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\begin{document}
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\section{Introduction}
\label{section:Introduction} Given a point $p\in {\mathbb C}^n$
and a real number $r>0$ we denote by $B^{2n}(p,r)$ the open ball
of radius $r$ centered at $p$ in ${\mathbb C}^{n}$. The
corresponding closed ball is denoted by $B^{2n}[p,r]$ and its
boundary sphere by $S^{2n-1}(p,r)= \partial B^{2n}[p,r]$.  We also
write $B^{2n}(1)=B^{2n}(0,1)$, $B^{2n}[1]=B^{2n}[0,1]$ and
$S^{2n-1}(1)=\partial B^{2n}[1]$. Let $\Omega$ be a germ of
holomorphic one-form with an isolated singularity  at the origin
$0\in \bc ^n, \, n\ge 3$. We address the problem of analytical
classification of $\Omega$ in the {\it non-integrable} case.
Motivated by the geometrical-analytical classification of
singularities in dimension $2$ we consider the case where the
kernel of $\Omega$ generates a germ of distribution $\Ker(\Omega)$
transverse to small spheres $S^{2n-1}(0,\epsilon)$. This is one,
though not the only, central motivation for this work. The problem
of existence of integral manifolds for germs of singularities of
integrable one-forms is an ancient problem already considered in
the work of Briot-Bouquet. The existence results in
\cite{Camacho-Sad} (for dimension $n=2$) and in
\cite{Cano-Cerveau} (for the non-dicritical case in dimension
$n=3$)  motivate the very basic question below:

\begin{Question}
\label{Question:first}  Is there a non-integrable germ of
holomorphic one-form $\Omega$ with an isolated singularity  at the
origin $0 \in \bc^{n}$ such that $\Ker(\Omega)$ is transverse to
the spheres $S^{2n-1}(0,\epsilon)$, \, for $\epsilon > 0$ small
enough and $\Ker(\Omega)$ admits no integral manifold through the
origin?
\end{Question}

Theorem~\ref{Theorem:nointegral manifold} gives a positive answer
to this question.  Other motivations are related to our previous
work  in \cite{Ito-Scardua1} and  \cite{Ito-Scardua2} where we
study the obstructions to the integrability of $\Omega$.
 Our first main result reads as:

\begin{Theorem}
\label{Theorem:main}  Let $n\ge 3$ and $\Omega$ be a holomorphic
one-form defined in a neighborhood of $B^{2n}[1]$  and such that
$\sing(\Omega) \cap S^{2n-1}(1)=\emptyset$. If  there exists a
holomorphic vector field $\xi$ in a neighborhood of $B^{2n}[1]$,
transverse to $S^{2n-1}(1)$, and such that $\Omega \cdot\xi= 0$,
then   $\Omega$ is not integrable.
\end{Theorem}

 Let us give examples of
 distributions as in  Theorem~\ref{Theorem:main}.
 Denote by $\mathcal A(2m)$ the set of all $2m\times2m$
skew-symmetric complex matrices and by $\ba(2m)$ the subset of
nonsingular elements in $\mathcal A(2m)$.
 In \cite{Ito-Scardua2} it is observed that if $A =
(a_{ij})_{i,j=1}^{2m}$  belongs to $\mathbb A(2m)$  then, for $m
\ge 2$,\,the one-form\, $\Omega_A = \sum\limits_{i,j=1}^{2m}
a_{ij}\, z_i\,dz_j$ defines a non-integrable holomorphic (linear)
distribution transverse to the spheres $S^{4m-1}(0,r) \subset \bc
^{2m}$, \, $r > 0$. Such a one-form will be called {\it linear}. A
particular case is the one-form $\Omega_{\bj(2m)} =
\sum\limits_{j=1}^m (z_{2j-1}\,dz_{2j} - z_{2_j}\,dz_{2j-1})$. One
may ask for non-linear examples. Given
$\ell=(\ell_1,...,\ell_m)\in {\mathbb N}^m$, we introduce the
corresponding {\it non-integrable Poincar\'e-Dulac normal form} as
$\Omega_{(\ell)} = \sum\limits_{j=1}^m [z_{2j-1}\,dz_{2j} -
(\ell_j z_{2j} + z_{2j-1}^{\ell_j})dz_{2j-1}]$ in coordinates
$(z_1,z_2,\dots,z_{2m}) \in \bc^{2m}$. We prove that also
$\Omega_{(\ell)}$ is  not integrable for $m \ge 2$, singular only
at the origin,  and $\Ker(\Omega_{(\ell)})$ is transverse to
$S^{4m-1}(0,r)$, \, $\forall\, r>0$ small enough (see
Example~\ref{Example:distributions}). The one-forms $\Omega_A,
\Omega_{\mathbb J(2m)}$ and $\Omega_{(\ell)}$ are our basic models
in the classification we pursue (see \S 4).

Let $\Ker(\Omega)$ be a codimension one holomorphic distribution
on a complex manifold $V^n$. Let  $p \in V^n$ be a {\it
singularity\/} of $\Ker(\Omega)$, that is, of $\Omega$. A germ of
codimension one analytic subset $\La_p$ at $p$ is  an {\it
integral manifold\/} of $\Ker(\Omega)$ through $p$ if any vector
$v_q\in T_q(V)$ which is tangent to $\Lambda_p$ at a point $q$
belongs to $\Ker(\Omega)(q)$. This means that if $\La$ is any
representative of $\La_p$ in a neighborhood $U$ of $p$ in $V^n$
and $\Lambda^*$ denotes the smooth part of $\Lambda$ then the
tangent bundle $T\La^*$ is a sub-bundle of
$\Ker(\Omega)\big|_{\Lambda^*}$ (see
Definition~\ref{Definition:invariant}). We shall always assume
$\La$ and $\La_p$ to be irreducible, nevertheless we do not
require that $\La\backslash\La^* = \sing(\La)$ is contained in
$\Sing(\Ker(\Omega))$. Regarding the existence of integral
manifolds for non-integrable distributions we have:

\begin{Theorem}
\label{Theorem:nointegral manifold} Let $m\ge 2$. Given $A\in
\mathbb A(2m)$ and $\ell\in {\mathbb N}^m$ the distributions
$\Ker(\Omega_{\mathbb J(2m)}), \, \Ker(\Omega_A)$ and
$\Ker(\Omega_{(\ell)})$ admit no integral manifold through the
origin.
\end{Theorem}

In the course of the proof of Theorem~\ref{Theorem:nointegral
manifold} we obtain the following Darboux's theorem type for (not
necessarily integrable) polynomial distributions. This is actually
a non-integrable version of the more precise Theorem 3.3 on page
102 of \cite{Jouanolou}:

\begin{Proposition}
\label{Proposition:Darboux}  Let $\Omega$ be a {\rm(}not
necessarily integrable{\rm)} polynomial one-form on $\bc^n$, \, $n
\ge 2$ and assume that $\cod \sing(\Omega)\ge 2$. If
$\Ker(\Omega)$ has infinitely many algebraic invariant
hypersurfaces then $\Omega$ is integrable. Indeed $\Omega = PdQ -
QdP$ for some polynomials $P,Q$ with no common factors and, in
particular, the leaves of the foliation $\fa_\Omega$ defined by
$\Omega$ are contained in the algebraic subvarieties $\{\la P -
\mu Q=0\}$ where $(\la,\mu) \in \bc^2-\{(0,0)\}$.
\end{Proposition}

As already mentioned, the examples  $\Omega_A, \,\Omega_{\bj(2m)}$
and $\Omega_{(\ell)}$ above constructed motivate the problem of
analytical classification of germs of non-integrable one-forms
defining distributions transverse to small spheres (see
Questions~\ref{Question:final} and \ref{Question:classify} in
\S\ref{section:classification}). In this direction we prove:

\begin{Theorem}
\label{Theorem:isotopy}   Let $\Om$ be a holomorphic one-form in a
neighborhood $U$ of the closed ball  $B^{4m}[1]\subset \bc^{2m}$
and such that  {\rm(1)} $\Om\cdot\vec R = 0$, where $\vec R$ is
the radial vector field in $\bc^{2m}$  and {\rm(2)} $\Sing(\Om)
\cap S^{4m-1}(1) = \emptyset$. Then $\Ker(\Om)$ is homotopic to
the linear distribution $\Ker(\Omega_{\bj(2m)})$ by distributions
$\Ker(\Om_s)$, $0 \le s \le 1$, such that $\Omega_0=\Omega$ and
$\Omega_1=\Omega_{\mathbb J_ {2m}}$, where $\Om_s$ is holomorphic
and satisfies $(1)$ and $(2)$ above.
\end{Theorem}

\noindent{\bf Acknowledgement}. We want to thank the anonymous
referee for  many valuable remarks, suggestions and for pointing
out gaps and unclear points in the first version. Specially for
suggesting clearer arguments and in particular for a drastic
simplification in the proof of Theorem~\ref{Theorem:nointegral
manifold}. Also mainly due to the referee is the discussion in
Remark~\ref{Remark:final}.

\section{Invariant manifolds}
\label{section:integral manifolds} In this section we discuss
Question~\ref{Question:first} and prove
Theorem~\ref{Theorem:nointegral manifold} and
Proposition~\ref{Proposition:Darboux}. First we prove the examples
mentioned in the introduction.

\begin{Example}
\label{Example:distributions}
 {\rm Let $A = (a_{ij})\in \mathbb A(2m)$ and
 $\Omega_A := \sum\limits_{i,j=1}^{2m} a_{ij}z_idz_j$ the
corresponding linear one-form in $\bc^{2m}$. Then $\Sing(\Omega_A)
= \{0\} \subset \bc^{2m}$ and $\Omega_A \cdot \vec R = 0$ for the
radial vector field $\vec R = \sum\limits_{j=1}^{2m}
z_j\,\dfrac{\po}{\po z_j}\cdot$ This implies that $\Ker(\Omega_A)$
is transverse to every sphere $S^{4m-1}(0,r)$, $r > 0$. The
non-integrability of $\Omega_A$ and $\Omega_{(\ell)}$ is a
straightforward computation (cf.\cite{Ito-Scardua1}).  For the
transversality of the non-integrable Poincar\'e-Dulac  normal form
$ \Omega_{(\ell)}$ with small spheres $S^{4m-1}(0,\epsilon)$  we
observe that if $\xi_{(\ell)} = \sum\limits_{j=1}^m
\big[(\ell_jz_{2j} + z_{2j-1}^{\ell_j})\, \frac{\po}{\po z_{2j}} +
z_{2j-1}\, \frac{\po}{\po z_{2j-1}}\big]$
 then $\Omega_{(\ell)} \cdot \xi_{(\ell)}=0$ and, as it is well-known,
 the vector field $\xi_{(\ell)}$ is transverse to the
spheres $S^{4m-1}(0,\epsilon)$ if $\epsilon>0$ is small enough.}
\end{Example}

 Now we shall prove Theorem~\ref{Theorem:nointegral manifold}.
 Let us first precise some notions involved.

\begin{Definition}
\label{Definition:invariant} {\rm Let $\Omega$ be a holomorphic
one-form on a complex manifold $V$. A {\it smooth} complex
immersed  submanifold $\La^* \subset V$ is an {\it integral
manifold\/} of $\Omega$ if  $T\La^* \subset
\Ker(\Omega)|\_{\La^*}$\,. In other words, any tangent vector to
$\La^*$ belongs to $\Ker(\Omega)$. If $\La \subset V$ is a
possibly singular complex analytic submanifold we say that $\La$
is an {\it integral manifold\/} of $\Omega$ if its regular part
$\La^*=\Lambda\setminus \sing(\Lambda)$ is an integral manifold of
$\Omega$. }
\end{Definition}

The following lemma is found in the algebraic setting in
\cite{Jouanolou}, Section 3.1, page 99.

\begin{Lemma}
\label{Lemma:Darboux} Let $\Ker(\Omega)$ be given by a holomorphic
one-form $\Omega$ in $V$ with $\cod \sing(\Omega) \ge 2$ and let
$\La \subset V$ be a codimension one analytic subset given by a
reduced equation $\La : \{f=0\}$ for some holomorphic $f\colon V
\to \bc$. The following conditions are equivalent:

\noindent {\rm (1)} $\La$ is $\Ker(\Omega)$-invariant. \, \, {\rm
(2)} $\Omega \wedge \frac{df}{f}$ is a holomorphic $2$-form on
$V$.

