# 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 ?

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Is $\phi$ surjective? Otherwise pick your favorite discrete valuation ring and look at the inclusion mapping into its field of fractions... – JSchlather Jan 31 '14 at 8:05
Not generally true. You have to assume (wlog) that $\mathfrak m$ is the kernel. – Karl Kronenfeld Jan 31 '14 at 8:08
Bah. ${}{}{}{}$ – anon Jan 31 '14 at 8:16

No, in general it is not true that a homomorphism from a local ring into a field has the maximal ideal as kernel. Just consider the inclusion ${\mathbb Z}_{(p)} \to {\mathbb Q}$ for some prime number $p$.

What is happening here is the following. Suppose you start out with a ring $A$ that is already local with maximal ideal ${\frak m}$ and a homomorphism $\phi \colon A \to L$. The kernel ${\frak p}$ of $\phi$ doesn't have to be ${\frak m}$, but you can still localize at ${\frak p}$, obtaining an other local ring $A_{\frak p}$.

So the claim by Lang that "$\phi$ and the canonical map $A \to A/{\frak m}$ have the same kernel" is true, but that's because earlier he localized exactly at the kernel. (It might have been clearer if Lang had written "we may assume that $A$ is a local ring with maximal ideal ${\frak p}$" (or "with maximal ideal ${\frak m} = {\frak p}$")).

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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.

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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 $\mathfrak m$.

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