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I believe that the answer is yes. Put $c:=g(x_1,\ldots, x_n)$, which is irreducible in the UFD $R:=k[x_1,\ldots, x_n]$. Assume $f$ is irreducible, but also assume by way of contradiction that $f(c)= …

As stated, I believe the answer is no. Set $q_1=x_1^2$, $q_2=x_2^4$, and consider $I=(q_1,q_2)$. Let $p_1=q_1$, and let $p_2=x_1p_1$. Then $p_1$ is (up to a constant) the only degree 2 polynomial i …

Suppose $F$ is a field, and $F_1, F_2$ are two extension fields of $F$. Is it always the case that there is a field $L$, containing three subfields $F, K_1, K_2$ and two ring isomorphisms $\varphi_{i …

The answer is no. In fact, even if $R$ is a (possibly noncommutative) right self-injective ring with no nontrivial idempotents, then $R$ is a local ring, and the unique maximal (left or right) ideal …

I assume that in condition (2) you meant to say that each monomial appearing in the $F$-support of $Red(f)$ is $\leq f$. (In other words, you should not assume reduction simply take monomials to mono …

Claim 1: Any injective module is the injective hull of a direct sum of cyclic modules. Proof: This is an easy Zorn's lemma type argument. Check out Lam's solution to exercise 3.22 in "Exercises in M …

(I would have made this a comment, but I don't yet have the 50 reps needed.) You can find Klein's 4 page paper by googling "Abraham Klein bounded index nilpotence" and clicking the first link (which …

I believe that the answer is yes. Let $A$ be a commutative ring with $1$, such that any non-empty set of ideals has a maximal element. For each ideal $I\leq A[[x]]$, and each $n\in \mathbb{N}$, defi …

Here is my, admittedly, ad hoc way of proving they are distinct. It comes from trying to make it concrete that the graded pieces have different sizes. First, you better assume that $n\geq 2$ as thes …

Edited due to mistakes pointed out in the comments: I think the answer to the problem might be yes. Here are some preliminary thoughts. First, you know $f_{\nu}(x)$ divides $f(x)$, since $f(x)$ is …

This is just an extended comment: If $Ra$ is essential in a direct summand of $R,$ then $a$ has this property. To see this, fix some idempotent $e\in R$ such that $Ra$ is essential in $Re.$ Then if …

On question 1: Let $A=\mathbb{F}_2[a,b,d\ :\ a^2=b^2,ad=a,bd=b]$. The element $a$ is not prime since $A/(a)$ has a nonzero nilpotent element $\overline{b}$. The element $a$ also is not irreducible s …

"Is there a property of an ideal $I$ that guarantees that $R/I$ is a principal ideal domain? A Bézout domain? Euclidean?" First, a clarification. Notice that the condition for an ideal $p$ to be prim …

Let $R$ be an integral domain. Given $a,b\in R$, then a division chain for $(a,b)$ is a sequence where we take $r_{-1}=a$, $r_0=b$, and for each $n>0$ we take $r_n=r_{n-1}s_n+r_{n-2}$ for some $s_n\i …

The answer to the question posed by Aaron Meyerowitz to darij grinberg in the comments is unfortunately negative, even in the integers, by taking $a=c=u=v=w=0$ but $b=d=1$. However, it has a positive …