The kernel of homomorphism of a local ring into a field is its maximal ideal? 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 ?

We can only infer $\mathfrak m=\ker\varphi$ when $\mathfrak m$ is the only prime ideal of $A$. In this case $A/\ker\varphi$ is a subring of $L$, which is necessarily an integral domain.
However, $A$ is assumed to be a subring of a field! So, $(0)\subset A$ is prime, and we can always consider the case of $\varphi$ being the canonical embedding of $A$ in the algebraic closure of its field of fractions. With this choice of $\varphi$ if $A$ is not a field, then $\ker\varphi$ is not the maximal ideal $\mathfrak m$.
