Factoring polynomials into prime prime polynomials

Question

How can I factorize $$p(x)=x^4+1 \in \mathbb{Z_5}$$ and $$p(x)=x^4+1 \in \mathbb{Z}_3$$ into prime polynomials?

Is $(x^2+2)(x^2+3)$ correct for $\mathbb{Z}_5$? Because $2 \cdot3 = 1$ and $2+3 = 0$ in $\mathbb{Z}_5$.

And how is it done for $\mathbb{Z}_3$?

Your answer for $\mathbb{Z}{5}$ seems correct, now just do the same for $\mathbb{Z}{3}$. — JohnColtraneisJC, Jun 22 '18 at 14:37
A generalization. Do notice that modulo a prime $\equiv1\pmod8$ this polynomial factors into a product of four linear polynomials. — Jyrki Lahtonen, Jun 22 '18 at 14:45

score 1 · Answer 1 · answered Jun 22 '18 at 14:40

1

Note that your factorisation in $\mathbf Z_5$ may as well be written as $$x^4+1=(x^2+2)(x^2-2),$$ which makes it more obvious.

Hint for $\mathbf Z_3$: $$ x^4+1=(x^2-1)^2+2x^2=(x^2-1)^2-x^2. $$

answered Jun 22 '18 at 14:40

Bernard

Even more clearly: $x^4+1=x^4-4=(x^2+2)(x^2-2)$. – lhf Jun 22 '18 at 14:42
That's what I meant, but I started from the O.P.'s factorisation. – Bernard Jun 22 '18 at 14:44
@Bernard For $\mathbb{Z}_3$ I ended up with $(x^2+x-1)(x^2-x-1)$. Is that correct? – user571590 Jun 22 '18 at 14:54
Absolutely correct. You can check performing the multiplication anyway. – Bernard Jun 22 '18 at 15:07

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