text:
\begin{document}

\title{\textbf{Decomposing Gorenstein Rings as Connected Sums}}

\author[H. Ananthnarayan]{H. Ananthnarayan}
\address{Department of Mathematics, I.I.T. Bombay, Powai, Mumbai 400076.}
\email{ananth@math.iitb.ac.in}

\author[E. Celikbas]{Ela Celikbas}
\address{Department of Mathematics, University of Missouri, Columbia, MO 65211.}
\email{celikbase@missouri.edu}

\author[Z. Yang]{Zheng Yang}
\address{Department of Mathematics, University of Nebraska, Lincoln, NE 68588.}
\email{s-zyang8@math.unl.edu}

\date{\today}

\subjclass[2010]{Primary 13A30, 13D40, 13H10}

\keywords{Gorenstein ring, local ring, associated graded ring, fibre product, connected sum, short ring, stretched ring,
Gorenstein up to linear socle.}

\begin{abstract}
In \cite{AAM}, Ananthnarayan, Avramov and Moore
give a new construction of Gorenstein rings from two Gorenstein local rings, called their connected sum.
Given a Gorenstein ring, one would like to know whether it can be decomposed as a connected sum and if so,
what are its components. We answer these questions in the case of an Gorenstein Artin local algebra over
a field, and give a characterization for such rings to be connected sums. We further investigate conditions
on the associated graded ring of a Gorenstein Artin ring which force it to be a connected sum, and as a
consequence, obtain results about its Poincar$\acute{\text{e}}$ series and minimal number of generators of its
defining ideal.
\end{abstract}

\maketitle

\section*{Introduction}

The main object of study in this paper is a construction of Gorenstein rings, called a connected sum, defined by
Ananthnarayan, Avramov and Moore in \cite{AAM}. Given Cohen-Macaulay local rings $R$, $S$ and $\sk$
of the same dimension, and ring homomorphisms
$R \overset{\varepsilon_R}\longrightarrow \sk \overset{\varepsilon_S}\longleftarrow S$,
the authors consider the fibre product (or pullback) $R\times_\sk S = \{(r,s) \in R \times S: \eR(r) = \eS(s)\}$
and define a connected sum of $R$ and $S$ over $\sk$ as an appropriate quotient of $R\times_\sk S$. They prove that
when $R$ and $S$ are Gorenstein, a connected sum is also a Gorenstein local ring of the same dimension.

In this paper, we focus on connected sums over a field in the Artinian case, i.e., when $R$ and $S$ are
Gorenstein Artinian local rings and $\sk$ is their common residue field. These objects have been studied
from different perspectives by various authors starting with Sah (cf. \cite{Sah}) in the graded case, and
in the local case, by Lescot (see Theorem \ref{CSGor}). A topologically influenced version was also studied
by Smith and Stong (cf. \cite[Section 4]{SS}), and quite a few authors approach this area via
{\it Macaulay's inverse systems}, (e.g., see \cite{BBKT}). Completely different techniques are used in this
article: we look at intrinsic properties of the ring and its defining ideal in Section 3, and in Section 4,
focus on conditions on the associated graded ring which force it to be a connected sum.

An important question is: Given a Gorenstein Artin ring $Q$, can it be decomposed as a connected sum?
This question was studied from a geometric point of view for {\it projective bundle ideals} by Smith and Stong
in \cite[Section 4]{SS}. It is also known (see \cite[Theorem 8.3]{AAM}) that if $Q$ has a non-trivial
decomposition as a connected sum over $\sk$ and if the embedding dimension of $Q$ is greater than 2, then
$Q$ cannot be a {\it complete intersection}. The question of decomposability has also been studied using the
inverse systems by the authors in \cite{BBKT} using polynomials which are {\it direct sums} and
corresponding {\it apolar} Gorenstein algebras (see Remark \ref{apolar}).

In Section 3, we study properties of fibre products and connected sums of algebras
over a field, and give two other necessary conditions for $Q$ to be decomposable as a non-trivial
connected sum: one concerning the second Hilbert coefficient $H_Q(2)$ of $Q$ (Theorem \ref{HilbertCoeff2}),
and the second in terms of the associated graded ring $\gr(Q)$ of $Q$ (Theorem \ref{GrFP}). In particular,
if $Q$ is a {\it compressed Gorenstein algebra} over $\sk$ with Loewy length at least 4, then $Q$ is
indecomposable as a connected sum.

Finally, we identify conditions under which a Gorenstein Artin local algebra over a field is a connected sum
in Proposition \ref{MainTheorem} and obtain a characterization for the same in terms of the defining ideals
(see Theorem \ref{MainTheorem1}). A secondary question is: If $Q$ is a connected sum, what are its components?
In general, it is not clear how to extract the components of a connected sum, but this characterization allows
us to do so in this case (see Proposition \ref{Prop3}).

In Section 4, we study the associated graded ring of a Gorenstein Artin ring and look for conditions which
force the given Gorenstein ring to be a connected sum. The motivation comes from the following:
In \cite{Sa1}, Sally proves a structure theorem for stretched Gorenstein rings and in \cite{ER}, Elias and
Rossi give a similar structure theorem for short Gorenstein rings with some assumptions on the residue field.
When such a short or stretched Gorenstein ring $Q$ is an algebra over a field, these structure theorems show
that $Q$ can be written as a connected sum of a graded Gorenstein Artin ring $R$ with the same Loewy length
as $Q$, and a Gorenstein Artin ring $S$ with Loewy length less than three.

In either case, using a construction of Iarrobino, we see that the associated graded ring of $Q$ has the
property that its socle in degrees two or higher is one-dimensional. We call such graded $\sk$-algebras
{\it Gorenstein up to linear socle}. Using the characterization proved in Section 3, we show that this
property allows us to decompose $Q$ as a connected sum, where one of its components has Loewy length
less than three and the other has a Gorenstein associated graded ring. This gives us results regarding
the Poincar$\acute{\text{e}}$ series of $Q$ and minimal number of generators of its defining ideal.
(See Theorem \ref{Structure Theorem} and its corollaries).

In Section 5, we use Theorem \ref{Structure Theorem} to give applications to short and stretched
Gorenstein $\sk$-algebras. In particular, we show that these rings, when they are not graded, are
non-trivial connected sums, and derive some consequences without any restrictions on the residue field.

The first two sections contain results regarding the main tools that are used in the rest of the paper.
In Section 1, we list some properties of associated graded rings and Poincar$\acute{\text{e}}$ series.
Section 2 contains some information on fibre products and connected sums, including their interactions
with the objects introduced in Section 1. The results in Section 1 are well-known and Section 2 includes
known results rephrased in our notation.

\section{Preliminaries}

In this section, we first introduce some notation and terminology, and recall the definitions and some basic
properties of the main objects that appear in this article. We also list the previously known results
and some consequences that are needed in this paper.

\subsection{Notation}\hfill{}

\noindent
a)  If $T$ is a local ring and $M$ is an $T$-module, $\lambda(M)$ and $\mu(M)$ respectively denote the {\it length} and the
{\it minimal number of generators} of $M$ as a $T$-module.
\vskip 2 pt

\noindent
b) Let $(T,\m,\sk)$ be an Artinian local ring. Then $\edim(T)$
denotes the {\it embedding dimension} of $T$ which is equal to $\mu(\m)$, and
the {\it socle} of $T$ is $\soc(T) = \ann_T(\m)$. Moreover, the {\it type} of $T$ is $\type(T) = \dim_{\sk}(\soc(T))$,
and the \emph{Loewy length} of $T$ is $\ll(T) = \max\{n: \m^n \neq 0\}$.\footnote{Loewy length is also referred to as
{\it socle degree} in the literature.}
\vskip 2 pt

\noindent
c) If $\sk$ is a field, a {\it graded $\sk$-algebra} $G$ is a graded ring $G = \oplus_{i \geq 0} G_i$, where
$G_0 = \sk$, with unique homogeneous maximal ideal $G_+ = \oplus_{i \geq 1} G_i$. Furthermore,
we say $G$ is {\it standard graded} if $G_+$ is generated by $G_1$.
\vskip 2 pt

\noindent
d)  Let $m$ and $n$ be positive integers. Then $\Y$ and $\Z$ denote the sets of indeterminates $\{Y_1, \ldots, Y_m\}$ and
$\{Z_1,\ldots,Z_n\}$ respectively, and $\Y \cdot \Z$ denotes $\{Y_iZ_j: 1 \leq i \leq m, 1 \leq j \leq n\}$.
\vskip 2 pt

\noindent
e) Let $\iota_Y: \sk[\Y] \hookrightarrow \sk[\Y,\Z]$ and $\iota_Z: \sk[\Z] \hookrightarrow \sk[\Y,\Z]$ be the natural inclusions.
For an ideal $I \subseteq \sk[\Y,\Z]$, $I \cap \sk[\Y]$ and $I \cap \sk[\Z]$ denote the respective contractions
$\iota_Y^{-1}(I)$ and $\iota_Z^{-1}(I)$.

Moreover, for an ideal $J$ of $\sk[\Y]$ or $\sk[\Z]$, $J^e$ denotes its extension to $\sk[\Y,\Z]$ via the natural inclusions.

\subsection{Associated Graded Rings}
\begin{definition} Let $(T,\m,\sk)$ be a Noetherian local ring.\hfill{}\\
i) The {\it graded ring associated to the maximal ideal} $\m$ of
$P$, denoted $\gr_{\m}(T)$ (or simply $\gr(T)$), is defined as $\gr(T) \simeq \oplus_{i=0}^{\infty} \m^i/\m^{i+1}$.

\noindent
ii) If $T = \oplus_{i \geq 0} T_i$ is a finitely generated graded $\sk$-algebra, where $T_0 = \sk$ and $T_i$
consist of the elements in $T$ of degree $i$,  then we define the {\it Hilbert function} of $T$ as
$H_T(i) = \dim_{\sk}(T_i)$ for $i \geq 0$. If $T$ is not graded, we define $H_T(i) = H_{\gr(T)}(i)$.

\end{definition}

In the following remark, we list some notation and facts about associated graded rings.

\begin{remark}\label{GR}  Let $(T,\m,\sk)$ be a Noetherian local ring.\hfill{}\\
\noindent
a) For any $n \geq 0$, a minimal generating set of $\gr(T)_n = \oplus_{i=n}^{\infty} \m^i/\m^{i+1}$, the
$n$th power of the homogeneous maximal ideal $\gr(T)_+$ of $\gr(T)$, lifts to a minimal generating set of $\m^n$.

In particular, if $T$ is Artinian local, then so is $\gr(T)$, and we have $\lambda(T)  = \lambda(\gr(T))$.

\noindent
b) Let $x \in T$ be such that $x \in \m^i \setminus \m^{i+1}$. We define the
{\it initial form} of $x$ to be the element $x^* \in \gr(T)$ of degree $i$ that is the image of $x$ in $\m^i/\m^{i+1}$.

