# Are there any examples of rings $R$ such that $\mathrm{End}(R,+,0)\not\cong R$?

In a handful of examples, I've noticed that the endomorphism ring $\mathrm{End}(R,+,0)$ is isomorphic to the ring $R$ itself. For instance, $\mathrm{End}(\mathbb{Z},+,0)\cong\mathbb{Z}$ and $\mathrm{End}(\mathbb{Z}/n\mathbb{Z},+,0)\cong\mathbb{Z}/n\mathbb{Z}$.

Is this true in general, or are there examples of rings which are not isomorphic to the endomorphism ring as above? If not, is it at least always true for $R$ a field? Thanks.

• If you ever feel like working in rings without identity, a rng $R$ without identity is never isomorphic to its endomorphisms (as an $R$ module or as a $Z$ module) since endomorphism rings always have identity. – rschwieb Aug 12 '12 at 16:45
• The finite field of order 4 is another easy example: the endomorphism ring of its underlying additive group has order 16. This is a finite analogue of $\mathbb{Q}[i]$. – Jack Schmidt Aug 12 '12 at 17:49

## 1 Answer

The correct statement is that a ring $R$ is precisely the endomorphism ring of $R$ as a right $R$-module ("Cayley's theorem for rings"). When $R$ is any quotient of $\mathbb{Z}$, a morphism of right $R$-modules is just a morphism of abelian groups, which gives the examples you describe.

In general, preserving right $R$-module structure is stronger. For example, if $R = \mathbb{Z} \times \mathbb{Z}$ then the endomorphisms of the additive group of $R$ are given by $2 \times 2$ integer matrices. For a field counterexample, if $R = \mathbb{R}$ then $R$, as an abelian group, is a vector space over $\mathbb{Q}$ of uncountable dimension and therefore it admits an enormous endomorphism ring. (For a simpler field counterexample, if $R = \mathbb{Q}(i)$ then $R$, as an abelian group, is $\mathbb{Q}^2$, so its endomorphism ring is $2 \times 2$ rational matrices.)