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Let $P \in \mathbb{Z}[x], P(x) = \displaystyle\sum\limits_{j=0}^n a_j x^j, a_n \neq 0$ and $a_0 \neq 0$; if $p/q$ is a root of P (with p and q coprimes) then $p|a_0$ and $q|a_n$

I've managed to prove the first part ($p|a_0$) and I suppose I'm not far from proving the second, though I'd really like some feedback since I'm just starting with making proofs of my own.

Proof:

$P(x) = a_n(x-p/q)\displaystyle\prod\limits_{j=2}^n (x-r_j)$, with $r_j$ being the other n-1 roots of P(x).

It follows that $a_0 = a_n(-p/q)\displaystyle\prod\limits_{j=2}^n (-r_j)$

Then, $-p/q|a_0$ and obviously $p/q|a_0$. Rephrasing, $a_0 = l\frac{p}{q} = \frac{l}{q} p$ with $l \in \mathbb{Z}$.

This implies $p|a_0$ if $l/q \in \mathbb{Z}$, but this is trivial since $q|lp$ and q and p are coprimes, so $q|l$.

Therefore, $p|a_0$.

As for the second part, we want to see that $q_i|a_n \forall i \leq n$.

We define $d$ as the least common multiple of $\{q_1, q_2,...,q_n\}$. Then, $q_i|a_n \forall i \leq n \iff d|a_n$.

Also, it follows that $d|\displaystyle\prod\limits_{j=1}^n q_j$, so we want to see that $a_n = l \displaystyle\prod\limits_{j=1}^n q_j$ with $l \in \mathbb{Z}$. Here's where I have my doubts with the proof as I have no way to show that l is indeed an integer.

Rearranging the previously given equation for $a_0$:

$a_n = a_0 \displaystyle\prod\limits_{j=1}^n \frac{-q_j}{p_j}$

Using the previous reasoning, as $p_i|a_0 \forall i \leq n$, then $a_0 = k \displaystyle\prod\limits_{j=1}^n p_j$.

Replacing $a_0$:

$a_n = k \displaystyle\prod\limits_{j=1}^n -q_j$, which is equivalent to $q|a_n$ as shown earlier.

Bill Dubuque
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Fernando Martin
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    The proof that $p|a_0$ isn't complete since it assumes without justification that $a_n\ \prod_{j=2}^n(-r_j)\in \mathbb Z$. But one can prove $p|a_0$ the same way as in my answer - see my comment to Mariano. – Bill Dubuque Nov 03 '10 at 05:13
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    I hadn't seen this attributed to Gauss — it's often called the "rational root test". – Dylan Moreland May 11 '12 at 01:57
  • @DylanMoreland: Over the last few years I've noticed that a handful of theorems have alternative names here in Latin America (not just my professors, but books, and even Wikipedia, offer this alternative names). This is one example. The intermediate value theorem is called Bolzano's theorem, the rank-nullity theorem is called the 'dimension theorem' and this corollary: http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Corollary is called Barrow's rule, just to name a few. – Fernando Martin May 11 '12 at 03:58

2 Answers2

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Hint $\displaystyle\ \ 0 = q^n P(\frac{p}q) = a_n\ p^n + \color{#c00}q\, (\overbrace{\!\cdots^{\phantom :\!}}^{\!\large \in\, \Bbb Z}\!)\ \Rightarrow\ \color{#c00}q\mid a_n\,p^n\ \!\overset{(q,\,p)\,=\,1}\Longrightarrow\! q\mid a_n\ $ by Euclid's Lemma.

Note $ $ This result is usually called RRT = Rational Root Test / Theorem, not Gauss' polynomial theorem. It is special case of Gauss's Lemma for polynomials, i.e. if $\,\alpha\,$ is an algebraic number with primitive minimal polynomial $\, f(x)\in \mathbb Z[x]\,$ and if $\, g(\alpha) = 0\,$ for $\, g(x)\in \mathbb Z[x]\,$ then $\,f\mid g\,$ in $\,\Bbb Q[x]\,$ hence $\,f\mid g\,$ in $\,\mathbb Z[x]\,$ by Gauss. So the leading coefficient of $\,f\,$ divides the leading coefficient of $\,g,\,$ and ditto for the constant coefficients. RRT is just the special case when $\,f\,$ has degree $= 1.\,$ Indeed $\ \alpha = p/q\in \mathbb Q\,$ with $\,(p,q)=1\,$ has primitive minimal polynomial $\, f(x) = q\,x - p\,$. Therefore if $\,g(\alpha) = 0\,$ above $\Rightarrow (q\,x-p)\,|\,g\ \ in\ \ \mathbb Z[x]\,$ thus $\,q\,|\,g_n,\ p\,|\,g_0\,\ {\rm in}\,\ \mathbb Z,\,$ which is precisely RRT.

Bill Dubuque
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  • Great answer! It was really that simple. I'll keep this open for a little while to see if someone can spot where I went wrong with my proof. – Fernando Martin Nov 03 '10 at 03:15
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If you know how to prove the first part, just apply it to the polynomial $t^n P(t^{-1})$.

Mariano Suárez-Álvarez
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