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seen Apr 15 at 21:11

Apr
15
comment Isomorphisms between finite abelian groups and cyclic groups
@LokiClock No problem.
Apr
15
comment Isomorphisms between finite abelian groups and cyclic groups
@LokiClock: actually, if $f\colon A \to B$ is an injective map between two finite sets of the same cardinality, then $|f(A)|=|A|=|B|$ so that $f(A)=B$.
Apr
15
comment Isomorphisms between finite abelian groups and cyclic groups
Regarding the surjective homomorphism $C_5\times C_5 \to C_{25}$, recall that for finite sets of the same cardinality, surjectivity is equal to injectivity.
Apr
15
awarded  Yearling
Apr
15
comment Isomorphisms in finite abelian groups
You're welcome!
Apr
15
comment Show that if B is simply-connected, then p is a homeomorphism.
I think this works fine!
Apr
15
answered Isomorphisms in finite abelian groups
Apr
15
comment Show that $E[(X-E[X])^2] = E[X^2]-E^2[X]$
Linearity also implies that you can bring constants out of the expectation, and the expectation of a constant is the constant.
Apr
15
comment Is the relative ideal of two affine curves $C\subset Z$ a finite dimensional vector space?
Anyway, if $I$ is homogeneous, then you can compute the Hilbert function of $J$ as the difference of the Hilbert functions of $(x,y)$ and $I$, and from this it is easy to see whether $J$ has finite dimension or not.
Apr
15
comment Is the relative ideal of two affine curves $C\subset Z$ a finite dimensional vector space?
Ok, but if $I$ is not homogeneous to begin with, you could get strange things. For example, if $I=(x+1)$ then $I\cap A_k=0$ for every $k$.
Apr
15
comment Fixed Point Involutions
For the first point, you could use the fact that every automorphism of $\mathbb{P}^2$ is linear. For the second point you can use again the strategy of user8268 and see what happens with the induced covering.
Apr
15
comment How to compute Lipschitz Constant for multivariate function $f(x,y)=1-xy$?
There is also the definition for a general metric space. In your case, a function $f\colon \mathbb{R}^2 \to \mathbb{R}^2$ is said to be Lipschitz continuous if there exists a $C>0$ such that $|f(x)-f(y)|\leq C\cdot |x-y|$ for all $x,y\in \mathbb{R}^2$. By $|\cdot |$ you could take the standard euclidean norm on $\mathbb{R}^2$.
Apr
15
comment How to compute Lipschitz Constant for multivariate function $f(x,y)=1-xy$?
en.wikipedia.org/wiki/Lipschitz_continuity
Apr
15
comment Is the relative ideal of two affine curves $C\subset Z$ a finite dimensional vector space?
Is your $I$ homogeneous? Otherwise I would not know the definition of Hilbert function.
Apr
15
comment Prove that if $C=A+iB$ is invertible, then so is $A+\lambda B$ for some $\lambda$
$\det(A+\lambda B)$ is a polynomial in $\lambda$.
Feb
11
awarded  Nice Question
Feb
9
comment a “paradox” regarding regular and complete intersection rings
yes, that's precisely the problem with the argument, as it has been clearly explained by user115654 below.
Feb
7
comment a “paradox” regarding regular and complete intersection rings
I think that the problem is that $S/(x_1,\dots,x_\nu)$ will not be regular, so that it is not true that the $x_1,\dots,x_\nu$ can be extended to a system of parameters. For example, if $S=k[X]_{(X)}$ and $I=(X^2)$ then $S/(X^2)$ is not regular since it is not reduced. I think that your argoment proves the general fact by contradiction.
Feb
7
asked Vanishing of global sections with very negative twists.
Jan
18
comment Asymptotics of partitions in at most n parts, bounded by r
@Wouter: yes, I had seen something related to the q-Pochammer function as well, but I did not manage to obtain a result from it. Thanks for the segnalation of the typo, it should be fixed now.