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I have a problem which could not be solve by using Eisenstein's criterion.

Let $f(x)=\sum_{i=0}^n a_i x^i(a_n\neq 0, n\geq 2010)$. Suppose there exists a prime $p$ such that

(1) $p\nmid a_n$;

(2) $p\mid a_i, i=0,1,\cdots,2008$;

(3) $p^2\nmid a_0$.

Then $f$ should have an irreducible polynomial factor with degree $\geq 2009$.

How to do then? I was pardoned.

XLDD
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1 Answers1

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In the (finally!) modified form, the question can be solved using the approach giving the Schönemann-Eisenstein theorem (please allow this small bit of publicity for a nearly forgotten mathematician (guess which one) who actually found and proved the theorem first).

Factor your polynomial into irreducible factors. Since the constant term is divisible by$~p$ but not by$~p^2$, exactly one irreducible factor has a constant term divisible by$~p$; I will call this one $f_0$, so one has $f=f_0f_1\ldots f_k$ where the constant term of $f_i$ is not divisible by$~p$ for $i>0$. As $p$ does not divide the leading coefficient of$~f$, it does not divide the leading coefficient of any of the$~f_i$ either. Reducing the decomposition modulo$~p$ then gives $\bar f=\bar{f_0}\bar{f_1}\ldots\bar{f_k}$, a product of (possibly reducible) factors$~\bar{f_i}\in\Bbb Z/p\Bbb Z[x]$, with $\deg\bar{f_i}=\deg f_i$ for all$~i$. Now the hypotheses give that $\bar f$ is divisible by$~x^{2009}$, while $x$ does not divide any $\bar f_i$ with $i>0$. Since $x$ is an irreducible element of the UFD $\Bbb Z/p\Bbb Z[x]$, all those $2009$ factors$~x$ must then divide$~f_0$. This gives $\deg\bar{f_0}\geq2009$, and since $\deg\bar{f_0}=\deg f_0$ we get$f_0$ as our irreducible factor of degree${}\geq2009$.

Marc van Leeuwen
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  • +1, excellent argument. Only, I do not think it is really necessary to argue by way of contradiction: I believe you are showing that the irreducible factor $f_{0}$ whose constant term is divisible by $p$ has degree at least $2009$. – Andreas Caranti Sep 13 '13 at 07:50
  • @AndreasCaranti: You are right, thank you for noting that. I'll try to rewrite this without going for a contradiction, but I will have to assume a decomposition into irreducible factors at the beginning, something I did not do initially, to avoid even suggesting that after reduction I would get a factorisation into irreducibels. But your suggestion is better anyway. – Marc van Leeuwen Sep 13 '13 at 08:26