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.

The well-known simple proof of the 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 $=1$).

More generally, we can present the proof in a form that works for both gcds and cancellable ideals by using only universal laws common to both (commutative, associative, distributive etc). Below I 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 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 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.
