**Claim:**is irreducible over but has factors modulo

**every**prime .

**Proof**

**:**

**must**have a solution, because for a prime , if is not a quadratic residue, then is. [Proof of this fact: If then where is the Legendre symbol.]

Concrete Nonsense

Thanks for visiting my blog. The labels may help you to find possibly interesting articles. Do leave your comments if you feel so.

Many a times, to prove that a polynomial in one indeterminate with integer coefficients is irreducible, we prove that it is irreducible modulo some prime, ; i.e. irreducible in the ring . Unfortunately, the converse is not true, and this can be seen by a simple counterexample:

The symmetry in the coefficients of the polynomial suggests that whenever in is a root, then so is in where . (This small result is also interesting in its own right! ).

So it is enough to consider the factorization of into two quadratics.

Case i:

Here, there exists a solution, say a for the congruence . (This is a standard result in number theory, first proved by Fermat.) Now, and so there are factors of .

Case ii:

Then, the factorization is possible provided there exist integers satisfying . This reduces to the four congruences-

(1)

(2)

(3)

(4)

(1) and (3) imply that . Together with (4) gives so that OR . Substituting in (2) gives or

But one of the two equations **must** have a solution, because for a prime , if is not a quadratic residue, then is. [Proof of this fact: If then where is the Legendre symbol.]

Hence the congruences have at least one solution for for modulo every prime and thus$latex x^4+1$ is reducible modulo every prime. ∎

Advertisements

%d bloggers like this:

## Leave a comment

Comments feed for this article