\end{Lemma}

\noindent{\bf Proof}. Since all objects involved are analytic we
may consider  the local case also at a generic (and therefore
non-singular) point $p \in \La^*$. In suitable local coordinates
$(z_1,\dots,z_n) = (z_1,\dots,z_{n-1},f)$ we have $p=0$ and $\La$
given by $\{f=z_n=0\}$. Also we may write $\Omega =
\sum\limits_{j=1}^n a_j\,dz_j$\,.  Suppose $\La$ is $\Ker(\Omega)$
invariant. Then, since $\La$ is given by $\{z_n=0\}$ we have
$\Omega \cdot \frac{\po}{\po z_j}\bigg|_{\{z_n=0\}} = 0$ for all
$j \in \{1,\dots,n-1\}$. In other words $z_n$ divides $a_j$ in
$\bc\{z_1,\dots,z_n\}$ for every $j \in \{1,\dots,n-1\}$ and
therefore $\Omega = \sum\limits_{j=1}^{n-1} z_n\, \tilde a_j\,
dz_j + a_n\, dz_n$ for some holomorphic $\tilde a_1,\dots,\tilde
a_{n-1} \in \bc\{z_1,\dots,z_n\}$. Thus $\Omega \wedge
\frac{dz_n}{z_n} = \sum\limits_{j=1}^{n-1} \tilde a_j\, dz_j
\wedge dz_n$ is holomorphic. Conversely, if  $\Omega \wedge
\frac{dz_n}{z_n}$ is holomorphic then $\Omega =
\sum\limits_{j=1}^{n-1} z_n\,\tilde a_j\,dz_j + a_n\,dz_n$ as
above therefore $\Omega \cdot\frac{\po}{\po z_j}$ vanishes on
$\{z_n=0\}$ for every $j=1,\dots,n-1$. Since $\{\frac{\po}{\po
z_1},\dots, \frac{\po}{\po z_{n-1}}\}$ generate $T\La$ in a
neighborhood of 0 we obtain that $T\La \subset \Ker(\Omega)$ in a
neighborhood of $p$. Therefore $\La$ is $\Ker(\Omega)$-invariant.
\qed

\vglue.1in

\noindent{\bf Proof of Proposition~\ref{Proposition:Darboux}}.
Once we have Lemma~\ref{Lemma:Darboux}, the proof is essentially
the same given in \cite{Jouanolou}, thus we will omit it and refer
to \cite{Jouanolou}.  \qed \vglue.1in

\subsection{Proof of
Theorem~\ref{Theorem:nointegral manifold}}
 Let $\Omega_{\bj(2m)}=\sum\limits_{j=1}^m (z_{2j-1}dz_{2j} - z_{2j}
dz_{2j-1})$ be given in $\bc^{2m}$. Then $d\Omega_{\bj(2m)}=2
\sum\limits_{j=1}^{m}dz_{2j-1}\wedge dz_{2j}$ is a non-degenerate
$2$-form on $\bc^{2m}$ and $(d\Omega_{\bj(2m)})^m$ is a non-zero
$2m$-form. Let $L$ be a complex manifold and $\zeta \colon L \to
\bc^{2m}$ a smooth embedding.

\begin{Lemma}
\label{Lemma:nointegral manifold} The following statements are
equivalent:

\noindent{\rm (1)} $T(\zeta(L))\subset
\Ker(\Omega_{\bj(2m)})\big|_{\zeta(L)}.$

\noindent{\rm (2)} $\zeta^*(\Omega_{\bj(2m)})=0$.
\end{Lemma}

\noindent{\bf Proof}. To prove that (1) implies (2) we take any
vector $v\in T(L)$, then $\zeta_*(v)\in T(\zeta(L))\subset
\Ker(\Omega_{\bj(2m)})\big|_{\zeta(L)}$ means
$\Omega_{\bj(2m)}(\zeta_*v)=0$. Therefore we get
$\zeta^*(\Omega_{\bj(2m)})(v)=\Omega_{\bj(2m)}(\zeta_*v)=0$. To
prove that (2) implies (1) take any $\tilde v = \zeta_* v \in
T(\zeta(L)), \, v \in T(L)$. Then $\zeta^*(\Omega_{\bj(2m)})=0$
means that $\Omega(\tilde
v)=\Omega_{\bj(2m)}(\zeta_*v)=\zeta^*(\Omega_{\bj(2m)})(v)=0$.
Thus $\tilde v$ belongs to
$\Ker(\Omega_{\bj(2m)})\big|_{\zeta(L)}$. \qed

\begin{Proposition}
\label{Proposition:dimensionless} Under the above notations,
assume that $T(\zeta(L))\subset
\Ker(\Omega_{\bj(2m)})\big|_{\zeta(L)}$, then the complex
dimension of $L$ is less than $m$ or equal to $m$, i.e., $\dim
_{\bc} L \leq m$.
\end{Proposition}

\noindent{\bf Proof}. First we observe that
$\zeta^*(d\Omega_{\bj(2m)})= d(\zeta^*\Omega_{\bj(2m)})=0$ on $L$.
Assume that $\dim_\bc L \geq m+1$. Take $m+1$ linearly independent
vectors $v_1,...,v_{m+1}$ in $T(L)$. Moreover, we take $m-1$
vectors $u_1,...,u_{m-1}$ in $T\bc^{2m}$ such that $\zeta_*
v_1,...,\zeta_* v_{m+1}, u_1,...,u_{m-1}$ are linearly independent
in $T\bc^{2m}$. Then we get
$(d\Omega_{\bj(2m)})^m(\zeta_*v_1,...,\zeta_*v_{m+1},u_1,...,u_{m-1})=0$
because $\zeta^*(d\Omega_{\bj(2m)})=0$. It is contradictory with
the fact that $(d\Omega_{\bj(2m)})^m$ is a non-zero $2m$-form.
\qed

\begin{Corollary}
\label{Corollary:nointegral manifold} Let $m\ge 2$. The one-form
$\Omega_{\bj(2m)}$ has no integral manifold through the origin.
\end{Corollary}

\noindent{\bf Proof}. Assume that $\Lambda$ is an integral
manifold of $\Omega_{\bj(2m)}$. Then $\Lambda^*= \Lambda -\{0\}$
is a complex hypersurface, i.e., $\dim_\bc \Lambda ^*=2m-1$. By
Proposition~\ref{Proposition:dimensionless}, we get $\dim_\bc
\Lambda^* \leq m$. We have a contradiction with the hypothesis
$m\ge 2$.

\vglue.2in This proves Theorem~\ref{Theorem:nointegral manifold}
for $\Omega_{\mathbb J(2m)}$.
 By the same argument,
$\Omega_A$ and $\Omega_{(\ell)}$ have no integral manifold through
the origin of $\bc^{2m}$. \qed

\vglue.1in Regarding the smoothness of invariant hypersurfaces of
holomorphic
 distributions with an isolated singularity we have:

\begin{Proposition}
\label{Proposition:smooth}
 Let $\Omega$ be a holomorphic one-form in a
complex manifold $M$ and $p\in M$ and $\Lambda\subset M$ an
integral manifold of $\Omega$ point. Given a point $p\in M$  where
$\Omega$ is nonzero then $\Lambda$  is smooth at the point $p$.
\end{Proposition}

\noindent{\bf Proof}. In suitable local coordinates
$(z_1,...,z_n)$ centered at $p$ the Pfaffian equation $\Omega=0$
is equivalent to an equation $dz_n=\sum\limits_{j=1}^{n-1} g_j
dz_j$ for some holomorphic functions $g_j$. Any integral manifold
then writes as $z_n=f(z_1,...,z_{n-1})$ for a holomorphic function
$f$ satisfying $\po f (z_1,...,z_{n-1})/ \po
z_j=g_j(z_1,...,z_{n-1},f(z_1,...,z_{n-1}))$. This implies that
$f$ cannot develop any singularities. \qed

\section{Proof of Theorem~\ref{Theorem:main}}
\label{section:proofmain} By hypothesis,  $\xi$ defines  a
one-dimensional holomorphic foliation  in a neighborhood of the
closed ball $B^{2n}[1]$, transverse to the sphere $S^{2n-1}(1)$.
 According to \cite{Ito}   $\xi$ has a unique
singularity $p$ in the open ball $B^{2n}(1)$ and this singularity
is in the Poincar\'e domain: the  vector field $\xi$ has a
non-singular linear part at $p$ having eigenvalues
$\la_1,...,\lambda_n$ such that the origin does not belong to the
convex hull of the set $\{\la_1,...,\la_n\}$ in $\re^2$. On the
other hand we have  the following result from \cite{Ito-Scardua2}:

\begin{Theorem} \label{Theorem 8.1}  Let $\omega$ be a holomorphic one-form in
an open subset $U \subset \bc^n$, $n \ge 2$. Suppose that the
distribution $\Ker(\omega)$ is transverse to a sphere
$S^{2n-1}(p,r) \subset U$, with $B^{2n}[p,r] \subset U$, $p \in
U$, $r>0$, then $n$ is even, $\omega$ has a single singular point
in the ball $B^{2n}(p,r)$ and this is a simple singularity in the
following sense:  if we write $\omega=\sum\limits_{j=1}^n f_j
dz_j$ in local coordinates centered at the singularity then the
matrix $(\partial f_j /\partial z_k)_{j,k=1}^n$ is non-singular at
the singularity.
\end{Theorem}

 By the above result  $\Omega$ has a unique singularity $q$ in $B^{2n}(1)$
 and this is a simple singularity.

\begin{Claim}
We have $p=q$.
\end{Claim}
\noindent{\bf Proof}.
 Let $\Omega =
\sum\limits_{j=1}^n f_j\,dz_j$  in a neighborhood $U$ of
$B^{2n}[1]$ on $\bc^n$. Write $\xi = \sum\limits_{i=1}^n
A_i\,\frac{\po}{\po z_i}$ then $0 = \Omega\cdot\xi =
\sum\limits_{j=1}^n f_jA_j$ and therefore $0 = \frac{\po}{\po z_k}
(\sum_{j=1}^n f_jA_j) = \sum_{j=1}^n (\frac{\po f_j}{\po z_k}\,
A_j + f_j\, \frac{\po A_j}{\po z_k}). $ Thus $0 =
\sum\limits_{j=1}^n \frac{\po f_j}{\po z_k}(p)\cdot A_j(p)$, \,
$\forall\,k$ and since the matrix $\big(\frac{\po f_j}{\po
z_k}(p)\big)_{j,k=1}^n$ is non-singular we have $A_j(p)=0$, \,
$j=1,\dots,n$. By the uniqueness of the singularity of $\xi$ we
get $p=q$. This proves the claim.

\vglue.1in Let us finish the proof.  Suppose by contradiction that
$\Omega$ is integrable. By a theorem of Malgrange
\cite{[Malgrange]}, since $\codim \sing(\Omega)=n \ge 3$ at $p$,
the one-form $\Omega$ admits a holomorphic (Morse-type) first
integral in a neighborhood of $p$, say $f\colon (W,p) \to
(\bc,0)$. Then $\Omega\cdot \xi=0$ implies that $\xi(f)=0$.
Because the germ of $\xi$ at $0$ is in the Poincar\'e domain, this
implies that $f$ is constant in a neighborhood of $0$,
contradiction with the fact that $f$ is of Morse type at $0$. This
ends the proof of Theorem~\ref{Theorem:main}. \qed

\section{On the analytical classification}
\label{section:classification}

The problem of  analytic classification of singularities of
 holomorphic one-forms in dimension two is a very well-developed
 topic.  Recently (cf. \cite{Cerveau-Mozo}) the analytic classification
 was obtained for germs of reduced {\it integrable} one-forms at
 the origin $0\in \bc^3$. As far as
 we know, nothing is found regarding the non-integrable case.
In the follow-up,  "to classify"   means to give a description in
terms of objects which are completely understood.
 Obviously  the class of germs of singular (non-integrable) holomorphic
 one-forms is too wide in order to be classified at a first
 moment and we also lack of geometric ingredients. This remark
 is one of the motivations for our  approach in this section. Other motivation is
 given by the well-known results for holomorphic foliations with singularities
 in dimension two collected in the
 following omnibus theorem:

\begin{Theorem}[\cite{[Brjuno]},
\cite{[Camacho-Kuiper-Palis]}, \cite{[Dulac]}, \cite{Ito}]
Given a germ of singular holomorphic one-form $\Omega$ at $0 \in
\bc^2$ the following conditions are equivalent:
\begin{enumerate}
\item[{\rm(1)}] $\Omega$ is in the {\it Poincar\'e domain}:
$\Omega = Ady - Bdx$ where the vector field $X = A\,\frac{\po}{\po
x} + B\,\frac{\po}{\po y}$ has a singularity at $(0,0)$ whose
linear part $DX(0,0)$  has non-zero eigenvalues $\la$, $\mu$ with
quotient $\la/\mu \in \bc\backslash\re_-$\,. \item[{\rm(2)}]
$\Ker(\Omega)$ is transverse to some {\rm(}and therefore to
every{\rm)}   sphere $S^3(0,\epsilon)$, \, for $\epsilon > 0$
small enough. \item[{\rm(3)}] There are local analytic coordinates
$(x,y) \in (\bc^2,0)$ such that $\Omega(x,y)$ is either linear
$\Omega = \la xdy - \mu ydx$ with $\la/\mu \in \bc\backslash
\re_-$\,, or it is of the form $\Omega = xdy - (ny+x^n)dx$ where
$n \in \bn$ {\rm(}called Poincar\'e-Dulac normal form{\rm)}.
\end{enumerate}
\end{Theorem}

