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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 $A\cong B$?

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I think this MO question could be useful. –  Zev Chonoles Apr 27 '11 at 6:08

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up vote 17 down vote accepted

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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That's a nice counterexample. An obvious follow-up would then be, is there a nice characterisation of rings for which the statement does hold? –  jerome d Apr 27 '11 at 7:25
    
$A\ncong B$ as rings, since the number of elements $f$ with $f^2=0$ is $p^2$ in $A$ and $p$ in $B$. But where is $p\neq2$ actually needed? –  Leon Lampret Jan 25 at 23:28
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@LeonLampret Dear Leon, Actually, I don't think it's needed. Regards, –  Matt E Jan 26 at 1:12
    
Also, how do you see that $A^\times\cong B^\times$? I know that $A^\times=\{a+bx+cy;a\neq0\}$ and $B^\times=\{a+bz+cz^2;a\neq0\}$ with $$(a+bx+cy)(d+ex+fy)=ad+(ae+bd)x+(af+cd)y~~~\text{ and}$$ $$(a+bz+cz^2)(d+ez+fz^2)=ad+(ae+bd)z+(af+be+cd)z^2$$ But what are the isomorphisms to $C_{p−1}\times C_p\times C_p$? –  Leon Lampret Jan 28 at 18:39
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Dear 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, –  Matt E Jan 28 at 21:49

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