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I asked this question on Mathematics Stackexchange (link), but got no answer. Let $K$ be a field, let $x_1,x_2,\dots$ be indeterminates, and form the $K$-algebra $A:=K[[x_1,x_2,\dots]]$. Recall th …

This question was asked, but not answered, on Mathematics Stackexchange. [In this post "ring" means "commutative ring with one".] Let $A$ be a noetherian ring, and let $f:A\to A$ be an endomorphism …

I asked this question on Mathematics Stackexchange, but got no answer. Let $A$ and $B$ be noetherian commutative rings with one, and let $f:A\to B$ and $g:B\to A$ be epimorphisms. Are the rings …

Edit of Feb. 14, 2019. After Laurent Moret-Bailly's accepted answer, only Questions 4 and 5 remain open. I don't care that much about Question 4, but I'm very curious about Question 5, which is Do …

I asked this question on Mathematics Stack Exchange but got no answer. Here is the question: Let $A$ be a domain (that is, a commutative ring with one in which the condition $ab=0$ implies $a=0$ or …

I asked this question on Mathematics Stackexchange, but got no answer. Is the product $$ \prod_{i\in I}A_i $$ of a family $(A_i)_{i\in I}$ of Jacobson rings a Jacobson ring? (Here "ring" means "c …

I asked this question on Mathematics Stackexchange, but got no answer. Let $B$ be a commutative ring with $1$, let $A$ be a subring such that any unit of $B$ which belongs to $A$ is a unit of $A$, an …

EDIT OF JULY 26, 2017 Proposition 2.4 page 21 reads: Let $M$ be a finitely generated $A$-module, let $\mathfrak a$ be an ideal of $A$, and let $\phi$ be an $A$-module endomorphism of $M$ such tha …

EDIT OF AUG. 31, 2010. The proof of the Cayley-Hamilton Theorem I like best (among the ones I know) is on page 21 (proof of Proposition 2.4) of Introduction to Commutative Algebra by Atiyah and MacDon …

Consider the following Condition (C) on a positive integer $k$: (C) If $R$ is a commutative ring, if $F$ is a free $R$-module, and if $f$ is an endomorphism of $F$, then $f$ is an $R$-linear combinat …

In view of Mariano Suárez-Alvarez's answer I see how badly phrased my question was, and decided to rewrite it. The drawback is that some comments of Martin Brandenburg are now incomprehensible, but I …

The Chinese Remainder Theorem gives a way to compute matrix exponentials. Indeed, let $A$ be a complex square matrix, put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algeb …