3

Let $k$ be a field, $A$ a local $k$-algebra with maximal ideal $\mathfrak m$. Suppose furthermore that the residue field $A/\mathfrak m$ is isomorphic to $k$ as a ring. Can we then deduce that

$$ k\to A \to A/\mathfrak m $$

is an isomorphism, i.e. that $A/\mathfrak m$ has dimension $1$ over $k$ with respect to the algebra structure inherited from $A$?

  • $A$ is a ring. Take any $a \in A$ such that $ a+ \mathfrak{m}= 1_{A/\mathfrak{m}}$ then $\rho(x)= ax$ is a ring morphism $k \to A$ such that $\phi(x) = \rho(x)+\mathfrak{m}$ is an isomorphism $k \to A/\mathfrak{m}$. What you want is to know is if local means for any $k$-algebra structure on $A$, it is compatible with $\rho,\phi$ ? – reuns Nov 11 '17 at 16:17

1 Answers1

6

It might appear disappointingly trivial, but you can view any non-automorphism $\varphi: k\to k$ as an example for this by setting $k := A$ and defining the $k$-algebra structure on $A$ via $\varphi$. As a concrete example, take $k(t)\to k(t), t\mapsto t^2$.

Hanno
  • 19,510
  • 1
    This is a nice answer, but has not been upvoted for two days now. I try to simplify the notation, maybe this will raise the deserved appreciation. We have an inclusion $k(t^2) \subset k(t)$, hence $k(t)$ is a $k(t^2)$-algebra. The residue field of $k(t)$ is $k(t)/0=k(t)$ itself and this isomorphic to $k(t^2)$. But the map $k(t^2) \hookrightarrow k(t) \to k(t)/0$ is not an isomorphism, since it is just the (proper) inclusion. – MooS Nov 13 '17 at 19:23