Proposition. Let $R$ be a ring and $\{Q_i\}_{i\in I}$ a family of injective $R$ -modules. Then the product $$Q=\prod_{i\in I}\ Q_i$$ is injective.

Proof. Let $B$ be an arbitrary $R$ -module, $A\subseteq B$ a submodule and $f:A\to Q$ a homomorphism. It is enough to show that $f$ can be extended to $B$ . For $i\in I$ denote by $\pi_i:Q\to Q_i$ the projection. Since $Q_i$ is injective for any $i$ , then the homomorphism $\pi_i \circ f:A\to Q_i$ can be extended to $f'_i:B\to Q_i$ . Then we have $$f':B\to Q;$$ $$f'(b)=\big( f'_i (b)\big)_{i\in I}.$$ It is easy to check, that if $a\in A$ , then $f'(a)=f(a)$ , so $f'$ is an extension of $f$ . Thus $Q$ is injective. $\square$

Remark. Unfortunetly direct sum of injective modules need not be injective. Indeed, there is a theorem which states that direct sums of injective modules are injective if and only if ring $R$ is Noetherian. Note that the proof presented above cannot be used for direct sums, because $f'(b)$ need not be an element of the direct sum, more precisely, it is possible that $f'_i(b)\neq 0$ for infinetly many $i\in I$ . Nevertheless products are always injective.