I'm trying to study algebra at IUM and I stuck on this question:
Let $\mathbb{F_4} := \mathbb{F_2}[\alpha]/(\alpha^2+\alpha+1)$ -- a field that consists of 4 elements $\{0,1,\alpha, \alpha+1\}$. Find all zero divisors and inverses in quotient ring $\mathbb{F_4}[x]/(x^2+[\alpha]x+1)$.
What I do: first, let's find zero divisors: let $a_1x+a_2$ and $b_1x+b$ $ \in \mathbb{F_4}[x]/(x^2+[\alpha]x+1)$, such that $(a_1x+a_2)(b_1x+b) \equiv 0$, then:
$a_1b_1x^2+(a_1b_2+a_2b_1)x+a_2b_2 \equiv 0$
Since ideal is generated by $x^2+[\alpha]x+1$, then $x^2=[\alpha]x+1$. So:
$(a_1b_1[\alpha]+a_1b_2+a_2b_1)x+a_1b_1+a_2b_2=0$
And here I stuck. What should I do next? Solve this equation for x? Or make a system of linear equations and solve it? In both cases I don't see how I'll be able to find zero divisors. The same thing is for inverses.
You want $$a_1b_1[\alpha]+a_1b_2+a_2b_1=a_1b_1+a_2b_2=0$$ Assume without loss of generality that $a_1=b_1=1$. Then $a_2b_2=1$ as well, and $$[\alpha]+b_2+a_2=0$$ We would then have to have $a_2+b_2=\alpha$, but this is incompatible with $a_2b_2=1$. Thus there are no zero divisors. You can see this by noting that $x^2+[\alpha]x+1$ is irreducible, which you can find out by testing for roots, of which there are none.
To find inverses, it would help first to enumerate the elements. It is sufficient to look first at the monic elements $x$, $x+1$, $x+[\alpha]$, $x+1+[\alpha]$. We have $$x(x+[\alpha])=[\alpha]x+1+[\alpha]x=1$$ $$(x+1)(x+1+[\alpha])=[\alpha]x+1+x+[\alpha]x+x+1+[\alpha]=[\alpha]$$ We see that the inverse of $x$ is $x+[\alpha]$, and the inverse of $x+1+[\alpha]$ is $(1+[\alpha])(x+1)$ (we had to multiply by the inverse of $[\alpha]$ since $(x+1)(x+1+[\alpha])=[\alpha]$). The remaining elements have inverses that are scalar multiples of these (except of course the elements of $\mathbb{F}_4$, which have their usual inverses).