Three non-isomorphic rings whose additive group is the Klein four-group.

Question

I was looking at OEIS sequence A037291, which gives the number of rings with identity that have $n$ elements. That page linked me to a website written by C. Noebauer which claims that there are three non-isomorphic rings with identity that have an underlying additive group that is isomorphic to $(\mathbb{Z}_2)^2$.

One of them is the direct product of rings $\mathbb{Z}_2 \times \mathbb{Z}_2$ with the usual (componentwise) multiplication: \begin{array}{c|ccc} \times_1 & (0,1) & (1,0) & (1,1) \\ (0,1) & (0,1) & (0,0) & (0,1) \\ (1,0) & (0,0) & (1,0) & (1,0) \\ (1,1) & (0,1) & (1,0) & (1,1) \\ \end{array}

Another is given by the field $\mathbb{F}_4$. Let $\alpha$ be a root of the polynomial $x^2 + x + 1 \in \mathbb{Z}_2[x]$ so that $\alpha^2 = \alpha + 1$. Then $\mathbb{F}_4 = \{0, 1, \alpha, 1 + \alpha\}$, and we can consider the ring isomorphism which sends $1 \mapsto (1,0)$ and $\alpha \mapsto (0,1)$. This gives us the multiplication table \begin{array}{c|ccc} \times_2 & (0,1) & (1,0) & (1,1) \\ (0,1) & (1,1) & (0,1) & (1,0) \\ (1,0) & (0,1) & (1,0) & (1,1) \\ (1,1) & (1,0) & (1,1) & (0,1) \\ \end{array}

What is the third such ring? And does it arise in any kind of natural setting?

Answer 1

So we have a ring with four elements $\{0,1,x,1+x\}$ such that the sum of every element with itself is $0$. Then we can determine the multiplication from $x^2$ alone because $x(1+x)=(1+x)x = x+x^2$ and $(1+x)^2 = 1 + x + x + x^2 = 1 + x^2$.

What is $x^2$? We have four possibilities.

1. $x^2 = 0$ implies $x(1+x) = (1+x)x = x + x^2 = x$ and $(1+x)^2 = 1+x^2=1$

2. $x^2 = 1$ implies $x(1+x)=(1+x)x = x+x^2 = 1+x$ and $(1+x)^2 = 1+x^2 = 0$. So this is really isomorphic to the previous case by sending $x \mapsto 1+x$.

3. $x^2 = x$ implies $x(1+x) = x + x^2 = x + x = 0$ and $(1+x)^2 = 1 + x^2 = 1+x$. This is $(\mathbb{Z}/2)^2$.

4. $x^2 = 1+x$ implies $x(1+x)=(1+x)x = x + x^2 = 1 + x + x = 1$ and $(1+x)^2 = 1 + x^2 = 1 + 1 + x = x$. This is $\mathbb{F}_{2^2}$.

So the missing ring is one in which there is an element squaring to zero, cases 1 and 2 above. It arises as the quotient $\mathbb{Z}[x]/\langle 2, x^2 \rangle = \mathbb{F}_2[x]/\langle x^2 \rangle$.

Interesting to note that all of these are commutative.