# Prove that the image of a ring via functor is a bimodule

I'm studying Morita theory on "Rings and Categories of Modules", Anderson. I have some problems with a Theorem about equivalent rings.

Let $R$ and $S$ be equivalent rings via inverse equivalences $F: {}_{R}Mod \to {}_{S}Mod$ and $G: {}_{S}Mod \to {}_{R}Mod$. Set $P=F(R)$, $Q=G(S)$. Then $P$ and $Q$ are naturally bimodules $_{S}P_{R}$ and $_{R}Q_{S}$.

The theorem states other results but I can't demonstrate this first assertion. In the proof he uses two isomorphisms $R \cong End(_{R}R)$ via right multiplication $\lambda$ for a scalar in $R$ and $End(_{R}R) \cong End(_{S}P)$ via $F$. So there is a ring isomorphism $r \mapsto F(\lambda(r))$ and I have to use this one to define a right multiplication of an element of $P$ for a scalar in $R$ to prove that $P$ is a right $R$-module. He also says that we use the fact that $_{R}R_{R}$ is a bimodule and that $F$ is an additive functor.

I have thought of: $P\times R \to P$ sending $(p,r) \mapsto pF(\lambda(r))$ but i'm not sure of this definition and if the result is in P. How can I prove that P is a right R-module? If someone has some suggestions I'd really appreciate!

• Actually it's $R^{op}$ that is isomorphic to $End(_R R)$, so if $f:R^{op}\to End(_S P)$ is your isomorphism, can't you define $x\cdot \lambda = f(\lambda)(x)$ ? – Max Aug 12 '18 at 13:57
• Sorry, I don't understand what you are saying. Can you try to explain? – robbis Aug 12 '18 at 13:59
• An endomorphism of $_R R$ is of the form $x\mapsto xa$ for some $a$: indeed put $a=f(1)$, then for $x\in R$, $f(x) = f(x 1) = xf(1)$. But $g:a\mapsto (x\mapsto xa)$ is not a morphism: indeed $g(ab) : x\mapsto xab = (xa)b = g(b)(xa) = g(b)\circ g(a)(x)$. It's rather a morphism from the opposite ring (that has multiplication defined as $x\times^{op}y= yx$) to $End(_R R)$. So you have an isomorphism $f:R^{op}\to End(_S P)$. Use this to define the right action of $R$ on $P$ by $x\cdot \lambda= f(\lambda)(x)$, for $x\in P, \lambda \in R$ – Max Aug 12 '18 at 14:03
• This will indeed be a bimodule structure because each $f(\lambda)$ is $S$-linear – Max Aug 12 '18 at 14:03
• Are the 2 isomorphisms in the text right? Isn't the composition of two isomorphisms an isomorphism? – robbis Aug 12 '18 at 14:25