This problem has been bugging me for a while. As was stated in the title, I wish to prove:
$A$ is an integrally closed domain $\Rightarrow$ $A[t]$ is integrally closed domain
Here's what I have so far…
Suppose $f \in k(t) = Frac(A[t])$ is integral over $A[t]$. Then, trivially, it is integral over $k[t]$. However, $k[t]$ is a PID, hence a UFD, and is thus integrally closed. So $f \in k[t]$.
So the problem is reduced to:
$A$ is an integrally closed domain $\Rightarrow$ $A[t]$ is integrally closed in $k[t]$.
Now, $f$ is integral. We may (with no loss of generality) assume $f$ is monic by adding a high power of $t$. (Since $t^n$ is integral, $t^n + f$ is integral if and only if $f$ is.) We write $$f^n + a_{n-1}f^{n-1} + \ldots + a_1 f + a_0 = 0$$
$$f (f^{n-1} + a_{n-1} f^{n-2} + \ldots + a_2 f + a_1 ) = -a_0$$
Now I'd like to be able to use the Gauss lemma and win, but every time I try to do that it beats me.
Attempt 1: $f$ is monic, so I'd like to make $f^{n-1} + a_{n-1} f^{n-2} + \ldots + a_2 f + a_1$ monic as well. However, if I divide by the leading coefficient $q$, I have to divide $-a_0$ by $q$. Unfortunately, there's no guarantee (that I know of) keeping $-a_0/q$ in $A[t]$.
Attempt 2: Add a really high power of $t$ to $f$. But there is no bound on the degree of the coefficients, so again there's no guarantee this will be monic.
I think I'm in the right direction, but this problem is getting into my head. Any thoughts are welcome! Thanks
(For those who are interested, this is a question from Neukirch's Algebraic Number Theory Chapter 1, Section 2, Question 2)
Best Answer
Assume that $f\in k[t]$ is integral over $A[t]$ and $f^n + a_{n-1}(t)f^{n-1} + \cdots + a_1(t)f + a_0(t) = 0$, where $a_i(t)\in A[t]$. Let $m$ be an integer greater than the degree of $f$ and all the degrees of $a_i$. Set $f_1(t)=f(t)-t^m$. If $q(x)=x^n+a_{n-1}(t)x^{n-1}+\cdots+a_1(t)x+a_0(t)$, then $f_1$ is a root of $q_1(x)=q(x+t^m)$. Note that $q_1(x)=x^n+b_{n-1}(t)x^{n-1}+\cdots+b_1(t)x+b_0(t)$ and $b_0(t)=q(t^m)$ is monic. Since $f_1(f_1^{n-1}+b_{n-1}f_1^{n-2}+\cdots +b_1(t))=-b_0(t)$ and $f_1$ and $b_0$ are monic, it follows that $f_1^{n-1}+b_{n-1}f_1^{n-2}+\cdots +b_1(t)$ is also monic, and now you can apply Gauss Lemma.