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The rational root theorem states that, given $P \in \mathbb{Z}[x], P = \displaystyle\sum_{j=0}^n a_j x^j$ with $a_0 \neq 0$ and $a_n \neq 0$, if $P(p/q)=0, p\perp q$ then $p|a_0$ and $q|a_n$.

As a follow-up to this question, I'd like to know how can I prove the following:

$$a_0 = a_n \displaystyle\prod_{j=1}^n (-r_j)$$, with $r_j$ being roots of P(x)

This is the same as saying:

$$a_0 = a_n \frac{-p}{q} \displaystyle\prod_{j=2}^n (-r_j)$$

Rearranging this, we get:

$$a_n = q\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)}$$

I want to prove that $$\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)} \in \mathbb{Z}$$ without using the rational root theorem (actually, I want to prove the rational root theorem using this identity).

So far I've tried to prove it by contradiction, assuming $a_0 = k(-p)\displaystyle\prod_{j=2}^n (-r_j) + \epsilon$ with $\epsilon \in (0, (-p)\displaystyle\prod_{j=2}^n (-r_j))$ but I don't seem to find anything weird rearranging that equation.

If you manage to find a proof please just post a small hint. Thanks in advance.

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Fernando Martin
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    @fmartin: You are missing a minus sign; if $r_1 = \frac{p}{q}$, then you factor out $-\frac{p}{q}$, or put a minus sign somewhere... – Arturo Magidin Nov 05 '10 at 19:20
  • @Arturo: You're right, I'll edit that. – Fernando Martin Nov 05 '10 at 19:27
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    Note that generally the roots $ r_j $ are not rationals but, rather, algebraic numbers. Why do you desire to pass to algebraic numbers to try prove a result that has a simple proof in integers via Bezout's theorem? – Bill Dubuque Nov 05 '10 at 19:56
  • @Bill: On the other hand, the full product of the "other" roots will be a rational... – Arturo Magidin Nov 05 '10 at 20:20
  • @Arturo. Of course but that's of no help for the matter at hand. – Bill Dubuque Nov 05 '10 at 20:23
  • @Bill: I know the theorem can be proved that way thanks to your response in the other thread. I was trying to prove the theorem using the naive approach posted above and I found myself unable to prove it. I just wanted to know how it can be proved, but now I know that's more difficult than I thought since, as you mentioned, not all roots are rationals. Thanks. – Fernando Martin Nov 05 '10 at 20:25
  • @Bill: Yes, I agree; but my point is that he's not necessarily "passing to the algebraic numbers", since the $r_j$, $j\gt 1$, are not treated separately but only "bundled up" into the product, which is a rational. (I agree that this seems a rather strange way to try to prove the result, though). – Arturo Magidin Nov 05 '10 at 20:26
  • @Arturo: But if you "bundle them up" into a rational then you're back to the usual rational methods of proof, so there was no reason to introduce algebraic integers in the first place. – Bill Dubuque Nov 05 '10 at 20:30
  • There are other generalizations of Gauss's Lemma, e.g. Kronecker's Lemma (a.k.a Dedekind's Prague Theorem) that can be proved by way of algebraic numbers. But that would be extreme overkill here and, furthermore, probably beyond the OP's knowledge of algebra and number theory. – Bill Dubuque Nov 05 '10 at 20:33

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Those are just Vieta's formulas for the coefficients in terms of the roots.

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