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.
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,
---------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 $R$.