\noindent
c) For an ideal $K $ of $T$, $K^*$ is the ideal of $\gr(T)$ defined by $\langle x^*: x \in K \rangle$.
Note that, if $R \simeq T/K$, then $\gr(R) \simeq \gr(T)/K^*$.
\end{remark}
We now recall a construction due to Iarrobino (cf. \cite{Ia}):

\begin{remark}[Iarrobino's Construction]\label{Iarrobino}
Let $(Q,\m_Q,\sk)$ be a Gorenstein Artin local ring where $\ll(Q)=s$ and $G = \gr(Q)$ be its associated graded ring.
Iarrobino showed that
$$C = \bigoplus_{i \geq 0}  \frac{(0:_Q \m_Q^{s-i})\cap \m_Q^i}{(0:_Q \m_Q^{s-i})\cap \m_Q^{i+1}}$$
is an ideal in $G$. He also proved that $Q_0 = G/C$ is a graded Gorenstein quotient of $G$ with
$\deg(\soc(Q_0)) = s$.

Note that $H_{Q_0}(i) = H_G(i)$ for $i \geq s-1$ since $C_i = 0$ for $i \geq s-1$.
\end{remark}

\subsection{Poincar$\mathbf{\acute{\text{e}}}$ Series}

\begin{definition}\label{PSdef} For a local ring $(T, \m, \sk)$, the {\it Poincar$\acute{\text{e}}$ series} of $T$,
denoted ${\p}^T(t)$, is the formal power series
$$\p^T(t)=\sum_{i\geq 0} \beta_i^Tt^i, \;\;\; \text{with}\;\; \beta_i^T=\dim_{\sk}\left( \Tor_i^T(\sk, \sk)\right).$$
\end{definition}

The Poincar$\acute{\text{e}}$ series of $T$ can also be written as
$$\p^T(t) = \prod_{i \geq 1}\frac{(1+t^{2i-1})^{\varepsilon_{2i-1}^T}}{(1-t^{2i})^{\varepsilon_{2i}^T}},$$
 (e.g., cf. \cite[Remark 7.1.1]{Av}), where $\varepsilon_i^T$ is called the $i$th {\it deviation} of $T$.
 \vskip 5pt

If $T$ is an Artinian ring, then it is isomorphic to a quotient of a regular local ring by Cohen's Structure Theorem.
If $T$ is also a $\sk$-algebra, then we can write $T$ as a quotient of a polynomial ring over $\sk$. In the following remark,
we compute the minimal number of generators of its defining ideal.

\begin{remark}[Minimal Number of Generators]\label{PS2}
Let $(T,\m,\sk)$ be an Artinian local ring with $\edim(T) = d$.

\noindent
a) By Cohen's Structure Theorem, there is a regular local ring $(\widetilde T, \m_{\widetilde T}, \sk)$ such that
$T \simeq \widetilde T/I_T$, where $I_T \subseteq \m_{\widetilde T}^2$.
By \cite[Corollary 7.1.5]{Av}, $I_T \subseteq \m_{\widetilde T}^2$ implies that $\varepsilon_1^T= \edim(T)$ and
$\varepsilon_2^T=\mu(I_T)$.

In particular, by comparing the coefficients of $t$ and $t^2$, we get that
$$\varepsilon_1^T = \beta_1^T=d\quad\text{ and }
\quad\mu(I_T) = \varepsilon_2^T =\beta_2^T - \binom{\beta_1^T}{2} = \beta_2^T - \binom{d}{2}.$$

\noindent
b) Suppose $T$ is also a $\sk$-algebra. We can write $T \simeq \sk[X_1,\ldots,X_d]/I_T$, where
$I_T \subseteq \langle \underline X\rangle^2$.
Since $T$ is Artinian, $T \simeq \widetilde T/I_T \widetilde T$, where $\widetilde T = \sk[|\underline X|]$.
Hence, by (a), $\mu(I_T) = \varepsilon_2^T =\beta_2^T - \binom{d}{2}$ in this case too.
\end{remark}

Next we list some properties of the Poincar$\acute{\text{e}}$ series of a Gorenstein Artin local ring.

\begin{remark}[Poincar$\mathbf{\acute{\text{e}}}$ Series of Gorenstein Rings]\label{PS} \hfill{}\\
Let $(T,\m,\sk)$ be a Gorenstein Artin local ring and $\overline T$ represent the quotient $T/\soc(T)$.

\noindent
a) If $\edim(T) \leq 3$, then $\p^T(t)$ is rational. The same holds for $\edim(T) = 4$ if we assume $\char(\sk) \neq 2$.

Indeed, if $\edim(T) = 1$, then $T$ is a quotient of a discrete valuation ring. In particular,
every ideal in $T$ is principal. This forces $\frac{1}{\p^{T}(t)} = 1- t$.
When $\edim(T) = 2$, it is a well-known result of Serre that $T$ is a complete intersection.
In particular, $\p^T(t)$ is rational. If $\edim(T) = 3$, then $\p^T(t)$ is rational, for example, by \cite{We}.
Finally, by \cite{JKM}, if $\char(\sk) \neq 2$, then $\p^T(t)$ is rational if $\edim(T) = 4$.

\noindent
b) If $\edim(T) \geq 2$, then, by \cite[Theorem 2]{LA}, we see that
$\frac{1}{\p^{T}(t)} = \frac{1}{\p^{\overline T}(t)} + t^2$.

\noindent
c) If $\ll(T) = 2$ and $\edim(T) = n$, it is clear that $\frac{1}{\p^{\overline T}(t)} = 1 - nt$
since $\m_T \overline T = \soc(\overline T) \simeq \sk^{\oplus n}$. Therefore, by (b),
$\frac{1}{\p^{T}(t)} = 1 - nt + t^2$ if $n \geq 2$. Moreover, by (a), if $\edim(T) = 1$, then $\frac{1}{\p^{T}(t)} = 1- t$.
In particular, $\p^T(t)$ is a rational function of $t$.
\end{remark}

\section{Fibre Products and Connected Sums}

We begin with the definitions and some basic properties of fibre products and connected sums. For more details,
see \cite[Sections 1 and 2]{AAM} and \cite[Chapter 4]{HAthesis}.

\begin{definition}\label{FPdef}
Let $(R,\m_R,\sk)$ and $(S,\m_S,\sk)$ be local rings with the same residue field $\sk$.
We define the {\it fibre product} $R$ and $S$ over $\sk$ to be
$R\times_{\sk} S = \{(r,s) \in R \times S: \pi_R(r) = \pi_S(s)\}$,
where $\pi_R$ and $\pi_S$ are the natural projections from $R$ and $S$ respectively onto $\sk$.

This construction is also referred to as the {\it pullback} of the diagram
$R \overset{\pi_R}\longrightarrow \sk \overset{\pi_S}\longleftarrow S$.
\end{definition}

Next we give some observations regarding fibre products.

\begin{remark}\label{FP} With the notation as in
Definition \ref{FPdef}, set $P = R\times_{\sk} S$. Then, we see
that:

\noindent
a) $I = \{(r,0): r \in \m_R\}$ and $J = \{(0,s): s \in \m_S\}$ are ideals of $P$.
Identifying $\m_R$ with $I$ and $\m_S$ with $J$, we can write
$R \simeq P/\m_S$, $S \simeq P/\m_R$, $\sk \simeq P/(\m_R + \m_S)$ and $\m_R \cap \m_S = 0$ in $P$.

\noindent
b) $P$ is a local ring with unique maximal ideal $\m_P = \m_R \times \m_S$.
Furthermore, if $R$ and $S$ are both different from $\sk$, then $\soc(P) = \soc(R) \oplus \soc(S)$. In particular,
$\edim(P) = \edim(R) + \edim(S)$ and $\type(P) = \type(R) + \type(S)$.

\noindent
c) If $R$ and $S$ are both Artinian local, then $\m_P = \m_R \times \m_S$ shows that $\lambda(P ) = \lambda(R) + \lambda(S) - 1$.
\end{remark}

As a consequence of Remark \ref{FP}(b), when $R$ and $S$ are Artinian local rings with $R \neq \sk \neq S$,
$P = R \times_\sk S$ is Artinian, but not Gorenstein. In \cite{AAM}, the authors construct a suitable quotient of $P$
which is a Gorenstein ring. They define:

\begin{definition}\label{CSdef}
Let $(R,\m_R,\sk)$ and $(S,\m_S,\sk)$ be Gorenstein Artin local rings different from $\sk$.
Let $\soc(R) = \langle\delta_R\rangle$, $\soc(S) = \langle\delta_S\rangle$. Identifying $\delta_R$ with
$(\delta_R, 0)$ and $\delta_S$ with $(0, \delta_S)$, a {\it connected sum} of $R$ and $S$ over $\sk$,
denoted $R\#_\sk S$, is the ring $R\#_\sk S = (R\times_\sk S)/\langle\delta_R - \delta_S\rangle$.
\end{definition}

Connected sums of $R$ and $S$ over $\sk$ depend on
the generators of the socle $\delta_R$ and $\delta_S$ chosen as can be seen from the following example.

\begin{example}
Let $R = \Q[Y]/\langle Y^3\rangle$ and $S = \Q[Z]/\langle Z^3\rangle$. Let $y$ and $z$ denote the respective
images of $Y$ and $Z$ in $R$ and $S$. Then $\soc(R) = \langle y^2 \rangle$ and $\soc(S) = \langle z^2\rangle$.
The connected sums $Q_1 = (R \times_\sk S) / \langle y^2 - z^2\rangle$ and
$Q_2 = (R \times_\sk S) /\langle y^2 - p z^2\rangle$ are not isomorphic as rings where $p$ is a prime number not
congruent to $3$ modulo $4$. For a proof of this fact, see \cite{AAM}.
\end{example}

\begin{remark}\label{CS} With notation as in Definition \ref{CSdef}, set $P = R \times_{\sk} S$ and let
$Q = P/\langle\delta_R - \delta_S\rangle$ be a connected sum of $R$ and $S$ over $\sk$.

\noindent
a) Note that $\lambda(Q) = \lambda(P) - 1 = \lambda(R) + \lambda(S) - 2$ since $0 \neq \delta_R - \delta_S \in \soc(P)$.

\noindent
b) If $\ll(R) \geq 2$ and $\ll(S) \geq 2$, i.e., $\soc(R) \subseteq \m_R^2$ and $\soc(S) \subseteq \m_S^2$, then
$\edim(Q) = \edim(P) = \edim(R) + \edim(S)$.

