Lemma 8, Section 4.3 of Hungerford’s Algebra

Question

Let $R$ be a ring with identity. A unitary $R$-module $J$ is injective if and only if for every left ideal $L$ of $R$, any $R$-module homomorphism $L\to J$ may be extended to an R-module homomorphism $R\to J$.

Sketch of proof: …… To show that $J$ is injective we must find a homomorphism $h:B\to J$ with $hg=f$ Let $\mathcal{S}$ be the set of all $R$-module homomorphisms $h:C\to J$, where $\text{Im}g \subseteq C\subseteq B$. $\mathcal{S}$ is nonempty since $fg^{-1}:\text{Im}g\to J$ is an element of $\mathcal{S}$ ($g$ is a monomorphism). Partially order $\mathcal{S}$ by extension: $h_1 \leq h_2$ if and only if $\text{Dom} h_1\subseteq \text{Dom } h_2$ and $h_2| \text{Dom } h_1 = h_1$. Verify that the hypotheses of Zorn's Lemma are satisfied and conclude that $\mathcal{S}$ contains a maximal element $h : H\to J$ with $hg = f$. We shall complete the proof by showing H = B ……

I showed poset $\mathcal{S}$ satisfy hypotheses of Zorn’s lemma. To be specific, if $\{h_i \mid i\in I\}$ is a chain in $\mathcal{S}$, then $h:\bigcup_{i\in I}\text{Dom }h_i\to J$ defined by $h(x)=h_i(x)$, where $x\in \text{Dom }h_i$, is an upper bound in $\mathcal{S}$ for the given chain.

I don’t understand why maximal element $h:H\to J$ of $\mathcal{S}$ satisfy $hg=f$. If possible, please give some intuition for defining $\mathcal{S}$.

Answer 1

Hungerford omits a critical part of the definition of $\mathcal{S}$ in his sketch.

We have an embedding of unitary $R$-modules $g\colon A\to B$, and a morphism $f\colon A\to J$. We wish to find a morphism $h\colon B\to J$ such that $hg=f$.

We define $\mathcal{S}$ to be the set of all $R$-module homomorphisms of the form $h\colon C\to J$, satisfying the following properties:

1. $\mathrm{Im}(g)\subseteq C\subseteq B$; and
2. $hg=f$ (this is the part missing in the sketch).

We then partially order $\mathcal{S}$ by setting $h_1\colon C_1\to J$ to be less than or equal to $h_2\colon C_2\to J$ if and only if $C_1\subseteq C_2$ and $h_2|_{C_1}=h_1$.

We know $\mathcal{S}$ is not empty, because have $fg^{-1}\colon \mathrm{Im}(g)\to J$ in $\mathcal{S}$, as $(fg^{-1})g = f$. Then as you do you establish that a chain has an upper bound (verifying that $hg=f$ for the $h$ is straighforward). By Zorn's Lemma, $\mathcal{S}$ contains maximal elements. We let $h\colon H\to J$ be a maximal element. It trivially satisfies $hg=f$. What we need to show is that $H=B$. The next paragraph in Hungerford's sketch establishes this equality, using the property being assumed.