TITLE: Global Dimension of a Ring and its Localizations
QUESTION [6 upvotes]: Why is the following true?

The global dimension of a noetherian ring $A$ is the supremum of the global dimension of its localizations at its maximal ideals:
  $$\operatorname{gldim}(A)=\sup_{\mathfrak m\in\operatorname{SpecMax}A} \operatorname{gldim}(A_{\mathfrak m}).$$

Could one approach by the definition of global dimension that is the supremum of projective dimensions of all $A$-modules?
Thanks for any help!

REPLY [3 votes]: If $S\subset A$ is a multiplicative set then $$\operatorname{gldim}(S^{-1}A)\le\operatorname{gldim}(A)$$ since $\operatorname{pd}_{S^{-1}A}(S^{-1}M)\le\operatorname{pd}_AM$ and every $S^{-1}A$-module is of the form $S^{-1}M$ with $M$ an $A$-module.
Suppose now $$\sup_{\mathfrak m\in\operatorname{SpecMax}A} \operatorname{gldim}(A_{\mathfrak m})<\operatorname{gldim}(A)=n<\infty.$$
Let $M$ be a finitely generated $A$-module with $\operatorname{pd}_AM=n$. Then $\operatorname{pd}_{A_{\mathfrak m}}M_{\mathfrak m}\le n-1$. (Recall that $\operatorname{gldim}(A)=\sup\operatorname{pd}_A(M)$ with $M$ finitely generated.) Consider a projective (free) resolution of $M$ of length $n$, and denote by $K_i$ the $i$th syzygy of $M$, that is, the kernels of the maps in a projective resolution. Since $\operatorname{pd}_{A_{\mathfrak m}}M_{\mathfrak m}\le n-1$ we get that all localizations $(K_{n-1})_{\mathfrak m}$ are projective (free). It follows that $K_{n-1}$ is finitely generated and flat, so it is projective. This entails $\operatorname{pd}_{A}M\le n-1$, a contradiction.


TITLE: I don't imagine how to make a cartesian equation of the parametric one: $x=2 \sin (t), y=-\frac{4\cos (t)^2 (2 + 1.3\cos(t))}{3 + \sin (t)^2}$
QUESTION [0 upvotes]: I know it's a downvoting question but I'm not good at math. I know that $x=a \sin(t)$ , $y=\frac{a \cos^2(t)(2+ \cos(t))}{(3- \sin^2(t))}$ is $y^2(a^2-x^2)=(x^2+2ay-a^2)^2$ but what it could be as a cartesian equation with the negative $y$ part I can't even imagine. Sorry. Could you help me?

REPLY [2 votes]: $$y=-\frac{4 \cos ^2(t) \left(\frac{13 \cos (t)}{10}+2\right)}{\sin ^2(t)+3}$$
$$\sin t=\frac{x}{2}$$
thus
$$y=-\frac{4 \left(1-\frac{x^2}{4}\right) \left(\frac{13 \cos (t)}{10}+2\right)}{\left(\frac{x}{2}\right)^2+3}$$
with some algebra we get
$$\cos t=\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}$$
and then
$$\cos^2 t=\left(\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}\right)^2$$
as $\cos^2 t= 1-\sin^2 t$ we get
$$1-\frac{x^2}{4}=\left(\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}\right)^2$$
and finally
$$676 \left(x^2-4\right)^2=25 \left(x^2 y-8 x^2+12 y+32\right)^2+169 x^2 \left(x^2-4\right)^2$$


TITLE: Using L'Hôpital's rule on $\lim_{x \to 0}(xe^{\frac{1}{x}}-x) $
QUESTION [1 upvotes]: As stated, we are supposed to use L'Hôpital's rule on:
$$\lim_{x \to 0}(xe^{\frac{1}{x}}-x) $$
We can write this function as:
$$\frac{e^{\frac{1}{x}}-1}{\frac{1}{x}}=\frac{e^{\frac{1}{x}}-1}{x^{-1}}$$
I'm a bit uncertain of the derivative of $e$ to the power of a fraction but I guess the numerator can be written as:
$$e^{x*\frac{1}{x^2}}$$
So the derivative would be:
$$\frac{1}{x^2}e^{x*\frac{1}{x^2}}=\frac{1}{x^2}e^\frac{1}{x}$$
The derivative of the denominator should simply be:
$$-1*x^{-2}=-\frac{1}{x^2}$$
Thusly
$$\frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}=\frac{\frac{1}{x^2}e^\frac{1}{x}}{-\frac{1}{x^2}}=-\frac{x^2}{x^2}*e^\frac{1}{x}$$
As x tends to 0, $e^{\frac{1}{x}}$ tends to infinity, and $-\frac{x^2}{x^2}$ should tend to...-1 I guess since the denominator will always equal the numerator. In any case, this term should always be negative when $x \neq 0$, so x should equal negative infinity.
That's not the answer however, $x$ should equal positive infinity, how do I reach this result?

REPLY [0 votes]: We have that for $x\to 0^+$ by $y=1/x \to \infty$ by l’Hopital rule we have
$$\lim_{x\to 0^+} xe^{1/x}-x=\lim_{y\to +\infty} \frac{e^y-1}{y}=\lim_{y\to +\infty} e^y= \infty$$
but for $x\to 0^-$ by $y=1/x \to -\infty$
$$\lim_{x\to 0^-} xe^{1/x}-x=\lim_{y\to -\infty} \frac{e^y-1}{y}=0$$
therefore the given limit doesn’t exist.


TITLE: Connection between results of two SVDs
QUESTION [1 upvotes]: Consider SVD of $M$:
$$
M = U \Sigma V^\top
$$
And SVD of $N= \ln M$:
$$
N = U^\prime \Sigma^\prime V^{^\prime\top}
$$
Anyone knows/has seen/can think of any interesting connection/relation between $U$ and $U^\prime$? (Or $\Sigma$ and $\Sigma^\prime$? )
Update: Let's suppose that both decompositions exist.
Update2: By relation, I mean not the trivial relation of $\ln( U \Sigma V^\top) \approx U^\prime \Sigma^\prime V^{^\prime\top}$.
[1] http://en.wikipedia.org/wiki/Singular_value_decomposition

REPLY [1 votes]: Assume that $\log$ denotes the principal logarithm and $M$ has no $\leq 0$ eigenvalues. Then $\log(M)$ is defined and the columns of $U,V$ are the eigenvectors of $MM^T,M^TM$ and the columns of $U',V'$ are the eigenvectors of $\log(M)\log(M)^T,\log(M^T)\log(M)$.
If $M$ is symmetric with eigenvalues $>1$, then $U=U',V=V',\Sigma'=\log(\Sigma)$. If $M=diag(1/2,3,5)$ then the condition $U=U',V=V'$ is not satisfied.
EDIT: Vedran is right. More precisely, let $f:U\subset M_n(\mathbb{R})\rightarrow M_n(\mathbb{R})$ be a matrix function (cf. the Higham's book) s.t. $f$ is defined on $spectrum(M)$. Then $f(M)$ is a polynomial in $M$. If $M$ is diagonalizable ($M=PDP^{-1}$) , then $f(M)=Pf(D)P^{-1}$. If $M$ is symmetric $\geq 0$, then $PDP^T$ ($P$ orthogonal) is a SVD of $M$ and $f(M)=Pf(D)P^T$ is a SVD of $f(M)$ only if $f(D)\geq 0$.