Factoring in $Z_3[x]$ I need to factor $x^6+x^4+x^2+1$ into irreducible parts in $Z_3[x]$. Obviously this polynomial reduces to $(x^4+1)(x^2+1)$ which is irreducible in $Z[x]$, but I'm not sure how to confirm that it's irreducible in $Z_3[x]$. I've tried trial and error, and haven't found anything but would love some suggestions if anyone has ideas. 
Just a heads up, I'm working ahead of my class since I have to work later this week and won't have time to wait for the lecture notes before finishing the assignment. Apologies if I've missed something obvious. 
 A: 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.
A: Hint $\rm\ mod\ 3\!:\,\ x^4\!+1\, \equiv\, x^4\!-3x^2\! + 1\,\equiv\, (x^2\!-1)^2 - x^2 =\, $ difference of squares
$\rm\  mod\,\ 2n\!+\!1\!:\,\ x^4\!+n^2\equiv\,x^4\!-(2n\!+\!1)\,x^2\! + n^2\equiv\, (x^2\!-\!n)^2\! - x^2$
