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Let $A$ be an integral domain and $S \subset A$ be an multiplicative closed set. Let $\alpha$ be an element in some extension of $A$ which is integral over $A$. Then I want to know whether $S^{-1}A[\alpha]\cong(S^{-1}A)[\alpha]$ holds or not. Actually the natural map $\phi:S^{-1}A[\alpha]\to(S^{-1}A)[\alpha]$ is surjective ring homomorphism, but can't prove that it is injective.

This question comes from here. Makoto Kato used this in the proof. I failed to prove it. Need some help. Thanks.

user371231
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There's a subtle but important point of ambiguity in the way this question is phrased; suppose that $B\supseteq A$ is the extension that contains $\alpha$. Talking about $(S^{-1}A)[\alpha]$ makes an assumption that $S^{-1}A$ is a subring of $B$, or at least that the two rings can be "amalgamated" in a coherent way: ie that there exists a ring $C$ containing both $S^{-1}A$ and $B$ as subrings. But this may not possible in general; for instance, take $A=\mathbb{Z}$, fix a prime $p\in A$, and let $S=\{p^n:n\in\mathbb{N}\}$. Now take $B$ to be the ring $\mathbb{Z}\times\mathbb{F}_p$. This ring has characteristic $0$, so it indeed contains $A$ as a subring. Furthermore, the element $\alpha=(0,1)\in B$ is integral over $A$, since it satisfies the monic polynomial $t^2-t\in A[t]$. But it is impossible for a ring $C$ to contain both $B$ and $S^{-1}A$ as subrings; indeed, if $C$ contains $S^{-1}A$, then $p$ is invertible in $C$, so $C$ contains no elements of additive order $p$.

So, it is possible for the question to not even make sense. However, this kind of pathological behavior only arises because $B$ is not an integral domain. If we do demand that $B$ be an integral domain, which is the case in the post that you link to, then the theorem as stated is correct. In that case, we can work in the field of fractions $K=Q(B)$. We can then safely consider everything as taking place in $K$, and both $(S^{-1}A)[\alpha]$ and $S^{-1}(A[\alpha])$ will be isomorphic to the subring $A[\alpha,1/s:s\in S]$ of $K$, as desired.

Atticus Stonestrom
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  • How can I show that the natural map $\phi:S^{-1}A[\alpha]\to(S^{-1}A)[\alpha]$ is injective? This is exactly what I wanted to know. – user371231 Sep 02 '21 at 10:01
  • hi @user371231; to reiterate, the notion of the ring $(S^{-1}A)[\alpha]$ only makes sense if we imagine $\alpha$ and $S^{-1}A$ as being embedded in some larger ring; eg in the field $K$ that I mention above. can you see why $S^{-1}A$ is isomorphic to the subring $A[1/s:s\in S]$ in $K$? so then we can consider $(S^{-1}A)[\alpha]$ as literally being the subring $A[\alpha,1/s:s\in S]$ of $K$. now suppose $\phi$ maps some element $x\in S^{-1}(A[\alpha])$ to $0$. by definition of $S^{-1}(A[\alpha])$, we have $x=f/s$ for some $f\in A[\alpha]$ and $s\in S$, and then $\phi(x)$ is just $f/s$ ... – Atticus Stonestrom Sep 02 '21 at 13:23
  • ... considered as an element of $A[\alpha,1/s:s\in S]$. in particular, if it is zero, this forces $f=0$ (since we can multiply through by $s$), and this means that $x$ itself is $0$, as desired. does that make sense? – Atticus Stonestrom Sep 02 '21 at 13:24
  • But $\alpha$ is not necessarily in $K$. One can always extend the natural injection $A\to S^{-1}A $ to $A[\alpha]\to (S^{-1}A)[\alpha]$, which inverts the elements of $S$. Thus by universal property we have a map $S^{-1}A[\alpha]\to(S^{-1}A)[\alpha]$ given by $\frac{\sum a_i \alpha^i}{s} \mapsto \sum \frac{a_i}{s}\alpha^i$. This map is surjective. I just want to show that this is injective as well. – user371231 Sep 02 '21 at 14:03
  • dear @user371231, I have constructed $K$ so that it contains both $A$ and $B$ as subrings; since $\alpha\in B$, also $\alpha\in K$. the argument you give does not quite work; again, what do you mean by $(S^{-1}A)[\alpha]$? (for example, what is $(S^{-1}A)[\alpha]$ when $A$ is $\mathbb{Z}$, $S={p^n:n\in\mathbb{N}}$, and $\alpha$ is the element $(0,1)$ of $\mathbb{Z}\times \mathbb{F}_p$?) – Atticus Stonestrom Sep 02 '21 at 14:08
  • (more precisely, I mean for $A$ here to be the subring of $\mathbb{Z}\times\mathbb{F}_p$ generated by $(1,1)$; this is isomorphic to $\mathbb{Z}$, hence my abuse of notation. if you prefer, you can think of $\alpha$ as being the element $\bar{x}$ of the ring $B:=\mathbb{Z}[x]\big/\langle x^2-x,px\rangle$ and $A$ as being the subring $\mathbb{Z}\subset B$) – Atticus Stonestrom Sep 02 '21 at 14:17
  • $A, S^{-1}A$ are subsets of $S^{-1}B$ and $\alpha \in B$, thus $S^{-1}A[\alpha]$ and $(S^{-1}A)[\alpha]$ are both subrings of $S^{-1}B$. – user371231 Sep 02 '21 at 14:33
  • @user371231 so, my point is that the localization map $B\to S^{-1}B$ may not necessarily be injective. for instance, in the example that I give, we have $p\alpha=0$, so if we localize at $p$ then the element $\alpha$ will be in the kernel of the localization map $B\to S^{-1}B$. apologies if it seems like I'm being pedantic, but I think it's important to draw this distinction – Atticus Stonestrom Sep 02 '21 at 14:36
  • @user371231 but if we assume that $B$ is an integral domain, then things do work out fine: since $(S^{-1}A)[\alpha]$ is a subring of $S^{-1}B$, the element $\sum\frac{a_i}{s}\alpha^i$ that you mention above is just equal to the element $\frac{1}{s}\sum a_i\alpha^i$ of $S^{-1}B$... if this is zero, then also $\sum a_i\alpha^i$ is $0$, and so the element $\frac{\sum{a_i\alpha^i}}{s}$ of $S^{-1}(A[\alpha])$ was zero to begin with. does that make sense? :) – Atticus Stonestrom Sep 02 '21 at 14:38
  • But $A$ is sitting inside $S^{-1}A$ and $S^{-1}A$ is sitting inside $S^{-1}B$. – user371231 Sep 02 '21 at 14:41
  • @user371231 indeed, but $A[\alpha]$ may not be sitting inside $S^{-1}B$. again, think about what happens when you localize the ring $\mathbb{Z}[x]\big/\langle x^2-x,px\rangle$ at the multiplicative set ${p^n:n\in\mathbb{N}}$, when $A=\mathbb{Z}$ and $\alpha=\bar{x}$ – Atticus Stonestrom Sep 02 '21 at 14:42
  • You are right. I messed up completely. I'll take some time to get over it. – user371231 Sep 02 '21 at 14:43
  • @user371231 no problem at all! :) it's a very easy mistake to make; sorry for all of the confusion and the back-and-forth!! – Atticus Stonestrom Sep 02 '21 at 14:44