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Let $R,S,T$ be rings, $_RM_S$ be a $R$-$S$-bimodule and $_RN_T$ a $R$-$T$-bimodule. Besides we have $$H=\text{Hom}_R(M,N).$$ We can turn $H$ into a $S$-$T$-bimodule via $$(s.f)(m):=f(m.s)\quad\text{and}\quad (f.t)(m):=f(m).t \qquad \forall f\in H,m\in M,s\in S,t\in T.$$

This is a calculation I have already done. Now I want to show, that $$\text{Hom}_R(_RR_R,_RN_T)~\cong~_RN_T$$ as $R$-$T$-bimodules.

My idea is, to show that $$\phi:\text{Hom}_R(_RR_R,_RN_T)~\rightarrow~_RN_T,~f\mapsto f(1_R)$$ is a bimodule-isomorphism.

At first, we have to show, that this is a bimodule-homomorphism:

  • $\phi(f+g)=f(1_R)+g(1_R)=\phi(f)+\phi(g)\quad\forall f,g\in H$
  • $\phi(f.t)=(f.t)(1_R)=f(1_R).t=\phi(f).t\quad\forall f\in H,t\in T$
  • $\phi(r.f)=(r.f)(1_R)=f(1_R.r)=...=r.f(1_R)=r.\phi(f)\quad\forall f\in H,r\in R$

My problem is, that I can't fill the $\dots$ in the third point. How would this work?

1 Answers1

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We have $$1_R\cdot r=r=r\cdot 1_R$$ Since $f$ is a left module homomorphism, $$f(r\cdot 1_R)=r\cdot f(1_R)$$ If I understand correctly this is what you are trying to prove.

Matt Samuel
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  • Yes thank you! Do you also have an idea for an inverse map of $\phi$? –  Oct 29 '16 at 15:15
  • @user114179 my guess is it'd be easier to prove injectivity by showing the kernel is trivial and proving surjectivity directly. – Matt Samuel Oct 29 '16 at 15:17
  • This is what I've tried first, but I have failed. –  Oct 29 '16 at 15:18
  • @user114179 inverting the homomorphism is easy. There is always a unique left module homomorphism from $R$ that sends $1_R$ to a given element. – Matt Samuel Oct 29 '16 at 16:11