\noindent Thus we have the following problem:

\begin{Problem}
To obtain the local analytical  classification  of germs of
non-integrable holomorphic one-forms $\Omega$ at $0\in \bc^n$
under the hypothesis of transversality with small spheres
$S^{2n-1}(0,\epsilon)$.
\end{Problem}

One may work with the following notions and  models:

\begin{Definition} {\rm
 We shall say that a germ of singular non-integrable
one-form $\Omega$ at the origin $0 \in \bc^{2m}$ is {\it in the
Poincar\'e domain\/} if $\Ker(\Omega)$ is transverse to small
spheres $S^{4m-1}(0,\epsilon)$, \, $\epsilon
> 0$. A germ $\Omega$ will be called  {\it analytically linearizable\/}
if $f^*\Omega=\Omega_A$  for some germ of biholomorphism $f \in
\text{Bih}(\bc^{2m},0)$ fixing the origin and some $A\in \mathbb
A(2m)$. Finally we shall say that $\Omega$ is ({\it analytically
conjugate to}) {\it a non-integrable Poincar\'e-Dulac  normal
form\/} if $f^*\Omega = \Omega_{(\ell)}$  for some $f \in \text{
Bih}(\bc^{2m},0)$ and some $\ell \in \bn^m$.}
\end{Definition}

\begin{Question}
\label{Question:final}
 Are conditions {\rm (i)}, {\rm (ii)}  and {\rm (iii)}  below equivalent?

\noindent {\rm (i)}  $\Omega$ is the Poincar\'e-domain.

\noindent {\rm (ii)}  $\Omega$ is analytically linearizable or
conjugate to a non-integrable Poincar\'e-Dulac  normal form.

\noindent{\rm (iii)} There is a   holomorphic vector field $\xi$
transverse to the spheres $S^{4m-1}(0,r)$ for $r>0$ small enough
and such that $\Omega \cdot \xi=0$?
\end{Question}

\begin{Remark}
\label{Remark:final} {\rm The non-integrable examples we have
given, do admit holomorphic sections $\xi$ transverse to small
spheres $S^{4m-1}(0,r)$ (see Example~\ref{Example:distributions}).
Let us observe that there are examples of non-integrable one-forms
$\Omega$ with isolated singularity at the origin in $\bc ^n, \, n
\ge 3$, which admit integral manifold. For example, take
$f=\frac{1}{2}\sum\limits_{j=1}^{2m} z_j^2$ and $\Omega=df + f\nu$
for some holomorphic one-form $\nu=\sum\limits_{j=1}^{2m}\nu_j\,
dz_j$ such that $d\nu(0)$ is nondegenerate. Then $\{f=0\}$ is an
integral manifold of $\Omega$ and also $\Omega(z)=0$ if and only
if $z_j+ \sum\limits_{k=1}^{2m} z_k^2 \nu_j(z)=0$ for every $j$.
This shows, because the Jacobian of the left hand side at $z=0$ is
the identity, that $\Omega$ has an isolated singularity at the
origin $z=0$. Finally, if $X_j\in \Ker(\Omega)$, then
$(d\Omega)(X_1,X_2)=f(d\nu)(X_1,X_2)$, we obtain that near the
origin $\Omega$ is non-integrable at every point where $f\ne 0$.
This example suggests that a more interesting question might be :
\begin{Question}
\label{Question:classify} Classify the non-integrable germs of
holomorphic one-forms $\Omega$ with an isolated singularity at the
origin and $\Ker(\Omega)$ transverse to all small spheres centered
at the origin, which admit an integral manifold through the
origin.
\end{Question}

The above construction suggests that if the integral manifold is
taken in the form $\{f=0\}$ for a holomorphic function $f$, then
maybe we can write $\Omega=g(df + f\nu)$ for some function $g$ and
a one-form $\nu$, and we have to study the kernels of $df + f\nu$.
}
\end{Remark}

\subsection{Proof of Theorem~\ref{Theorem:isotopy}}

Now we shall  prove Theorem~\ref{Theorem:isotopy}. We need:
\begin{Lemma} \label{Lemma:arcwise}   The open subset $\ba(2m)\subset \mathcal A(2m)$
is arcwise-connected.
\end{Lemma}

\noindent{\bf Proof}.  First we recall that $\mathcal A(2m)$ is a
 complex vector space. Denote by $f\colon \mathcal A(2m) \to \bc$
the determinant function, i.e., $f(A) = \Det(A)$, $\forall\,A \in
\mathcal A(2m)$. Then $f$ is holomorphic and $f \not= 0$ because
if $\bj(2m) = \left[\begin{smallmatrix}
\big[\begin{smallmatrix}0 &-1\\ 1 &0\end{smallmatrix}\big]_{\ddots} &\\
&\big[\begin{smallmatrix}0 &-1\\ 1
&0\end{smallmatrix}\big]\end{smallmatrix}\right]$ then $\bj(2m)
\in \mathcal A(2m)$ and $f(\bj(2m)) = \Det(\bj(2m)) = (-1)^m \ne
0$.

\noindent Now $\ba(2m) = \big\{A \in \mathcal A(2m) : f(A) \ne
0\big\} = \mathcal A(2m)\backslash f^{-1}(0)$. Since $\mathcal
A(2m)$ is an affine space and $f^{-1}(0)$ is a thin subset of
$\mathcal A(2m)$ it follows from \cite{[Gunning-Rossi]} Corollary
4 page 20 that $\ba(2m)$ is (arcwise) connected.\qed

\vglue.1in We shall adopt the following natural definition:

\begin{Definition} \label{Definition 8.4} {\rm Two codimension one holomorphic
distributions $\Delta_1$, $\Delta_2$ on a complex manifold $M^n$
are {\it homotopic\/} if there is a $C^\infty$ family
$\{P_s\}_{s\in [0,1]}$ of holomorphic distributions in $M$ such
that $P_0=\Delta_1$ and $P_1=\Delta_2$. }
\end{Definition}

\begin{Lemma} \label{Lemma:linearhomotopy}  Let $\Omega_A$ be a linear one-form in
$\bc^{2m}$ with $A \in \ba(2m)$. Then the distribution
$\Ker(\Omega_A)$ is homotopic to the {canonical distribution}
$\Omega_{\bj(2m)} = \sum\limits_{j=1}^{2m} \big(z_{2j} dz_{2j-1} -
z_{2j-1} dz_{2j}\big)$.
\end{Lemma}

\noindent{\bf Proof}.  This is an immediate consequence of
Lemma~\ref{Lemma:arcwise}. Given a smooth path $\al\colon [0,1]
\to \ba(2m)$ connecting $\al(0) = A$ to $\al(1) = \bj(2m)$ we
define a homotopy  by setting $P_s=\Ker(\Omega_{\alpha(s)})$
(recall that $\al(s)\in \mathbb A(2m)$). \qed

\vglue.2in \noindent Now we consider a holomorphic one-form $\Om$
defined in a neighborhood $U$ of the closed unit ball $B^{4m}[0,1]
\subset \bc^{2m}$ and such that $\Om \cdot\vec R = 0$.  Write $\Om
= \sum\limits_{\nu=1}^{+\infty} \omega_\nu$ where $\omega_\nu$ is
a homogeneous one-form of degree $\nu \ge 1$ in $U$ (assume $U$ is
a ball). Then $0 = \Om\cdot\vec R = \sum\limits_{\nu=1}^{+\infty}
\omega_\nu\cdot\vec R$ and since $\omega_\nu\cdot \vec R$ is a
homogeneous polynomial of degree $\nu+1$ we conclude that
$\omega_\nu\cdot\vec R = 0, \quad \forall\,\nu \ge 1.$  In
particular, $\omega_1\cdot\vec R = 0$ and therefore $\omega_1 =
\Omega_A$ for some $A \in \mathcal A(2m)$. Assume now that
$\Ker(\Om)$ is transverse to $S^{4m-1}(1)$ then it follows from
Theorem~\ref{Theorem 8.1} that $\Omega$ has only one singularity
which is simple. Because the group of holomorphic transformations
of the unit ball acts transitively, we can assume by a holomorphic
change of coordinates that the origin is the only singularity of
$\Om$ in $B^{4m}[0,1]$ and, since it is a simple singularity, $A$
is non-singular, i.e., $A \in \ba(2m)$.

\begin{Lemma} \label{Lemma:deformation}  Under the above hypothesis there is a real
analytic deformation $\big\{\Ker(\Om^t)\big\}_{t\in[0,1]}$ of
$\Ker(\Om)$ into $\Ker(\Omega_A)$ by holomorphic distributions
$\Ker(\Omega^t)$ transverse to $S^{4m-1}(1)$ outside the
intersection $S^{4m-1}(1) \cap \Sing(\Omega^t)$.
\end{Lemma}

\noindent{\bf Proof}.  We define $\Om^t$ by $\Om^t
:=t^{-1}\Omega(tz)$. Then $\Om^t$ converges to a holomorphic
one-form for each $t \in [0,1]$ by canonical convergence criteria,
indeed, $\Om^t$ defines a holomorphic one-form in $U$ for each $t$
in the closed unit disc $\ov\bd \subset \bc$.  We have  $\Om^1 =
\Om$ and an easy computation shows that $\Omega^t$ converges for
$t\rightarrow 0$ to $\Omega_A$ where $A$ is the Jacobi-matrix of
$\Omega$ at the origin. Furthermore, because $\vec R(tz)=t\vec R
(z)$, one has $\Omega^t(z)\cdot \vec R(z)=t^{-1}\Omega(tz) \cdot
\vec R(z)= t^{-2} \Omega(tz) \cdot \vec R(tz) =0$ for every $t\ne
0$. Thus $\Ker(\Om^t)$ is transverse to $S^{4m-1}(1)$ outside
$\Sing(\Om^t) \cap S^{4m-1}(1)$.\qed

\vglue.1in

\noindent Now we study the singular set $\Sing(\Om^t)$.
\begin{Lemma}
\label{Lemma:singularset} We have $\sing(\Om^t) = \{0\}$,
$\forall\, t \in \ov\bd$.
\end{Lemma}
\noindent{\bf Proof}.  Take any $z\in S^{4m-1}(1)$. Consider the
complex line $\ell(z)=\{tz : t \in \bc\}$ through $z$. If $0<
|t|<1$, then $tz \ne 0$ belongs to $B^{4m}[1]$. Therefore
$\Omega^t(z)= \frac{1}{t}\Omega(tz)$ is non-zero. If $t=0$, then
$\Omega^0(z)=\Omega_A(z)$ is non-zero on $S^{4m-1}(1)$. This
proves the lemma. \qed

\vglue.1in \noindent{\bf Proof of Theorem~\ref{Theorem:isotopy}}:
For the proof we just have to apply
Lemmas~\ref{Lemma:linearhomotopy}, ~\ref{Lemma:deformation} and
~\ref{Lemma:singularset} above.\qed

\bibliographystyle{amsalpha}

config: arxiv


TITLE: Why $C_0^\infty$ is dense in $L^p$?
QUESTION [11 upvotes]: Why $C_0^\infty$ is dense in $L^p$?
Would you give me a simple proof or the outline of the proof?

