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