Is $x^4+1$ irreducible in $F_3[x]$?

Question

I don't know any irreducibility criteria over $F_p[x]$ - so I tried to decompose it using $1=-2-3n\pmod 3$ and make $2+3n = p^2$ so that $x^4+1$ turns into $x^4-p^2$ - but didn't succeed.

Easy to see there are no roots. Try to factor it as the product of two quadratics. — lulu, May 17 '18 at 18:00
Background information: over $\Bbb R$ it is $(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$. There is no square root of $2$ in $\Bbb F_3$, which suggests the answer should be no. One obtains this factorization by writing it as a product of quadratics and proceeding, which is how you'll get a contradiction here — anon, May 17 '18 at 18:18
$x^4+1 = x^4+4,$ factors by completing the square, as in the linked dupe (put $,k=1,$ there). $\ \ $ — Bill Dubuque, Mar 27 '24 at 07:15

score 4 · Accepted Answer · answered May 17 '18 at 18:00

No. One reason is that you can use a nice factorization of $a^4+4b^4$ usually seen in mathematical competitions (eg here https://artofproblemsolving.com/wiki/index.php?title=Sophie_Germain_Identity). This is relevant in your case because $x^4+1=x^4+4$ in your field.

score 1 · Answer 2 · answered May 17 '18 at 18:00

Hint If $X^4+1$ would be reducible, it would either be the product of a first degree and a cubic, or of two quadratics.

The first case happens if and only if $X^4+1$ has a root in $\mathbb F_3$, which is easy to check.

In the second case $$X^4+1=(X^2+aX+b)(X^2+cX+d)$$

Now open the brackets and solve.

Is $x^4+1$ irreducible in $F_3[x]$?

2 Answers2