REPLY [0 votes]: Maybe I am missing something, but rather than using convolution and Lusins Lemma to approximate continuous finitely supported functions I would use the functions
$$
f(x) :=
\begin{cases}
e^{1-\frac{1}{1-x^2}} & \text{ for } \vert x\vert <1 \\
0,& otherwise
\end{cases}
$$
and
$$
g_n(x) := \frac{1}{n}\tan(\frac{x ~\pi }{2})
$$
to approximate indicator functions $\chi_{[a,b]}(x)$ by the functions $\frac{a+b}{2}+ \frac{b-a}{2} f\circ g_n (x)$.


text:
\begin{document}

\title [$EM-$Graded Rings]{$EM-$Graded Rings}

\author[{{T. Alraqad}}]{\textit{Tariq Alraqad}}

\address
{\textit{Tariq Alraqad, Department of Mathematics, University of Hail, Saudi Arabia.}}
\bigskip
{\email{\textit{t.alraqad@uoh.edu.sa}}}

 \author[{{H. Saber}}]{\textit{Hicham Saber}}

\address
{\textit{Hicham Saber, Department of Mathematics, University of Hail, Saudi Arabia.}}
\bigskip
{\email{\textit{hicham.saber7@gmail.com}}}

 \author[{{R. Abu-Dawwas }}]{\textit{Rashid Abu-Dawwas }}

\address
{\textit{Rashid Abu-Dawwas, Department of Mathematics, Yarmouk
University, Jordan.}}
\bigskip
{\email{\textit{rrashid@yu.edu.jo}}}

 \subjclass[2010]{13A02, 16W50}

\date{}

\begin{abstract}
The main goal of this article is to introduce the concept of $EM-G-$graded rings. This concept is an extension of the notion of $EM-$rings. Let $G$ be a group and $R$ be a $G-$graded commutative ring.  The $G-$gradation of $R$ can be extended to $R[x]$ by taking the components $(R[x])_{\sigma}=R_{\sigma}[x]$. We define $R$ to be $EM-G-$graded ring if every homogeneous zero divisor polynomial has an annihilating content. We provide examples of $EM-G-$graded rings that are not $EM-$rings and we prove some interesting results regarding these rings.
\end{abstract}

\keywords{Graded rings, EM-rings,, homogeneous, polynomial ring, annihilating content.

 }
 \maketitle

\section{Introduction}\label{int}
Throughout this article, all rings are commutative rings with one. The set of all zero divisors of a ring $R$ is $Z(R)$, the set of all regular elements is denoted by $reg(R)$, and the set of all idempotent elements is denoted by $E(R)$. A zero divisor polynomial $f(x)\in R[x]$ is said to have an annihilating content if there are $c_f\in Z(R)$ and $f_1\in reg(R[x])$ such that $f(x)=c_ff_1(x)$. Annihilating contents simplify computations of annihilator of polynomials. For instance, in \cite{ab17} annihilating contents are used to obtain some results related to the zero divisors graph of $R[x]$, where $R$ is a principal ideal rings. As a generalization, Abuosba and Ghanem \cite{ab19} introduce the concept of $EM-$rings as follows: A ring $R$ is called $EM-$ring if every zero divisor polynomial in $R[x]$ has an annihilating content. These rings were studied extensively in \cite{ab19} and \cite{gh18}.

Our goal in this article is to extend the notion of $EM-$rings by using the concept of graded rings. Let $G$ be a group. A ring $R$ with unity $1$, is said to be  $G$-graded if there exist additive subgroups $\{R_{\sigma}\mid \sigma\in G\}$ such that $\displaystyle R=\oplus_{\sigma\in G}R_{\sigma}$ and $R_{\sigma}R_{\tau}\subseteq R_{\sigma\tau}$ for all $\sigma,\tau\in G$. This gradation is denoted by $(R,G)$. The elements of $R_{\sigma}$ are called homogeneous of degree $\sigma$. The set of all homogeneous elements is denoted by $h(R)$, and the set of homogenous zero divisors is denoted by $hZ(R)$.  If $x\in R$, then $x$ can be written uniquely as $\displaystyle\sum_{\sigma\in G}x_{\sigma}$, where $x_{\sigma}$ is the component of $x$ in $R_{\sigma}$.

If $R$ is a $G-$graded ring, then $R[x]$ is $G-$graded with gradation defined by $(R[x])_\sigma=R_\sigma[x]$, $\sigma\in G$. Clearly, $R[x]=\oplus_{\sigma\in G}R_{\sigma}[x]$, and $R_{\sigma}[x]R_{\tau}[x]\subseteq R_{\sigma\tau}[x]$, for all $\sigma,\tau\in G$. Furthermore, $f(x)=\sum_{i=0}^{n}a_ix^i\in h(R[x])$ if and only if there exists $\sigma\in G$ such that $a_i\in R_{\sigma}$ for all $i$. i.e. all coefficients of $f$ are homogenous of the same degree. We will say that $R$ is an $EM-G-$graded ring if every zero divisor homogeneous polynomial in $R[x]$ has an annihilating content. It is evident that every $EM-$ring (that is $G-$graded) is an $EM-G-$graded ring, however the converse is not true as we will see in Examples \ref{e1} and \ref{e2}.

The next section includes some preliminary results on graded rings and $EM-$rings that will be needed in the following section. Section \ref{sec1} is devoted to the concept of $EM-G-$graded rings and its properties. We show that if $R_e$ is an $EM-$ring and the other components satisfy some certain condition then $R$ is $EM-G-$graded ring. We deduce that if $(R,G)$ is a cross product and $R_e$ is an $EM-$ring then $R$ is an $EM-G-$graded ring. Some extensions of $EM-G-$graded rings are also investigated. Some of these results are similar to results obtained about $EM-$ring in \cite{ab19}. We show that if $R$ is $EM-G$-graded ring then so is $R[x]$. We also prove that if $R$ is an $EM-G$-graded ring and  $S\subseteq h(R)$ is multiplicatively closed then $S^{-1}R$ is $EM-G-$graded ring. In this section we also obtain a nice result related to the idealization $R(+)R$ that generalizes \cite[Theorem 5.1]{gh18}. We show that $R$ is an $EM-$ring if and only if $R(+)R$ is an $EM-Z_2-$graded ring with gradation $H_0=R\oplus 0$ and $H_1=0\oplus R$.

\section{Preliminaries}\label{sec0}
In this section we list some results on graded rings and $EM-$rings that will be needed in the squeal. For a general reference on graded rings the reader is referred to \cite{Nastasescue}.

\begin{defn}\label{def0}
Let $G$ be a group. A ring $R$ with unity $1$, is said to be  $G$-graded if there exist additive subgroups $\{R_{\sigma}\mid \sigma\in G\}$ such that $\displaystyle R=\oplus_{\sigma\in G}R_{\sigma}$ and $R_{\sigma}R_{\tau}\subseteq R_{\sigma\tau}$ for all $\sigma,\tau\in G$.
\end{defn}
When $R$ is $G$-graded we denote that by $(R,G)$. The support of $(R,G)$ is defined as $supp(R,G)=\{\sigma\in G:R_{\sigma}\neq0\}$. If $x\in R$, then $x$ can be written uniquely as $\displaystyle\sum_{\sigma\in G}x_{\sigma}$, where $x_{\sigma}$ is the component of $x$ in $R_{\sigma}$. It is well known that $R_{e}$ is a subring of $R$ with $1\in R_{e}$. An ideal $A$ of a $G-$graded ring $R$ is called $G-$graded ideal provided that $A=\oplus_{\sigma\in G}(A\cap R_{\sigma})$. In case $A$ is a $G-$graded ideal of a $G-$graded ring $R$ then the factor ring $R/A$ is a $G-$graded ring with gradation defined by $(R/A)_{\sigma}=R_{\sigma}+A/A$.

For a polynomial $f(x)\in R[x]$, we denote by $C(f)$ the ideal generated by the coefficients of $f$. Abuosba and Ghanem \cite{ab19} noted that if $c_f$ is an annihilating content of $f$ then $Ann_{R[x]}(f(x))=Ann_{R[x]}(c_f)$ and $Ann_R(C(f))=Ann_R(c_f)$. Next we show that if $f$ is homogenous then $C(f)$ is a $G-$graded ideal.

\begin{lem}\label{l2} Let $R$ be a $G-$graded ring and consider the $G-$grading on $R[x]$ whose components are $R([x])_{\sigma}=R_{\sigma}[x]$, $\sigma\in G$. If $f(x)=\sum_{i=0}^na_ix^i\in h(R[x])$ then $C(f)$ is a $G-$graded ideal of $R$. \end{lem}

\begin{proof}
Since $f$ is homogeneous, there is $\tau\in G$ such that $\{a_0,\dots,a_n\}\subseteq R_{\tau}$. Let $w=\sum_{\sigma\in G}w_{\sigma}\in C(f).$ Then we can write $w$ as $w=\sum_{i=o}^nr_ia_i$. In addition, each $r_i$ can be written as $r_i=\sum_{\sigma\in G}r_{\sigma,i}$. So, we have
$$\sum_{\sigma\in G}w_{\sigma}=\sum_{i=0}^n\left(\sum_{\sigma\in G}r_{\sigma,i}\right)a_i=\sum_{\sigma\in G}\left(\sum_{i=0}^nr_{\sigma,i}a_i\right).$$
Now, for each $\sigma\in G$ we have $r_{\sigma}$ and $ia_i$ belong to $R_{\sigma}R_{\tau}\subseteq R_{\sigma\tau}$. Therefore $w_{\sigma}\in C(f)$, for all $\sigma\in G$. Hence $C(f)$ is $G-$graded ideal of $R$.
\end{proof}

\begin{lem}\label{l1} Let $R$ be a $G-$graded ring. If for every $\sigma\in supp(R,G)$, there exists $u_{\sigma}\in R_{\sigma}$ such that $R_{\sigma}=R_eu_{\sigma}$ and $Ann_{R_e}(u_{\sigma})=\{0\}$, then $reg(R_e[x])\subseteq reg(R[x])$.
\end{lem}
\begin{proof}
Let $h(x)=\sum_{i=0}^na_ix^i\in reg(R_e[x])$. Then $Ann_{R_e}(a_0,a_1,\dots,a_n)=\{0\}$. Let $t=\sum_{\sigma\in G}t_g\in Ann_R(a_0,a_1,\dots,a_n)$. Then for each $i$, we have $0=\sum_{\sigma\in G}t_{\sigma}a_i$. Since $t_{\sigma}a_i\in R_{\sigma}$, by the unique representation of $0$, we get $t_{\sigma}\in Ann_R(a_0,a_1,\dots,a_n)$ for each ${\sigma}\in G$. On the other hand, we have that for every ${\sigma}\in G$ there exists $s_{t_{\sigma}}\in R_e$, such that $t_{\sigma}=u_{\sigma}s_{t_{\sigma}}$.  Since $u_{\sigma}s_{t_{\sigma}}a_i=t_{\sigma}a_i=0$ and $Ann_{R_e}(a_0,\dots,a_n)=\{0\}$, we obtain that $s_{t_{\sigma}}a_i=0$ for each $i$. So, $s_{t_{\sigma}}\in Ann_{R_e}(a_0,\dots,a_n)=\{0\}$, for all ${\sigma}\in G$. This implies that $t=0$. Hence $h(x)\in reg(R[x])$.
\end{proof}

\begin{defn}\label{d22} A grading $(R,G)$ is called crossed product over $supp(R,G)$ if for every $\sigma\in supp(R,G)$, $R_\sigma$ contains a unit.\end{defn}

\begin{prop}\label{p2} \cite[Proposition 1.7]{Da10} If $(R,G)$ is crossed product over $supp(R,G)$, then for each $\sigma\in supp(R,G)$, $R_{\sigma}=R_eu$ for some unit $u\in R_{\sigma}$
\end{prop}

\begin{defn}\label{ae} Let $R$ be a ring with two gradations $(R,G)$ and $(R,G')$. Then $(R,G)$ is almost equivalent to $(R,G')$ if there exists an automorphism $\phi:R\longrightarrow R$ such that for each $\tau\in G'$ the exists $\sigma\in G$ such that $\phi(R_{\sigma})=R_{\tau}$.
\end{defn}

\begin{prop}\label{p3}\cite[Proposition 8.1.2]{Nastasescue} Let $R$ be a $G-$graded ring. If $S\subseteq h(R)$ is a multiplicatively closed subset of $R$, then the ring $S^{-1}R$ is $G-$graded by the gradation $(S^{-1}R)_{\lambda}=\{\frac{a}{s}\mid s\in S, a\in R\text{ such that } \lambda=(deg(s))^{-1}deg(a)\}$.
\end{prop}

