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Let $R$ be an integrally closed domain. Then I want to show $R[x]$ is integrally closed.

Let $K$ be the field of fractions of $R$ and if I choose a function $a(x)\in K(x)$ which is integral over $R[x]$, then it is also integral over $K[x]$. Since a UFD is integrally closed, we have $a(x)=a_nx^n+\cdots+a_1x+a_0\in K[x]$. So it is enough to show each $a_i\in K$ is contained in $R$.

Since $a(x)$ is integral over $R[x]$, we have an equation $a(x)^m+c_1(x)a(x)^{m-1}+\cdots+c_m(x)=0$. Looking at the zeroth degree term, and since $R$ is integrally closed, we have $a_0\in R$. Thus if we can show that $a_1\in R$ as well, the remaing $a_i\in R$ by the same method. Since both $a(x)$ and $a_0$ are integrally closed, their difference $a'(x):=a(x)-a_0=a_nx^n+\cdots+a_1x$ is integrally closed. Then we have a new equation \begin{equation} \label{new_eqn} a'(x)^k+d_1(x)a'(x)^{k-1}+\cdots+d_k(x)=0 \, . \tag{1} \end{equation} Here, since $a'(x)$ has no constant term, it is deduced that $d_k(x)$ has no constant term, so a multiple of $x$. So, the equation (\ref{new_eqn}) can be divided by $x$ and we get $a_1\in R$ once more by looking at the constant terms and using the condition that $R$ is integrally closed.

There are somewhat complicated proofs (e.g. Trouble with proving $A$ is an integrally closed domain $\Rightarrow$ $A[t]$ is integrally closed domain) for the same statement, but I don't know why the above proof is not given in any reference. Does this proof has some flaw?

Viktor Vaughn
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user190964
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1 Answers1

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I like your approach. You just need to patch the hole in your proof by showing

If $f \in K[x]$ is integral over $R[x]$ with lowest nonzero coefficient $f_0$, then $f_0$ is integral over $R$.

Badam Baplan
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    Doesn't the claim in the quote simply follow from evaluating the integral relation at $x=0$? – soomakan. Feb 26 '22 at 19:57
  • Yes, good point. I edited the answer because it was overcomplicating something trivial. – Badam Baplan Feb 27 '22 at 17:15
  • Sorry, I think I misunderstood some part in this, this only works in degree 0 term. – soomakan. Feb 27 '22 at 21:22
  • @lyuinnhe. once we know that $f \in K[x]$ is integral over $R[x]$ and $f_0 \in K$ is integral over $R$, we also get that $f - f_0 \in K[x]$ is integral over $R[x]$ (integral elements are closed under addition and multiplication). It's easy to check that $(f-f_0)/x$ is also integral over $R[x]$ and then we can proceed by induction. – Badam Baplan Feb 28 '22 at 18:56
  • @BadamBaplan Why it is easy to check that $(f-f_0)/x$ is integral over $R[x]?$ I don't see why this is even true – rowcol Nov 29 '22 at 07:14
  • @rowcol you're right, this part is really the meat of the proof, this was a terrible answer from many years ago. – Badam Baplan Nov 29 '22 at 16:22
  • @rowcol that said, it certainly is true that $fx$ integral implies $f$ integral. One way to do it is to pick a finite type $\mathbb{Z}$ algebra $R'$ generated by the elements of $R$ that are needed to express the integral dependence for $fx$. From the integral dependence for $fx$, you can find a polynomial $g \in R'[x]$ such that $(fx)^n g \in R'[x]$ for all $n$. – Badam Baplan Nov 29 '22 at 16:29
  • @rowcol This implies $f^n g \in R'[x]$ for all $n$, too. Then $fI \subseteq I$ for the ideal $I$ generated by $f^n g$. Since $R'$ is Noetherian, $I$ is a finitely generated ideal. Therefore (using the determinant trick), $f$ is integral over $R'[x]$, and moreover $f$ is integral over $R[x]$. – Badam Baplan Nov 29 '22 at 16:29