\noindent
c) By the definition of a connected sum, it is clear that $Q/\soc(Q) \simeq R/\soc(R) \times_\sk S/\soc(S)$.
\end{remark}

A special case of Theorem 2.8 in \cite{AAM} is the following, see also \cite[Proposition 4.4]{Le}:

\begin{theorem}\label{CSGor} Let $(R,\m_R,\sk)$ and $(S,\m_S,\sk)$ be Gorenstein Artin local rings different from $\sk$.
Then a connected sum of $R$ and $S$ over $\sk$ is also a Gorenstein Artin local ring.
\end{theorem}

The following remark tells us what the Poincar$\acute{\text{e}}$ series of fibre products and
connected sums over $\sk$ are, in terms of the
Poincar$\acute{\text{e}}$ series of the components.

\begin{remark}[Poincar$\mathbf{\acute{\text{e}}}$ Series of Fibre Products and Connected Sums]\label{PS3} \hfill{}\\
Let $(R,\m_R,\sk)$ and $(S,\m_S,\sk)$ be Artinian local rings.

\noindent
a) If $P = R \times_{\sk} S$, then, by \cite[Theorem 1]{DK},
\begin{equation*}\tag{\ref{PS3}.1}
\frac{1}{\p^{P}(t)}=\frac{1}{\p^{R}(t)} + \frac{1}{\p^{S}(t)} -1.
\end{equation*}

\noindent
b) Furthermore, if $R$ and $S$ are Gorenstein Artin, where $\ll(R)$, $\ll(S) \geq 2$, then a connected sum
$Q = R \#_{\sk} S$ is Gorenstein, $\edim(Q) = \edim(R) + \edim(S) \geq 2$ and $\overline Q \simeq \overline R \times_\sk
\overline S$. Hence $\frac{1}{\p^{Q}(t)}=\frac{1}{\p^{\overline R}(t)} + \frac{1}{\p^{\overline S}(t)} - 1 + t^2$.
In particular, we get
\begin{equation*}\tag{\ref{PS3}.2}
\frac{1}{\p^{Q}(t)}=\frac{1}{\p^{R}(t)} + \frac{1}{\p^{S}(t)} - 1 + \phi(t) = \frac{1}{\p^P(t)} + \phi(t),
\end{equation*}
where $\phi(t) = - t^2$ when $\edim(R) \geq 2$ and  $\edim(S) \geq 2$, $\phi(t) = t^2$ when $\edim(R) = 1 = \edim(S)$,
and $\phi(t) = 0$ otherwise.
\end{remark}

\section{Connected Sums of {$\mathbf \sk$}-Algebras}

When $R$ and $S$ are $\sk$-algebras, we can say more about their fibre products and connected sums over $\sk$.
We begin with the following observations:

\begin{remark}\label{kalg}
Let $R = \sk[Y_1,\ldots,Y_m]/I_R$ and $S = \sk[Z_1,\ldots, Z_n]/I_S$ be $\sk$-algebras (not equal to $\sk$) with
$I_R \subseteq \langle \Y\rangle^2$ and $I_S \subseteq \langle\Z\rangle^2$. Note that this implies $\edim(R) = m$ and $\edim(S) = n$.
\vskip 2 pt

\noindent
a) Let $P$ denote the fibre product $R \times_\sk S$. Then $P \simeq \sk[\Y, \Z]/I_P$ where
$I_P = I_R^e + I_S^e+ \langle\Y \cdot \Z\rangle.$ In particular, if $R$ and $S$ are standard graded, then so is $P$ and,
for $i \geq 1$, $H_P(i) = H_R(i) + H_S(i)$.
\vskip 2 pt

\noindent
b) Now suppose $R$ and $S$ are Gorenstein and $Q = R \#_\sk S$ is their connected sum over $\sk$. Then, by
Definition \ref{CSdef}, there exist $\Delta_R \in \sk[\Y]$ and $\Delta_S \in \sk[\Z]$ such that their respective
images $\delta_R \in R$ and $\delta_S \in S$ generate the respective socles and
$Q \simeq  (R \times_\sk S)/\langle\delta_R - \delta_S\rangle \simeq \sk[\Y, \Z]/I_Q$, where
$I_Q= I_P + \langle \Delta_R-  \Delta_S\rangle =
I_R^e + I_S^e+ \langle \Y \cdot \Z\rangle+ \langle \Delta_R-  \Delta_S\rangle$.
In particular, if $R$ and $S$ are standard graded, then $Q$ is standard graded if and only if $\ll(R) = \ll(S)$.
\end{remark}

A different point of view to study Gorenstein Artin $\sk$-algebras is via {\it Macaulay's inverse systems}.
The next remark tells us what connected sums are in this light.

\begin{remark}[Connected Sums and Inverse Systems]
Let $\sk$ be a field of characteristic zero. One can study Gorenstein Artin $\sk$-algebras via Macaulay's inverse systems.
In this correspondence, such rings correspond to polynomials. For example, the ring
$\sk[X_1,\ldots,X_n]/I$, where $I = \langle X_iX_j, X_i^2 - X_j^2: 1 \leq i < j \leq n\rangle$ corresponds to the polynomial
$Z_1^2 + \cdots + Z_n^2$. For more details, see \cite[Section 1.4]{HAthesis} or \cite[Section 2]{ER}.

It can be seen that if $R$ and $S$ are Gorenstein Artin $\sk$-algebras corresponding to polynomials $F(\underline Y)$ and
$G(\underline Z)$ respectively, then the Gorenstein Artin $\sk$-algebra corresponding to $F + G$ is a connected sum of $R$ and
$S$ over $\sk$. (For example, see \cite[Remark 4.24]{HAthesis}).

In particular, whenever the polynomial corresponding to a Gorenstein Artin $\sk$-algebra $Q$ is of the form
$F(\underline Y) + Z_1^2 + \cdots + Z_n^2$, then $Q$ is a connected sum, over $\sk$, of the Gorenstein Artin rings
corresponding to the polynomials $F(\underline Y)$ and $Z_1^2 + \cdots + Z_n^2$, say $R$ and $S$ respectively.
As seen above, $S$ is a Gorenstein Artin local ring with $\ll(S) = 2$. This observation plays an important role
in Section 5.
\end{remark}

The main question we would like to address is:

\smallskip
\noindent
{\bf Main Question:} When is a given Gorenstein Artin local ring $Q$ decomposable as a connected sum over $\sk$?
\smallskip

\begin{remark}\label{apolar}
In terms of the inverse systems, answering the main question amounts to the following:
Given a polynomial $F$ corresponding to $Q$, write $F = F_1 + F_2$, where
$F_1$ and $F_2$ are polynomials in a disjoint set of variables.

When $F$ is homogeneous, the above property has been studied by the authors in \cite{BBKT}. They define such a polynomial
to be a {\it direct sum} and the corresponding Gorenstein algebra to be {\it apolar}.
\end{remark}

Before we proceed to answer the main question, observe the following trivial cases:

\begin{remark}[Trivial Fibre Products and Connected Sums]\label{trivial}
Let $(R,\m_R,\sk)$ be an Artinian local ring different from $\sk$ and $S = \sk$. Then it is easy to see that
$R \times_{\sk} S \simeq R$. We call this the trivial fibre product.
Note that $R \times_\sk S$ is a non-trivial fibre product if and only if $\ll(R)$ and $\ll(S)$ are both at least $1$.

Similarly, let $(R,\m_R,\sk)$ be a Gorenstein Artin local ring different
from $\sk$ and $S$ be a ring of length two. Note that this forces $S$ to be Gorenstein.
One can check that $R \#_\sk S \simeq R$.
This is referred to as the trivial connected sum decomposition of $R$. Thus $R \#_\sk S$ is a not a trivial connected
sum if and only if $\ll(R)$, $\ll(S) \geq 2$. (A simple length count proves the converse).
\end{remark}

In light of this, we make the following key definition about connected sums. Note that a similar terminology is also used
for fibre products in this article.

\begin{definition}
Let $Q$ be a Gorenstein Artin local $\sk$-algebra. We say that $Q$ can be {\it decomposed non-trivially as a connected sum}
over $\sk$ if there exist Gorenstein Artin $\sk$-algebras $R$ and $S$ such that
$Q \simeq R \#_\sk S$ and $R \not\simeq Q \not\simeq S$.

If this happens, we say that $R$ and $S$ are the {\it components} in a non-trivial connected sum decomposition of $Q$.

If this does not happen, we say that $Q$ is {\it indecomposable} as a connected sum over $\sk$.
\end{definition}

It is known (see \cite[Theorem 8.3]{AAM}) that if $Q$ is a {\it complete intersection} with $\edim(Q) \geq 3$, then $Q$ is
indecomposable as a connected sum over $\sk$.

In this section, we give two other necessary conditions: one concerning the second Hilbert coefficient $H_Q(2)$ of $Q$, and
the second in terms of the associated graded ring $\gr(Q)$ of $Q$. Finally, we give a characterization of connected sums of
$\sk$-algebras over $\sk$ in terms of the defining ideals. The notation in the rest of this section is as in Remark \ref{kalg}.

\subsection{Hilbert Functions}
We first obtain a numerical criterion that is satisfied by connected sums.

\begin{proposition}\label{HilbertCoeff1}
Let $R$ and $S$ are Gorenstein Artin $\sk$-algebras, with $\edim(R) = m$ and $\edim(S) = n$. If $Q \simeq R\#_\sk S$ is a
non-trivial connected sum, then $H_Q(2) \leq \binom{m+n+1}{2} - mn$.
\end{proposition}

\begin{proof}
With notation as in Remark \ref{kalg}, since $\underline Y \cdot \underline Z \in I_Q$, we have $Y_i^* Z_j^* \in I_Q^*$
for all $i$ and $j$. Hence, $H_{\sk[\underline Y, \underline Z]}(2) = \binom{m+n+1}{2}$ implies that
\vskip 2pt
$H_Q(2) =  H_{\gr(Q)}(2) = \binom{m+n+1}{2} -
\dim_{\sk}((I_Q^* + \langle\underline Y \cdot \underline Z\rangle^3)/ \langle\underline Y \cdot \underline Z\rangle^3)
\leq \binom{m+n+1}{2} - mn$.
\end{proof}

Note that for positive integers $m$ and $n$, if $d = m+n$ is fixed, the minimum value of $mn = m(d-m)$ is obtained when
$m = 1$ or $m = d-1$. Hence, an immediate corollary to the above proposition is:

\begin{theorem}\label{HilbertCoeff2}
Let $Q$ be a Gorenstein Artin local $\sk$-algebra with $\edim(Q) = d$. If $H_Q(2) \geq \binom{d}{2} + 2$, then $Q$ is
indecomposable as a connected sum over $\sk$.
\end{theorem}

We say $Q$ is a {\it compressed Gorenstein $\sk$-algebra} if it has a maximum possible Hilbert function given the
embedding dimension $d$ and Loewy length $s$, i.e., if the Hilbert function of $Q$ is
$H_Q(i)=\text{min}\{\binom{d+i-1}{i}, \binom{d+s-i-1}{s-i}\}$.

