Factorization over $\mathbb{Q}$ and $\mathbb{Z_{41}}$

Question

Factor $f(x) = x^4+1$ over $\mathbb{Q}$ and over $\mathbb{Z_{41}}$.

1)I can't factor $f(x)$ over $\mathbb{Q}$ because $f(x+1)$ is irreducible by Eisenstein's criterion.

2)I don't know where to start: If there was a root $a$ then $a^4 \equiv -1 \,(\mod 41)$, how can I solve this congruence?

Answer 1

The polynomial splits over $\Bbb F_{41}$ into linear factors as $$ x^4+1=(x^2-9)(x^2+9)=(x + 3)(x -3)(x + 14)(x -14), $$ because $-9^2=1$ and $-14^2=9$ in $\Bbb F_{41}$.

Answer 2

(Online/free) SAGE software is a wonderful tool that I advise you to use in such questions ; it will help to build conjectures or, in the present case, give the solution (which shouldn't prevent you to construct proofs and have a thorough reasoning...)

You can invoke SAGE in this way : https://sagecell.sagemath.org/.

and make your request like this:

 F=GF(41)
 R.<x>=PolynomialRing(F)
 factor(R(x^4+1))

(GF = Galois Field, $R.\!\langle x\rangle$ for $R[x]$ where $x$ is an indeterminate variable).

Just press button "Evaluate" to obtain the following result :

$$(x + 3) * (x + 14) * (x + 27) * (x + 38)$$

which is the same $\pmod{41}$ as the one given by @Dietrich Burde.

Remark : what happens is you replace the first line by

 F=RationalField() ?

Answer : (your first question) The answer is ... plainly $x^4+1$, a way to say that it isn't factorizable over $\mathbb{Q}$, but in this case, if you don't know or don't think to Eisenstein criterion, it will be difficult to find a track to prove it !