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Let $u,v\in\mathbb{Z},$ and let $f=X^4+uX^2+v.$ Let $p$ be a prime number, and let $r\geq 1.$

In a paper I'm reading, one can find the following result.

Proposition. If $f$ is reducible modulo $p^r,$ then it is the product of two monic polynomials of $\mathbb{Z}/p^r[X]$ of degree $2$.

The authors pretend that the proof of this result is similar to the case where we consider polynomials in $\mathbb{Z}[X]$.

For the case of polynomials in $\mathbb{Z}[X]$, this is easy: since $\mathbb{Z}$ is an integral domain, the sum of the degrees of two factors of $f$ is $4$, and each of the leading coefficients is invertible, so $f$ has either a monic linear factor or a monic quadratic factor. In the first case, $f$ has a root $m$, but $-m$ is also a root, and we are done if $m\neq 0$ (if $m=0$ the result is obviously true).

Unfortunately, weird things may happen in $\mathbb{Z}/p^r[X].$ The degree of a product is not the sum of the degrees, so the proof above does not work anymore. Worse: for example, if $u=v=0,p=2,r=2$, then for any polynomial $g\in \mathbb{Z}/4[X]$, we have $f=X^4=(X^2+2g)^2,$ so $f$ may have divisors of arbitrary degree.

However, note there is still a decomposition as in the statement of the proposition, since $f=X^2\cdot X^2.$

I'm trying to find a proof of the result, but for the moment, I'm struggling. If $f$ is a product of two coprime polynomials modulo $p$, then by Hensel's lemma, the decomposition may be lifted to $\mathbb{Z}/p^r[X],$ but honestly if I can avoid the use of Hensel's lemma, I would be happier.

If $f$ is the square of an irreducible polynomial modulo $p$, I don't know...

Question. Am I missing something obvious? Is the proposition true (I believe so...)? Does somebody have some arguments to fill the gap, or a complete proof?

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2 Answers 2

You should check out Theorem 2 in this 2011 monthly paper. (Irreducible Quartic Polynomials with Factorizations modulo p Author(s): Eric Driver, Philip A. Leonard, Kenneth S. Williams )

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1  
This is exactly the paper I'm reading. Proof of Theorem 2 is left to the reader, because the proof is similar to Theorem 1. However, I'm pointing out in my first message why this is not that clear. So my initial question remains... –  GreginGre Mar 20 at 9:36
    
@GreginGre Ah, I see... have you tried emailing the authors? –  Igor Rivin Mar 20 at 14:33

Ok, I finally found how to fill the gap.

In fact, a polynomial of $\mathbb{Z}/p^r[X]$ which is not $0$ modulo $p$ may be written in a unique way as a product of a unit of $\mathbb{Z}/p^r[X]$ by a monic polynomial.

Hence, if $f=gh$ where $g,h$ are both non units, then we write $g=ug_0,h=vh_0$, so $f=uvg_0h_0.$ Now $g_0h_0$ is monic and $uv$ is a unit, so $uv=1$ and $f=g_0h_0.$

But now since $g_0$ and $h_0$ are monic, one may conclude as for the case of $\mathbb{Z}[X]$ by degree considerations.

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It's nice that you found the solution. Yet you should accept your answer so that the post won't pop up in the list. –  Loïc Teyssier Apr 25 at 15:23

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