How to find all invertible elements in $F_3/(f(x))$

Question

Is there a algorithm or smt like this? For example, $F_3/f(x) = \{ax^3 + b x^2+c x+d | a,b,c,d \in F_3\}$ where $f(x)=x^4+2x^2+x+2$ and $F_3=Z/3Z$. Am I right that there are 8 zero-divisors

Answer 1

Hint:

In a finite commutative ring every non-zero element is either a zero-divisor or a unit. So instead of finding a unit we can aim at finding the zero divisors (which might be easier).

Let $R=\mathbb{F}_3[x]/\langle f(x) \rangle$. First observe that in $\mathbb{F}_3$ we have $$f(x)=x^4+2x^2+x+2=(x+2)^2(x^2+2x+2)$$ Say $g(x)$ is a zero divisor. Then $\exists \, \text{ nonzero } h(x)$ such that $g(x)h(x) \equiv 0 \pmod{f(x)}$. This also means we want \begin{align*} g(x)h(x) &\equiv 0 \pmod{(x+2)^2}\\ g(x)h(x) &\equiv 0 \pmod{x^2+2x+2} \end{align*} So we need to find zero divisors in $\mathbb{F}_3[x]/\langle x^2+2x+2 \rangle$ and also in $\mathbb{F}_3[x]/\langle (x+2)^2 \rangle$.

Since $x^2+2x+2$ is irreducible in $\mathbb{F}_3[x]$, so this ring is a field. Hence all non-zero elements are units in this ring. So no zero divisors.

But for $\mathbb{F}_3[x]/\langle (x+2)^2 \rangle$, $x+2$ is an obvious zero divisor. Check if there are others!!

Hopefully now you can complete it.