Factoring in $Z_3[x]$

The roots of any polynomial over $\mathbb{F}_3$ lie in some finite extension field $\mathbb{F}_{3^n}$.

The roots of $x^4 + 1$ are all primitive eighth roots of unity. Therefore, we know they live in any extension field such that $8 \mid 3^n - 1$.

The smallest such field is the one with $n=2$. However, this is merely a degree 2 extension! The minimal polynomial of an eighth root of unity in characteristic 3, then, is a quadratic polynomial. The irreducible factorization of $x^4 + 1$ over $\mathbb{F}_2$, therefore, must be a product of two quadratic polynomials.

We can say more: if $\zeta$ is a primitive eighth root of unity, then $\zeta$ and $\zeta^3$ are roots of one factor, and $\zeta^5$ and $\zeta^{15} = \zeta^7$ are roots of the other factor. We have a factorization

$$ x^4 + 1 = (x^2 - (\zeta+\zeta^3) x + \zeta^4) (x^2 - (\zeta^5 + \zeta^7) x + \zeta^4) $$

If you construct a realization of the field $\mathbb{F}_9$, you could do the arithmetic to simplify the coefficients to elements of $\mathbb{F}_3$.

But we can cheat: from ordinary arithmetic in characteristic 0, we know the standard eighth roots of unity

• $\zeta = \frac{\sqrt{2}}{2} (1 + \mathbf{i})$
• $\zeta^3= \frac{\sqrt{2}}{2} (-1 + \mathbf{i})$
• $\zeta^5 = \frac{\sqrt{2}}{2} (-1 - \mathbf{i})$
• $\zeta^7 = \frac{\sqrt{2}}{2} (1 - \mathbf{i})$

and so the factorization should simplify to

$$ x^4 + 1 = (x^2 - \frac{\mathbf{i}\sqrt{2}}{2} x + (-1))(x^2 - \frac{-\mathbf{i}\sqrt{2}}{2} + (-1))$$

In characteristic 3, $(\mathbf{i} \sqrt{2})^2 = -2 = 1$, so if we pick $1$ as the product of square roots, we should have

$$ x^4 + 1 = (x^2 - \frac{1}{2} x + (-1))(x^2 - \frac{-1}{2} + (-1)) = (x^2 - 2x - 1)(x^2 + 2x - 1)$$

and if we didn't believe that this "cheat" was actually a valid derivation, we can verify this factorization is correct directly.