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)


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – user135520
    Jan 3, 2019 at 18:46
  • 3
    $\begingroup$ 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 $h = f_1^{n-1} + ... + b_1(t)$. $\endgroup$
    – vacant
    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 $K$.


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