Let $R$ be a discrete valuation ring with a uniformizer $\pi$ and $(A, \mathfrak{m}_A)$ a local $R$-algebra that is essentially of finite type (i.e., is a localization of a finite type $R$-algebra), such that the image of $\pi$ is in $\mathfrak{m}_A$, and such that $A/\mathfrak{m}_A$ is a finite extension of $R/\pi$. Choose $x_1, \dotsc, x_n \in \mathfrak{m}_A$ such that $(\pi, x_1, \dotsc, x_n) = \mathfrak{m}_A$ in $A$ and let $f\colon R[X_1, \ldots, X_n] \rightarrow A$ be the $R$-algebra morphism satisfying $f(X_i) = x_i$. Let $B$ be the localization of $R[X_1, \ldots, X_n]$ at $(\pi, X_1, \dotsc, X_n)$ and let $g\colon B \rightarrow A$ be the morphism induced by $f$. Does $g$ make $A$ a finite type $B$-algebra?
$\begingroup$
$\endgroup$
12
-
1$\begingroup$ Please give more motivation for your questions, so people can recognize the context before spending time to try to answer them. For example, is your question related to some step in the proof of Artin approximation as presented in the book "Neron Models"? $\endgroup$– user27920Commented Nov 5, 2014 at 7:47
-
$\begingroup$ @user52824: Maybe Question Mark is just trying to figure out what is correct and what is incorrect in those sources. There are mistakes, and it is good for students to try to find them and correct them. $\endgroup$– Jason StarrCommented Nov 5, 2014 at 13:32
-
2$\begingroup$ @JasonStarr: Sure, but I have responded to many of QM's questions and often only find out afterwards what the motivation was. It would be nice to know beforehand (e.g., "I am reading book X and on page Y the following thing happens which is unclear..."). In the end if I or you are doing the actual "correcting" then it can help in making a more informed response if we know where the question is coming from. For example, maybe the question is about an "error" which you or I can see does not affect the intended application but QM doesn't realize that. $\endgroup$– user27920Commented Nov 5, 2014 at 16:12
-
$\begingroup$ OK, I agree that this question is a bit contrived. This is a step in the proof of 3.6/16 in "Neron models." Namely, this concerns paragraph 3 on p. 92 and its sentence "it is easily seen that $Spec(A_0) \rightarrow Spec(S_0)$ is quasi-finite at..." My question is why that map is of finite type to begin with (as is part of the definition of quasi-finite). The reason why I sometimes avoid giving context is because I don't like answers of the sort "oh, but that doesn't matter, since in the main case of interest so-and-so this can be seen directly..." (the case of interest depends on taste...). $\endgroup$– Question MarkCommented Nov 5, 2014 at 16:44
-
$\begingroup$ This is slightly off-topic, but I think there is one further inaccuracy in section 3.6: in Proposition 7 one should assume that $\#I = 1$. Surely, they must've intended $I$ to be at least nonempty (why no rat'l points over $K^{sh}$ implies the same over $K^{'sh}$?), but also the first sentence of the proof doesn't justify how to "easily reduce to the case $\# I = 1$," since there is no reason why separated glueing be smooth, or (if one also does the smoothening afterwards) why all of its $R'$-points factor through the open immersions of the blowups of the $X_i$'s. $\endgroup$– Question MarkCommented Nov 5, 2014 at 16:51
|
Show 7 more comments