In an algebra course, whenever we are asked to prove that a certain monic polynomial is irreducible in , it is a useful to trick to reduce the coefficients of this polynomial modulo for some prime and show that the resulting polynomial is irreducible in (the unique field with with elements). Proof of this result can be found in any standard introductory algebra course. Here is an application. If we are asked to show the irreducibility of in , we could reduce the coefficients mod to get the polynomial in . Now we can easily show that is irreducible in . Since is cubic, it suffices to show that has no root in . This is clear, because for . This shows that is irreducible in , and as a result is irreducible in , as desired.
What if we reduced coefficients of mod 5? Then our reduced polynomial would be in . But wait! This polynomial is now reducible, namely in . Does this mean that our original function is also reducible in ? Well, of course not, for we just proved above that is irreducible in . Moral of the story: if, for some prime , reducing coefficients of a polynomial modulo renders it reducible in , we cannot conclude that polynomial is reducible in .
Things are about to get more interesting. There exist some polynomials (and I will give an example) that are irreducible in but reducible in for every prime . Basically, these special polynomials fully destroy any possible converse for the theorem “if is irreducible in , then it is irreducible in .” One such polynomial is . First, let’s show that is irreducible in . It suffices to show is irreducible in . But we have
which is irreducible in by Eisenstein’s Criterion for prime . I will hopefully write about Eisenstein’s Criterion in one of the future blog posts. So we have established that is irreducible in .
It remains to show that is reducible for every prime . Rotman’s Advanced Modern Algebra has two different proofs of this fact. First one relies on the following argument. For it is clear that is reducible, since it has a root in . Now, for odd prime , one can show that can be written as a product of two quadratic polynomials in . To find coefficients of these polynomials, one simply writes down two arbitrary monic quadratics, and multiplies it out, and the result must equal to . The coefficients can then be analyzed, and some very special cases of quadratic reciprocity allows one to conclude that such two quadratics exist.
Second approach given in the book is the use of field extensions, and elementary property of cyclic groups. Here is the full proof:
Proposition. The polynomial is reducible in for every prime .
Proof. If , then we have , and we are done. Let be an odd prime. Then, it is easy to check that . Thus, . If denotes the group of units for the unique finite field containing elements, then it is easy to see that , and so . By a well known theorem is cyclic. Thus, the group contains a cyclic group of order , and so it contains all -th roots of unity; in particular, all the roots of are contained in (Here we use the fact that 8-th cyclotomic polynomial is ). Thus, the splitting field of over is
. Now, . So if were irreducible in , then would get , contradiction.