4

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?

user26857
  • 52,094
Clade
  • 235
  • 5
    Hint to get started. You know the ring is ${0,1,a,b}$. What are the possible values for $1+1$. – Ethan Bolker Apr 01 '17 at 15:41
  • 1+1 could be 0 or a correct? – Clade Apr 01 '17 at 15:45
  • 3
    Correct. Now explore those two possibilities with further questions about possible values for other sums and products. The decision tree you build will have four leaves. – Ethan Bolker Apr 01 '17 at 15:48
  • If 1+1=0 then we are in $\mathbb{Z}_4$ but I am not sure about what it would look like if it were a... – Clade Apr 01 '17 at 15:56
  • 2
    No. In $\mathbb{Z}_4$, $1+1=2 \ne 0$, so you can call that $a$. When $1+1=0$ you can ask other questions that lead to three possibilities. You can figure out those questions by writing out the addition and multiplication tables for the four rings you know you have to end up with. (The SE system will complain soon about carrying on a back and forth conversation in comments.) – Ethan Bolker Apr 01 '17 at 16:02
  • Duplicate of http://math.stackexchange.com/questions/279388/there-are-at-least-three-mutually-non-isomorphic-rings-with-4-elements, but I've preferred to vote for closing this as off-topic. – user26857 Apr 02 '17 at 09:46
  • @EthanBolker why is it necessary that the ring has an identity? Rings need not have an identity. – GraduateStudent Dec 04 '19 at 18:29
  • @SunShine It's conventional to assume there's a multiplicative identity. If you want to pursue this two yaer old question further without that assumption, feel free. – Ethan Bolker Dec 04 '19 at 18:37

1 Answers1

8

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.

egreg
  • 238,574
  • 1
    It seems like the first case can be identified with $\mathbb{Z}/4\mathbb{Z}$, third case with $\mathbb{F}_4$ and fourth case with $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. What about the second one? It is not exactly the $L$ in the list? Maybe I'm making a mistake in my calculation. Do you think there is a theoretic way of classifying? Some rep theory way? – nekodesu Dec 17 '17 at 18:32
  • @octoberbear $\mathbb{Z}/4\mathbb{Z}$ is not among the list, because it is not an algebra over $\mathbb{Z}/2\mathbb{Z}$. – egreg Dec 17 '17 at 18:56
  • Ah you are right! So how do we identify the first and second cases again? I'm not familiar with anything that are isomorphic to them. Also the $L$ in Clade's list is not present in the cases we calculated, is this true? @egreg – nekodesu Dec 17 '17 at 20:12
  • @octoberbear Case 1 is exactly $L$, where $a=\left(\begin{smallmatrix}0&1\0&0\end{smallmatrix}\right)$. The third case is not $\mathbb{F}_4$, because it has zero divisors. The field case is the fourth one. Note that only $L$ has nilpotent elements, so also case 2 is $L$. – egreg Dec 17 '17 at 21:05