Computing $\dim_K(K\otimes_A K)$ where $K$ is field of fractions of discrete valuation ring $A$

Question

I have a discrete valuation ring $A$ with field of fractions $K$. For each $A$-module $M$, we regard the tensor product $K\otimes M$ as a $K$-vector space by giving it the unique scalar multiplication satisfying $a(b\otimes m)=(ab)\otimes m$ for all $a,b\in K$ and $m\in M$.

I am trying to compute $f(M):=\dim_K(K\otimes_A M)$ in some simple cases, and feel that I may have misunderstood the tensor products in some way. Let's look at $f(K)$ (the value should be $1$).

Each element of $K\otimes K$ is a finite sum of elementary tensors $a\otimes b$, where $a,b\in K$. Using the property of the $K$-multiplication listed above, we have $a\otimes b=a(1\otimes b)$. Then $b=p^{\nu(b)}u$ for some $u\in A^\times$ (where $p\in A$ is irreducible and $\nu$ is the valuation), so $a\otimes b=au(1\otimes p^{\nu(b)})$. Yet since $b\in K$, we could have $\nu(b)<0$ i.e. $b\notin A$.

My understanding is that whereas we can take factors of $K$ out of the left component of the elementary tensor, we can only take out factors of $A$ from the right component, since $M$ is an $A$-module and not a $K$-vector space (however $K$ of course is also a $K$-vector space, but is considered as an $A$-module here). I'm guessing $K\otimes K$ has basis $1\otimes 1$ over $K$, but for some reason I can't see how $1\otimes p^{-n}=p^{-n}(1\otimes 1)$ is meant to work here.

Where am I going wrong?

Answer 1

If $n\geq 1$, then $$1\otimes p^{-n}=(p^np^{-n})\otimes p^{-n}=p^n(p^{-n}\otimes p^{-n})=p^{-n}\otimes p^np^{-n}=p^{-n}\otimes 1=p^{-n}(1\otimes 1).$$ Since each $b\in K^\times$ can be expressed in the form $p^ku$ for some $k\in\mathbb{Z}$ and $u\in A^\times$, we have that $1\otimes b=b(1\otimes 1)$ for all $b\in K$. Then $a\otimes b=ab(1\otimes 1)$ for all $a,b\in K$ and we see that $K\otimes_A K$ has basis $1\otimes 1$ over $K$ and finally $f(K)=1$.