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.
The proof follows exactly like the proof of the Rational Root Theorem.
Let $\alpha\in K$ be a root. We can express $\alpha$ as $\frac{a}{b}$ with $a,b\in A$, and using unique factorization we may assume that no irreducible of $A$ divides both $a$ and $b$.
If $f(x) = x^n + c_{n-1}x^{n-1}+\cdots+c_0$, then plugging in $\alpha$ and multiplying through by $b^n$ we obtain $$a^n + c_{n-1}ba^{n-1}+\cdots + c_0b^n = 0.$$ Now, $c_{n-1}ba^{n-1}+\cdots+c_0b^n$ is divisible by $b$, hence $a^n$ is divisible by $b$. Since no irreducible of $A$ divides both $a$ and $b$, it follows that $b$ must be a unit by unique factorization. Hence $\alpha\in A$.
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.
The well-known simple proof of RRT = Rational Root Test remains valid in a UFD or GCD domain. The sought result is simply the special case when $\:\!\rm f(x)\:\!$ is monic (lead coef $=1$).
Generalization: $ $ we prove RRT in a form that works for both $\rm\color{#0a0}{gcds}$ and cancellable $\rm\color{#c00}{ideals}$ by using only universal laws common to both (commutative, associative, distributive etc). Below we give such a universal proof for degree $3\,$ (to avoid notational obfuscation). It should be clear how this generalizes to any degree.
If $\rm\:D\:$ is a $\rm\color{#0a0}{gcd}$ domain and monic $\rm\:f(x)\in D[x]\:$ has root $\rm\:a/b,\ a,b\in D\,$ then
$\rm\qquad f(x)\, =\, c_0 + c_1 x + c_2 x^2 + x^3,\:$ and $\rm\:b^3\:\! f(a/b) = 0\:$ yields
$\rm\qquad c_0 b^3 + c_1 a b^2 + c_2 a^2 b\, =\, -a^3\ $
$\rm\qquad\qquad\ \ \,\Rightarrow\ (b^3, a b^2,\, a^2 b)\mid \color{#c00}{a^3},\ $ by the gcd divides LHS of above so also RHS
$\rm\qquad\ (b,a)^3 = \, (b^3,\, a b^2,\, a^2 b,\ \color{#c00}{a^3}),\ $ so, by the prior divisibility
$\rm\qquad\qquad\quad\:\! =\, (b^3,\, a b^2,\, a^2 b) $
$\rm\qquad\qquad\quad =\, b\, (b,a)^2,\ $ so cancelling $\rm\,(b,a)^2$ yields
$\rm\qquad\ \, (b,a) =\, b\:\Rightarrow\: b\:|\:a,\ $ i.e. $\rm\: a/b \in D.\ \ $ QED
The degree $\rm\:n> 1\:$ case has the same form: we cancel $\rm\:(b,a)^{n-1}$ from $\rm\,(b,a)^n = b\,(b,a)^{n-1}.$
The $\rm\color{#c00}{ideal}$ analog is the same, except replace "divides" by "contains", and assume that $\rm\,(a,b)\ne 0\,$ is invertible (so cancellable), e.g. in any Dedekind domain. Thus the above yields a uniform proof that PIDs, UFDs, GCD and Dedekind domains satisfy said monic case of the Rational Root Test, i.e. that they are integrally closed.
The proof is more concise using fractional gcds and ideals. Now, with $\rm\:r = a/b,\:$ we simply cancel $\rm\:(r,1)\:$ from $\rm\:(r,1)^n = (r,1)^{n-1}$ so $\rm\:(r,1) = (1),\:$ i.e. $\rm\:r \in D.\:$ See this answer for further details.