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Is there an example of two groups $G_{1}, G_{2}$ such that there are two non isomorphic ring $R_{1}$ and $R_{2}$ such that the additive group of both rings is isomorphic to $G_{1}$ and their unit groups is isomorphic to $G_{2}$?

Lets generalize this question as follows:

Is there an example of two groups $G_{1}$ and $G_{2}$ such that there are infinite number of non isomorphic rings which their additive and unit groups are isomorphic to $G_{1}$ and $G_{2}$, respectively?

this question is motivated by the following post:

A basic question about rings

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  • $\begingroup$ Likely you can riff on the following example, although G_2 is empty: Let G_1 be a countable direct sum of the two element group, and define multiplication as in the two element ring, except the nth version "zeroes out" the first n coordinates. Gerhard "Willing To Be Backup Singer" Paseman, 2014.01.17 $\endgroup$
    – Gerhard Paseman
    Jan 17, 2014 at 23:36
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    $\begingroup$ For all squarefree integers $d > 1$, the ring ${\mathbf Z}[\sqrt{d}]$ has additive group isomorphic to ${\mathbf Z}^2$ and unit group isomorphic to $\{\pm 1\} \times \mathbf Z$. (Same result when $d < 0$ with unit group being $\{\pm 1\}$ except for $d = -1$ and $d = -3$.) For different squarefree integers $d_1$ and $d_2$, the rings ${\mathbf Z}[\sqrt{d_1}]$ and ${\mathbf Z}[\sqrt{d_2}]$ are nonisomorphic. More generally, there are lots of infinite families of these examples if you look at the ring of integers in number fields having a common degree over $\mathbf Q$. $\endgroup$
    – KConrad
    Jan 18, 2014 at 0:30
  • $\begingroup$ @KConrad Thank you, could you plese consider my same question but replacing "Ring" with "field"? $\endgroup$
    – Ali Taghavi
    Jan 18, 2014 at 0:33
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    $\begingroup$ Here are lots of examples. For every prime number $p$ let $K$ and $L$ be two non-isomorphic quadratic extension fields of the $p$-adic numbers ${\mathbf Q}_p$ that are ramified. (There are always at least three such extensions of ${\mathbf Q}_p$.) As additive groups, $K$ and $L$ are isomorphic to ${\mathbf Q}_p^2$. For $p \geq 5$, the unit groups $K^\times$ and $L^\times$ are both isomorphic to ${\mathbf Z} \times {\mathbf Z}/(p-1) \times {\mathbf Z}_p^2$. The proof I can think of for this description of $K^\times$ and $L^\times$ as abstract groups uses the $p$-adic logarithm. $\endgroup$
    – KConrad
    Jan 18, 2014 at 0:55
  • $\begingroup$ If you don't know about $p$-adics, then use the field example in Todd's comment to his answer, which can be extended to ${\mathbf Q}(x_1,\dots,x_n)$ for any $n \geq 1$. $\endgroup$
    – KConrad
    Jan 18, 2014 at 0:57

1 Answer 1

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Take $\mathbb{R}$ and $\mathbb{R}[x]$. All units in the latter are constant polynomials, so the unit groups are isomorphic. The additive groups of both have continuum dimension as rational vector spaces, so they are isomorphic. But clearly they are not isomorphic.

It seems the same argument shows that we could take polynomial rings over $\mathbb{R}$ in finitely many variables, and that gives an infinite class of nonisomorphic rings.

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  • $\begingroup$ thank you for the answer.what about if we replace "ring" with "field"? $\endgroup$
    – Ali Taghavi
    Jan 17, 2014 at 23:53
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    $\begingroup$ For fields, I think we could take $\mathbb{Q}(x)$ and $\mathbb{Q}(x, y)$. Both additive groups have countable dimension as $\mathbb{Q}$-vector spaces. The multiplicative groups are both isomorphic to $\mathbb{Q}^\ast \times F$ where $F$ is a free abelian group on countably many generators. In each case the generators of $F$ are given by monic irreducible polynomials; the fact that both $\mathbb{Z}[x]$ and $\mathbb{Z}[x, y]$ are UFD's means that the multiplicative monoids of nonzero elements are free abelian monoids (modulo sign), and the result for the fields follows. $\endgroup$
    – Todd Trimble ♦
    Jan 18, 2014 at 0:34

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