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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.