Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $p\in\mathbb{N}$ is a prime, is $x^n+px+p^2$ irreducible in $\mathbb{Z}[x]$?

I've proved that any non-unit factor in $\mathbb{Z}[x]$ must have degree at least 2.

Eisenstein's criterion doesn't hold, but perhaps we can make some minor change so that it does? (E.g. a linear substitution, as with cyclotomic polynomials??)

Or could we start from scratch, assuming $x^n+px+p^2 = (b_kx^k+...+b_0)(c_{n-k}x^{n-k}+...+c_0)$ and trying to get conditions on the $b$'s and $c$'s that give a contradiction. I've got that $b_0=\pm p$, $c_0=\pm p$, $b_1=0$, $c_1=\pm 1$, and $c_2=\pm b_2$, but can't see how this helps!

A hint given in the question says 'Consider powers of $p$ dividing coefficients.' Does this suggest some variant of Eisenstein, or an unusual way of implementing Eisenstein?

Thanks for any help with this!

share|improve this question
    
HINT: Consider that is factorizable as $fg$, the reduce mod p, and you will have $\bar{f}\bar{g}=x^n$, which by unique factorization in $\mathbb{F}_p[x]$ gives that all coefficients except the leading ones are multiple of $p$. Use that fact for obtaining a contradiction when considereng the terms of dregrees zero and one of those polynomials. –  Josué Tonelli-Cueto Apr 28 '12 at 10:55
    
Example 2.2.4 from the book Prasolov: Polynomials gives a solution of this problem using Newton diagram. (However, I do not know this technique and, based on the hint you wrote, this is obviously not the solution you're after.) –  Martin Sleziak Apr 28 '12 at 11:13
add comment

1 Answer 1

up vote 2 down vote accepted

As you write $$x^n+px+p^2=(b_kx^k+\ldots+b_0)(c_{n-k}x^{n-k}+\ldots+c_0) $$

First, we may assume $b_k=c_{n-k}=1$, and $k>0,n-k>0$. Then $\mod p$, it will give $x^n=polynomial\times polyonoimal$ in $\mathbb{Z}/p[x]$, this forces that $p\mid b_i,p\mid c_j$ for $i\neq k,j\neq n-k$. Now consider the term of the degree one of the original polynomial, we obtain $px=(b_0c_1+c_0b_1)x$, this gives a contradiction if $n-k>1,k>1$.

Now if in the case $$ x^n+px+p^2=(x^{n-1}+b_{n-2}x^{n-2}+\ldots+b_0)(x+c_0) $$ We have $c_0=p,-p$, thus $p$ or $-p$ is a root of $x^n+px+p^2$, but this is not true.

share|improve this answer
3  
A little plug for newton polygons: Looking at the newton polygon of the polynomial, we see that over $\mathbb{Q}_p,$ the polynomial is either irreducible or factors as a product of a degree $n-1$ eisenstien polynomial times a linear factor with a root of valuation 1. So if the polynomial factors over $\mathbb{Z}$ a linear factor with a root of valuation 1 must appear. Now evaluate at $p$ and $-p$ to obtain the result. –  jspecter Apr 28 '12 at 11:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.