Let $A$ and $B$ be finite commutative rings with unity. Denote the additive group structure of each to be $A^{(+)}$ and $B^{(+)}$, and the multiplicative group of units of each to be $A^{(\times)}$ and $B^{(\times)}$ respectively. Supposing that $A^{(+)}\cong B^{(+)}$ and $A^{(\times)}\cong B^{(\times)}$, does it follow then that $A\cong B$?
Does a finite ring's additive structure and the structure of its group of units determine its ring structure?
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abstract-algebra
ring-theory
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1I think [this MO question](http://mathoverflow.net/questions/7133/classification-of-finite-commutative-rings) could be useful. – 2011-04-27
1 Answers
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No. Suppose that $p$ is an odd prime, and consider $A = \mathbb F_p[x,y]/(x^2,xy,y^2)$ and $B = \mathbb F_p[z]/(z^3)$.
In each case the addive group is isomorphic to a direct sum of 3 copies of a cyclic group of order $p$, while the multiplicative group is isomorphic to a direct product of a cyclic group of order $(p-1)$ and two cyclic groups of order $p$.
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1Dear Leon, Each of these rings has a natural projection to $\mathbb F_p$, which induces a surjection of the units onto $\mathbb F_p^{\times}$; call the kernel $U$. It's easy to see in each case that $U^p = 1$ (actually, maybe *here* is where I needed $p$ odd!). So $U$ is an ab. group of order $p^2$ and killed by $p$, hence a product of two cyclic groups of order $p$. So each unit group is an extension of $C_{p-1}$ by $C_p \times C_p$, and any such extension splits (easily checked). Regards, – 2014-01-28