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I would like to know if somewhere the number of non-isomorphic rings with additive group $(\mathbb{Z}_{16})^2$ is mentioned. If not, is someone able to calculate it?

And (easier) the commutative case? A341548(16) in OEIS.

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It is not an answer but rather a method, outlined in https://arxiv.org/abs/1903.01623v1 . We adapt it to the ring $Z_{16}$.

We say an algebra is curled if every element $x$ of order 16 is linearly dependent with $x^2$. An algebra is straight otherwise.

If the algebra is curled, it has a basis $x,y$ such that $x^2=kx$ and $y^2=ly$. WLOG, $k,l\in\{0,1,2,4,8\}$. Writing $xy=ax+by$ and $yx=cx+dy$, we get a collection of 6 elements satisfying equations (3), (4) and (5) from the paper. Any collection will give an algebra. The number of collections is $5^2\cdot 16^4 = 1638400$. Such search is within computer possibilities but I suspect that it can be narrowed down much further.

If the algebra is straight, it has a basis $x,y$ such that $x^2=ky$. Again, we can restrict to $k\in\{1,2,4,8\}$. I would choose the smallest such $k$ possible among all the bases. If $k=1$, then $xy=yx=bx+cy$ and $y^2$ is determined. In this case, a straight algebra is commutative and is determined by 2 elements: $16^2 = 258$ possibilities. The analysis in the paper goes through.

The cases of higher $k$ require considerations, not covered in the paper. I suspect they are not difficult with $k$ being the smallest possible quite helpful: write $xy$ and $yx$, making sure that no $x+ay$ yields a smaller $k$.

To determine the isomorphism types, I would the criterion on page 4 and computer. The group $GL_2(Z_{16})$ has 4 generators and $6\cdot 16^3=393216$ elements. One can apply the generators to the $2\times 2$-matrices determining algebras: this gives all the orbits. There is probably a simpler method of just looking for

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  • $\begingroup$ I have to reflect on it. Can you tell me how many of them are commutative? $\endgroup$ Commented Apr 9, 2021 at 12:01
  • $\begingroup$ I don't really know how rewards work. In this case the answer is not entirely satisfactory. $\endgroup$ Commented Apr 14, 2021 at 10:28

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