Let $R$ be a commutative ring with unity and let $M$ be a finitely generated noetherian R-module. Can someone tell me how the isomorphism $$ Hom_R(R\oplus R, M)\simeq Hom_R(R,M)\oplus Hom_R(R,M)\simeq M\oplus M $$ is given? I know how the second isomorphism is given but I don't know about the first. Where does it send a map $R\oplus R\to M$ to?


2 Answers 2


The other answer is nice but only if you're comfortable with the language of categories and universal properties. More concretely, if we have a homomorphism

$$\phi:R\oplus R\to M,$$

we get two homomorphisms $\phi_1,\phi_2:R\to M$ by defining


You can check this sending $\phi\mapsto(\phi_1,\phi_2)$ gives an isomorphism $$\hom(R\oplus R,M)\cong\hom(R,M)\oplus\hom(R,M).$$

  • $\begingroup$ Where is the assumption that $M$ is a finitely generated Noetherian $R$-module exploited in this proof? $\endgroup$
    – gen
    Commented Mar 6, 2022 at 22:23
  • 1
    $\begingroup$ @gen it's not, those properties don't matter for this fact. $\endgroup$ Commented Mar 7, 2022 at 0:27

The universal property of the biproduct is precisely the existence of the first isomorphism (assuming it is natural in $M$).

More generally, the coproduct of $X \amalg Y$ of two objects in any category is the object that satisfies an isomorphism natural in $Z$: $$ \hom(X \amalg Y, Z) \cong \hom(X, Z) \times \hom(Y, Z) $$ In the case of $R$-modules, finite products and coproducts are both given by the biproduct; i.e. $X \amalg Y \cong X \oplus Y$ and $M \times N \cong M \oplus N$.

The projection $$ \hom(X \amalg Y, Z) \to \hom(X, Z) $$ is precisely the map induced by the insertion $i_1 : X \to X \amalg Y$. That is, it sends a function $g : X \amalg Y \to Z$ to the function $g \circ i_1 : X \to Z$.

The same is true for the other projection. The isomorphism from the natural property is the map $g \mapsto (g \circ i_1, g \circ i_2)$.


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