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 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 $\alpha\in A$.

The well-known simple proof of the Rational Root Test immediately generalizes to any UFD or GCD domain. The sought result is simply a monic special case. 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},\ $ since the gcd divides the LHS of above so also the 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 if one knows about fractional gcds and ideals. Now, with $\rm\:r = a/b,\:$ one simply cancels $\rm\:(r,1)\:$ 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.