Proving $L(R,R) \otimes L(R,R) \cong L(R \otimes R , R \otimes R)$.

Question

Lang wrote in page 627, Algebra, proof of Corollary, that the isomoprhism

$L(R,R) \otimes L(R,R) \rightarrow L(R \otimes R, R \otimes R)$

is clear. Where given $R$-modules $E,F$, $L(E,F)$ is the set of $R$-module homomorphisms. What I did was

• $R \otimes R \cong R$ given by $\pi: x \otimes y \mapsto xy$, inverse by $i: x \mapsto x \otimes 1$. Given $f \in RHS$,$$f \mapsto (\pi \circ f \circ i ) \otimes id $$

• Conversely, a map from LHS to RHS is given by $f \otimes g \mapsto T(f,g)$ where $$T(f,g): x \otimes y \mapsto f(x) \otimes g(y)$$

• The map are inverses. $$f \mapsto T( (\pi \circ f \circ i), id ) : x \otimes y \mapsto (\pi \circ f \circ i)(x) \otimes y= f(x \otimes y)$$ $$f \otimes g \mapsto (\pi \circ T(f,g) \circ i) \otimes id = f \otimes g$$

• Last equality holds as $$ (\pi \circ T(f,g) \circ i) \equiv \alpha_f \alpha_g \, id $$ which holds because $f \equiv \alpha_f \, id$, $g \equiv \alpha_g \, id $

It took me some playing to figure out the first map from RHS to LHS. Is there are higher perspective which explains this map?

Answer 1

It's even simpler than that. There is an isomorphism $L(R,R) \cong R$ given by $f \mapsto f(1)$. Its inverse is $x \mapsto f_x$ where $f_x(y) = xy$. Since Therefore $L(R,R) \otimes_R L(R,R) \cong R \otimes_R R \cong R$.