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Yes: this is an old chestnut. Let me write $\oplus_n\mathbf{Z}$ for what you call $\mathbf{Z}[x]$ and $\prod_n\mathbf{Z}$ for what you call $\mathbf{Z}[[x]]$ (all products and sums being over the set …

Every few years this question re-appears in my life (as it just did this week) and I have to rediscover the Euler Characteristic proof which I half-remember. Let me just write my thoughts down here so …

In this question, all rings are commutative with a $1$, unless we explicitly say so, and all morphisms of rings send $1$ to $1$. Let $A$ be a Noetherian local integral domain. Let $T$ be a non-zero $ …

The question. Let $R$ be a commutative ring. Let $M$ be an $R$-module with the property that there exists an $R$-module $N$ such that $M\otimes_R N\cong R$. Does there always exist an ideal $I$ of $R$ …

For Dedekind domains, like the integers of a number field, PID iff UFD. There's definitely a quantitative statement relating the class number to failure of PIDness: the higher the class number, the sm …

I am surprised that Brian got to this one first without making what I thought was another obvious comment: affinoids are Jacobson rings! A function which is zero at all points of an affinoid rigid spa …

Your proof seems wrong to me. I might be misunderstanding some things you wrote, but surely $\mathbf{Q}{}_p=\mathbf{Z}_p[X]/(pX-1)$ is finite type over $\mathbf{Z}_p$, and contains many elements which …

If U is an open set in X, but U isn't X, then there are non-zero sheaves on X whose support lies outside U. Now add O_X to one of these to get a counterexample.

Look at the subspace of $\mathbf{A}^{n+1}$ cut out by your polynomials. This set is invariant under the diagonal action of $k^\times$. So the functions that vanish on it will be an ideal $I$ (the radi …

I computed the nimbers of a few rings, for what it's worth. I don't see any sensible pattern so perhaps the general answer is hopelessly hard. This wouldn't be surprising, because even for very simple …

Every matrix with trace zero over a PID is a commutator, according to the MR review of Rosset, Myriam(IL-BILN); Rosset, Shmuel(IL-TLAV) Elements of trace zero that are not commutators. Comm. Algebra …

I now believe a-fortiori's argument: translations are a problem, but, as a-fortiori observed, they are the only problem. Let me spell it out. Say $f:Sym(M)\to Sym(N)$ is an isomorphism. For $m\in M$ …

Let me have a punt at this. $F$ alg closed inside $F'$ iff $\overline{F}\otimes_FF'$ is a field, right? So now it's easy because $\overline{L}$ is an algebraic closure of $F$, and I don't think I even …

Here is an example where representability fails. If $R$ is an $A$-algebra representating $\otimes_AM$ on $A$-algebras, and if $B\to C$ is an injective map of $A$-algebras, then $R(B)\to R(C)$ will be …

Contrary to what I guessed initially, I now think the question has a great answer: the functor is representable if and only if $M$ is locally free, and the proof is EGA I, 9.4.10. Edit: this is an an …