\begin{corollary}\label{compressed}
If $Q$ is a compressed Gorenstein $\sk$-algebra with $\ll(Q) \geq 4$, then $Q$ is
indecomposable as a connected sum over $\sk$.
\end{corollary}

\begin{remark}\label{generic}
Since a {\it generic Gorenstein $\sk$-algebra} is compressed (see \cite{Ia84}), we see that they are
indecomposable as a connected sum over $\sk$.
This has also been proved in \cite[Proposition 4.4]{SS}, which the authors obtain by using different techniques.
\end{remark}

\subsection{Associated Graded Rings}
We first prove a basic property of the associated graded ring of a connected sum of $\sk$-algebras to obtain a
necessary condition. We need the following:

\begin{lemma} If both $R$ and $S$ are Artinian $\sk$-algebras, then
$$\gr(R\times_\sk S)\simeq \gr(R)\times_\sk \gr(S).$$
\end{lemma}

\begin{proof} Let $P = R\times_\sk S$. By Remark \ref{FP}(a), we have $R\simeq P/J$, $S\simeq P/I$, and
$\sk=P/(I+J)$. Thus if $I=\langle\underline y \rangle$ and $J=\langle \underline z \rangle$, we see that
$\m_P=\langle \underline y, \underline z \rangle$
is the maximal ideal of $P$. Hence $\m_P^*=\langle y_1^*,\ldots, y_m^*, z_1^*,\ldots, z_n^* \rangle$ is the maximal ideal
$P_+$ of $\gr(P)$. Thus $\langle y_1^*,\ldots, y_m^* \rangle \subseteq I^*$ and
$\langle z_1^*,\ldots, z_n^* \rangle \subseteq J^*$
force $I^*+J^*=P_+$. Since $\lambda(P)=\lambda(R)+\lambda(S)-1$ and $\lambda(\gr(P))=\lambda(P)$, we have
$\lambda(\gr(P))=\lambda(\gr(R)\times_\sk \gr(S))$.

Now, by Remark \ref{GR}(b), $R = P/J$ and $S = P/I$ imply that $\gr(R) \simeq \gr(P)/J^*$ and
$\gr(S) \simeq \gr(P)/I^*$, respectively. In particular, the natural projection $\gr(P) \twoheadrightarrow \sk$ factors through the
surjective maps $\gr(P) \twoheadrightarrow \gr(R)$ and $\gr(P) \twoheadrightarrow \gr(S)$. Hence $\gr(P)$ maps onto
$\gr(R) \times_\sk \gr(S)$. Since $\lambda(\gr(P))=\lambda(\gr(R)\times_\sk \gr(S))$, we get the desired isomorphism.
\end{proof}

\begin{proposition} \label{Prop2} Let $R$ and $S$ be Gorenstein Artin $k$-algebras with $\ll(R) \neq \ll(S)$. Then
the associated ring of $R\#_\sk S$ is a fibre product.

Moreover, if $\ll(R)$ and $\ll(S)$ are at least two, the non-trivial connected sum $R\#_\sk S$ is not a standard
graded $\sk$-algebra.
\end{proposition}

\begin{proof} Let $P = R \times_\sk S$ and $Q = R \#_\sk S$. Let $\soc(R)=\langle \delta_R \rangle$ and
$\soc(S)=\langle \delta_S \rangle$.
Since $Q\simeq P/\langle \delta_R - u\delta_S \rangle$ for some unit $u$ in $S$, by Remark \ref{GR}(b), we have
$\gr(Q) \simeq \gr(P)/\langle \delta_R-u\delta_S \rangle^*$. Without loss of generality, we may assume that $\ll(R) > \ll(S)$.
Hence $\langle \delta_R- u\delta_S \rangle^* = \langle u\delta_S \rangle^*$. Thus we see that
$\gr(Q) \simeq (\gr(R)\times_\sk \gr(S))/\langle u \delta_S \rangle^*\simeq \gr(R) \times_\sk \gr(S/\langle \delta_S \rangle)$.

Finally, if $\ll(R) > \ll(S) \geq 2$, then $\gr(R) \neq \sk \neq
\gr(S/\langle \delta_S \rangle)$. Hence $\gr(Q)$ is not Gorenstein, by Remark
\ref{FP}(b). Thus $Q \not\simeq \gr(Q)$, hence $Q$ is not standard
graded.
\end{proof}

The condition $ll(R) \neq \ll(S)$ is a necessary assumption in the above proposition. Firstly, note that
as observed in Remark \ref{kalg}, if $R$ and $S$ are graded with $\ll(R) = \ll(S) \geq 2$, then
$Q \simeq R \#_\sk S$ is graded. In particular, $Q$ is non-trivial connected sum, but $Q \simeq \gr(Q)$ is
indecomposable as a fibre product over $\sk$.

The graded nature of $R$ and $S$ does not play a role in $\gr(Q)$ being indecomposable as a fibre product over
$\sk$, as can be seen in the following example.

\begin{example} Let $R \simeq \sk[Y_1,Y_2]/\langle Y_1^2Y_2, Y_1^3 - Y_2^2\rangle$,
$S = \sk[Z]/\langle Z^5 \rangle$ and $Q \simeq R \#_\sk S$.
Then $G = \gr(Q) \simeq \sk[Y_1,Y_2,Z]/\langle Y_1Z, Y_2Z, Y_1^2Y_2,Y_2^2,Y_1^4 - Z^4\rangle$.
Note that in this case, $R$ is not standard graded, and $\ll(R) = \ll(S) = 4$.
We now show that $G$ is indecomposable as a fibre product over $\sk$.

Suppose $G \simeq A \times_\sk B$ is a non-trivial fibre product, for some $\sk$-algebras $A$ and $B$.
Since $3 = \edim(G) = \edim(A) + \edim(B)$, we may assume that $\edim(A) = 2$ and $\edim(B) =1$.
Furthermore, $2 = \type(G) = \type(A) + \type(B)$ forces $A$ and $B$ to be Gorenstein, and $\soc(G) = \soc(A) \oplus \soc(B)$
implies that $\ll(A) = 4$ and $\ll(B) = 2$ or vice versa.
Since the Hilbert function of $B$, $H_B(i) = 1$ for each $i \leq \ll(B)$, one can check that if $\ll(B) = 4$, then $\ll(A) = 3$.
As this cannot happen, we must have $\ll(A) = 4$ and $\ll(B) = 2$.

Thus, if $G \simeq A \times_\sk B$ is a non-trivial fibre product, we may assume that $A$ and $B$ are Gorenstein,
$\edim(A) = 2$, $\ll(A) = 4$, and $B \simeq \sk[V]/\langle V^3\rangle$. In particular, we can write
$A \simeq \sk[U_1,U_2]/\langle f_1,f_2\rangle$. Note that $U_1$, $U_2$ and $V$ are indeterminates over $\sk$.

Thus $G \simeq \sk[U_1,U_2,V]/\langle U_1V,U_2V,V^3,f_1,f_2 \rangle$. Let small letters denote the respective images of the
indeterminates in $G$ and $$\phi: \sk[U_1,U_2,V]/\langle U_1V,U_2V,V^3,f_1,f_2 \rangle \longrightarrow
\sk[Y_1,Y_2,Z]/\langle Y_1Z, Y_2Z, Y_1^2Y_2,Y_2^2,Y_1^4 - Z^4\rangle$$ be an isomorphism.

Write $\phi(u_1) = a_{11}y_1 + a_{12}y_2 + a_{13}z$, $\phi(u_2) = a_{21}y_1 + a_{22}y_2 + a_{23}z$ and
$\phi(v) = a_{31}y_1 + a_{32}y_2 + a_{33}z$. The fact that $\phi(v^3) = 0$ forces $a_{31} = 0 = a_{33}$. Hence $a_{32} \neq 0$ and
therefore, $a_{11} = 0 = a_{21}$ since $\phi(u_1v) = 0 = \phi(u_2v)$. This gives us a contradiction as $a_{i1} = 0$ for all $i$
implies that $y_1 \not\in \im(\phi)$. Hence $G$ cannot be written as a non-trivial fibre product.
\end{example}

Proposition \ref{Prop2} gives us a strategy to answer the main question, one may expect that if the associated graded ring is
indecomposable as a fibre product over $\sk$, then the Gorenstein ring itself is indecomposable as a connected sum over $\sk$.
However, the above example shows that this is not necessarily true. Hence, the best possible answer to the main question using
this approach is the following application of Proposition \ref{Prop2}:

\begin{theorem}\label{GrFP}
Let $Q$ be a Gorenstein Artin $\sk$-algebra such that its associated graded ring
$\gr(Q)$ is indecomposable as a fibre product over $\sk$. If $R$ and $S$ are Gorenstein Artin $\sk$-algebras such that
$Q \simeq R \#_\sk S$ is a non-trivial connected sum over $\sk$, then:\\
{\rm a)} $\ll(R) = \ll(S)$ and\\ {\rm b)} if $Q$ is not graded, then at least one of $R$ and $S$ is not a graded Gorenstein ring.
\end{theorem}

\begin{proof}
Suppose $Q \simeq R \#_\sk S$ is a non-trivial connected sum over $\sk$. By Proposition \ref{Prop2}, since
$\gr(Q)$ is indecomposable as a fibre product over $\sk$, we must have $\ll(R) = \ll(S)$, proving (a).

If both $R$ and $S$ are graded, then $\ll(R) = \ll(S)$ forces $Q$ to be graded, which contradicts the hypothesis in (b).
Hence (b) holds.
\end{proof}

\subsection{A Characterization in terms of the Defining Ideals}

The following theorem gives us a criteria to determine whether a given Gorenstein Artin local $\sk$-algebra
is a connected sum.

\begin{proposition}\label{MainTheorem}
Let $Q = \sk[\underline Y, \underline Z]/I_Q$ be a Gorenstein Artin local ring. Let $R = \sk[\underline Y]/I_R$ and
$S = \sk[\underline Z]/I_S$, where $I_R = I_Q \cap \sk[\underline{Y}]$ and $I_S = I_Q \cap \sk[\underline{Z}]$. Suppose
$Y_i \cdot Z_j \in I_Q$ for $1 \leq i \leq m$, $1 \leq j \leq n$. Then \\
a) $R$ and $S$ are Gorenstein Artin and\\
b) $Q \simeq R \#_\sk S$.
\end{proposition}

\begin{proof}
The inclusions $\sk[\underline Y]$, $\sk[\underline Z] \hookrightarrow \sk[\underline Y, \underline Z]$ induce inclusions
$R \hookrightarrow Q$ and $S \hookrightarrow Q$. Let $y$ and $z$ denote the respective images of $Y$ and $Z$ in the quotient rings
$Q$, $R$ and $S$.

