Field of fractions of $(E \otimes_K F)/p$ and $(E \otimes_K F)/q$ are not isomorphic.

Question

Let $E$ and $F$ are two field extensions of a field $K.$ Now suppose $p$ and $q$ are two distinct prime ideals of $E \otimes_K F.$ Then how can I show that the field of fractions of the two domains $(E \otimes_K F)/p$ and $(E \otimes_K F)/q$ are not isomorphic ? I don't know even how to approach. Help me.

Answer 1

Let $\mathbb Q(s,t)$ be the rational field of functions in two algebraically independent indeterminates $s,t$ .
The tensor product $A:=\mathbb Q(s)\otimes_ \mathbb Q \mathbb Q(t)\subset \mathbb Q(s,t)$ consists of all fractions of the form $\frac {P(s,t)}{Q(s)R(t)}$ where $P(s,t)$ is an arbitrary polynomial in the indeterminates $s,t$ while $Q(s),R(t)$ are nonzero polynomials in one indeterminate.
Of course $A$ is a domain, since it is a subring of the field $\mathbb Q(s,t)$ but not a field because $s+t$, for example, is not invertible in $A$.
Consider the two prime ideals of $A$: $\mathfrak p=(0)\subset A$ and $\mathfrak q=( s^2+t^2-1)A \subset A$ .
We have on the one hand $\operatorname {Frac}A/\mathfrak p=\operatorname {Frac}A=\mathbb Q(s ,t).$
But on the other hand $\operatorname {Frac}A/\mathfrak q=\operatorname {Frac}(\mathbb Q[s,t]/(s^2+t^2-1))$.
These fields are "very" non isomorphic because they have transcendence degree over $\mathbb Q$ respectively $2$ and $1$.