In $ \mathbb{Z}[x]$, root $a\in \mathbb{Q}$ $\Rightarrow$ $a\in \mathbb{Z}$

Question

Let monic $f(x)\in \mathbb{Z}[x]$. Let $a\in \mathbb{Q}$ where $f(a)=0$. Prove that $a\in \mathbb{Z}$

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Thinking the answer below looks alot like the proof of rational root test.

Suppose $a \not \in \mathbb{Z}$ we can say (a is imaginary or irrationa or rational,not integer...)

Let us say $a\in \mathbb{Q}$. So, $a=\frac{r}{s}$ where $gcd(r,s)=1$ and $s>1$ is a root.

Thanks to the Rational root test
$$s\mid (a_n=1) \text{ and } r\mid a_0 $$ $s\mid 1 \Rightarrow s \leq 1$ But $s>1$. Contradition

Answer 1

Suppose $a\notin\mathbb{Z}$; then we can write $$ a=\frac{p}{q},\qquad \gcd(p,q)=1, q>0 $$ Now, if $f(x)=c_0+c_1x+\dots+c_{n-1}x^{n-1}+x^n$, from $f(a)=0$ we get $$ c_0q^n+c_1pq^{n-1}+\dots+c_{n-1}p^{n-1}q+p^n=0 $$ that can be written $$ p^n=q(-c_0q^{n-1}+c_1pq^{n-2}+\dots+c_{n-1}p^{n-1}) $$ Can you derive a contradiction if you assume $q>1$?

Hint: if $q>1$, then there exists a prime $q_0$ dividing $q$.

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This is a variation on the theme of the rational root test. If a polynomial with integer coefficients $c_0+c_1x+\dots+c^nx^n$ (where $c_n\ne0$) has $p/q$ as a root, where $\gcd(p,q)=1$, then $p$ is a divisor of $c_0$ and $q$ is a divisor of $c_n$.

The technique of proof is essentially the same and you could try your hand at it.

Your case is the special one where $c_n=1$, so $q=\pm1$ (which does not contradict the above proof, because one can assume $q>0$).

Answer 2

HINT: Suppose I have a quadratic monic polynomial with integer coefficients, $$f(x)=x^2+\mbox{[linear stuff]}.$$ I tell you that ${1\over 2}$ is a root of this polynomial. Well, $({1\over 2})^2={1\over 4}$, so we must have $$\mbox{[linear stuff]}({1\over 2})=-{1\over 4};$$ if the coefficients of my linear stuff are integers, why is this a problem?

(If you don't see where I'm going with this, try plugging $x={1\over 2}$ into some specific examples, e.g. $x^2+3$, $x^2-7x+1$, $x^2+13x+12345346$, . . . )