When I try to prove that $\operatorname{End}_R(R):=\text{Hom}_R(R,R)\text{(left)}\cong R^{\text{op}}$, I used the map $f \mapsto f(1)$, which requires $R$ to have an identity. Is it really necessary to assume that $R$ has an identity? If not, is there an counterexample for this?

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    $\begingroup$ $\operatorname{End}_R(R)$ has an identity, so for an isomorphism $End_R(R)=R^{op}$ to hold, $R$ must have an identity $\endgroup$ – Roland Feb 18 '17 at 22:10

The endomorphism ring always has an identity. If the ring $R$ hasn't one, you're doomed.

For a more spectacular failure, consider $R$ to be the Prüfer $p$-group, with trivial multiplication. Its endomorphism ring is the ring of $p$-adic integers, which has even larger cardinality.

  • $\begingroup$ oh~~~sorry for this stupid question... $\endgroup$ – No One Feb 18 '17 at 22:12
  • $\begingroup$ By the way, what do you mean by trivial multiplication here? $\endgroup$ – No One Feb 18 '17 at 22:15
  • $\begingroup$ @TiWen $ab=0$, for all $a,b\in R$. $\endgroup$ – egreg Feb 18 '17 at 22:18

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