# Modified version of Eisenstein's irreducibility criterion

I have an assignment to extend/modify (and of course prove it) Eisenstein's criterion as follows:

Let $f(x)=\sum a_ix^i\in\mathbb{Z}[x]$ with $n\ge 2$ and let $p$ be a prime such that $p\mid a_i$ for $i=0,\ldots n-2$ but NOT for $i=n-1, n$ and $p^2\not\mid a_0$.

My first thought was that irreducible polynomials in a polynomial ring fill the role of prime numbers in the ring of integers (in some sense) and then I wondered under what condition will the sum of two primes be a prime. Now I need really special primes to have their sum as a prime. Since their parity has to be different one of them has to be $2$ and the second has to be a lower twin of a twinprime-pair. I was thinking about sums since if I define $$f_1(x)=a_nx^n+a_{n-2}x^{n-2}+\ldots+a_0$$ and $$f_2(x)=a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_0$$ will be irreducible polynomials by Eisenstein's irreducibility criterion (the original one) and the sum of these two will be $$f'(x)=f_1(x)+f_2(x)=a_nx^n+a_{n-1}x^{n-1}+2a_{n-2}x^{n-2}+\ldots+2a_0$$ and this looks fairly much like $f(x)$ and the $2$'s won't disturb much (this is just a bold feeling) since all these terms are already divisible by a prime namely $p$ so we could write $f'(x)$ as $$f'(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}'x^{n-2}+\ldots+a_0'$$ and as long as $p\neq 2$ this fulfills the criterion of the modification. And from here try to "cook up" some criteria on the $a_i's$. But I feel stuck and I don't even know whether this is a good approach or I am just wasting time. Could anyone give me some hints how to continue from here or if the approach is wrong, give me a hint how to try to tackle the problem. Thanks.

• I suppose you mean to prove that f is irreducible. But in that case it's false: I'll leave it to you to cook up a degree-2 polynomial that factors even though its first two coefficients are odd and its constant term is even and not a multiple of 4. – John Brevik Feb 26 '15 at 17:48
• it also follows from the very awesome theory of the newton polygon – oxeimon Feb 26 '15 at 18:55

The assertion is true for every $n \geq 2$, if you additionaly require that $f$ has no rational root (if $f$ is monic, this means no integer root, if $f$ is not monic the denominator of a rational root divides the coefficient of the leading term).

You can always construct a polynomial $x^n+(p-1)x^{n-1}-px^{n-2}+px^{n-3} - \dotsb \pm p$, which fulfills the requirements but has the root $1$, hence is reducible.

So let us further assume that $f$ has no root (in particular we only have to deal with the case $n \geq 4$).

Modulo $p$ we have $f = x^n+ax^{n-1}=x^{n-1}(x+a)$. Let $f=gh$ with $g,h$ polynomials of degree at least $2$ (since we have no roots). Modulo $p$ we have $g=x^r$ and $h = x^s(x+a)$, with $r \geq 2, s \geq 1$. We deduce $f(0) = g(0)\cdot h(0) = 0 \mod p^2$, the same contradiction as in the prove of the original criterion.

• hey! I don't understand where $g=x^r$ and $h=x^s(x+a)$ comes from and why can we deduce a contradiction from them. Could you explain a little bit further? thanks – Vinyl_cape_jawa Mar 2 '15 at 7:52
• This is just unique factorization in $\mathbb Z/p\mathbb Z[X]$. – MooS Mar 2 '15 at 7:57
• I am sorry but I still don't see how unique factorization in $\mathbb{Z}_p[x]$ helps here. I mean the map $\mathbb{Z}[x]\to\mathbb{Z}_p[x]$ preserves reducibility/irreducibility when $p$ does not divide the highest coefficient but I am unsure that this works the other way around, or am I totally misunderstanding you? – Vinyl_cape_jawa Mar 2 '15 at 8:02
• $g$ and are $h$ are some factors of degree $\geq 2$ of the polynomial $x^{n-1}(x+a)$ in $\mathbb Z/p\mathbb Z[X]$. So they obviously have the form $x^r$ and $x^s(x+a)$ for some $r$ and $s$ respectively. The lower bounds for $r,s$ just come from the fact that $g,h$ have degree at least $2$. This is the same as I would say in integers: $a$ and $b$ are factors of the number $p^n \cdot q$, hence they have the form $p^r$ and $p^sq$. – MooS Mar 2 '15 at 8:05
• I see. But then how do we get $f(0)=g(0)h(0)$ when we reduce mod $p^2$. This whole exercise is just weird. I thought I am going to get some further assertions on the coefficients (some extra besides the ones for Eisenstein's) so it will be iredducile if ..., but I can't see this from the answer you gave, I have the feeling that I misinterpreted the question. – Vinyl_cape_jawa Mar 2 '15 at 8:18