The following lemma can be found in \cite[Lemma 2.3]{an85}
\begin{lem}\label{l3}
If $f\in R[x]$ with $C(f)=Ra$ then there exists $g\in R[x]$ with $C(g)=R$ and $f(x)=ag(x)$
\end{lem}

\begin{defn}\label{d2} Let $H$ be a subset (res. subring, ideal) of a ring $R$. Then $H$ is called $EM-$subset (res. $EM-$subring, $EM-$ideal) of $R$ if for each $f(x)\in H[x]\cap Z(R[x])$, there exists $c_f\in Z(R)$ and $g(x)\in reg(R[x])$ such that $f(x)=c_fg(x)$.
\end{defn}

\section{$EM-G-$graded rings}\label{sec1}

We start this section by introducing the concept of $EM-G-$graded ring.

\begin{defn}\label{d1} Let $R$ be a $G-$graded ring and consider the $G-$grading on $R[x]$ whose components are $R([x])_{\sigma}=R_{\sigma}[x]$, $\sigma\in G$. Then $R$ is called $EM-G-$graded ring if every non-zero polynomial in $hZ(R[x])$ has an annihilating content.
\end{defn}

We can see that every $EM-$ring that is $G-$graded is an $EM-G-$graded ring. The next example shows that the converse is not true.

\begin{exa}\label{e1} The ring $R=\mathbb{Z}_4[y]/(y^2)=\{a+bY: a,b\in \mathbb{Z}_4\text{ and } Y^2=0\}$ is not an $EM-$ring beacuse the polynomial $f(x)=2+Yx\in Z(R[x])$ does not have annihilating content (see \cite[Example 2.6]{ab19}). Now $R$ is $\mathbb{Z}_2-$graded by $R_0=\mathbb{Z}_4$, $R_1=\mathbb{Z}_4Y$. Let $f(x)=a_0+a_1x+\dots+a_nx^n\in hZ(R[x])$. Then either $\{a_0,\dots,a_n\}\subseteq \{0,2\}$ or  $\{a_0,\dots,a_n\}\subseteq \mathbb{Z}_4Y$. If $\{a_0,\dots,a_n\}\subseteq \{0,2\}$, then $a_n=2$, and hence $2$ is an annihilating content of $f$. Assume $\{a_0,\dots,a_n\}\subseteq \mathbb{Z}_4Y$. Then there is a polynomial $f_1(x)$ whose one of the coefficients is $1$ or $3$ such that $f(x)=Yf_1(x)$ or $f(x)=2Yf_1(x)$. Hence either $Y$ or $2Y$ is an annihilating content of $f$. Therefore $R$ is an $EM-\mathbb{Z}_2-$graded ring.
\end{exa}

\begin{thm}\label{t1} Let $R$ be $G-$graded ring. Then $R$ is an $EM-G-$graded ring if and only if $R_{\sigma}$ is an $EM-$subset of $R$ for all $\sigma\in G$.
\end{thm}
\begin{proof}Straightforward because $hZ(R[x])=\cup_{g\in G}(R_g[x]\cap Z(R[x]))$. \end{proof}

\begin{cor}\label{c1}
If $R$ is an $EM-G-$graded ring then $R_e$ is an $EM-$subring of $R$.
\end{cor}

\begin{thm}\label{t2} Let $R$ be a $G-$graded ring such that $R_e$ is an $EM$-ring. If for every $\sigma\in supp(R,G)$, there exists $u_\sigma\in R_\sigma$ such that $R_{\sigma}=R_eu_{\sigma}$ and $Ann_{R_e}(u_{\sigma})=\{0\}$, then $R$ is an $EM-G-$graded ring.
\end{thm}
\begin{proof}
Let $f(x)\in hZ(R[x])$. Then these exists $\sigma\in G$ such that $a_i\in R_{\sigma}$ for all $i$. Therefore $f(x)=u_{\sigma}h(x)$ for some $h(x)\in R_e[x]$. If $h(x)\in reg(R_e[x])$, then by Lemma \ref{l1}, $h(x)\in reg(R[x])$, and hence $u_{\sigma}$ is an annihilating content of $f$. Suppose $h(x)\in Z(R_e[x])$. Then there exist $c\in Z(R_e)$ and $h_1(x)\in reg(R_e[x])$ such that $h(x)=ch_1(x)$. So we have $f(x)=u_{\sigma}ch_1(x)$ where $u_{\sigma}c\in Z(R)$ and $h_1(x)\in reg(R[x])$. Hence $u_{\sigma}c$ is an annihilating content of $f(x)$.
\end{proof}

\begin{exa}\label{e2} The ring $R=\mathbb{Z}_6[x,y]/(xy)$ is not an $EM-$ring (see \cite[Example 3.11]{ab19}). Now $R$ is $\mathbb{Z}\times \mathbb{Z}-$graded by $R_{(0,0)}=\mathbb{Z}_6$, $R_{(i,0)}=\mathbb{Z}_6x^i$, $R_{(0,i)}=\mathbb{Z}_6y^i$ ($i\geq 1$), and $R_{(i,j)}=0$ otherwise. For each i,  $R_{(i,0)}=R_{(0,0)}x^i$, $R_{(0,i)}=R_{(0,0)}y^i$, and $Ann_{R_{(0,0)}}(x^i)=Ann_{R_{(0,0)}}(y^i)=\{0\}$. So by Theorem \ref{t2}, we get that $R$ is an $EM-\mathbb{Z}\times \mathbb{Z}-$graded ring.
\end{exa}

Let $R$ be any commutative ring (need not to be $G-$graded) and $n\geq2$. Then the ring $H=R[x]/(x^n)$ is $\mathbb{Z}_n-$graded by the gradation $H_k=Rx^k$, $k\in \mathbb{Z}_n$. For each $k\in \mathbb{Z}_n$, we have $H_k=Rx^k=H_0x^k$, and $Ann_R(x^k)=\{0\}$. So by Theorem \ref{t2}, we obtain following result

\begin{cor}\label{c6} Let $R$ be an $EM-$ring and $n\geq2$ be an integer. Then $H=R[x]/(x^n)$ is $EM-\mathbb{Z}_n-$graded by the gradation $H_k=Rx^k.$
\end{cor}

Let $R$ be a ring and $G$ be a group. Then the group ring $R[G]$ is $EM-G-$graded ring with gradation  $H_{\sigma}=R\sigma$, $\sigma\in G$. The following corollary also follows directly from Theorem \ref{t2}.

\begin{cor}\label{c66} Let $R$ be an $EM-$ring and $G$ be a group. Then $R[G]$ is $EM-G-$graded ring with gradation  $H_{\sigma}=R\sigma$, $\sigma\in G$.
\end{cor}

\begin{cor}\label{c2}
Let $R$ be a $G-$graded ring. If $(R,G)$ is crossed product over $supp(R,G)$ and $R_e$ is an $EM-$ring then $R$ is an $EM-G-$graded ring.
\end{cor}
\begin{proof} The result follows directly from Proposition \ref{p2} and Theorem \ref{t2}. \end{proof}

\begin{thm} Let $R$ be a ring with two gradations $(R,G)$ and $(R,G')$ such that $(R,G)$ is almost equivalent to $(R,G')$. If $R$ is $EM-G-$graded ring then $R$ is $EM-G'-$graded ring.
\end{thm}
\begin{proof}
Let $f(x)=\sum_{i=0}^{n}a_ix^i$ be a homogenous zero divisor of $R[x]$ under the gradation $(R,G')$. So $\sum_{i=0}^{n}\phi^{-1}(a_i)x^i$ a homogenous zero divisor of $R[x]$ under the gradation $(R,G)$. Hence there exists $c\in Z(R)$ and $g(x)\in reg(R[x])$ such that $\sum_{i=0}^{n}\phi^{-1}(a_i)x^i=cg(x)$. This implies that $\phi(c)$ is an annihilating content of $f$.
\end{proof} .

A ring $R$ is called Bezout ring if every finitely generated ideal is principal. $R$ is called Armendariz if the product of two polynomials on $R[x]$ is zero if and only if the product of their coefficients is zero. Abuosba and Ghanem \cite{ab19} showed that every Bezout ring is $EM-$ring and every $EM-$ring is Aremndariz. We extend these results into the notion of graded rings. We shall call a $G-$graded ring $R$, Bezout$-G-$graded if each finitely generated $G-$graded ideal is principal. In addition, we will say $R$ is  Armendariz$-G-$graded ring if the product of two polynomials in $hZ(R[x])$ is zero if and only if the product of their coefficients is zero.

\begin{cor}\label{c3} Every Bezout$-G-$graded ring is an $EM-G-$graded ring.
\end{cor}
\begin{proof} The result follows directly from Lemma \ref{l2} and Lemma \ref{l3}.
\end{proof}

\begin{thm}\label{t3} Every $EM-G-$graded ring is Aremndariz$-G-$graded.
\end{thm}
\begin{proof}Suppose $R$ is an $EM-G-$graded ring. Let $f(x)=\sum_{i=0}^na_ix^i$ and $g(x)=\sum_{i=0}^mb_ix^i$ be two polynomials in $h(R[x])$ such that $f(x)g(x)=0$. Since $R$ is an $EM-G-$graded, we get $f(x)=c_f\sum_{i=0}^{n_1}r_ix_i$ and $g(x)=c_g\sum_{j=0}^{m_1}s_jx^j$ where $c_fc_g=0$. Hence, for each $i,j$, we have $a_ib_j=c_fr_ic_gs_j=0$.
\end{proof}

\begin{thm}\label{t4} Let $R$ be an $EM-G-$graded ring and $S\subseteq h(R)$ be a multiplicatively closed subset of $R$. Then $S^{-1}R$ is $EM-G-$graded ring by the gradation $(S^{-1}R)_{\lambda}=\{\frac{a}{s}\mid s\in S, a\in R\text{ such that } \lambda=(deg(s))^{-1}deg(a)\}$.
\end{thm}
\begin{proof} Assume $f(x)\in hZ(S^{-1}R[x])$. Then $f(x)=\frac{h(x)}{t}$ for some $t\in S$ and some $h(x)\in hZ(R[x])$. Since $R$ is an $EM-G-$graded ring, there exist $c\in Z(R)$ and $g(x)=\sum_{i=0}^nb_ix^i\in reg(R[x])$ such that $h(x)=cg(x)$. So $f(x)=c\frac{g(x)}{t}$. Suppose that there exists $\frac{d}{k}\in S^{-1}R$ such that $\frac{d}{k}\frac{g(x)}{t}=0$. Then there exists $u\in S$ such that $udg(x)=0$. Since $g(x)$ is regular in $R[x]$ and $u\in S$, we get $d=0$. Hence $\frac{g(x)}{t}$ is regular in $S^{-1}R[x]$. Therefore $S^{-1}R[x]$ is an $EM-G-$graded ring.
\end{proof}

If $R$ is a commutative ring with unity, then $reg(R)$ is muliplicatively closed subset of $R$, and the localization $T(R)=(reg(R))^{-1}R$ is called the total quotient ring of $R$. Abuosba and Ghanem \cite{ab19} deduced that if $R$ is an $EM-$ring then so is $T(R)$. A similar result is obtain for $EM-G-$graded rings. Let $hreg(R)$ be the set of regular homogenous elements of $R$ and define $hT(R)=(hreg(R))^{-1}R$.

\begin{cor}\label{c4} If $R$ is $EM-G-$graded ring then so is $hT(R)$.\end{cor}

\begin{thm}\label{t5} Let $R$ be a $G-$graded ring. If $hT(R)$ is $EM-G-$graded ring then for every $f(x)\in hZ(R[x])$ there exists $c\in R$ such that $Ann_{R[x]}(f)=Ann_{R[x]}(c)$.
\end{thm}

\begin{proof} Let $f(x)\in hZ(R[x])$. Then $\frac{f(x)}{1}\in hZ(hT(R)[x])$. So there exist $\frac{c}{t}\in Z(hT(R)[x])$, and $g(x)\in reg(hT(R)[x])$ such that $\frac{f(x)}{1}=\frac{k}{t}g(x)$. So we have $u(tf(x)-kg(x))=0$ for some  $u\in hreg(R[x]).$ Since $u$, $t$, and $g(x)$ are regular, we get $Ann_{R[x]}(f)=Ann_{R[x]}(k)$.
\end{proof}

