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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.

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If you are familiar with The Rational Root Theorem, I should think you could recast it in your more general setting. – Gerry Myerson May 27 '12 at 23:55
You can find a proof of your theorem, here – Makoto Kato May 28 '12 at 0:22
up vote 9 down vote accepted

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 $\alpha\in A$.

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The well-known proof of the Rational Root Test immediately generalizes to any UFD or GCD domain. The sought result is simply a monic special case. One can present the proof in a form that works for both gcds and cancellable ideals by using only universal laws common to gcds and ideals, e.g. commutative, associative, distributive laws. Below I give such a universal proof for degree $3$ (to avoid notational obfuscation). It is clear from this how it generalizes to any degree.

Let $\rm\:D\:$ be a domain and suppose monic $\rm\:f(x)\in D[x]\:$ has root $\rm\:a/b,\ a,b\in D.\:$ The notation $\rm\:(d,e,\ldots)\:$ denotes $\rm\:gcd(d,e,\ldots)\:$ (or the ideal $\rm\:dD + eD +\:\!\ldots\:$ in the ideal case).

$\rm\qquad f(x)\, =\, c_0 + c_1 x + c_2 x^2 + x^3,\:$ 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}),\ \ $ hence, 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: one cancels $\rm\:(b,a)^{n-1}$ 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. $\rm\:r \in D.\:$ For more, see my posts in a 2009/5/22 sci.math thread (mathforum or Google groups) which includes discussion of how most elementary irrationality proofs are simply unwindings of the elegant one-line proof employing Dedekind's notion of conductor ideal (universal denominator ideal).

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