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For fields it is well known, that all fields with finite cardinality $n$ are isomorphic. Does a similar result also hold for rings, i.e. are rings that have the same finite cardinality isomorphic ?

EDIT: Sorry, user Keenan Kidwell got it right, I mean "cardinality", not characteristic.

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    $\begingroup$ I know that isn't nice to answer to a question with another question, but I'll take the risk. Show that the ring $R=\mathbb{Z}[X]/(2,X^2+1)$ is not a field, has $4$ elements and characteristic $2$, but is not isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ which is not a field, has $4$ elements and characteristic $2$. $\endgroup$
    – user26857
    Commented Dec 2, 2012 at 21:04
  • $\begingroup$ Using the fact that fields are a subset of rings, we have a field of size $p^n$ for any pair $(n,p)$, but we also have a ring that is not a field of size $p^n$. Specifically, that is $\mathbb{Z}_p \oplus \ldots \oplus \mathbb{Z}_p$. $\endgroup$ Commented Dec 2, 2012 at 22:42
  • $\begingroup$ @andybenji If you have looked on this page would have seen that there already were such examples. $\endgroup$
    – user26857
    Commented Dec 3, 2012 at 9:03

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This is not true for fields. $\mathbb{F}_{p^n}$ is a field (in particular, a ring) of characteristic $p$, and so is $\mathbb{F}_{p^m}$, but these fields are isomorphic if and only if $n = m$.

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Surely not. For example, $\mathbb Z_2$ and $\mathbb Z_2\oplus\mathbb Z_2$ have both characteristic $2$. They cannot be isomorphic, as they have different cardinality.

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I guess that you're mixing up characteristic with cardinality. Any two fields of the same finite cardinality are isomorphic. On the other hand, the characteristic of a field, defined as the smallest positive integer $n$ such that $1+\cdots+1=0$ ($1$ added to itself $n$ times) in the field, or else if no such integer exists, as $0$, is either $0$ or $p$ with $p$ a prime, so always "finite." But there is no relationship between characteristic and cardinality beyond the fact that a field with finite cardinality must have positive characteristic.

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A more dramatic example would be $\mathbb{F}_p$ and $\mathbb{F}_p(t)$ where $t$ is an indeterminate, and $p$ is prime. They both have characteristic $p$, but one of them is finite while the other is not. You can also take an algebraic closure of $\mathbb{F}_p$, instead of $\mathbb{F}_p(t)$.

Later edit: As a counterexaple to your revised statement take $\mathbb{Z}/8\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. They both have the same cardinality, but cannot be isomorphic. One has characteristic $4$, while the other has characteristic $8$.

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  • $\begingroup$ What happens if the rings have also the same characteristic? $\endgroup$
    – user26857
    Commented Dec 2, 2012 at 20:40
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    $\begingroup$ Take $\mathbb{F}_{2^3}$ and $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. $\endgroup$
    – Rankeya
    Commented Dec 2, 2012 at 20:44
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    $\begingroup$ I've thought about $\mathbb{F}_{2^2}$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ :) $\endgroup$
    – user26857
    Commented Dec 2, 2012 at 20:51
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Even with your edit it is not true.

Consider $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. Both have $4$ elements. The first one has an element of (additive) order $4$, the second does not, so they cannot be isomorphic.

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