Suppose $\alpha$ is a rational root of a monic polynomial in $\mathbb{Z}[x]$. Prove $\alpha \in \mathbb{Z}$ .

Question

Suppose $\alpha$ is a rational root of a monic polynomial in $\mathbb{Z}[x]$. Prove $\alpha \in \mathbb{Z}$ .

attempt:

Suppose $p(x)$ is monic. Then the leading coefficient $a_n = 1.$

Thus, $p(x) = x^n + a_{n-1}x^{n-1}+ ...+ a_1x + a_0$. Then let $\alpha = r/s \in \mathbb{Q}$ be in the lowest term, that is $(r,s) = 1$. And assume $\alpha = r/s$ is a root of $p(x)$, then $r|a_0$ and $s|(a_n= 1)$.

Then since $\alpha $ is a root, we have :

$p(\alpha) = \alpha^n + a_{n-1}\alpha^{n-1}+ ...+ a_1\alpha + a_0 = (r/s)^{n} + a_{n-1}(r/s)^{n-1}+ ...+ a_1(r/s) + a_0 = \frac{r^n}{s^n} + a_{n-1}\frac{r^{n-1}}{s^{n-1}} + ....+ a_1\frac{r}{s} + a_0 = 0$.

Then multiply both sides by $s^n$.

Thus $ s^np(\alpha) = s^n(\frac{r^n}{s^n} + a_{n-1}\frac{r^{n-1}}{s^{n-1}} + ....+ a_1\frac{r}{s} + a_0) = r^n + a_{n-1}r^{n-1}s + ...+ a_1rs^{n-1} + a_0s^n = 0$.

But $(r,s) = 1$ , so $s = 1$. Thus, $s | a_n$ , so $\alpha $ is an integer.

Can someone please verify this? any better suggestion would help. Thank you !

Answer 1

There are two proofs in your write-up that are intermingled to make a confusing reading. All-in-all, all the necessary arguments are given and every single statement is correct, but the structure could be better.

In the first part you prove the claim in the task, you use the rational root theorem to find that $s=1$ and thus $α=r$ is an integer. At this point the claim has been proven. There is no need to later come back to it.

Then you have a second part that is not really cleanly separated from the first part where you state and prove the rational root theorem (specialized to $a_n=1$) in that $gcd(r,s)=1$ and $$ a_nr^n+s(a_{n-1}r^{n-1}+...+a_1rs^{n-2}+a_0s^{n-1})=0 $$ implies $s|a_n$ and similarly $r|a_0$.

From your question it is not clear if that second part is necessary for your solution write-up. That depends on if the rational root theorem was previously provided in the lecture or otherwise assumed to be generally known.

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It is a sometimes somewhat hard to apply advice to structure your text cleanly in claims (lemma, proposition) and proofs wherever possible, so that the text in-between is mostly historical remarks and motivation. This might give the text a rigid feel, but it is much easier for the reader to orient themselves in it.

Answer 2

This is another proof using Ring theory and Field theory.

Let $f(x)\in \mathbb{Z}[x]$ be a monic polymonial having $\alpha$ as a root.

Since $\mathbb{Z}[x]$ is an U.F.D., there is an monic irreducible polynomial $p(x)\in \mathbb{Z}[x]$ having $\alpha$ as a root ($p(x)$ is a factor of $f(x)$).

By Gauss's lemma, the polynomial $p(x)$ is also irreducible over the faction field $\mathbb{Q}$ of $\mathbb{Z}$. We have $$\mathbb{Q} = \mathbb{Q}(\alpha) \cong \mathbb{Q}[x] / (p(x))$$ hence the degree of $p(x)$ is $1$ (since the field extension $\mathbb{Q}(\alpha)$ has degree $1$), so $p(x) = x - \alpha$, therefore $\alpha\in \mathbb{Z}$.

Answer 3

Given a monic polynomial $f(x) = x^n + \cdots + a_1 x + a_0$ in $\mathbb{Z}[x],$ we may consider $f(x)$ as a monic polynomial in $\mathbb{Q}[x].$ Given that $\alpha$ is a rational root of $f(x),$ the Rational Roots Theorem implies that we have $\alpha = \frac{r}{s}$ for some integers $r$ and $s$ such that $\gcd(r,s) = 1,$ $r \,|\, a_0,$ and $s \,|\, 1.$ We must have therefore that $s = 1,$ from which it follows that $\alpha = r$ is an integer, as desired.