\noindent a) Let $f \in \soc(R)$. Then $Y_i \cdot F  \in I_R \subseteq
I_Q$ for each $i$, where $F \in \sk[\underline Y]$ is a preimage in
$\sk[\underline Y]$ of $f$. Moreover, since $Y_i Z_j \in I_Q$ for
each $i$ and $j$, $Z_j F \in I_Q$. Hence $f \in \soc(Q)$. Therefore
$0 \neq \soc(R) \subseteq \soc(Q)$ which is a one-dimensional
$\sk$-vector space. Thus $\dim_{\sk}(\soc(R)) = 1$, i.e., $R$ is
Gorenstein Artin.

We can show that $S$ is also a Gorenstein Artin local ring by a similar argument.

\noindent
b) Let $P = R \times_\sk S$. Then $P \simeq \sk[\underline Y, \underline Z]/I_P$, where
$I_P = I_R^e + I_S^e + \langle \underline Y \cdot \underline Z \rangle$. By the hypothesis, $I_P \subseteq I_Q$ and hence
induces a natural surjective map $\pi: P \longrightarrow Q$.

Let $\delta_R$ and $\delta_S$ be the generators of $\soc(R)$ and $\soc(S)$ respectively, with their respective
preimages in $\sk[\underline Y, \underline Z]$ being $\Delta_R$ and $\Delta_S$. Note that if $\Delta_R \in I_Q$,
then $\Delta_R \in I_R$, which is not true. Similarly, $\Delta_S \not\in I_Q$, i.e.,
$\pi(\delta_R) \neq 0 \neq \pi(\delta_S)$ in $Q$. But $\soc(P) = \langle \delta_R, \delta_S\rangle$ maps into
$\soc(Q)$, which is one-dimensional. Hence $\pi(\delta_R) = u \pi(\delta_S)$ for some unit $u \in \sk$. Thus
$\delta_R - u \delta_S \in \ker(\pi)$ for some $u \in \sk$.

Finally, we show that $\ker(\pi) \subseteq \soc(P)$. Let $f \in \ker(\pi)$ and $F$ be a preimage in
$\sk[\underline Y, \underline Z]$.
In order to prove $f \in \soc(P)$, it is enough to prove that $Y_i F$, $Z_j F \in I_P$ for all $i$ and $j$.

We show that $Y_1 F \in I_P$. The others follow similarly. Write
$F = F_1(\underline Y) + F_2(\underline Y, \underline Z)$, where every term
of $F_2$ is a multiple of some $Z_j$. Then $Y_1 F_2 \in I_P \subseteq I_Q$.

Now $f \in \ker(\pi)$ implies $F \in I_Q$. Thus $Y_1 F_1 = Y_1 F - Y_1 F_2 \in I_Q \cap \sk[\underline Y] = I_R$. Therefore,
$Y_1 F_1$ is also in $I_P$, which implies that  $Y_1 F \in I_P$. This proves that $\ker(\pi) \subseteq \soc(P)$.

Thus, we have
$$0 \subsetneq \langle \delta_R - u \delta_S\rangle \subseteq \ker(\pi) \subsetneq \langle \delta_R, \delta_S\rangle = \soc(P).$$
Since $\lambda(\soc(P)) = 2$, this forces $\ker(\pi) = \langle \delta_R - u \delta_S\rangle$. Therefore,
$Q \simeq P/\langle \delta_R - u \delta_S\rangle$ for some unit $u \in \sk$, and hence is a connected sum of
$R$ and $S$ over $\sk$.
\end{proof}

The requirement that $\underline Y \cdot \underline Z \in I_Q$ is necessary in the above proposition,
otherwise $R$ or $S$ may not be Gorenstein as can be seen from the following example.

\begin{example}
Let $Q=\sk[Y_1,Z_1,Z_2]/I_Q$, where $I_Q=\langle Y_1Z_1-Z_2^2, Y_1^2, Z_1^2\rangle$. Then
$I_Q \cap \sk[Z_1,Z_2]=\langle Z_1^2, Z_1Z_2^2, Z_2^4\rangle$.
(One can verify this using a computer algebra package. For example, we use the elimination package in Macaulay2).
Note that $Q$ is Gorenstein, but $S=\sk[Z_1,Z_2]/(I_Q\cap \sk[Z_1,Z_2])$ is not Gorenstein because it has a
two dimensional socle.
\end{example}

It follows from Remark \ref{FP}(a) that, if $P = R \times_\sk S$, then we can identify $R$ and $S$ with appropriate quotients of
$P$. On the other hand, if $Q = R \#_\sk S$, in general, it is not clear how one can recover the components $R$ and $S$ from
$Q$. Proposition \ref{Prop3} shows that $R$ and $S$ can be identified with subrings in the case of $\sk$-algebras.

\begin{proposition} \label{Prop3}
Let $P$, $Q$, $R$ and $S$ be $\sk$-algebras with $P \simeq \sk[\underline Y,\underline Z]/I_P$,
$Q \simeq \sk[\underline Y,\underline Z]/I_Q$, $R \simeq \sk[\underline Y]/I_R$ and $S \simeq \sk[\underline Z]/I_S$. \\
a) If $P \simeq R \times_\sk S$, then $I_R = I_P \cap \sk[\underline{Y}]$ and $I_S = I_P \cap \sk[\underline{Z}]$.\\
b) If $R$ and $S$ are Gorenstein Artin with $Q \simeq R \#_\sk S$,
then $I_R = I_Q \cap \sk[\underline{Y}]$ and $I_S = I_Q \cap \sk[\underline{Z}]$.
\end{proposition}

\begin{proof} \hfill{}\\
a) By Remark \ref{kalg}(a), since $I_P = I_R^e + I_S^e + \langle \underline Y \cdot \underline Z\rangle$,
it is enough to show $I_P \cap \sk[\underline Y] \subseteq I_R$.

Let $F(\underline Y) \in I_P \cap \sk[\underline{Y}]$.
We can write $F = F_1(\underline Y) + F_2(\underline Y, \underline Z)$, where $F_1 \in I_R$ and every term
of $F_2$ is a multiple of some $Z_j$. But $F-F_1\in \sk[\underline Y]$, hence $F_2 = F - F_1 = 0$.

\noindent
b) Let $I_{R'} = I_Q \cap \sk[\underline{Y}]$ and $I_{S'} = I_Q \cap \sk[\underline{Z}]$. Since
$I_R^e + I_S^e \subseteq I_Q$, we have $I_R \subseteq I_{R'}$ and $I_S \subseteq I_{S'}$, which induce natural surjective maps
$\pi_1: R \longrightarrow R'$ and $\pi_2: S \longrightarrow S'$, where $R' = \sk[\underline Y]/I_{R'}$, $S' = \sk[\underline Z]/I_{S'}$.
In particular, $\lambda(R') \leq \lambda(R)$ and $\lambda(S') \leq \lambda(S)$. In order to prove that
$I_R = I_{R'}$ and $I_S = I_{S'}$, it is enough to prove that $\pi_1$ and $\pi_2$ are isomorphisms, in particular, it is enough to
show that $\lambda(R') = \lambda(R)$ and $\lambda(S') = \lambda(S)$.

Since $Y_i \cdot Z_j \in I_Q$ for $1 \leq i \leq m$, $1 \leq j \leq n$, by Theorem \ref{MainTheorem},
$R'$ and $S'$ are Gorenstein Artin and $Q \simeq R' \#_\sk S'$.
Hence $\lambda(R') + \lambda(S') = \lambda(Q) +2 = \lambda(R) + \lambda(S)$. Since $\lambda(R') \leq \lambda(R)$
and $\lambda(S') \leq \lambda(S)$, we get $\lambda(R') = \lambda(R)$ and $\lambda(S') = \lambda(S)$, completing the proof.
\end{proof}

We conclude this section with a result on the minimal number of generators of the defining ideals of $P$ and $Q$.

\begin{proposition}\label{Prop10}
With notation as in Remark \ref{kalg}, we have the following:\\
{\rm a)} $\mu(I_P) = \mu(I_R) + \mu(I_S) + mn$. \\
{\rm b)} $\mu(I_Q) = \mu(I_R) + \mu(I_S) + mn + \psi = \mu(I_P) + \psi$, \\
where $\psi = 1$ when $m$, $n \geq 2$, $\psi = -1$ when $m = 1 = n$,
and $\psi = 0$ otherwise.
\end{proposition}

\begin{proof}\hfill{}\\
a) Comparing the coefficients of $t$ and $t^2$ in Equation \ref{PS3}.1, we see that $\beta_1^P = \beta_1^R + \beta_1^S$
and $\beta_2^P = \beta_2^R + \beta_2^S + 2 \beta_1^R\beta_1^S$. Hence (a) follows by Remark \ref{PS2}(b).

\noindent b) First of all, note that if $m = 1 = n$, then $\mu(I_R)
= 1 = \mu(I_S)$. Since $Q$ is Gorenstein Artin with $\edim(Q) = m +
n = 2$, it is a well known result of Serre that $\mu(I_Q) = 2$. Hence, without loss of
generality, we may assume that $m \geq 2$.

It is clear from Equation \ref{PS3}.2 that $\beta_2^Q = \beta_2^P$ when $n = 1$ and $\beta_2^Q = \beta_2^P + 1$
when $n \geq 2$. Since $\beta_1^Q = \beta_1^P$ in either case, the proof of (b) is complete by Remark \ref{PS2}(b).
\end{proof}

We use Proposition \ref{MainTheorem} and Proposition \ref{Prop3}(b) to give equivalent conditions for a
Gorenstein Artin $\sk$-algebra to be a connected sum over $\sk$. We can summarize the results obtained in this subsection
in the following:

\begin{theorem}\label{MainTheorem1} Suppose $R$, $S$, and $Q$ are Artin $\sk$-algebras of the form $R \simeq \sk[\underline Y]/I_R$,
$S \simeq \sk[\underline Z]/I_S$ and $Q \simeq \sk[\underline
Y,\underline Z]/I_Q$. Then the following are equivalent:

\noindent
i) $R$ and $S$ are Gorenstein Artin, and $Q=R\#_{\sk} S$.

\noindent
ii) $Q$ is Gorenstein Artin, $I_R = I_Q \cap \sk[\underline{Y}]$, $I_S = I_Q \cap \sk[\underline{Z}]$, and
$Y_iZ_j\in I_Q$ for all $i,j$.

