Necessity (⇒): If $I$ is injective, then by definition, the inclusion $\mathfrak{a} \hookrightarrow R$ allows us to extend any $\phi: \mathfrak{a} \to I$ to $\tilde{\phi}: R \to I$.

Sufficiency (⇐): Assume the extension property holds for ideals. We must show that for any submodule $A \subseteq B$ and homomorphism $f: A \to I$, there exists an extension $\tilde{f}: B \to I$.

Step 1: Consider the partially ordered set $\mathcal{P}$ of pairs $(C, g)$ where $A \subseteq C \subseteq B$ and $g: C \to I$ extends $f$. Order by $(C_1, g_1) \leq (C_2, g_2)$ if $C_1 \subseteq C_2$ and $g_2$ extends $g_1$.

Step 2: Apply Zorn's Lemma. Every chain in $\mathcal{P}$ has an upper bound (take the union of the submodules and the extension is unique). Thus $\mathcal{P}$ has a maximal element $(M, h)$ with $A \subseteq M \subseteq B$.

Step 3: We claim $M = B$. Suppose not, and let $b \in B \setminus M$. We will extend $h$ to $M + Rb$, contradicting maximality.

Step 4: Define the ideal: $\mathfrak{a} = \{r \in R : rb \in M\}$

This is an ideal of $R$. Define $\phi: \mathfrak{a} \to I$ by: $\phi(r) = h(rb)$

This is well-defined: if $r \in \mathfrak{a}$, then $rb \in M$, so $h(rb)$ makes sense.

Step 5: By hypothesis, $\phi$ extends to $\tilde{\phi}: R \to I$. Now define $\tilde{h}: M + Rb \to I$ by: $\tilde{h}(m + rb) = h(m) + r \tilde{\phi}(1)$

for $m \in M$ and $r \in R$.

Well-definedness: If $m + rb = m' + r'b$, then $(r - r')b = m' - m \in M$, so $r - r' \in \mathfrak{a}$. Thus: $h(m) + r\tilde{\phi}(1) = h(m) + r'\tilde{\phi}(1) + (r-r')\tilde{\phi}(1)$ $= h(m) + r'\tilde{\phi}(1) + \phi(r-r')$ $= h(m) + r'\tilde{\phi}(1) + h((r-r')b)$ $= h(m + (r-r')b) + r'\tilde{\phi}(1)$ $= h(m') + r'\tilde{\phi}(1)$

Step 6: The extension $\tilde{h}$ contradicts the maximality of $(M, h)$, so we must have $M = B$. Therefore, $I$ is injective.