$R$-module maps $R\times R \to R$

Question

I can't figure out exercise 4 from chapter V, 2 of MacLane's Algebra book. The question is:

Show that any $R$-module map $R\times R \to R$ is of the form $(x_1,x_2) \mapsto \lambda_1x_1 +\lambda_2x_2$ for suitable scalars $\lambda_1$ and $\lambda_2 \in R$.

$R$-modules are left modules in this case, and the ring $R$ need not to be commutative. If $f: R^2 \to R,\, f(x_1,x_2) = \lambda_1x_1 +\lambda_2x_2$ is a map of $R$-modules, then for every $\mu \in R$ we get $f(\mu,0) = \lambda_1\mu$ by definition and $f(\mu,0) = \mu f(1,0) = \mu(\lambda_1 1) = \mu\lambda_1$ by properties of $R$-module maps. So $\lambda_1$ (and $\lambda_2$) need to be in the center of $R$.

For this reason I feel like the exercise is mistaken and it should rather be "$(x_1,x_2) \mapsto x_1\lambda_1 +x_2\lambda_2$". In this case everything works fine. Given a morphism $f$ of $R$-modules, set $\lambda_1 = f(1,0)$ and $\lambda_2 = f(0,1)$. What am I missing? Is there a way of solving the exercise in its original form?

Answer 1

For the original form, $R$ needs to be considered as a right $R$-module, not a left $R$-module.