If the above conditions are satisfied, then $\p^{Q}(t)$ is given by Equation (\ref{PS3}.2), $\mu(I_Q)$ is given by the formula in
Proposition \ref{Prop10}(b) and  $I_Q = I_R^e + I_S^e + \langle \Y \cdot \Z\rangle+ \langle \Delta_R-  \Delta_S\rangle$, where
$\Delta_R \in \sk[\Y]$ and $\Delta_S \in \sk[\Z]$ are such that their respective
images $\delta_R \in R$ and $\delta_S \in S$ generate the respective socles.
\end{theorem}

\begin{proof} Note that (ii) implies (i) by Proposition \ref{MainTheorem}. If (i) holds, we first observe
that $Q$ is Gorenstein by Theorem \ref{CSGor}. The rest of the
theorem follows from Proposition \ref{Prop3}(b), Remark \ref{PS3}(b), Proposition \ref{Prop10}(b) and Remark
\ref{kalg}(b).
\end{proof}

\section{Associated Graded Rings and Connected Sums}

In this section, we explore the connections between associated graded rings and connected sums.
In particular, we study conditions on the associated graded ring of a Gorenstein Artin
$\sk$-algebra which force it to be a connected sum.
We note that as an immediate consequence of Proposition \ref{Prop2}, we get the following:

\begin{corollary}\label{Cor1}
Let $R$ and $S$ be Gorenstein Artin $\sk$-algebras where $\ll(S) = 2$ and $\ll(R) \geq 3$. If $Q\simeq R\#_\sk S$,
then there is a surjective map $\pi: G=\gr(Q) \twoheadrightarrow A = \gr(R)$ such that $\ker(\pi)_i=0$, $i\geq 2$.
\end{corollary}

The main theorem we prove in this section (Theorem \ref{Structure Theorem}) shows that
 the converse of Corollary \ref{Cor1} is true when $A$ itself is a graded
Gorenstein ring.

With $A$ and $G$ as in Corollary \ref{Cor1}, if $A$ is further
assumed to be graded Gorenstein, one can see that $\soc(G)\cap
(G_+)^2 \simeq \soc(A)$, which is a one-dimensional $\sk$-vector
space. In fact, as we show in Proposition \ref{GLS}, these properties are equivalent.
Hence, we introduce the following:

\begin{definition} We say that a graded Artinian $\sk$-algebra $G$ is \emph{Gorenstein up to linear socle}
if the socle of $G$ in degree two and higher is a one-dimensional vector space over $\sk$, i.e.,
$\dim_{\sk} (\soc(G)\cap (G_+)^2)=1$.
\end{definition}

In Theorem \ref{Structure Theorem}, we prove that if the associated graded ring of a
Gorenstein Artin local ring $Q$, with Loewy length at least 3, is Gorenstein up to linear socle, then $Q$ can be decomposed as a
connected sum.

Before we state the main theorem, we explore what it means to be Gorenstein up to linear socle. We begin with the
following proposition which characterizes this property.

\begin{proposition}\label{GLS}
Let $G$ be a graded $\sk$-algebra with $\ll(G) \geq 2$. Then the following are equivalent:\\
i) $G$ is Gorenstein up to linear socle.\\
ii) There exist graded $\sk$-algebras $A$ and $B$, where $A$ is graded Gorenstein and $B_+^2 = 0$ such that
$G \simeq A \times_\sk B$.   \\
iii) There exists a graded Gorenstein ring $A$ and a surjective ring homomorphism $\pi: G \twoheadrightarrow A$
such that $\ker(\pi) \cap (G_+)^2 =0$.
\end{proposition}

\begin{proof}
First of all, note that if $G$ itself is Gorenstein, then the above statements are all trivially true.
$G$ satisfies (i), we can take $A = G$ in (iii), and write $G$ as the trivial fibre product $G \times_{\sk} \sk$ in (ii).
Hence, we may assume that $G$ itself is not Gorenstein, i.e., $(\soc(G) + (G_+)^2)/(G_+)^2 \neq 0$.

\noindent
(i) $\Rightarrow$ (ii): Let $G$ be Gorenstein up to linear socle. Then we can find a
$\sk$-basis, say $\{\overline{z_1},\ldots,\overline{z_n}\}$, for $(\soc(G) + (G_+)^2)/(G_+)^2$, where
$n = \type(G) - 1$. Extend this to a $\sk$-basis
$\{\overline{y_1}, \ldots, \overline{y_m}, \overline{z_1},\ldots,\overline{z_n}\}$ of
$G_+/(G_+)^2$, and lift it to a minimal generating set $\{\underline y, \underline z\}$ of $G_+$ in $G_1$.
Since $\langle \underline y\rangle \cap \langle \underline z\rangle = 0$ and
$\langle \underline y\rangle + \langle \underline z\rangle = G_+$, it is easy to see that $G \simeq A \times_\sk B$,
where $A = G/\langle \underline z\rangle$ and $B = G/\langle \underline y\rangle$ are graded $\sk$-algebras.

Since $z_i \in \soc(G)$ for each $i$, their images in $B$, which are the generators of $B_+$, are in $\soc(B)$, in
particular, $B_+ = \soc(B)$. Thus $B_+^2 = 0$. Furthermore, $\type(B) = n$ and $\type(G) = \type(A) + \type(B)$
implies $\type(A) = 1$, i.e., A is a graded Gorenstein $\sk$-algebra, proving (ii).

\noindent
(ii) $\Rightarrow$ (iii): Suppose (ii) is true.
By Remark \ref{FP}(a), $A \simeq G/J$, where $J = \{(0,b) : b \in B_+\}$. Since $B_+^2 = 0$, (iii) holds.

\noindent
(iii) $\Rightarrow$ (i): Assuming (iii), we see that $\pi |_{(G_+)^i}: (G_+)^i \longrightarrow (A_+)^i$ is an isomorphism for each
$i \geq 2$. In particular, if $\soc(A) = (A_+)^s$, then $s = \ll(A) = \ll(G) \geq 2$ and $\dim_{\sk}((G_+)^s) = 1$.

Suppose $z \in \soc(G) \cap (G_+)^2$. Then $\pi(z) \in \soc(A) = (A_+)^s$. Since $s \geq 2$, this forces $z \in (G_+)^s$.
Thus $(G_+)^s \subseteq \soc(G) \cap (G_+)^2 \subseteq (G_+)^s$, proving (i).
\end{proof}

The following remark lists observations from Proposition \ref{GLS} which we use repeatedly.

\begin{remark}\label{TG-1} Let $(Q, \m_Q, k)$ be a Gorenstein Artin local ring such that $G=\gr(Q)$ is Gorenstein up to
linear socle. Let the notation be as in Proposition \ref{GLS}.

Firstly, note that $\edim(Q) = \edim(G)=\edim(A)+\edim(B)$ and
$\lambda(Q) = \lambda(G) = \lambda(A) + \lambda(B) - 1$. Hence,
if $n= \type(B)= \type(G) - \type(A) = \type(G)-1 $, then since $\edim(B)=\type(B) = \lambda(B) - 1$, we get
$\lambda(A) = \lambda(Q) - n$ and $\edim(A) = \edim(Q) - n$.

Furthermore, by the proof of (i) $\Rightarrow$ (ii), we note that $A \simeq G/\langle \soc(G)\cap G_{1}\rangle$.
Hence, $\ll(A) = \ll(Q)$.
\end{remark}

\noindent
{\bf Key Note:} With notation as above, note that $G$ itself is Gorenstein if and only if $n = \type(G) - 1= 0$. For the
next two results, we will assume that $G$ is Gorenstein up to linear socle but not Gorenstein, i.e.,
$\type(G) = n + 1 \geq 2$.\\

In the following lemma, we record some properties of Gorenstein rings whose associated graded rings are Gorenstein up to
linear socle.

\begin{lemma} \label{ST}
Let $(Q, \m_Q, k)$ be a Gorenstein Artin local ring with $\ll(Q)= s$ such that $G=\gr(Q)$ is Gorenstein up to linear socle
but not Gorenstein. If $s\geq 3$, then the following hold:\\
{\rm a)} $(0 :_Q \m_Q^2) \cap \m_Q^2 = \m_Q^{s-1}$.\\
{\rm b)} $(0:_Q \m_Q^2)/\m_Q^{s-1}$ is a $\sk$-vector space of dimension $\type(G) -1$.\\
{\rm c)} If $w\in (0:_Q \m_Q^2)\setminus \m_Q^{s-1}$, then $w\in \m_Q\setminus \m_Q^2$,
$w\cdot \m_Q=\soc(Q)$ and $w^* \in \soc(G) \setminus (G_+)^2$.
\end{lemma}
\begin{proof} By Proposition~\ref{GLS}, there exists a graded Gorenstein ring $A$ and a surjective ring homomorphism
$\pi: G \twoheadrightarrow A$ such that $\ker(\pi) \cap (G_+)^2 =0$.
Note that the induced map $\pi: (G_+)^i \longrightarrow (A_+)^{i}$ is an isomorphism for each $i \geq 2$.

\noindent
a) Let $w \in (0 :_Q \m_Q^2)$. Then $\pi(w^*) \in (0:_A (A_+)^2)$. Since $A$ is graded Gorenstein, we have
$(0:_A (A_+)^2) = (A_+)^{s-1}$ and hence $\pi(w^*)\in (A_+)^{s-1}$. Suppose further $w \in \m_Q^2$. Then
$\deg(w^*) \geq 2$ in $G$. Since $\pi: G_i \longrightarrow A_{i}$ is an isomorphism for $i \geq 2$,
$\pi(w^*) \in (A_+)^{s-1}$ forces $w^* \in (G_+)^{s-1}$, i.e., $w \in \m_Q^{s-1}$. Thus
$(0 :_Q \m_Q^2) \cap \m_Q^2 \subseteq \m_Q^{s-1}$. The other inclusion is clear since $\m_Q^s = \soc(Q)$.

\noindent
b) By (a),
$$\frac{(0:_Q \m_Q^2)}{\m_Q^{s-1}} \simeq \frac{(0:_Q \m_Q^2)}{(0 :_Q \m_Q^2) \cap \m_Q^2} \simeq
\frac{(0:_Q \m_Q^2) + \m_Q^2}{\m^2_Q}$$ which is a $\sk$-vector space since $\m_Q^2 \neq 0$.

Let $n = \lambda(\ker(\pi)) = \edim(Q) - \edim(A)$. Note that since $(0:_Q \m_Q^2)$ is the canonical module of
$Q/\m_Q^2$, $\lambda(0:_Q \m_Q^2) = \lambda(Q/\m_Q^2)$. Also, $\ker(\pi)_i = 0$ for $i \geq 2$ gives
$\lambda(\m_Q^i) = \lambda((G_+)^i) = \lambda((A_+)^i)$ for $i \geq 2$. Hence
$\lambda((0:_Q \m_Q^2)/\m_Q^{s-1})= \lambda(Q/\m_Q^2) - \lambda((A_+)^{s-1}) = \lambda(Q/\m_Q^2) - \lambda(A/(A_+)^2)$,
where the last equality follows from $\lambda((A_+)^{s-1}) = \lambda(A/(A_+)^2)$, which holds since $A$ is a graded
Gorenstein ring.

