# Trouble with proving $A$ is an integrally closed domain $\Rightarrow$ $A[t]$ is integrally closed domain

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)

• In the Stacks Project, this is 030A. Apr 12 at 7:42

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.

• What is the statement of Gauss's lemma here? I've seen a version for polynomials with integer coefficients and wanted to see the punchline here. Jan 3, 2019 at 18:46
• I think they are using the statement of Theorem 2 here. The link says "Let $A$ be an integrally closed domain with fraction field $K$. Let $f(t)$ be a monic polynomial in $A[t]$ and suppose that $f(t)$ factors in $K[t]$ as $g(t)h(t)$ with $g$ and $h$ monic. Then $g$ and $h$ have coefficients in $A$." Which gives the desired result when $g = f_1$ and $h = f_1^{n-1} + ... + b_1(t)$. Feb 10, 2019 at 5:02

Here is another way of doing it. We denote the field of fractions of $$A$$ by $$K$$. Now we first note that $$K[t]$$ is integrally closed in $$K(t)$$ (this follows readily from the fact that UFD's are integrally closed). So now all we have to show is that $$A[t]$$ is integrally closed in $$K[t]$$. To do this we choose a polynomial $$f \in K[t]$$ that is integral over $$A[t]$$ such that $$f$$ has degree $$n$$ in the variable $$t$$. We also assume that we have shown the result (that is, if $$f \in K[t]$$ is integral over $$A[t]$$ then it is in $$A[t]$$) for all $$f$$ with degree less than $$n$$. We first recall the mantra "integral + finite type = finite". Now we consider the extension

$$A[t] \subset (A[t])[f]$$

It is finite as it is integral and finite type. Now as one can check, the ring of leading coefficients of polynomials in $$(A[t])[f]$$, denoted by $$L$$, is finite over $$A$$. In particular, if

$$f = k_nt^n+\dots+k_0$$

we have that $$A[k_n] \subset L$$, and hence $$k_n$$ is integral over $$A$$. Since $$A$$ is integrally closed in its field of fractions, we have that $$k_n \in A$$. We now note that $$f-k_nt^m$$ is integral over $$A[t]$$ and has degree less than $$n$$. We finish the proof by noting that the result for $$n=0$$ is equivalent to the integral closure of $$A$$ in $$K$$.