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

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?

• You need to prove that $a\otimes b = 1\otimes ab$. Aug 8, 2022 at 19:01
• I can see how this works for $a\in A$, $b\in K$ but if $a\notin A$ how can you multiply by $a$ when right component is $A$-module? Aug 8, 2022 at 19:30
• $a\otimes b = a \pi^r \otimes b \pi^{-r}$ where $r$ is chosen such that $a\pi^r \in A$ (where $\pi$ is your $p$). Note that it is non necessary that $A$ is a DVR, only that $K = Frac(A)$. Aug 8, 2022 at 19:34
• More generally, if $A\to B$ is an epimorphism of commutative rings, then $B\otimes_A B\cong B.$ (Localizations of rings are epimorphisms.) Aug 8, 2022 at 20:41

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 $$f(K)=1$$.
• Let us note that the fact that $A$ is a valuation ring is not actually very relevant, it would work for any domain (just by replacing $p^n$ by any denominator in $A$). Aug 8, 2022 at 21:37