Thus $\dim_{\sk}((0:_Q \m_Q^2)/\m_Q^{s-1}) = \lambda((0:_Q \m_Q^2)/\m_Q^{s-1}) = \edim(Q)-\edim(A)= \type(G)-1= n$ by
Remark~\ref{TG-1}.

\noindent
c) If $w\in \m_Q^2$, then $w\in \m_Q^2 \cap (0:_Q \m_Q^2)=\m_Q^{s-1}$ by (a). Hence $w\notin \m_Q^2$ and
so $w^*\notin (G_+)^2$. Now $w\cdot \m_Q^2 = 0$ implies that $w\m_Q \subseteq \soc(Q)$. Since $Q$ is Gorenstein and
$s \geq 3$, $\soc(Q) \subseteq w \m_Q$ proving $w\m_Q=\soc(Q) =\m_Q^s$. In particular, since $s \geq 3$,
$w^*\in \soc(G)\setminus (G_+)^2$.
\end{proof}

We now prove the following proposition, which is crucial in our proof of the main theorem.

\begin{proposition} \label{basisprop} Let $(Q, \m_Q, k)$ be a Gorenstein Artin local ring such that
$G=\gr(Q)$ is Gorenstein up to linear socle but not Gorenstein. If $\ll(Q) \geq 3$, there is an ideal $J$ in $Q$ such that the
following hold:\\
{\rm a)} $J\m^2_Q = 0$, and hence $J\m_Q \subseteq \soc(Q)$,\\
{\rm b)} $\mu(J) = \type(G)-1$,\\
{\rm c)} $0:_Q J + J = \m_Q$, \\
{\rm d)} $\m^r_Q= (0:_Q J)^r$ for $r\geq 2$ and\\
{\rm e)} $\mu(0:_Q J) + \mu(J) = \edim(Q)$, and hence $J^2 \neq 0$.
\end{proposition}

\begin{proof} By Lemma \ref{ST}(b), there exist elements $z_1,\ldots,z_n \in
(0:_Q\m_Q^2)\setminus \m_Q^{s-1}$, where $n = \type(G)-1$,
such that their images form a basis for $(0:_Q \m_Q^2)/\m_Q^{s-1}$.
By Lemma \ref{ST}(c), $\underline{z} \in \m_Q \setminus \m_Q^2$. Now
(a) follows by setting $J = \langle z_1,\ldots,z_n\rangle$.

Notice that $\underline z$ is a part of a minimal generating set of
$\m_Q$, i.e., $\underline z$ are linearly independent modulo
$\m_Q^2$. Indeed, suppose $\sum_{i=1}^{n} a_i z_i\in\m_Q^2$. Then
$\sum_{i=1}^{n} a_i z_i\in \m_Q^{s-1}$ by Lemma \ref{ST}(a), and
hence $\sum a_i \overline{z_i}=0$ in $(0:_Q \m_Q^2)/\m_Q^{s-1}$. Since
$\{\overline{z_1}, \ldots, \overline{z_n}\}$ is linearly independent, $a_i\in
\m_Q$ for all $i$. In particular, $\underline z$ forms a minimal
generating set for $J$ showing that $\mu(J) = \type(G)  - 1$, which proves (b).

Let $I = 0:_QJ$. If $w \in J \subseteq (0:_Q \m_Q^2)$, then $w \cdot
\m_Q^{s-1} = 0$. Hence, if $w \in J \cap I$, then $w \in (0:_Q (J+
\m_Q^{s-1})) = (0:_Q(0:_Q \m_Q^2))= \m_Q^2$ since $Q$ is Gorenstein
Artin. Thus $w \in\m_Q^{s-1}$ by Lemma \ref{ST}(a). Write $w =
\sum_{i=1}^n a_iz_i$. Since $w \in \m_Q^{s-1}$ and $\underline z$
are linearly independent modulo $\m_Q^{s-1}$, $a_i \in \m_Q$. Thus,
$w \in J \cdot \m_Q \subseteq \soc(Q)$. Therefore $J \cap I \subseteq
\soc(Q)$. Since $Q$ is Gorenstein and $J \neq 0 \neq I$, the other
inclusion is clear, giving $J \cap I = \soc(Q)$. Finally, since
$0:_Q I = (0:_Q(0:_Q J)) = J$, we see that $I + J = \m_Q$ by taking
annihilators of both sides in $Q$, proving (c).

In order to prove (d), let $A$ and $\pi:G \longrightarrow A$ be as in Proposition \ref{GLS}.
Since $z_i \in 0:_Q\m_Q^2$,
$\pi(z_i^*) \in 0:_A(A_+)^2 = (A_+)^{s-1}$. But $z_i \in \m_Q \setminus \m_Q^2$ implies that
$\deg(z_i^*) = 1$ in $G$. Hence either $\deg(\pi(z_i^*)) = 1$ or $\pi(z_i^*) = 0$ in $A$. Since
$\pi(z_i^*) \in (A_+)^{s-1}$, $\deg(\pi(z_i^*)) \neq 1$, forcing $\pi(z_i^*) = 0$ in $A$. Counting lengths, we see
that $\ker(\pi) = \langle z_1^*, \ldots, z_n^*\rangle$.

Let $\{y_1, \ldots, y_m\}$ be a minimal generating set for $I$. Since $A_+ = \pi(G_+)$ and $\pi(z_i^*) = 0$, we get
$A_+ = \langle \pi(y_1^*), \ldots, \pi(y_m^*)\rangle$. Thus $(A_+)^r = \langle \pi(y_1^*), \ldots, \pi(y_m^*)\rangle^r$
for each $r$. Therefore, the fact that $\pi: (G_+)^r \longrightarrow (A_+)^r$ is an isomorphism for $r \geq 2$ forces
$(G_+)^r = \langle\underline y^*\rangle^r$ and hence $\m_Q^r = I^r$ for $r \geq 2$.

Finally, observe that $J^2 = 0$, implies $J \subseteq 0:_Q J$ and hence $0:_Q J = \m_Q$. Now, $\mu(J) = n \neq 0$ by assumption.
Hence, $J^2 \neq 0$ follows if we prove $\mu(I) + \mu(J) = \mu(\m_Q) = \edim(Q)$.

Now $J\m_Q= \soc(Q)$ implies that $\lambda(J)=\lambda(J/J\m_Q) + 1= n+1$, and
$I^2 \subseteq I\m_Q \subseteq \m_Q^2 = I^2$ forces $I\m_Q = \m_Q^2$. Since $I \simeq 0:_QJ$ is isomorphic to the canonical module of
$Q/J$, we see that
\begin{align*}
\mu(I) = \lambda(I/\m_Q^2)& =\lambda(I)-\lambda(\m_Q^2)=\lambda(Q/J)-\lambda(\m_Q^2) =\lambda(Q/\m_Q^2)-\lambda(J)\\
& =1+\lambda(\m_Q/\m_Q^2)-(n+1)= \edim(Q) - n = \edim(A),
\end{align*}
which proves (e).
\end{proof}

\begin{remark}\label{Imp} The following is an important observation which is hidden in the proof of (d) above:
With hypothesis and notation as in Proposition \ref{basisprop},
$\langle \soc(G) \cap G_1\rangle = \langle z_1^*, \ldots, z_n^*\rangle$.
\end{remark}

We are now ready to state and prove the main theorem of this section.

\begin{theorem}\label{Structure Theorem}
Let $(Q, \m_Q,\sk)$ be a Gorenstein Artin $\sk$-algebra such that $G=\gr(Q)$ is Gorenstein up to linear socle.
If $\ll(Q) \geq 3$, then there exist Gorenstein Artin local $\sk$-algebras $R$ and $S$ such that
$Q \simeq R \#_\sk S$, where\\
a) $\ll(S) \leq 2$ and $\gr(R)\simeq G/\langle \soc(G)\cap G_{1}\rangle$.\\
b) $\p^Q(t)$ is rational in $t$ if and only if $\p^R(t)$ is so.\\
Furthermore, if $G$ is not Gorenstein, then $\ll(S) = 2$.
\end{theorem}

\begin{proof} If $G$ is Gorenstein, then one can take $Q = R$ and $S = \sk[Z]/\langle Z^2\rangle$. Then the conclusions of
the theorem are trivially true. Hence, we now assume that $G$ is not Gorenstein.

By Proposition~\ref{basisprop}, we
see that $\m_Q = \langle y_1,\ldots,y_m, z_1,\ldots,z_n\rangle$ where $n = \type(G) - 1 \neq 0$, $\underline y
\cdot \underline z=0$ and $\langle \underline z  \rangle^2 \neq 0$. Hence we can write $Q \simeq \widetilde{Q}/I_Q$
where $\widetilde Q = \sk[Y_1,\ldots,Y_m,Z_1,\ldots,Z_n]$, $I_Q \subseteq
\langle \underline Y, \underline Z\rangle^2$, $\underline Y \cdot \underline Z \in I_Q$,
$\langle \underline Z  \rangle^3 \subseteq I_Q$ and $\langle \underline Z  \rangle^2 \not\subseteq I_Q$.

Hence, if $I_R = I_Q \cap \sk[\underline Y]$ and $I_S = I_Q \cap \sk[\underline Z]$, then by
Theorem \ref{MainTheorem1}, $R = \sk[\underline Y]/I_R$ and
$S= \sk[\underline Z]/I_S$ are Gorenstein Artin such that $Q\simeq R\#_\sk S$.

Now, $\langle \underline Y, \underline Z \rangle^2\cdot Z_j \subseteq I_Q$ for each $j$
implies $\langle \underline Z  \rangle^3 \subseteq I_S$ proving $\m_S^3 = 0$. On the other hand,
$\langle \underline Z  \rangle^2 \not\subseteq I_Q$ and hence, $\m_S^2 \neq 0$. Thus $\ll(S) = 2$.

Since $\ll(Q) \geq 3 > \ll(S)$, we see that $\ll(R) = \ll(Q) \neq \ll(S)$. Hence, by (the proof of) Proposition \ref{Prop2},
$G \simeq \gr(R) \times_\sk \gr(S/\soc(S))$. In particular, $\gr(R) \simeq G/\langle z_1^*,\ldots, z_n^*\rangle$. By
Remark \ref{Imp}, we get $\gr(R) \simeq G/\langle \soc(G)\cap G_{1}\rangle$, proving (a).

