1

I was solving this question - "How many non-isomorphic rings are there of order 4 elements?"

I know there are two groups of order $4$, i.e., $\mathbb{Z}_2\times \mathbb{Z}_2$ (Characteristic $2$) and $\mathbb{Z}_4$ (Characteristic $4$). With help from Non isomorphic rings of order 4, I got that there are $3$ non-isomorphic ring structures on $\mathbb{Z}_2\times \mathbb{Z}_2$. I also know that the there are only $4$ non-isomorphic rings of order $4$. Thus there is only $1$ non-isomorphic ring structure possible on $\mathbb{Z}_4$.

Let $R=\{0,1,a,b\}$ ( $\cong \mathbb{Z}_4$ as a group) and characteristic of $R$ is $4$. Note $1+1=a$ and $b=1+a$. Then $b^2=(1+a)^2=1+a^2+2a=1+a^2$ and $ab=a(1+a)=a+a^2$ and $ab=ba$.

Then let us study these possible cases:

Case $a^2=0$. Then $b^2=1$ and $ab=a$

Case $a^2=1$. Then $b^2=a$ and $ab=b$

Case $a^2=a$. Then $b^2=b$ and $ab=0$

Case $a^2=b$. Then $b^2=0$ and $ab=1$

Now Case $a^2=0$ is the usual ring structure on $\mathbb{Z}_4$. And the case $a^2=b$ is not valid since $ab=1$ implies $ab^2=b$, since $b^2=0$ this implies $b=0$. I don't know how to eliminate the cases $a^2=1$ and $a^2=a$.

Some help would be appreciated.

Sampah
  • 177
Ramandeep Singh Arora
  • 361
  • 1
  • 12
  • 1
    With this kind of things, it's useful to take not of what you are and what you aren't using. So far in your case study I haven't seen you use the additive structure of a ring. – take008 Dec 05 '19 at 13:24
  • @take008 Thanks, I got it. Since $2b=a$, then $2ab=a^2$ and since $2a=0$ this implies $a^2=0$. Thus only first case is valid. – Ramandeep Singh Arora Dec 05 '19 at 13:40

1 Answers1

1

There's no need to go to all this trouble. Notice that $$a^2=(1+1)(1+1)=1+1+1+1=0$$ $$b^2=(1+1+1)(1+1+1)=9\cdot 1=1$$ $$ab=6\cdot 1=a$$

Matt Samuel
  • 59,287