Let $(A,m)$ and $(B,n)$ be local commutative rings that are also $\mathbb{k}$-algebras. Let $\phi :B \rightarrow A$ be a local $\mathbb{k}$-algebra homomorphism.
Suppose that $A/m \cong \mathbb{k}$. I want to show that $B/n \cong \mathbb{k}$.
If we call the maps
$\tau : \mathbb{k} \rightarrow B \twoheadrightarrow B/n$
$\overline{\phi}: B/n \rightarrow A/m$,
$\rho : \mathbb{k} \rightarrow A \twoheadrightarrow A/m$
I have shown that $\overline{\phi}\circ\tau = \rho$. $\textbf{If we suppose further that $\rho$ is an isomorphism}$, as $\overline{\phi}$ is injective, it follows that $\tau$ must be surjective and hence an isomorphism (as it is already a nonzero ring homomorphism from a field).
But I don't see why $\rho$ has to be an isomorphism?
I don't think that the fact $A/m \cong \mathbb{k}$ implies that $\rho$ is an isomorphism, as the inclusion $k(t^2) \hookrightarrow k(t)$ gives a counterexample (see Local $k$-algebra with residue field $k$). So I am not sure that my approach is even on the right track!
This problem is related to Hartshorne Chapter 2, exercise 2.15.
Any help would be appreciated :).
Best Answer
If you want $A/m\cong k$ as a $k$-algebra then that is saying that $\rho$ is an isomorphism, at least in my book.
But one can have $A/m\cong k$ just as a ring isomorphism and then $B/n$ need not be isomorphic to $k$. As an example, take a tower of field extension $k\subset L\subset K$ with $k\cong K$ and $k\not\cong L$. Then let $(A,m)=(K,0)$ and $(B,n)=(L,0)$.
Why do such fields exist? For example, take $k=\Bbb C$, $L=\Bbb C(t)$ and $K$ the algebraic closure of $L$.