Finally, note that $\ll(R) = \ll(Q) \geq 3$. Thus, the proof of (b) in this case follows from Remark \ref{PS}(c) and
Remark \ref{PS3}(b).
\end{proof}

\begin{corollary}
Let the hypothesis and notation be as in Theorem \ref{Structure Theorem} and its proof. If we further assume that
$G$ is not Gorenstein, then $\mu(I_Q) = \mu(I_R) + \binom{n+1}{2} + mn$ when $m \geq 2$ and
$\mu(I_Q) = \binom{n+1}{2} + n$ when $m = 1$.
\end{corollary}

\begin{proof} By Theorem \ref{Structure Theorem}, since $G$ is not Gorenstein, we have $\ll(S) = 2$. Thus,
if $n = \edim(S) \geq 2$, then by Remark \ref{PS}(c) and Remark \ref{PS2}(b),
$\mu(I_S) = \binom{n+1}{2} - 1$. Hence the proof follows from Proposition \ref{Prop10}(b).
\end{proof}

\begin{corollary}\label{Corr}
Let $(Q, \m_Q,\sk)$ be a Gorenstein Artin $\sk$-algebra such that
$G=\gr(Q)$ is Gorenstein up to linear socle. If $\ll(Q) \geq 3$, $\p^Q(t)$ is a rational function of $t$ in the
following situations:\\
{\rm i)} $\char(\sk) \neq 2$ and $\edim(Q) - \type(G) = 3$, or $\edim(Q) - \type(G) \leq 2$.\\
{\rm ii)} $\lambda(Q) - \type(G) \leq 10$.
\end{corollary}

\begin{proof}
Note that by Theorem \ref{Structure Theorem}(a), there are Gorenstein Artin local rings $R$ and $S$
such that $Q \simeq R \#_\sk S$, where $\ll(S) \leq 2$ and $\gr(R) \simeq G/\langle \soc(G)\cap G_{1}\rangle$.

Let $A = G/\langle \soc(G)\cap G_{1}\rangle$. By Remark \ref{TG-1}, $A$ is a graded Gorenstein Artin
$\sk$-algebra such that $\ll(A) = \ll(Q)$, $\edim(A) = \edim(Q) - \type(G) + 1$ and $\lambda(A) = \lambda(Q) - \type(G) + 1$.

\noindent
(ii) The assumption that $\lambda(Q) - \type(G) \leq 10$ implies that $\lambda(A) \leq 11$. Since $A$ is graded Gorenstein, and
hence has a palindromic Hilbert function, the hypothesis that $\ll(Q) \geq 3$ forces $\edim(A) \leq 4$. Thus (ii) reduces
to (i).

\noindent (i) In this case, $\char(\sk) \neq 2$ and $\edim(R) = \edim(A) = 4$, or $\edim(R) = \edim(A) \leq 3$.
Hence, by Remark \ref{PS}(a), $\p^R(t)$ is
rational. Thus, by Theorem \ref{Structure Theorem}(b), $\p^Q(t)$ is rational.
\end{proof}

\section{Applications: Short and Stretched Gorenstein Rings}

\begin{definition} Let $(Q,\m_Q,\sk)$ be a Gorenstein Artin local
ring with Hilbert function $H_Q$.

\noindent i) We say that $Q$ is a \emph{short Gorenstein ring} if
$H_Q = (1,h,n,1)$, i.e., if $\m_Q^3 = \soc(Q)$.

\noindent ii) We say $Q$ is a \emph{stretched Gorenstein ring} if
$H_Q =  (1,h,1,\cdots,1)$, i.e., $\m_Q^2$ is principal and $\m_Q^3
\neq 0$.
\end{definition}

In her paper on stretched Gorenstein rings, Sally proved a structure
theorem (\cite[Corollary 1.2]{Sa1}) for a stretched Gorenstein
local ring $(Q,\m_Q,\sk)$ when $\char(\sk) \neq 2$. Elias and Rossi
proved a similar structure theorem (\cite[Theorem 4.1]{ER})
for a short Gorenstein local ring $(Q,\m_Q,\sk)$ when $\char(\sk) =
0$ and $\sk$ is algebraically closed.  In particular, if $Q$ is a
$\sk$-algebra where $\char(\sk) = 0$, then, if $Q$ is either a short (with $\sk$ algebraically closed) or a
stretched (with $\ll(Q) \geq 3$) Gorenstein Artin $\sk$-algebra, the structure theorems of Elias-Rossi and Sally respectively
show that $Q$ corresponds to a polynomial of the form $F(\underline Y) + Z_1^2 + \cdots + Z_n^2$ in terms of Macaulay's
inverse systems. Thus, in the respective cases, this shows that $Q$ is a connected sum, where one of the components is
a Gorenstein ring with Loewy length equal to two.

Theorem \ref{Structure Theorem} generalizes these two results, which can be seen as follows:

\begin{proposition}\label{maincor}
Let $Q$ be either a short or a stretched Gorenstein ring. Then $G = \gr(Q)$
is Gorenstein up to linear socle.
\end{proposition}

\begin{proof}
Let $Q_0$ be the quotient of $G = \gr(Q)$ as defined by Iarrobino
(see Remark \ref{Iarrobino}). Since the last two Hilbert
coefficients of $G$ and $Q_0$ are the same by Remark
\ref{Iarrobino}, and the
Hilbert function of a graded Gorenstein $\sk$-algebra is palindromic, we see the following:\\
i) Let $Q$ be a short Gorenstein ring with $H_Q = (1,h,n,1)$. Then $H_{Q_0} = (1,n,n,1)$.\\
ii) Let $Q$ be a stretched Gorenstein ring with $H_Q =
(1,h,1,\ldots,1)$. Then $H_{Q_0} = (1,1,1,\ldots,1)$.

Thus if we take $A$ to be $Q_0$ in Proposition \ref{GLS}(ii), we see that
$G$ is Gorenstein up to linear socle.
\end{proof}

\begin{remark}\label{SSGedim}
From the above proof and Remark \ref{TG-1}, it is clear that if $Q$ is a short Gorenstein ring with $H_Q = (1,h,n,1)$, then
$\edim(Q) - \type(G) = n - 1$ and if $Q$ is a stretched Gorenstein ring with $H_Q = (1,h,1,\ldots,1)$, then $\edim(Q) = \type(G)$.
\end{remark}

\begin{theorem}
Let $(Q, \m_Q,\sk)$ be a Gorenstein Artin $\sk$-algebra with $\ll(Q) \geq 3$. Then $Q$ is stretched if and only if
$G = \gr(Q)$ is Gorenstein up to linear socle with $\edim(Q) = \type(G)$. In particular, $Q$ is a connected sum with
$\frac{1}{\p^Q(t)} = 1 - \edim(Q) t + t^2$ when $\edim(Q) \geq 2$.
\end{theorem}
\begin{proof}
Assume $G = \gr(Q)$ is Gorenstein up to linear socle with $\edim(Q)
= \type(G)$. Then Proposition \ref{GLS} and Remark \ref{TG-1} give
that $G \simeq A \times_\sk B$ with $\edim(B)=\edim(Q)-1$ and
$\edim(A)=1$. Since $\ll(A) = \ll(Q)$, we have $Q$ is stretched by
Theorem \ref{Structure Theorem}. The converse is a consequence of Proposition \ref{maincor} and Remark \ref{SSGedim}.

Finally, the formula for $\frac{1}{\p^Q(t)}$
follows from Remark \ref{SSGedim} and Remark \ref{PS3}.
\end{proof}

\begin{remark}
In \cite{CENR}, the authors show that if $(Q, \m_Q,\sk)$ is a short Gorenstein Artin $\sk$-algebra with
Hilbert function $H_Q = (1,h,n,1)$, then $\p^Q(t)$ is rational with the assumption that $\sk$ is an algebraically closed field of
characteristic zero. In Theorem \ref{n=4}, we show the same is true for $n \leq 3$ without any restrictions on $\sk$ and for
$n = 4$ with the extra assumption that $\char(\sk) \neq 2$. \end{remark}

\begin{theorem}\label{n=4}
Let $(Q, \m_Q,\sk)$ be a short Gorenstein Artin $\sk$-algebra with
Hilbert function $H_Q = (1,h,n,1)$. Then $Q$ is a connected sum.
Furthermore, if $\char(\sk) \neq 2$ and $n = 4$, or $n\leq 3$, then $\p^Q(t)$ is rational.
\end{theorem}
\begin{proof}
By Theorem \ref{Structure
Theorem} and Proposition \ref{maincor}, $Q$ is a connected sum. The fact that $\p^Q(t)$ is rational
follows from Remark \ref{SSGedim} and Corollary \ref{Corr}.
\end{proof}

\noindent
{\bf Acknowledgement}\\
We would like to thank L. L. Avramov for comments inspiring this work. We would also like to thank him,
A. A. Iarrobino and M. E. Rossi for insightful discussions and encouraging words along the way.


TITLE: Is the set of all $(x,y,z)$ such that $z^2-x^2-y^2-1 = 0$ open or closed?
QUESTION [3 upvotes]: As the title says, is the set of all $(x,y,z)$ such that $z^2-x^2-y^2-1 = 0$ open or closed? Moreover, how can I prove it?
I understand the definition of open and closed sets, but I don't get this exercise yet. I've been trying this for days.
Thanks for all help you can give me.
Regards!

REPLY [6 votes]: Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ be defined by $f(x,y,z)=z^2-x^2-y^2-1$. Obviously, $f$ is continuous. Your set is $f^{-1}(\{0\})$. Since $\{0\}$ is closed, your set is closed. Since $\mathbb{R}^3$ is connected, your set can't be open, since it is obviously not empty nor the entire $\mathbb{R}^3$.


TITLE: Can someone help check if I evaluated my integral right?
QUESTION [1 upvotes]: The integral given to be evaluated is $\int 6x\cdot\mathrm{arctanh}(x)\,dx$.
I tried to evaluated and got the following answer:
$$3x^2\operatorname{arctanh}x + 3x -3\ln\,\bigl\lvert(1+x)/\sqrt{1-x^2}\bigr\rvert + c$$
I used some trig substitutions and integration by parts to find it. Did I get it right or no?
Thank you.

REPLY [2 votes]: Consider the function $y = \tanh^{-1} x$; that is to say, $$x = \tanh y = \frac{\sinh y}{\cosh y} = \frac{e^y - e^{-y}}{e^y + e^{-y}}.$$  Hence $$0 = x(e^y + e^{-y}) - e^y + e^{-y} = (x-1)e^y + (x+1)e^{-y},$$ or equivalently, $$e^{2y} = \frac{1+x}{1-x}.$$  Therefore, $$y = \tanh^{-1} x = \frac{1}{2} \log \frac{1+x}{1-x}, \quad |x| < 1.$$  Now observe that $$\begin{align*} \log \left| \frac{1+x}{\sqrt{1-x^2}} \right| &= \log \left| \frac{1+x}{\sqrt{1-x} \sqrt{1+x}} \right| = \log \left| \frac{\sqrt{1+x}}{\sqrt{1-x}} \right| = \log \left| \sqrt{\frac{1+x}{1-x}} \right| = \frac{1}{2} \log \left| \frac{1+x}{1-x} \right| \\ &= \tanh^{-1} x. \end{align*}$$  This immediately leads to the desired simplification.