Let $R_{\alpha}$, $\alpha\in I$ be a family of $G-$graded rings. Then $R=\prod_{\alpha\in I}R_{\alpha}$ is $G-$graded by the gradation $R_g=\prod_{\alpha\in I}(R_{\alpha})_g$, $g\in G$. Next we generalize \cite[Theorem 3.12]{ab19}.

\begin{thm}\label{t6}Let $R_{\alpha}$, $\alpha\in I$ be a family of $G-$graded rings, and $R=\prod_{\alpha\in I}R_{\alpha}$. Then $R$ is $EM-G-$graded if and only if $R_{\alpha}$ is $EM-G-$graded for each $\alpha\in I$.
\end{thm}

\begin{proof} Suppose $R$ is $EM-G-$graded. Fix $j\in I$, and let $f(x)=\sum_{i=0}^na_ix^i\in hZ(R_j[x])$. Consider the polynomial $g(x)=\sum_{i=0}^nb_{\alpha,i}x^i\in R[x]$, where for each $i$, $b_{j,i}=a_i$, and $b_{\alpha,i}=0$, for all $\alpha\neq j$. Since all $a_i's$ are in the same component of the $G-$grading of $R_{\alpha}$, we get $g(x)\in hZ(R[x])$. Hence there exist $c=(c_{\alpha})\in Z(R)$ and $g_1(x)=\sum_{i=0}^m (d_{\alpha,i})x^i \in reg(R[x])$ such that $g(x)=cg_1(x)$, and $m\geq n$. Then $f(x)=c_j\sum_{i=0}^md_{j,i}x^i$. It is clear that $c_j\neq0,$ because $0\neq a_n=c_jd_{j,n}$. Assume $yd_{j,i}=0$ for all $i$. Let $q=(q_{\alpha})$ where $q_j=y$ and $q_{\alpha}=0,$ when $\alpha\neq j$. Then $(q_{\alpha})(d_{\alpha,i})=0$ for all $i$. So $y=0$ and hence $\sum_{i=0}^n d_{j,i}x^i \in reg(R_{\alpha}[x])$.

For the converse assume $R_{\alpha}$ is $EM-G-$graded for all $\alpha\in I$. Let  $f(x)=\sum_{i=0}^n(a_{\alpha,i})x^i\in hZ(R[x])$. Then there exists $(b_{\alpha})\in R[x]$ such that $(b_{\alpha})(a_{\alpha,i})=(0)$ for all $i$. For each $\alpha\in I$, let $f_{\alpha}(x)=\sum_{i=0}^na_{\alpha,i}x^i$ and let $J=\{\alpha\in I\mid f_{\alpha}\notin reg(R_{\alpha})\}$. So, for each $\alpha\in J$, we have $f_{\alpha}(x)\in hZ(R_j[x])$, and hence $f_{\alpha}=c_{\alpha}g_{\alpha}(x)$ for some $c_{\alpha}\in Z(R_{\alpha})$ and some $g_{\alpha}(x)\in reg(R_{\alpha}[x])$. Now for $\alpha\notin J$, let $c_{\alpha}=1$ and $g_{\alpha}(x)=f_{\alpha}(x)$. Hence we have $f(x)=(c_{\alpha})(g_{\alpha}(x))$, where $(c_{\alpha})\in Z(R)$ and $(g_{\alpha}(x))\in reg(R[x])$ as desired.
\end{proof}

\begin{thm}\label{t7} If $R$ is an $EM-G-$graded ring then so is $R[x]$.
\end{thm}
\begin{proof} Let $f(x)=\sum_{i=0}^nf_i(x)y^i\in hZ(R[x,y])$. Then there exists $h(x)\in R[x]$ such that $h(x)f_i(x)=0$ for all $i$. Let
\[g(x)=f_0(x)+f_1(x)x^{deg(f_0)+1}+f_2(x)x^{deg(f_0)+deg(f_1)+2}+\dots+f_n(x)x^{\sum_{i=0}^{n-1}deg(f_i)+n}.\]
So $hg=0$. Moreover, since $f(x,y)\in h(R[x,y])$, there exists $\lambda\in G$ such that $R_{\lambda}$ contains all coefficients of $f_i(x)$, for all $i$. Hence $g(x)\in hZ(R[x])$. Thus there exists $c\in Z(R)$ and $g_1(x)=\sum_{i=0}^mb_ix^i\in reg(R[x])$ such that $g(x)=cg_1(x)$. So $f_0(x)=c\sum_{i=0}^{deg(f_0)}b_ix^i=cw_0(x)$, $f_1(x)=c\sum_{i=0}^{deg(f_1)}b_{i+deg(f_0)+1}x^i=cw_1(x)$, and so on. Thus $f(x,y)=c\sum_{i=0}^{n}w_i(x)y^i$, and since $\cap Ann(b_i)=0$, we have $\sum_{i=0}^{n}w_i(x)y^i\in reg(R[x,y])$. Therefore $R[x]$ is $EM-G-$graded.
\end{proof}

\begin{cor}\label{c5} Let $R$ be an $EM-G-$graded ring and $k$ be a positive integer. Then $R[x_1,x_2,\dots,x_k]$ is an $EM-G-$graded ring.
\end{cor}

\begin{rem}\label{r1} If $R$ is a $G-$graded ring then  $(x^2)$ is a $G-$graded ideal of $R[x]$. So $R[x]/(x^2)$ is a $G-$graded ring with gradation $(R[x]/(x^2))_g=R_g[x]+(x^2)/(x^2)$ i.e. $a+bX\in h(R[x]/(x^2))$ if and only if there exists $\sigma\in G$ such that $a,b\in R_\sigma$.
\end{rem}

\begin{thm}\label{t8} Let $R$ be $G-$graded such that $h(R)\cap Z(R)=\{0\}$. Then $H=R[x]/(x^2)$ is an $EM-G-$graded ring.
\end{thm}
\begin{proof}
Let $f(y)=\sum_{i=0}^n(a_i+b_iX)y^i\in hZ(H[y])$. Then by the argument in Remark \ref{r1} we get that $a_i,b_i\in h(R)$ and $a_i\in Z(R)$ for all $i$. Hence $a_i=0$ for all $i$. So we have $f(y)=X\sum_{i=0}^nb_iy^i$, which yields $X$ is an annihilating content for $f$. Thus $H$ is an $EM-G-$graded ring.
\end{proof}

\begin{thm}\label{t9} Let $R$ be a $G-$graded ring. If $H=R[x]/(x^2)$ is an $EM-G-$graded ring, then so is $R$.
\end{thm}
\begin{proof}
Let $f(y)=\sum_{i=0}^nr_iy^i\in hZ(R[y])$. Then $f(y)\in hZ(H[y])$. Therefore $f(y)=(c+dX)\sum_{i=0}^{m}(a_i+b_iX)y^i$, with $\cap_{i=0}^mAnn(a_i+b_iX)=\{0\}$. So we have $\cap_{i=0}^mAnn(a_i)=\{0\}$ and $r_i=ca_i$ for each $i$. Hence $f(y)=c\sum_{i=0}^na_iy^i$. Therefore $R$ is an $EM-G-$graded ring.
\end{proof}

\begin{thm}\label{t10} Let $R$ be $G-$graded ring such that for each $a\in h(R)$ there exists $b\in E(R)$ such that $ann(a)=bR$. Then $R$ is an $EM-G-$graded ring.
\end{thm}
\begin{proof} Let $f(x)=\sum_{i=0}^na_ix^i\in hZ(R[x])$. Then $a_i\in h(R)$ for all $i$. So for each $i$ there exists $b_i\in E(R)$ and $u_i\in reg(R)$ such that $a_i=u_ib_i$. Let $b=1-\prod_{i=0}^n(1-b_i)$. Clearly $1-b\in E(R)$, and so $b\in E(R)$. Since $b_i(1-b_i)=0$, we get $b(b_i+1-b)=bb_i=b_i-b_i\prod_{j=0}^n(1-b_j)=b_i$ and $1=\sum_{i=0}^n(b_i+1-b)-\prod_{i=0}^n(b_i+1-b)$. Hence $Ann(b_0+1-b,b_1+1-b,\dots,b_n+1-b)=\{0\}$ and for each $i$, $a_i=bu_i(b_i+1-b).$  Thus $f(x)=bg(x)$ where $b\in Z(R)$ and $g(x)=\sum_{i=0}^nu_i(b_i+1-b)x^i\in reg(R[x])$. Therefore $R$ is $EM-G-$graded ring.
\end{proof}

Let $R$ be a ring and $M$ be an $R-$module. Then the idealization $R(+)M$ is the ring whose elements are those of $R\times M$ equipped with addition and multiplication defined by $(r_1,m_1)+(r_2,m_2)=(r_1+r_2,m_1+m_2)$ and $(r_1,m_1)(r_2,m_2)=(r_1r_2,r_1m_2+r_2m_1)$ respectively. The annihilator of the module $M$ in $R$ is the set $Ann_R(M)=\{r\in R\mid rM=\{0\}\}$. It is well known that the idealization $R(+)M$ is $\mathbb{Z}_2-$graded by the gradation $(R(+)M)_0=R\oplus 0$ and $(R(+)M)_1=0\oplus M$.

\begin{thm}\label{t11} Let $R$ be a ring and $M$ be an $R-$module such that $Ann_R(M)=\{0\}$. If $R(+)M$ is an $EM-\mathbb{Z}_2-$graded by the gradation $H_0=R\oplus 0$ and $H_1=0\oplus M$, then $R$ is an $EM-$ring.
\end{thm}
\begin{proof} Suppose $R(+)M$ is an $EM-\mathbb{Z}_2-$graded by the gradation $H_0=R\oplus 0$ and $H_1=0\oplus M$. Let $f(x)=\sum_{i=0}^na_ix^i\in Z(R[x])$. Then $Ann_{R(+)M}\{(a_i,0)\}\neq\{0\}$. So $\sum_{i=0}^n(a_i,0)=x^i=(c,m)\sum_{i=0}^k(b_i,m_i)x^i$ with $k\geq n$ and $Ann_{R(+)M}\{(b_i,m_i)\}=\{0\}$. Suppose $0\neq w\in Ann_R\{\alpha_i\}$. Since $Ann_R(M)=\{0\}$, there exists $t\in M$ such that $wt\neq0$. However, since $(0,wt)(b_i,m_i)=(0,wtb_i)=(0,0)$ for all $i$, we get $(0,wt)\in Ann_{R(+)M}\{(b_i,m_i)\}=\{0\}$, a contradiction. Therefore $f(x)=c\sum_{i=0}^nb_ix^i$ with $Ann_R\{b_i\}=\{0\}$ as desired.
\end{proof}

Ganam and Abuosba in \cite{gh18} proved that if $R(+)R$ (equivalently $R[x]/(x^2)$) is an $EM-$ring then so is $R$. However the converse of this  result is not true. From Theorem \ref{t2} and Theorem \ref{t11} we obtain the following result.

\begin{cor}\label{c7} A ring $R$ is an $EM-$ring if and only if $H=R(+)R$ is an $EM-Z_2-$graded ring with gradation $H_0=R\oplus 0$ and $H_1=0\oplus R$.
\end{cor}

\section{Further Questions}\label{sec3}

In this section we highlight two problems that may be of interest for future research.

