UFDs are integrally closed Let $A$ be a UFD, $K$ its field of fractions, and $f$ an element of $A[T]$ a monic polynomial.
I'm trying to prove that if $f$ has a root $\alpha \in K$, then in fact $\alpha \in A$.
I'm trying to exploit the fact of something about irreducibility, will it help? I havent done anything with splitting fields, but this is something i can look for.

Overkill: It suffices to show that $A$ is integrally closed, which we prove using Serre's criterion. For this, we recall some of the definitions, including the definitions of the properties $R_n$ and $S_n$ that appear in Serre's criterion. We will assume $A$ is locally Noetherian (can one assume less for this approach to work?)
Background
Definition. A ring $A$ is said to satisfy the criterion $R_n$ if for every prime ideal $\mathfrak p$ of $A$ such that $\operatorname{ht}(\mathfrak p)\le n$, the localization $A_{\mathfrak p}$ is a regular local ring, which means that the maximal ideal can be generated by $\operatorname{ht}(\mathfrak p)$ elements.
Definition. A Noetherian ring $A$ is said to satisfy the criterion $S_n$ if for every prime ideal $\mathfrak p$ of $A$, we have the inequality $$\operatorname{depth}(A_{\mathfrak p}) \ge \min\{n,\operatorname{ht}(\mathfrak p)\}$$
This relies on the notion of depth, which is the length of a maximal regular sequence in the maximal ideal.
Exercise. Give a definition of the $S_n$ condition for modules. (Note: there are actually two distinct definitions in the literature, which only agree when the annihilator of the module is a nilpotent ideal.)
Exercise. Show that a Noetherian ring $A$ is reduced if and only if $A$ satisfies $R_0$ and $S_1$.
With these definitions out of the way, we now state the criterion of which we will benefit.
Theorem. (Serre's Criterion). A Noetherian integral domain $A$ is integrally closed if and only if $A$ has the properties $R_1$ and $S_2$.

Proof that UFDs are Integrally Closed
Firstly, localizations of UFDs are UFDs while intersections of integrally closed domains in the field of fractions of $A$ are integrally closed. Recalling that $A=\bigcap_{\mathfrak p\in\operatorname{Spec}A} A_{\mathfrak p}$, we may assume $A$ is local. Now, $A$ is $R_1$ because prime ideals $\mathfrak p$ of height $1$ are principal and thus $A_{\mathfrak p}$ is a DVR. Also, $A$ is $S_1$ because $A$ is an integral domain and thus reduced, while for any irreducible $f\in A$ we have $A/fA$ is an integral domain, so $A/fA$ is $S_1$. This implies
\begin{equation*}
    \operatorname{depth} A \ge \min\{2,\dim A\}
\end{equation*}
The argument works for any local UFD, in particular the localizations of $A$. So, $A$ is $S_2$. By Serre's criterion, $A$ is integrally closed.
