Very tough polynomial question.

andreas · #1 $Old$ December 1st, 2008, 04:49 AM

Let $f(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n$ be a polynomial with integer coefficients and assume that for a given prime integer $p$ that:

1. $p$ does not divide $a_0$ ,

2. $p$ divides $a_1,a_2,...,a_n$

3. $p^2$ does not divide $a_{n-1}$
Prove that either $f(x)$ has a rational root, or $f(x)$ is irreducible over the filed $Q$

What is the meaning of the last sentence? Do I have to show that there are two cases: case1 irreducible, case2 has rational root?

I assume that I should consider two cases. Case1 is easy, it occurs when $p^2$ does not divide $a_{n}$ . In this case Eisenstein's criterion gives answer. But what about case2 when $p^2$ divides $a_{n}$ ?

I managed to show the following in case2:

$f(x)=x(a_0x^{n-1}+a_1x^{n-2}+...+a_{n-1})+a_n$ where $a_0x^{n-1}+a_1x^{n-2}+...+a_{n-1}$ is irreducible according to Eisenstein's criterion. I have a feeling that $f(x)$ has a rational root, but do not know how to show it...

vincisonfire · #2 $Old$ December 1st, 2008, 05:19 AM

First you should have learned that roots in $\mathbb Q$ *are of the form $\frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0}$ If not tell me I'll show you.
Notice that it could be the case that none of them works. In this case the polynomial is irreducible over $\mathbb Q$ .
Else, let $*c_0$ be a root found using the criterion $\frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0}$ .
Then it is possible to rewrite the polynomial as $g(x)=(x-c_0)(b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1})$ . You can now simply factor the second polynomial with the criterion $\frac{\pm \text{ factors of }b_{n-1}}{ \text{ factors of }b_0}$ .
At this point you can use the Eisenstein's criterion.
$h(x)=b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1}$
Even though you have divided these coefficients by some number, you can again put it in the form of a polynomial with integer coefficient by multiplying the whole thing by the least common multiple of the denominator.
You still have that some $p|b_i$ but $p^2 \text{ doesn't divide } b_{n-1}$ that is some multiple of $a_{n-1}$ .
Therefore, this h(x) is irreducible and you thus are either one or no root.

andreas · #3 $Old$ December 1st, 2008, 05:22 AM

Quote:

=vincisonfire;229638]First you should have learned that roots in $\mathbb Q$ *are of the form $\frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0}$

I know that and I tried but did not find very helpful.

You did not use $a_n$ so you continued my case2 ?

vincisonfire · #4 $Old$ December 1st, 2008, 06:19 AM

The three conditions you post add up all together.
You have only one case with three conditions. Not three with one condition.
What you did in your "case 2" doesn't work.
Any polynomial of the form $ax^2 +bx$ is reducible. But you can't divide the polynomial in the way you did. You can get a factor out but not a term.
Remark that I used the three conditions in my answer. I'm pretty sure I'm right even though my explanation can seem a little rusty.

andreas · #5 $Old$ December 1st, 2008, 09:46 AM

Quote:

You still have that some $p|b_i$ but $p^2 \text{ doesn't divide } b_{n-1}$ that is some multiple of $a_{n-1}$ .
Therefore, this h(x) is irreducible and you thus are either one or no root.

How do you prove this?

vincisonfire · #6 $Old$ December 1st, 2008, 09:57 AM

When you make the division, you'll find that $b_{n-1} = \frac{a_n}{c_0}$ because $a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n = (x-c_0)(b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1})$
You thus have an integer that p does not divide, divided by an integer that p does divide.
Since p divides all $a_i$ expcept $a_0$ , when you multiply everything by the least common multiple you are sure you won't have a $p^2$ factor in the factorization of $b_{n-1}$ .
Although, you are sure that you will have p factor in some $a_i$ since $p$ doesn't divide $a_0$