$K\subset L$ is finite implies $\mathcal O_K/ \mathfrak{m}_K \subset \mathcal O_L/\mathfrak{m}_L$ is finite

Question

Given complete discrete valuation fields $K$ and $L$, finite extension $K\subset L$, corresponding d.v.r. $\mathcal O_K$ and $\mathcal O_L$ with maximal ideals $\mathfrak{m}_K$ and $\mathfrak{m}_L$ is it true that $\mathcal O_K/ \mathfrak{m}_K \subset \mathcal O_L/\mathfrak{m}_L$ is a finite field extension. I can see that one is subset of another, but why should this extension be finite?

Thank you.

Answer 1

You can show the following statement from which the result follows

If $\alpha_1,\dots,\alpha_n\in\mathcal O_L^\times$ are elements such that $\bar\alpha_1,\dots,\bar\alpha_n\in\mathcal O_L/\mathfrak m_L$ are linearly independent over $\mathcal O_K/\mathfrak m_K$, then $\alpha_1,\dots,\alpha_n$ are linearly independent over $K$.

Do this as follows:

1. Suppose $\bar\alpha_1,\dots,\bar\alpha_n$ are linearly independent over $\mathcal O_K/\mathfrak m_K$, and suppose you have an equation $c_1\alpha_1+\cdots+c_n\alpha_n=0$ in $L$ with $c_i\in K$.

2. Scale by the $c_i$ with largest absolute value so that, without loss of generality, you can assume some $c_i$ has absolute value $1$ and the rest $\le1$.

3. Consider the equation modulo $\mathfrak m_L$ and derive a contradiction.