Two equivalent definitions of injective modules

Question

According to my lecture notes, these two definitions of an injective module are equivalent:

(Let $R$ be a ring, $Q$ an $R$-module.)

1. For every injective $R$-module homomorphism $u:M’ \rightarrow M$ and every $R$-module homomorphism $f:M’ \rightarrow Q$, there is a $R$-module homomorphism $\tilde{f}:M \rightarrow Q$ such that $\tilde{f} \circ u = f$.

2. For every ideal $I$ in $R$ and every $R$-module homomorphism $f:I \rightarrow Q$ there is a $R$-module homomorphism $\tilde{f}:R \rightarrow Q$ such that $\tilde{f}|_I = f$.

I can see how $(2)$ is just a particular case of $(1)$, but I’m not sure how to prove the other direction. I’ve been told Zorn’s lemma can be used here but I don’t see how. Any help would be appreciated.

(I’ve seen there’s already a few questions regarding equivalent definitions of injective modules on this site, but I haven’t found any dealing with definition $(2)$.)

Answer 1

WLOG, take $u : M’ \to M$ to be the inclusion of a submodule.

Consider the partial order $(S, \leq)$ defined by $S = \{(W, g) \mid W$ is a submodule of $M$, $g : W \to Q$ is $R$-linear$\}$ and $(W, g) \leq (X, h)$ if and only if both $W \subseteq X$ and $h|_W = g$.

Note that every chain of $S$ has a least upper bound. In particular, if $C$ is a chain in $S$, then let $W = \{0\} \cup \bigcup\limits_{(X, h) \in C} X$ and define

$$g(x) = \begin{cases} h(x) & x \in X, (X, h) \in C \\ 0 & x = 0 \end{cases}$$

Then $W$ is a submodule of $M$, $g$ is a well-defined linear map $g : W \to Q$, and $(W, g)$ is an upper bound of $C$.

Now recall the following version of Zorn’s Lemma:

Let $(S, \leq)$ be a poset, and suppose every chain in $S$ has an upper bound. Then for all $x \in S$, there is a maximal element $x \leq m \in S$.

Applying this version of Zorn’s Lemma with $x = (M’, f)$ gives us some maximal $(W, g) \in R$ with $M’ \subseteq W$ and $g|_{M’} = f$. It now suffices to prove that $W = M$.

Consider an arbitrary $x \in M$. Let $I = \{r \in R \mid rx \in W\}$. Then $I$ is an ideal. Note that the map $h(r) = g(rx)$ is an $R$-module homomorphism $h : I \to Q$. Extend this to an $R$-module homomorphism $\tilde{h} : R \to Q$.

Now define $X = W + (x)$, and define $i : X \to Q$ by $i(w + rx) = g(w) + \tilde{h}(r)$ whenever $w \in W$, $r \in R$. Verify that $i$ is well-defined and $R$-linear. Then $(W, g) \leq (X, i)$. Since $(W, g)$ is maximal, we have $W = X$ and thus $x \in W$.

Therefore, $W = M$. So $g = \tilde{f}$ is the function we require.

Finally, note that we do require Zorn’s Lemma (or at least some form of the axiom of choice) to show (2) implies (1). This is because we can show without choice that $\mathbb{Q}$ is an injective $\mathbb{Z}$-module using definition (2), but we cannot show this using definition (1). See my answer here.