Question about fibre product of closed points. Given an integral smooth curve $X$ over an algebraically closed field $k$, with quotient field $K$. Let $V_p=\operatorname{Spec} \mathcal{O}_{X,p}$ for all closed points $p\in X(k)$.
Then it’s said that, $V_q\times V_p=V_p$ for $p=q$, and $V_q\times V_p=\operatorname{Spec} K$ for $p\neq q$.
I don’t know why this follows? If we may simplify it, for an integral domain $R$, let $m,n$ be two maximal ideals? Then the product of localizations $R_m\otimes R_n$ equals to the quotient field of $R$? Hope someone could help. Thanks!
 A: A correctly-stated version of the problem is to ask for $V_p\times_C V_q$ to be equal to $V_p$ if $p=q$ and $\operatorname{Spec} K$ for $p\neq q$. The choice of what space the fiber product is over really matters here: for instance, if you try to take the fiber product over $k$, the $p\neq q$ case won't give you $\operatorname{Spec} K$.
Your idea of turning this in to a tensor product computation will do what you want. Your smooth integral curve is birational to a smooth projective curve, so the complement of any finite number of points on it is affine, and we can find an affine open subset of the curve which contains $p,q$ for any two closed points $p,q$.
Now we have $R$ which is an integral domain with two distinct maximal ideals $m,n$ and we want to show that $R_m\otimes_R R_n\cong Frac(R)$. There's an obvious map from this tensor product to $Frac(R)$ given by $a\otimes b \mapsto ab$, and it's not hard to see that it's both injective and surjective. Alternatively, one can note that this is just the tensor product of the localizations $S^{-1}R\otimes_R T^{-1}R$ where $S=R\setminus m$ and $T=R\setminus n$, so as $m,n$ are comaximal, this gives that all nonzero elements of $R$ are invertible and thus the tensor product is $K$.
