I have difficulty understanding the proof of Theorem 3.2 in Lang's Algebra Chapter VII.
Let $A$ be a subring of a field $K$ and let $x\in K, x\neq 0$. Let $\phi:A \rightarrow L$ be a homomorphism of $A$ into an algebraically closed field $L$. Then $\phi$ has an extension to a homomorphism of $A[x]$ or $A[x^{-1}]$ into $L$.
The proof begins saying
We may first extend $\phi$ to a homomorphism of the local ring $A_{\mathfrak{p}}$, where $\mathfrak{p}$ is the kernel of $\phi$. Thus without loss of generality, we may assume that $A$ is a local ring with maximal ideal $\mathfrak{m}$.
After this, it seems that $\mathfrak{m}$ is considered to be the kernel of $\phi:A \rightarrow L$.
Next, it is proved that $\phi$ has an extension to a homomorphism of $A[x]$ into $L$ when $\mathfrak{m}A[x^{-1}] = A[x^{-1}]$. This part is clear to me.
When $\mathfrak{m}A[x^{-1}] \neq A[x^{-1}]$, it is shown that $\mathfrak{P} \cap A = \mathfrak{m}$, where $\mathfrak{P}$ is some maximal ideal of $A[x^{-1}]$. An embedding $\psi$ of $A/\mathfrak{m}$ into $L$ is defined so that $A \rightarrow A/\mathfrak{m} \stackrel{\psi}{\rightarrow} L$ is equal to $\phi$. Then,
We note that $A/\mathfrak{m}$ is canonically embedded in $B/\mathfrak{P}$ where $B=A[x^{-1}]$, and extend $\psi$ to a homomorphism of $B/\mathfrak{P}$ into $L$, which we can do whether the image of $x^{-1}$ in $B/\mathfrak{P}$ is transcendental or algebraic over $A/\mathfrak{m}$. The composite $B \rightarrow B/\mathfrak{P} \rightarrow L$ gives us what we want.
My first question is "what is a canonical embedding ?". From $\mathfrak{P} \cap A = \mathfrak{m}$, I know that the injection $A \rightarrow B$ induces an injection $A/\mathfrak{m} \rightarrow B/\mathfrak{P}$. So, surely $A/\mathfrak{m}$ is embedded in $B/\mathfrak{P}$. But, in what sense is it canonical ?
My second question is "why is it possible to extend $\psi$ to a homomorphism of $B/\mathfrak{P}$ into $L$. It is obvious that it is possible if $B/\mathfrak{P}$ is an algebraice extension of $A/\mathfrak{m}$. But, of course this is not the case.
Any help would be appreciated. Thanks in advance.