Several results on $EM-$rings can be extended to $EM-G-$graded rings if we can show that every homogenous zero divisor polynomial has at least one homogenous annihilating content. The rings discussed in Theorem \ref{t2}, Corollary \ref{c6}, and Corollary \ref{c66} have this property, however we don't know if this always the case. Based on this observation we ask the following question.

\begin{ques} In an $EM-G-$graded ring $R$, is it guaranteed that every polynomial in $hZ(R[x])$ has an annihilating content that belongs to $hZ(R)$? If not, then what kinds of graded rings have this property?
\end{ques}

The second problem is related to the notion of strongly $EM$ rings that was defined in \cite{ab19}. An $EM$ ring $R$ is called strongly $EM$ ring if every zero divisor power series has an annihilating content. If $R$ is $G-$graded ring then $R[[x]]$ is $G-$graded by the gradation $(R[[x]])_g=R_g[[x]]$, $g\in G$. We define an $EM-G-$graded ring $R$ to be strongly $EM-G-$graded ring if every nonzero homogeneous power series in $R[[x]]$ has an annihilating content. It is not guaranteed that if every power series in $hZ(R[[x]])$ has an annihilating content then every polynomial in $hZ(R[x])$ has an annihilating content. So similar to a question from \cite{ab19}, we ask the following question.

\begin{ques} What types of graded rings have the property that if every power series in $hZ(R[[x]])$ has an annihilating content then every polynomial in $hZ(R[x])$ has an annihilating content?
\end{ques}

The following theorem describes one of these types.

\begin{thm}\label{t12}  Let $R$ be a $G-$graded ring such that $Z(R[[x]])$ is an ideal of $R[[x]]$. Then $R$ is strongly $EM-G-$ring if and only if every homogeneous power series in $R[[x]]$ has an annihilating content.
\end{thm}
 \begin{proof} Suppose that every homogeneous power series in $R[[x]]$ has an annihilating content. Let $f(x)=\sum_{i=0}^{n}a_ix^i\in hZ(R[[x]])$. Then $f(x)=c_ff_1$ for some $c_f\in Z(R)$ and $f_1(x)=\sum_{i=0}^{\infty}b_ix^i\in reg(R[[x]])$. Thus we have, $a_i=c_fb_i$, for $i=0,1,\dots, n$, and $0=c_fb_i$, for $i\geq n+1$. So $\sum_{i=n+1}^{\infty}b_ix^i\in Z(R[[x]])$. Since $Z(R[[x]])$ is an ideal of $R[[x]]$ and $f_1(x)\in reg(R[[x]])$, we have $\sum_{i=0}^{n}b_ix^i\in reg(R[[x]])$. Thus $f(x)=c_f\sum_{i=0}^{n}b_ix^i$, where $\sum_{i=0}^{n}b_ix^i\in reg(R[x])$. Therefore $R$ is $EM-G-$graded.
\end{proof}


TITLE: Quotient of a PID by a prime ideal is a PID too.
QUESTION [1 upvotes]: I am trying to prove that if $R$ is a PID and $P$ is a prime ideal of $R$ then $R/P$ is also a PID.
This is question 8.2.3 from Dummit & Foote.
My thoughts and what I know : $P$ is also maximal. $R/P$ is hence a field. Also is an integral domain due $P$ being prime. $P$ is generated by by some element of $R$, say $a$, written as $P=(a)$. My conjecture is that an ideal $I$ of $R/P$ is generated by the coset $a+P$, ie. $I=(a+P)$.
I am having some trouble understanding that the set $I$ is equal to a set generated by a coset. I'm not sure how to show the inclusions to show that those two sets are the same. Any help is appreciated. Thanks for your time.

REPLY [0 votes]: As an alternative to the other given answer, you can split this into two parts.
First show that in any quotient of a PID all ideals are principal (just take the image of a generator under the canonical homomorphism).
Next, the quotient by a prime ideal is always an integral domain.
Putting these together, we see that if $R$ is a ring where all ideals are principal and $P$ is a prime ideal of $R$ then $R/P$ is a PID (so we could even relax the condition on the original ring a bit).


TITLE: Set Theory - Ordinal Exponents - Arithmetic
QUESTION [4 upvotes]: So I've run into an issue with simplifying the statement, with ω being the first infinite ordinal:
$(ω2)^ω$
Now in the case of a finite ordinal $a$ as an exponent, it is easy to inductively prove that
$(ω2)^a = (ω^a)2$
However, this is a little unclear as to what happens with the infinite exponent.
So which of the following would be true and why:
$(ω2)^ω = (ω^ω)2$
-or-
$(ω2)^ω = ω^ω$   ?
We could consider $(ω2)^ω$ as either
$ω(2ω)(2ω)...2$
(which in some way seems like nonsense as there is no "last" term) or as
$ω(2ω)(2ω)...$
The problem is that we are oscillating back and forth between $ω$ and $2$.  Does this even have a clearly defined value?

REPLY [0 votes]: In general. If $k$ is a limit ordinal then $\sup_{a<k}b^a=b^k.$
Proof: $b^a$ is defined as the order-type of the set of functions $f:a\to b$ such that $\{c\in b: f(c)\ne 0\}$ is finite. Denote this set by $(b^a)^*.$
The defined order $<^*$ on $(b^a)^*$ is reverse-lexicographic: For non-empty $a,b$ and distinct $f,g \in (b^a)^*$ let $c=\max \{d:f(d)\ne g(d)\}$. Define  $f<^*g \iff f(c)<g(c).$
For the non-trivial case $b\ne 0\ne k,$ let $U=\sup_{a<k}b^a=\cup_{a\in k}b^a.$
(i).  Clearly $x\in U\implies \exists a\in k\;(x\in b^a\leq b^k)\implies x\in b^k.$ So $U\leq b^k.$
(ii). To show that $U\geq b^k,$ let $\psi : [b^k,\epsilon]\to [(b^k)^*,<^*]$ be the (unique) order-isomorphism. For any $x\in b^k$ take any $a\in k$ such that $\sup \{c\in k:\psi(x)(c)\ne 0\}<a.$ Then $x$ is less than the order-type of $\{g\in (b^k)^*: \{c:g(c)\ne 0\}\subset a.\}$
But this last set is order-isomorphic to $(b^a)^*,$ which is order-isomorphic to $b^a.$  So $x<b^a$ for some $a\in k.$ That is, $x\in b^k\implies x\in U.$
In your Q, for $n<\omega$ we have $\omega^n \leq (\omega 2)^n<\omega^{\omega}.$ Therefore $\omega^{\omega}=\sup_{n<\omega}\omega ^n\leq \sup_{n<\omega}(\omega 2)^n=(\omega 2)^{\omega}\leq \omega^{\omega}.$


TITLE: Evaluating integral similar to $\sin^{-1}(3x)$
QUESTION [2 upvotes]: I was just doing revision for an upcoming exam and I came across a question I do not know how to answer. It seems pretty simple, however I am having a blank moment. If anyone knows how to solve that I would really appreciate the help.
The question:
It is given that y = $\sin^{-1}(ax)$, where a is a positive constant and $-\frac{1}{a}\leq x\leq \frac{1}{x}$.

a) Find $\frac{dy}{dx}$ in terms of y and a.
b) Hence find $\frac{dy}{dx}$ in terms of x and a.
c) Hence, or otherwise, find $\int \frac{1}{\sqrt{1-9x^{2}}}dx$.
d) Find the exact value of $\int_{0}^{1/6}\sqrt{1-9x^{2}} dx$

I have finished parts a to c (workings below) but cannot seem to solve part d (I do realize it's probably relatively simple, however I have just spent good 45 minutes trying to solve it or find the answer and I can't seem to solve it)
$$
y=\sin^{-1}(ax)\\\sin(y)=ax\\\frac{dy}{dx}\cos(y)=a\\\Rightarrow\frac{dy}{dx}=\frac{a}{\cos(y)}=\frac{a}{\sqrt{1-\sin^{2}(y)}}=\frac{a}{\sqrt{1-a^{2}x^{2}}}\\\int\frac{1}{\sqrt{1-9x^{2}}}dx = \frac{1}{3}\int{\frac{3}{\sqrt{1-9x^{2}}}}dx=\frac{1}{3}\sin^{-1}(3x)+c$$
Any help would be greatly appreciated.

REPLY [0 votes]: use this formula
$$\int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}\arcsin \frac{x}{a}+C$$
$$\int_{0}^{1/6}\sqrt{1-9x^{2}} dx$$
$$\left(\frac{3x}{6}\sqrt{1-9x^{2}}+\frac{1}{6}\arcsin\frac{3x}{1} \right)_{0}^{1/6}$$
$$\left(\frac{x}{2}\sqrt{1-9x^{2}}+\frac{1}{6}\arcsin\frac{3x}{1} \right)_{0}^{1/6}$$
$$(\frac{1}{12}\sqrt{1-\frac{9}{36}}+\frac{1}{6}\arcsin\frac{1}{2})-0$$
$$\frac{\sqrt{27}}{72}+\frac{\pi}{36}$$
$$\dfrac1{72} \left(2 \pi + 3 \sqrt{3} \right)$$


TITLE: Why is this map continuous? Group operation on étalé space of a sheaf
QUESTION [0 upvotes]: Let $X$ be some topological space and suppose we have a presheaf of abelian groups, $F\colon \mathrm{\textbf{Op}}(X) \to \textbf{Ab}$. Construct the corresponding étalé space as follows:
a) For each $x \in X$ set $\mathscr{F}_x \colon= \mathrm{Lim}_{U \ni x}
 \, F(U)$ and let $\phi_{x}^{U}$ be the homormorphism from $F(U)$ into the direct limit. Set $\mathscr{F} = \sqcup \mathscr{F}_x$.
b) For $f \in F(U)$ set $$[f,U] = \{ (\phi_{x}^{U}(f),x) | x \in U \}.$$ One can check that the sets $[f,U]$ for $U$ open in $X$ and $f \in F(U)$ form a basis for topology on $\mathscr{F}$, so give $\mathscr{F}$ the topology generated by them.
Let $\pi$ be the projection from $\mathscr{F} \to X$ that simply takes every element of $\mathscr{F}_x$ to $x$. One can check that $\pi$ is a local homeomorphism. One can also check that the map $(f,x) \mapsto (-f,x)$ is continuous from $\mathscr{F}$ to $\mathscr{F}$.

Now let $\mathscr{F} + \mathscr{F}$ represent the subspace of
  $\mathscr{F} \times \mathscr{F}$ consisting of pairs $((f,x), (g,x))$
  for some $x \in X$. How can we confirm that the map $((f,x),(g,x))
 \mapsto (f+g,x)$ is continuous from $\mathscr{F} + \mathscr{F}$ to
  $\mathscr{F}$?

For context, this construction is a functor from presheaves to category of  étalé spaces and is one half of the composition of functors that yields sheaffification, and the equivalence of categories. My construction above is verbatim as in Serre's \textit{FAC}. I can't figure out how to show that group operation is continuous, however.

REPLY [1 votes]: How about this, we will shoot for pointwise continuity.
Let $\psi$ be the map from $\mathscr{F} + \mathscr{F}$ to $\mathscr{F}$ that takes $((f,x),(g,x))$ to $(f+g,x)$.
Let $((f,x),(g,x)) \in \mathscr{F} + \mathscr{F}$ and let $[t,U]$ be any basis element of $\mathscr{F}$ containing the image $\psi((f,x),(g,x)) = (f+g,x)$. Usually we would consider an arbitrary open set $V$ containing $(f+g,x)$, but it is sufficient to show it for a basis element.
Now since $(f+g,x) \in [t,U]$ this means that $x \in U$ and $\phi_{x}^{U}(t) = f+g$. By properties of the direct limit we know there exists open $V$ containing $x$ with some $s_1 \in F(V)$ such that $\phi_{x}^{V}(s_1) = f$, likewise we have an open $W$ containing $x$ with an $s_2$ in $F(W)$ such that $\phi_x^{W}(s_2) = g$.
This means that
\begin{align*}
f+g &= \phi^{V \cap W}_{x} \circ \phi^{V}_{V \cap W}(s_1) + \phi^{V \cap W}_{x} \circ \phi^{W}_{V \cap W}(s_2)\\
&= \phi^{V \cap W}_{x}( \phi^{V}_{V \cap W}(s_1) + \phi^{W}_{V \cap W}(s_2)).
\end{align*}
Again by properties of the direct limit, this means that there exists some open $O$ containing $x$ such that $O$ is contained in $U$ and contained in $V \cap W$ such that
\begin{align*}
\phi^{U}_{O}(t) &= \phi^{V \cap W}_{O}(\phi^{V}_{V \cap W}(s_1)+\phi^{W}_{V \cap W}(s_2))\\
&= \phi^{V}_{O}(s_1) + \phi^{W}_{O}(s_2).
\end{align*}
This means that $[\phi^{V}_{O}(s_1) + \phi^{W}_{O}(s_2),O] = [\phi^{U}_{O}(t),O] \subset [t,U]$
So now take $[\phi^{V}_{O}(s_1),O] \times [\phi^{W}_{O}(s_2),O] \cap \mathscr{F} + \mathscr{F}$ which is open. As a set we have $$[\phi^{V}_{O}(s_1),O] \times [\phi^{W}_{O}(s_2),O] \cap \mathscr{F} + \mathscr{F} = \{(\phi^{O}_{x}(\phi^{V}_{O}(s_1)),x),(\phi^{O}_{x}(\phi^{W}_{O}(s_2)),x) : x \in O \}$$
The image of the set under $\psi$ is $[\phi^{V}_{O}(s_1) + \phi^{W}_{O}(s_2),O]$.


