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It is well-known that if we pass from a UFD to a new ring where we have factored one of the primes, it does not need to stay a UFD. The classic example is passing from $\mathbb{Z}$ to $\mathbb{Z}[\sq …

I have a (fairly large) system of polynomial equations, of the form $$ c_1d_1=0,\ c_1d_2+c_2d_1=0,\ldots $$ (In case it is relevant, all the polynomials are homogeneous of degree 2, except for exactly …

"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 …

Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)? Edited to add: As no answers are forthcoming, does anyone …

The following are just some partial thoughts about the new question which are too long for comments. Throughout, assume that $R_{\mathfrak{m}}$ is a UFD, $R_0$ is a field, etc... First, let $x\in \ma …

(Note: The original question had an incorrect premise. The ideal generated by elements of positive degree in an arbitrary graded ring is not generally maximal.) Here is the solution to the original p …

(This is only a comment.) If you instead look at the family $\{x^d-x+n\}$, then it is known that there are infinitely many irreducible elements (for each positive $n$). It suffices to show that $f(x …

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 …

I've figured out how to do the finite graph case, in the negative. The ordering of quadratically closed fields is not dense in that case either. Throughout let $qcl(F)$ denote the quadratic closure …

There are three answers. Throughout let $qcl(F)$ be the quadratic closure of a field $F$ inside $\mathbb{C}$. Part 1: Yes there is a quadratically closed field strictly between $qcl(\mathbb{Q})$ and …

The answer is no. Consider the set $\mathbb{N}^{<\omega}$, consisting of $\omega$-tuples with only finitely many nonzero entries. This set is totally ordered under the relation $\prec$, where $(a_0,a …

Factorization in the ring $\mathbb{Z}[x]/(x^2+1)\mathbb{Z}[x]\cong \mathbb{Z}[i]$ is well known. For instance, $5$ and $13$ (and any prime $\equiv 1\pmod{4}$) are no longer prime. The factorization o …

Fix $I_2=\langle x^2\rangle \vartriangleleft \mathbb Z[x]$. Let $f=x$. For each polynomial $g\in \mathbb{Z}[x]$ we have $h(1-gf)\in 1+I$ when taking $h=1+gf$. Your question: "Which polynomials $g\in …

Sketch: First, if $e$ is an idempotent in $A$, show that $B$ can be replaced with $B+Re$, and the hypotheses still holds. Use this to reduce to the case that $A$ and $B$ contain the same idempotents …

In rings: Let $R$ be a ring where idempotents lift modulo the Jacobson radical $J(R)$. Any countable set of orthogonal idempotents in $R/J(R)$ lifts to an orthogonal set of idempotents; but this fai …