Let $K$ be a complete discrete valued field with ring of integers $A$, maximal ideal $m$ and uniformizer $\pi_K$.
Let $L$ be a finite extension of $K$ with ring of integers $B$, maximal ideal $m'$ and uniformizer $\pi_K'$.
The ring map $A \to B$ ist injective and flat ($B$ is even a free $A$-module)
What happens after tensoring with $-\otimes_A A/(\pi_K)^n$? We get a ring map $$A/(\pi_K)^n \to B/(\pi_L)^{e\cdot n}$$
where $e$ is the ramification index. Is this map still injective and/or flat?
Best Answer
Since $\pi_K^n B = \pi_L^{en} B$, the question of whether $$A/(\pi_K^n) \to B/(\pi_L^{en})$$ is injective boils down to checking whether $(\pi_K^nB)\cap A=\pi_K^n A$. However, if $\pi_K^n b = a$ for some $b\in B$, $a\in A$, then $b=\frac{a}{\pi_K^n}\in K$, and $b$ is integral over $A$, which is normal. Thus $b\in A$. But then $\pi_K^n b \in \pi_K^n A$, as desired.
Thus, yes, this map remains injective.
Flatness is easier. It's not hard to check that if $M$ is a flat $R$-module, then if $S$ is any $R$-algebra, $S\otimes_R M$ is a flat $S$-module. Apply this with $R=A$, $S=A/(\pi_K^n)$, $M=B$. In fact, $B/(\pi_L^{en})$ is still free over $A/(\pi_K^n)$, since if $M$ is a free $R$-module, $M=R^n$, then $S\otimes_R M = S^n$.