# Why does the ring $A$ need to be commutative for $Hom_A(X,X')$ to be an $A$-module?

Refer to Lang's "Algebra" second paragraph from top, p. 122. Let $A$ be a ring and $X$, $X^'$ be $A$-modules. Let $Hom_A(X,X')$ be the set of $A$-homomorphisms from $X$ to $X'$. It is mentioned that if $A$ is commutative, then we can make $Hom_A(X,X')$ into an $A$-module by defining $(\alpha f)(x) = \alpha f(x)$ for $\alpha \in A$, $f \in Hom_A(X,X')$ and $x \in X$, whereas if $A$ is not commutative, then we can only regard $Hom_A(X,X')$ as an abelian group. The question is: why do we need the commutativity property to have a well-defined operation of $A$ on $Hom_A(X,X')$ as above? What goes wrong if we remove commutativity?

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But it is a module over the center of A. –  Frank Murphy Sep 15 '11 at 21:22

For the function $\alpha f$ to be a module homomorphism, we need it to satisfy $$(\alpha f)(x+y) = (\alpha f)(x) + (\alpha f)(y)$$ and $$(\alpha f)(\lambda x) = \lambda (\alpha f)(x).$$ The first condition is not a problem, but the second condition may fail if $R$ is not commutative: let $\alpha$ and $\lambda$ be two elements of $R$ such that $\alpha \lambda\neq \lambda\alpha$. Then $$(\alpha f)(\lambda x) = \alpha \Bigl(f(\lambda x)\Bigr) = \alpha\Bigl(\lambda f(x)\Bigr) = \alpha\lambda f(x);$$ but $$\lambda(\alpha f)(x) = \lambda\Bigl(\alpha f(x)\Bigr) = \lambda\alpha f(x).$$ We have no warrant for asserting that $\alpha\lambda f(x) = \lambda\alpha f(x)$ for all $x$, so $\alpha f$ need not be a module homomorphism when $R$ is not commutative.
Thank you very much, i had missed that! Something else that might be related: again from Lang p. 136 in the middle, it mentions that if $A$ is commutative and $\alpha$ a two-sided ideal, then $A/\alpha$ is ring. Is this not true even if $A$ is not commutative? –  Manos Jun 11 '11 at 6:12
If the ideal is two sided, you're fine even if the ring is noncommutative. But if $\alpha$ is a left-but-not-right ideal in a noncommutative ring (or a right-but-not-left ideal), then multiplication in $A/\alpha$ will not be well defined. Try it with $A$ the $2\times 2$ matrices over $\mathbb{R}$, and $\alpha$ the left ideal of matrices whose second column is all zeros (which is not a right ideal); try to spot where multiplication of congruence classes fails to be well-defined. –  Arturo Magidin Jun 11 '11 at 6:15