# Question about extensions of homomorphisms

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.

-

I guess that "canonical" means nothing else but the fact that the homomorphism is induced by $\mathfrak{P} \cap A = \mathfrak{m}$.

As for your second question: if $x^{-1}+\mathfrak{P}$ is transcendental over $A/\mathfrak{m}$, then $B/\mathfrak{P}$ is a polynomial ring over $A/\mathfrak{m}$, hence the existence of an extension follows from the universal property of polynomial rings.

H

-
Thank you for answering my question. Am I right in thinking that an extension of $\psi$ is defined as a unique homomorphism of $A/\mathfrak{m}$-algebras $\tilde{\psi}:B/\mathfrak{P} \rightarrow L$ such that $\tilde{\psi}(x^{-1}+\mathfrak{P})=\lambda$ for any $\lambda\in L$. – Aki Dec 7 '12 at 13:03