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Let $R$ be a commutative ring. It is well-known that if $b \in R$ and $c \in R$ are two nilpotent elements with $b^k = 0$ and $c^\ell = 0$ (where $k$ and $\ell$ are positive integers), then $b+c$ is n …

We have $K\left[x_1,...,x_m,y_1,...,y_n\right] \cong K\left[x_1,...,x_m\right] \otimes K\left[y_1,...,y_n\right]$ (where all tensor products are over $K$), and under this isomorphism, the ideal of $K\ …

This is a sidenote to Sasha's answer. The "yes" part can be proven in a completely elementary way without prime ideals and Krull dimension. Here is a sketch, as I have to prepare a talk for Monday and …

The following sounds so natural, I'm surprised I have never asked it before: Question 1. Let $R$ be a commutative ring. Let $P \in R\left[X\right]$ be a polynomial. Can we find a commutative ring $S$ …

You don't need most of your assumptions to ensure that your quotient is $0$. More generally, we have: (1) If $A$ is any commutative ring, and $f : M \to N$ is a surjective homomorphism of $A$-modules …

Yes. I claim that (1) $\prod\limits_{i=1}^s \left(M-u_i\right) = \prod\limits_{i=1}^{t} \left(m-u_i\right) \cdot \prod\limits_{i=t+1}^{s} \left(M-u_i\right) $ for every $t \in \left\lbrace 0,1,...,s …

Yes, it is. Let a tensor $t$ be in the equalizer of $f$ and $g$. Then, $f\left(t\right)=g\left(t\right)$. If we write $t$ in the form $\sum\limits_{j\in I} a_j\otimes b_j$ (with $I$ being a finite set …

This question is posed way too generically in order to obtain an answer that is useful to you by more than mere coincidence, but here are three things that I found of use: (1) Your algebra is graded, …

I tend to believe that the answer to your question, in the generality you want, is negative. Since I am not sure of the proof (and have not written it up in detail), I am making this answer community …

This looks extremely easy, but then again it's late at night... Let $k$ be a commutative ring with unity. An element $a$ of a $k$-algebra $A$ is said to be transcendental over $k$ if and only if ever …

For the sake of completeness, here is the proof I suggested in the comments, in some more detail. Lemma 1. Let $\mathbf{k}$ be a commutative ring. Let $A$ be a $\mathbf{k} $-algebra. Let $I$ be a two …

Fix a field $\mathbf{k}$ and an $\mathbb{N}$-graded commutative $\mathbf{k}$-algebra $A = \bigoplus\limits_{n = 0}^{\infty} A_n$ of finite type. ("Finite type" means that each $A_n$ is a finite-dimen …

All three questions are answered by Proposition (1.5) in Chapter PARTL ("Partially symmetric functions") in The LLPT Notes (by Dan Laksov, Alain Lascoux, Piotr Pragacz and Anders Thorup). (The notes a …

Solved. The key is that the set of polynomials $P$ for which there exist polynomials $P_d\in\mathbb{Z}\left[\Xi\right]$ for all divisors $d$ of $n$ such that $\displaystyle P=\sum_{d\mid n}dP_d^{n/d}$ …

EDIT: Proved it on my own. It easily follows from the Witt integrality theorem. Sorry for posting. Let $P\in\mathbb{Z}\left[\Xi\right]$ be a polynomial (where $\Xi$ is a family of symbols that we use …