Let $K$ be a field, $\Omega$ an algebraically closed field. Let $\Sigma$ be the set of all pairs $(A,f)$ where $A$ is a subring of $K$ and $F$ is a homomorphism of $A$ into $\Omega$. We partially order $\Sigma$ as follows: $$(A,f) \leq (A',f') \iff A \subseteq A' \land f'|A = f$$
Lemma 5.19 in Atiyah-Macdonald consists of showing that the maximal element $(B, g)$ of $\Sigma$ is a local ring by extending $g$ into $\overline g: B_{\ker g} \to \Omega$ with $\overline g (b/s) = g(b)/g(s)$, where $B_{\ker g}$ is the localization of $B$ at $\ker g$ and then using maximality to show $B = B_{\ker g}$.
My question is: doesn't this require that the condition $A \subseteq A'$ in the definition of our ordering not be taken literally, instead interpreting it as an injective homomorphism? Since to conclude $B = B_{\ker g}$ on the basis of maximality, we'd need to have $(B, g) \leq (B_{\ker g}, \overline g)$ by virtue of $B \subseteq B_{\ker g}$, but this last inclusion is false.
Full statement in Atiyah-Macdonald:
Lemma 5.19: B is a local ring and $\mathfrak m = \ker g$ is its maximal ideal. Proof: Since $g(B)$ is a subring of a field and therefore an integral domain, the ideal $\mathfrak m = \ker g$ is prime. We can extend $g$ to a homomorphism $\overline g: B_{\mathfrak m} \to \Omega$ by putting $\overline g(b/s) = g(b)/g(s)$ for all $b \in B$ and all $s \in B - \mathfrak m$, since $g(s)$ will not be zero. Since the pair $(B,g)$ is maximal, it follows that $B = B_{\mathfrak m}$, hence $B$ is a local ring and $\mathfrak m$ is its maximal ideal.
