# Is this true?

## Main Question or Discussion Point

Does anyone know if this is true and if so where they know it from?

Given a polynomial over the integers there exists a finite field K of prime order p, such that p does not divide the first or last coefficient, and the polynomial splits over K.

I realize this could be considered an abstract algebra question but I feel that this could easily have an elementary number theory proof. Of course I would love to see a reference in an algebra text of this result as well.

Thanks a lot,
Steven

## Answers and Replies

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matt grime
Homework Helper
Yes, it's true. Just adjoin the roots of the poly to the field, the extension is still finite, since the roots satisfy a polynomial.

But couldn't that be a field who's order is a prime power instead of just a prime?

matt grime
Homework Helper
oh, ok, i see your point now, sorry,

Does this argument sound reasonable?

A poly will split completely in some field if that field contains all it's roots.

Any algebraic extension of the rationals is a subset of a primitive extension.

Just consider the minpoly of that primitive element (using the algebraic extension where the original polynomial splits).

Every poly factors partially modulo some prime (a non constant f(x) can't equal +-1 for all values of x) so just consider the prime where that minpoly factors.

The original polynomial will completely split in that finite field since it will contain all the roots of the original polynomial.

Do any of those steps sound dubious to anybody (I know one of them sounds dubious to me but I don't want to say which because I don't want to bias anyone)

(I know this is a lot of algebra but number theory often is a lot of algebra)

Thanks,
Steven

shmoe
Homework Helper
snoble said:
Every poly factors partially modulo some prime (a non constant f(x) can't equal +-1 for all values of x) so just consider the prime where that minpoly factors.

The original polynomial will completely split in that finite field since it will contain all the roots of the original polynomial.
How are you concluding that this field contains all the roots? It seems like you're assuming the minimal polynomial of this primitive element will split in this field, which doesn't have to be the case. Sorry, can't offer much more help now. I'll try to think about it some more.

matt grime
Homework Helper
it is dubious - that minpoly can only at best partially factor - not factor - as you yourself state.

Ok, good. That was the part that I found dubious as well.

Here is how I justified that to myself (though I am no way convinced by this justification).

The roots of the original poly being in a primitive expansion means that for each root there is a polynomial (well rational function) such that the polynomial maps the primitive element to the root. ie $r_i\in \mathbb{Q}(p)$ implies that $\exists g_i$ rational function such that $g_i(p)=r_i$. But we don't really care about actually getting mapped to the right root - we just care that we get mapped to some root. So really what I'm saying is given this original polynomial f, and a primitive element p there is this set of g_i's such that f(g_i(p))=0.

So here's the nasty jump. It seems reasonable that any root of the minpoly of p should also be a root of every other poly that p is a root of (including polynomials that are numerators of rational functions). Doesn't it make sense that this sort of algebraic information of p should coded into its minpoly? I mean $f\circ g_i$ is just another poly (well a rational function) that p is a root of. So it shouldn't matter that the minpoly doesn't completely split in the prime field.

Again I'm not really happy with this explanation but maybe it will spark a thought in someone's head and they'll be able to say "oh hey I do recognize this... this is just a special case of Euler's product of Bernoulli integers... see this reference."

Oh... and I would like to apologize for posting so much in response to my own question (it seems like bad form to me somehow). I really haven't asked a question just so I could be a smart ass and try to answer it myself later. The comments so far have been greatly appreciated. I am looking forward to any other help that you guys can offer.

Thanks,
Steven

Hurkyl
Staff Emeritus
Gold Member
I'm 90% sure there exists a class of integer polynomials that are irreducible mod p for every prime p. IIRC, there's even a word for it!

shmoe
Homework Helper
A polynomial can be irreducible over Q, reducible mod some prime, yet not split in that prime order field. e.g. x^3+2x+1=(x+1)(x^2+x+1) mod 2.

Hurkyl, that's impossible no? As snoble pointed out any polynomial will have a root modulo some prime. If f is our poly, f(n) will have some prime divisor (for some large enough value of n to avoid f(n)=+/-1) so f is reducible mod that prime. There are however irreducible polynomials that are reducible modulo every prime, x^4+1 works I think. Locally irreducible seems to be the phrase.

Hurkyl
Staff Emeritus
Gold Member
I wish I could remember what it is I'm thinking. I can't find it in any of the texts I have handy.

I know of exactly what you are thinking about Hurkyl and I too have forgotten their names. However the key is that they are multivariate. I think they are important to a certain type of invariant theory. But integral polynomials over a single variable do factor. Of course the question is do they factor completely over some prime. Actually at this point that isn't even a question in my mind since I've now read both Mahler and van der Poorten state that fact - but without reference or proof. I just need to figure out why this is true or at least find a reference discussing it. Actually a reference would be more useful than a proof in this case.

I'm sure it's simple since they both just remarked it off the cuff. Unfortunately not so simple that I've been able to figure it out on my own

Steven

Ok guys thanks for thinking about this one but I finally found a reference that proves it by a guy named Cassels. Turns out you need a bit more care in picking your prime... you need the poly to factor and the prime not to divide the discriminant. Then you can use a corollary to Hensel's lemma. If anyone's invested some serious thought into this and is interested in the references please let me know. Also if you know of a simpler way to prove this I would love to see it.

Thanks again,
Steven

matt grime