Non isomorphic rings of order 4

Question

How do I show that $\mathbb{Z}_4$, $\mathbb{Z}_2 \times \mathbb{Z}_2$, $L=\left\{\left(\begin{smallmatrix} a & b \\ 0 & a \end{smallmatrix}\right)\mid a,b\in\mathbb{Z}_2\right\}$ and $\mathbb{F}_4$ are the only non isomorphic rings of order 4?

Answer 1

The characteristic is either $2$ or $4$. Characteristic $4$ is easily dealt with, because $\mathbb{Z}/4\mathbb{Z}$ is a subring of every such ring.

Thus we remain with characteristic $2$, hence an algebra over $\mathbb{F}_2$. Take a vector space basis $\{1,a\}$, so the elements are $0$, $1$, $a$ and $1+a$. The only products we need to consider are $a^2$, $a(1+a)=(1+a)a$ and $(1+a)^2$. However, $a(1+a)=a+a^2$ and $(1+a)^2=1+a^2$, so we need to look at $a^2$.

Case $a^2=0$. Here $a(1+a)=a$, $(1+a)^2=1$

Case $a^2=1$. Here $a(1+a)=1+a$, $(1+a)^2=0$

Case $a^2=a$. Here $a(1+a)=0$, $(1+a)^2=1+a$

Case $a^2=1+a$. Here $a(1+a)=1$, $(1+a)^2=a$

Now identify the four cases among the rings in the list.