Field Theory – Prove Polynomial Irreducibility Over Rational Numbers

field-theorygalois-theoryirreducible-polynomials

I've tried a few "criteria" to check if this is irreducible. According to Maple it only has one entirely real root which I suspect is not rational but I can't prove it so I'm attempting to check if $p$ is irreducible.

Eisenstein's Criterion doesn't work here and I'm yet to find a suitable transformation such that it could work. I also read that if a polynomial is irreducible over $\Bbb F_q$, with $q$ a prime not dividing the leading coefficient, then it is irreducible over $\Bbb Q$ so I reduced the polynomial modulo $2$ to obtain

$$p \equiv x^5 + x^3 + 1 \mod 2.$$

I think this is correct but then I need to know how to check the irreducibility of this new polynomial over $\Bbb F_2$. Do I simply need to check that neither $0$ nor $1$ are roots of this polynomial? (And am I applying this theorem correctly?)

If this polynomial IS irreducible over $\Bbb Q$, is the splitting field obtained by simply adjoining the roots to $\Bbb Q$?

Best Answer

.A little bit of scouting for nice irreducibility criteria throws up some very nice results:

Here is a lovely lemma by (Prof.) Ram Murty:

Let $f(x) = a_mx^m + ... + a_1x + a_0$ be a polynomial of degree $m$ in $\mathbb Z[x]$. Let $H = \displaystyle\max_{0 \leq i \leq m-1} \left|\frac{a_i}{a_m}\right|$. If $f(n)$ is prime for some $n \geq H+2$, then $f(x)$ is irreducible in $\mathbb Z[x]$.

I'll give the link : http://cms.dm.uba.ar/academico/materias/2docuat2011/teoria_de_numeros/Irreducible.pdf

In our case, $a_m = 1$, and the maximum of all the quantities in question is $2$. Hence, if $f(n)$ is prime for some $n \geq 4$, then we are done.

You can check that for $n=4$, the number $f(4) =919$, which is prime!

Hence, it follows that the polynomial is irreducible.

ASIDE : There is also a "shifted" base (base shifts from $0...n-1$ to $|b| < \frac n2$) version of Cohn's criteria, which will tell you that if $f(10)$ is prime, then the given polynomial is irreducible. This matches that description, since all coefficients are between $-5$ and $5$. Very interestingly, $f(10) = 98779$ is also prime! (Hence, another proof by another wonderful result).

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