# Very tough polynomial question.

• Dec 1st 2008, 02:49 AM
andreas
Very tough polynomial question.
Let $\displaystyle f(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n$be a polynomial with integer coefficients and assume that for a given prime integer $\displaystyle p$that:

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

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

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

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 $\displaystyle p^2$ does not divide $\displaystyle a_{n}$. In this case Eisenstein's criterion gives answer. But what about case2 when $\displaystyle p^2$ divides $\displaystyle a_{n}$ ?

I managed to show the following in case2:

$\displaystyle f(x)=x(a_0x^{n-1}+a_1x^{n-2}+...+a_{n-1})+a_n$ where $\displaystyle a_0x^{n-1}+a_1x^{n-2}+...+a_{n-1}$ is irreducible according to Eisenstein's criterion. I have a feeling that $\displaystyle f(x)$has a rational root, but do not know how to show it...
• Dec 1st 2008, 03:19 AM
vincisonfire
First you should have learned that roots in $\displaystyle \mathbb Q$*are of the form $\displaystyle \frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0}$ 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 $\displaystyle \mathbb Q$.
Else, let $\displaystyle *c_0$ be a root found using the criterion $\displaystyle \frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0}$.
Then it is possible to rewrite the polynomial as $\displaystyle g(x)=(x-c_0)(b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1})$. You can now simply factor the second polynomial with the criterion $\displaystyle \frac{\pm \text{ factors of }b_{n-1}}{ \text{ factors of }b_0}$.
At this point you can use the Eisenstein's criterion.
$\displaystyle h(x)=b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1}$
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 $\displaystyle p|b_i$ but $\displaystyle p^2 \text{ doesn't divide } b_{n-1}$ that is some multiple of $\displaystyle a_{n-1}$.
Therefore, this h(x) is irreducible and you thus are either one or no root.
• Dec 1st 2008, 03:22 AM
andreas
Quote:

=vincisonfire;229638]First you should have learned that roots in $\displaystyle \mathbb Q$*are of the form $\displaystyle \frac{\pm \text{ factors of }a_n}{ \text{ factors of }a_0}$
I know that and I tried but did not find very helpful.

You did not use $\displaystyle a_n$ so you continued my case2 ?
• Dec 1st 2008, 04:19 AM
vincisonfire
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 $\displaystyle ax^2 +bx$ 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.
• Dec 1st 2008, 07:46 AM
andreas
Quote:

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

How do you prove this?
• Dec 1st 2008, 07:57 AM
vincisonfire
When you make the division, you'll find that $\displaystyle b_{n-1} = \frac{a_n}{c_0}$ because $\displaystyle a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n = (x-c_0)(b_0x^{n-1}+b_1x^{n-2}+...+b_{n-2}x+b_{n-1})$
You thus have an integer that p does not divide, divided by an integer that p does divide.
Since p divides all $\displaystyle a_i$ expcept $\displaystyle a_0$, when you multiply everything by the least common multiple you are sure you won't have a $\displaystyle p^2$ factor in the factorization of $\displaystyle b_{n-1}$.
Although, you are sure that you will have p factor in some $\displaystyle a_i$ since $\displaystyle p$ doesn't divide $\displaystyle a_0$