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?
Does $\operatorname{End}_R(R)\cong R^{\text{op}}$ still hold when $R$ doesn't have an identity?
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abstract-algebra
ring-theory
modules
rngs
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3$\operatorname{End}_R(R)$ has an identity, so for an isomorphism $End_R(R)=R^{op}$ to hold, $R$ must have an identity – 2017-02-18
1 Answers
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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.
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0oh~~~sorry for this stupid question... – 2017-02-18
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0By the way, what do you mean by trivial multiplication here? – 2017-02-18
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0@TiWen $ab=0$, for all $a,b\in R$. – 2017-02-18