4 votes

If $G=AB$ and $|G|=|A||B|$, then $G=A^{-1}B$.

Maps between finite sets with the same cardinality are injective iff surjective. Because of $|G|=|A||B|=|A\times B|$, such a map is $\phi\colon A\times B\rightarrow G,(a,b)\mapsto ab$. It is ...
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4 votes

Finitely generated $R$-Algebra vs finitely generated $R$-Module for $R$-Algebra $S$

Consider a set $Q \subseteq S$. Suppose $Q$ generates $S$ as an $R$-module. Then I claim $Q$ also generates $S$ as an $R$-algebra. For suppose we have some sub-$R$-algebra $A \subseteq S$ such that $Q ...
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3 votes
Accepted

If $G=AB$ and $|G|=|A||B|$, then $G=A^{-1}B$.

The key here is the following property of maps on finite sets: Suppose $f : X \to Y$, where $X$ and $Y$ are finite with $|X| = |Y|$. Then $f$ is injective if and only if it is surjective. Let us ...
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  • 43.4k
2 votes

integrality under base change (Vakil 7.3.N)

Just to answer my own question here from @CraniumClamp 's hint, I should note that I originally wasn't being careful to show that $A \otimes_BC$ is integral over $C$ instead of $B$ — keeping track of ...
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2 votes

Every PID is hereditary

A further reference are the lecture notes by P. Clark. A ring is called hereditary if every submodule of a projective module is projective. Then he states the example, see page $57$: Example: A PID is ...
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1 vote
Accepted

Symmetric part of the inverse matrix

In case $\operatorname{Sym}(A)$ is positive definite, we have $\operatorname{Sym}(A^{-1})\preceq\operatorname{Sym}(A)^{-1}$. Let $A=P(I-K)P$ where $P=\operatorname{Sym}(A)^{1/2}\succ0$ and $K$ is skew-...
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  • 123k
1 vote

A problem involving prime ideals that are not maximal in $K[x,y]$

Yes, you can say this. Let us suppose $K[x] \cap P$ is nontrivial. Since $K$ is algebraically closed and $P$ is prime, there is some $a \in K$ such that $x - a \in P$. And if $K[y] \cap P$ is also ...
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1 vote

How to check if ring of formal power series is a discrete valuation ring?

My try (there maybe a chance that I am entirely off.. I'm still learning): So, in your definition we know that the ring is DVR if it's at least a PID. Consider the ideal generated by $(X^{\frac25} )$ ,...
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1 vote

A B-topological ring that is not semi-topological?

It is straightforward to adapt Sebastian Spindler's answer to your other question to this one. Let $R=\mathbb{R}$ with the usual notion of addition, and $$a\times_R b=\begin{cases}ab&a,b\in\...
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1 vote

Every PID is hereditary

There are a few ways to do this. The important fact is : $(*)$ a submodule of a free module $R^n$ over a PID $R$ is free. Now there are a few equivalent ways to proceed. $(*)$ implies that every ...
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1 vote
Accepted

Let $M = \operatorname{cok}(\phi)$ where $\phi: \mathbb{Z}^3 \to \mathbb{Z}^3$. Express $M$ as a direct sum of cyclic modules.

Smith Normal Form and isomorphism type of cokernel look good. You can do a sanity check by considering the rank of the same matrix over $\mathbb Q$. The first two columns are linearly independent so ...
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