TITLE: Is the Lebesgue decomposition only meaningful for measures which are supported on different sets?
QUESTION [1 upvotes]: Lately I've been trying to understand why we concern ourselves with the Lebesgue decomposition. The only satisfactory answer I can think of is that it allows us to work more easily with "weird" measures. For example, say we have a measurable space $(X,\mathcal{F})$, a positive measure $\nu$, a signed measure $\lambda$ and two signed measures $\lambda_{a}$ and $\lambda_{s}$ such that $\lambda=\lambda_{a}+\lambda_{s}$, $\nu \perp \lambda_{s}$ and $\lambda_{a} \ll \nu$. Say $\mu$ is the Lesbesgue measure and $X=\mathbb{R}$ and for $E \in \mathcal{F}$ we have
$$
\nu(E)=\int_{E}g \, d\mu
$$
and
$$
\lambda(E)=\int_{E} f \, d\mu
$$
where
$$
g(x)=
\begin{cases}
1 & \text{ if }x \in (-\infty,0] \\ 0 & \text{ otherwise}
\end{cases}
$$
and
$$
f(x)=\begin{cases}
x^{2} & \text{ if }x \in (-\infty,1] \\ 0 & \text{ if } x \in (1,\infty)
\end{cases}$$
which can be decomposed into $f_{1}+f_{2}$ given by
$$
f_{1}(x)=\begin{cases}
x^{2} & \text{ if }x \in (-\infty,0] \\ 0 & \text{ otherwise}
\end{cases}$$
and
$$
f_{2}(x)=\begin{cases}
x^2 & \text{ if }x \in (0,1] \\ 0 & \text{ otherwise}
\end{cases}
$$
then define
$$
\lambda_{a}(E)=\int_{E} f_{2}\, d\mu
$$
and
$$
\lambda_{s}(E)=\int_{E}f_{1} \, d\mu
$$
clearly we have that $\nu \perp \lambda_{s}$ and $\lambda=\lambda_{a}+\lambda_{s}$ and also that $\lambda_{a} \ll \nu$. So is this only useful when we consider decompositions w.r.t. measures with "weird" supports? For example consider the same example but where
$$
\nu(E)=\int_{X} \mathbb{1}_{E} \, d\mu
$$
or simply that $\nu(E)=\mu(E)$. Then absolute continuity for $f$ becomes easy since any set $E$ such that $\nu(E)=0$ we will have
$$
\lambda(E)=\int_{E} f \, d\mu=0
$$
and then we have that $\lambda=\lambda_{a}+0$ since the Lebesgue measure is supported over all of the real numbers. Would this also be true for any measure supported on $\mathbb{R}$? I'm just having a hard time seeing where this fits in to what I've been studying. I'm somewhat new to measure theory so maybe I just lack the experience but I also worry that I am missing a key piece of insight.

REPLY [1 votes]: For your first question:

If two measures share the same support then the singular component of the Lebesgue decomposition of one measure w.r.t. to another measure is always zero?

No, this is not necessarily true. Let $m$ denote Lebesgue measure on $[0,1]$ and $\mu$ be a Dirac measure on $[0,1]$ for the point $x = 1/2$. That is
$$ \mu(A) = \left\{ \begin{array}{cl}
1 & \text{ if } 1/2 \in A \\
0 & \text{ if } 1/2 \notin A
\end{array} \right.,\  A \in [0,1] \text{ Borel}. $$
Define $\nu := m + \mu$ on the Borel sets of $[0,1]$. What is the Lebesgue decomposition of $\nu$ with respect to $m$?
For for your second question:

If $\nu$ and $\mu$ are measures and we want to find the Radon Nikodym derivative of $\nu$ w.r.t. $\mu$ then the support of $\nu$ can be strictly smaller than that of $\mu$ and still have absolute continuity, right?

Yes. Suppose $\mu$ denotes Lebesgue measure on $\mathbb{R}$ and let $\nu(A) := \mu(A\cap [0,1])$ for all Borel $A\subseteq \mathbb{R}$. Then if $\mu(A) = 0$, $\nu(A) \leq \mu(A) = 0$. Hence $\nu \ll \mu$.


TITLE: Quasi-Linear PDE
QUESTION [0 upvotes]: I'm trying to solve the following PDE:
$$\begin{align}xu(x, y)\partial_xu(x, y) + yu(x, y)\partial_yu(x, y) &= x^2 + y^2 + (u(x, y))^2\\
u(1, y) &= y^2\end{align}$$
I wrote it in the form $$xv\partial_xv + yv\partial_yv + x^2+y^2+v^2 = 0$$
then I wrote
$$\begin{align}x'(t) &= x(t)v(t)\\
y'(t) &= y(t)v(t)\\
v'(t) &= x(t)^2 + y(t)^2+ v(t)^2\end{align}$$
I need two first integrals; one was easy by taking $x'y - y'x = 0$ and I got
$$\Phi_1(x,y) = -x(t)y(t)$$ but I'm not succeeding finding the second, can anyone help please? thank you.

REPLY [0 votes]: Calling $v = u^2$ we have
$$
xv_x+y v_y = 2(v+x^2+y^2),\ \ \ v(1,y) = y^4
$$
now changing variables according
$$
\cases{
r^2= x^2+y^2\\
\theta=\frac yx
}
$$
we have $v(x,y) = w(r,\theta)$
and
$$
 r w_r - 2w=2r^2
$$
this equation is solved as a ODE and has as solution
$$
w(r,\theta) = 2r^2\ln r+r^2\phi(\theta)
$$
now considering the condition
$$
v(1,y) = w\left(\sqrt{1+y^2}, y\right)=y^4\Rightarrow \phi\left(\frac yx\right) = \frac{y^4}{x^2 \left(x^2+y^2\right)}-\ln \left(1+\frac{y^2}{x^2}\right)
$$
and finally
$$
v(x,y) = u^2(x,y) = \frac{y^4}{x^2}+\left(x^2+y^2\right) \ln \left(x^2+y^2\right)-\left(x^2+y^2\right) \ln
   \left(1+\frac{y^2}{x^2}\right) = \frac{y^4}{x^2}+(x^2+y^2)\ln(x^2)
$$


TITLE: Is there always a non-abelian group of order $3q$ where $q$ is not squarefree?
QUESTION [6 upvotes]: Suppose $p = 3q$ and $q$ is not square-free, how do I show that there is always a non-abelian group of order p? Should I construct it using semi-direct product?

REPLY [5 votes]: Semidirect products will certainly do it.
It suffices to show that there are always nonabelian groups of order $3p^2$ with $p$ a prime. Indeed, if this is true and $q$ is not square free, then we can write $q=p^2r$ for some prime $p$ and some integer $r$; then taking $G\times C_r$, where $G$ is nonabelian of order $3p^2$ and $C_r$ is cyclic of order $r$ will give you a nonabelian group of order $3p^2r=3q$.
For $p=2$, we can take $A_4$, the alternating group of degree $4$, which has order $12$.
For $p=3$, we can take the Heisenberg group of $3\times 3$ upper triangular matrices with $1$s in the diagonal and coefficients in $\mathbb{F}_3$, which is a nonabelian group of order $27$.
For $p\gt 3$, we have two cases: $p\equiv 1\pmod{3}$ and $p\equiv 2\pmod{3}$. For the first one, we use the following Lemma:
Lemma. Let $n$ be a positive integer greater than $1$, and let $C_n$ be the cyclic group of order $n$. Then $C_n$ has an automorphism of order $3$ if and only if $3|\phi(n)$, where $\phi$ is Euler's $\phi$ function.
Proof. If we identify $C_n$ with the additive group of integers modulo $n$, then every automorphism corresponds to a map $1\mapsto r$ where $\gcd(r,n)=1$. The order is the smallest $k$ such that $r^k\equiv 1\pmod{n}$. This is equivalent to looking at the multiplicative group of units modulo $n$, and finding the order of $r$ in that group. The multiplicative group of units modulo $n$ has order $\phi(n)$, hence has an element of order $3$ if and only if $3$ divides $\phi(n)$, by Lagrange's Theorem (for the "only if" clause) and Cauchy's Theorem (for the "if" clause). $\Box$
If $p\equiv 1\pmod{3}$, then $C_p$ has an automorphism of order $3$ (since $\phi(p) = p-1$ is a multiple of $3$); this means that there is a group of order $3p$ that is nonabelian (a semidirect product $C_p\rtimes C_3$); hence we can construct a nonabelian group of order $3p^2$ as $(C_p\rtimes C_3)\times C_3$.
So we are reduced to the case in which $p\gt 3$ and $p\equiv 2\pmod{n}$. We cannot do what we did above, because all groups of order $3p$ are abelian (in fact, cyclic), which can be proven using the Sylow Theorems and a bit of extra work; neither can we get a nonabelian group of order $3p^2$ by using $C_{p^2}$, because, again by Sylow's Theorems it would have to be a semidirect product, but $C_{p^2}$ has no automorphism of order $3$, so again we get just the cyclic group of order $3p^2$.
So instead we will use a semidirect product with normal subgroup $C_p\times C_p$.  Consider the group $C_p\times C_p$; write $p=3k+2$; we are looking for an automorphism of order $3$.  Look at the map $C_p\times C_p\to C_p\times C_p$ given by $(a,b)\mapsto (a,b+(k+1)a)$. This is indeed an automorphism (easy check), and the order is exactly $3$ (again, easy check). This automorphism lets us construct a nonabelian group $(C_p\times C_p)\rtimes C_3$ of order $3p^2$.
So we have constructed a group of order $3p^2$ for every prime $p$, and we are done.


TITLE: A combinatorics problem refer to this problem?
QUESTION [0 upvotes]: If i define $f(m,n)=$ $$\sum_{1\leq k\leq mn}\left\{ \frac{k}{m}\right\} \left\{ \frac{k}{n}\right\} .$$
Then prove $$f(m+n,n) - f(m,n) =\frac{n^2-n}{4}$$
for all $m$ and $n$.
This question came from part of answer from this question: A sum of fractional parts.

REPLY [3 votes]: Since $k$ goes from $1$ to $mn$, then the pairs $((k \bmod m),(k \bmod n))$ will meet all cases once.
thus the product is
$$\sum_{1\le k\le m}\left\{\frac{k}{m}\right\}\sum_{1\le t\le n}\left\{\frac{t}{n}\right\}=\frac{m-1}{2}\frac{n-1}{2}\;.$$
Done.


TITLE: Give some example of linear functional?
QUESTION [3 upvotes]: Give  some example  of linear functional ?
My attempt : I know  the definition of linear functional
linear  functional  on $V$ is a  function $ f : V \to F$ such that
$f(a_1 v_1 + a_2v_2)=a_1f(v_1) + a_2f(v_1)$ for all $ a_1,a_2 \in F$ and all $v_1,v_2 \in V$
Here   i don't  know how to find the example of linear functional
But i can find the example of linear transformation  take  $f : \mathbb{R}^3 \to \mathbb{R}^3$ defined
$f(x,y,z)= (x+y,y+z,z+x)$

REPLY [1 votes]: Common examples are integrals.  Another (apparently different) example is differentiation.  Take as your space the collection of (once) differentiable functions on some open interval, say $(0,1)$.  Then use linearity of the derivative:
$$  \frac{\mathrm{d}}{\mathrm{d}x} \left( a f(x) + b g(x) \right) = a \frac{\mathrm{d}f(x)}{\mathrm{d}x} + b \frac{\mathrm{d}g(x)}{\mathrm{d}x}  \text{.}  $$
If you demand finite dimensionality, use polynomials of bounded degree.


TITLE: Almost hermitian manifolds with closed symplectic form.
QUESTION [4 upvotes]: An hermitian manifold $M$ is a manifold equipped with a Riemannian metric $g$, a non-degenerate two-form $\omega$, and an almost complex structure $J$ such that
$\omega(\cdot,\cdot) = g(J\cdot,\cdot)$
$(M,g,J,\omega)$ is a Kahler manifold if and only if $J$ is integrable, namely it is a complex structure on $M$, and $\omega$ is closed, namely it is a symplectic form on $M$. What happens if we relax the condition of $J$ being integrable but we still demand $\omega$ to be closed? Does this kind of manifolds have a name? Do they enjoy interesting geometric properties as the more constained Kahler manifolds do?
Thanks.

REPLY [5 votes]: The naming system is as follows:

$J$ not integrable, $\omega$ not closed: Almost Hermitian
$J$ integrable, $\omega$ not closed: Hermitian
$J$ not integrable, $\omega$ closed: Almost Kahler
$J$ integrable, $\omega$ closed: Kahler