I have a question about the proof of Theorem 3.2. of Algebra by Serge Lang.
In the theorem $A$ is a subring of a field $K$ and $\phi:A \rightarrow L$ is a homomorphism of $A$ into an algebraically closed field $L$.
In the beginning of the proof
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 assum that $A$ is a local ring with maximal ideal $\mathfrak{m}$.
Then, later
Since $\phi$ and the canonical map $A \rightarrow A/\mathfrak{m}$ have the same kernel,
I understand that $\mathfrak{m}$ is the kernel of the extension of $\phi$ to $A_\mathfrak{p}$. But in the proof, $\mathfrak{m}$ seems to be assumed to be the kernel of $\phi$ even if $A$ itself is a local ring.
In general, is the kernel of homomorphism of a local ring to a field its maximal ideal? If so, why is that ?
Best Answer
I believe, Lang meant that one can consider only the case of a local ring and the kernel being $\mathfrak{m}$ (i.e. $A:=A_{\mathfrak{p}}$). For the kernel: let $R$ be a ring of germs of infintely-differentiable functions, and take a homomorphism of $R$ to formal power series. That is embeddeble into the field of Laurent series, but the kernel is obviously not maximal ideal.