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Show that if an integral domain $A$ is integrally closed in its field of fractions $K$, then so is $A[T]$ in its ring of fractions, $K(T) := \mathrm{Frac}(A[T])$.

What I have done so far (albeit not much) is the following.

Pick any $p(T) \in K(T)$ that is integral over $A[T]$, that is, the following is true.

$p(T)^n + a_{n-1}p(T)^{n-1} + \cdots+ a_1p(T) + a_0 = 0$ with each $a_i \in A[T]$.

One hint I got was that I could try eliminating the denominator of $p(T) = \frac{f(T)}{g(T)}$ by multiplying out by $g(T)^n$. But that is about it, I'm stuck quite some time trying to do the next step.

So any help or insights regarding this is deeply appreciated.

user26857
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PIandpie
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1 Answers1

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You can refer Here for some reduction.

It is an exercise of Eisenbud's book (Ex.4.17 and Ex.4.18). Some hints are at the back of the book.

Actually, we can prove a strong assertion,

$R$ is integrally closed in $S$ $\iff$ $R[X]$ is integrally over $S[X]$

---------Edit: The following proof breaks unless reduce the ring to a Noetherian case, firstly-----------

Proof. If $f\in S[X]$ is integral over $R[X]$. Then $R[X][f(X)]\subseteq S[X]$ is a finitely generated $R[X]$-module. Let $M'$ be the submodule of $S$ generated of the coefficients of the polynomials in $R[X][f(X)]$, then $M'$ is finitely generated over $R$. Then for any coefficients of $f$, say $\alpha$, $R[\alpha]\subseteq M'$, is contained in a finitely generated $R$-module, thus $\alpha$ is integral over $R$, thus in $R$.

Cubic Bear
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    You used the result that for a ring extension $A\subseteq B$, $\alpha \in B$ is integral over $A$ if (and only if) $A[\alpha]$ is contained in a finitely generated $A$-module contained in $B$. Actually I think there is a subtle detail that you ignored, the correct condition to use is $A[\alpha]$ is contained in a subring $C$ of $B$ such that is $C$ is a finitely generated $A$-module. This invalidates your proof. – Mathematical Jan 26 '18 at 03:30
  • A possible workaround is to look at only the leading coefficients of the polynomials in $R[X][f(X)]$, which form a ring but not just an $R$-module. Then you figure out that the leading coefficients, and in particular, the leading coefficient of $f(X)$, is integral over $R$. Then use induction. I didn't think it through carefully though. – Mathematical Jan 26 '18 at 03:40
  • You are right, the hints reduce the problem into Noetherian case. I was not fully understood why he do so, thank you. – Cubic Bear Jan 27 '18 at 08:52