11
This is false.
Let $p$ be an odd prime, let $\ell$ be another prime, and let $m$ be a small prime divisor of $p^{\ell}-1$, that doesn't divide $p-1$. Let $n= 1 + \frac{ p^{\ell}-1}{m}$.
Then $n-1$ is a multiple of $p-1$, is not a multiple of $p^{\ell}-1$, and is not a multiple of $p^{k}-1$ for any other $k$ because $p^{\ell}-1$ is not a multiple of $p^{k}-...
9
For the module-finite (i.e. second) question:
Yes, it is true: The point is that $A \subseteq B$ is integral, and the claim is true for all integral extensions.
When $A \subseteq B$ is integral, every prime $\mathfrak{p} \subseteq A$ comes as $\mathfrak{P}\cap A$ for a prime $\mathfrak{P} \subseteq B$ ("lying over theorem"), and morever, if $\mathfrak{P}$ ...
8
No, $R$ is not necessarily Japanese. What follows is a explicit example from a note that Rankeya Datta and I wrote, now available on his website.
Example. We use Hochster's example [Hochster, Ex. 1]; see one of my other answers for the relevant results therein. Let $I$ be the set of positive integers, and set
$$R_i := k[x_i^2,x_i^3] \qquad\text{and}\qquad ...
7
A counterexample for the first question is any DVR $R$. Clearly, $R$ is not Jacobson. But if $\pi$ is the uniformizer, then $Q(R) = R[\frac{1}{\pi}]$ is a finitely generated $R$-algebra and a field, hence Jacobson.
2
As requested in the comments, here is Proposition 6.1.6.1 of Lurie's "Spectral Algebraic Geometry", specialized to the case of ordinary commutative rings:
Let $n\ge 0$ be an integer, let $A_0$ be a commutative ring and let $B_0$ be an $A_0$-algebra of finite presentation. Suppose we are given a diagram $\{A_\alpha\}_{\alpha\in I}$ of $A_0$-algebras ...
1
Aldo Conca and Matteo Varbaro have posted a preprint that purports to answer this question, at least for graded ASLs:
Squarefree Gröbner degenerations
It appears that if $A$ is Cohen-Macaulay, then $k[P]$ must also be Cohen-Macaulay! See Corollary 3.9.
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