Let $R$ and $T$ be commutative rings with unity. Let $Q$ be a prime ideal of $R$ and $\phi:R \to T$. Suppose $T \otimes_R (R_Q/Q R_Q)$ has only one prime ideal.
Then I would like to prove that the vertical map on the left hand side of
\begin{array}{cc}
T \otimes_R R_Q/Q R_Q & \leftarrow & T
\\
\uparrow & &\uparrow
\\
R_Q/Q R_Q & \leftarrow & R
\end{array}
is an isomorphism. How can I prove this?
I thought I could prove this by showing that given any $t \otimes r$, we have $t \otimes r = 1 \otimes s$ for some $s \in R_Q/Q R_Q$, but this seems to only work if $t$ is in the image of $\phi$...
Edit. The question as asked does not seem to be correct, as can be seen in the comment. What assumptions can I add to make this true? I am trying to understand the details of a proof in Mumford the fibre of $f$ over $y$ is $\operatorname{Spec} \kappa(y)$ given $f^{\#}(\mathfrak{m}_y) O_{X,x} = \mathfrak{m}_x$. Thank you
Lemma: Let $f:X\rightarrow Y$ be a morphism of schemes. Then $f^{-1}(p) \cong X\times_{Y} \kappa(p)$ as sets where $\kappa (p)$ is the residue field at $p\in Y$.
Proof: Assume $X=\operatorname{Spec}A,Y=\operatorname{Spec}B$ are affine and $p\in \operatorname{Spec} B$. Set $S=B\backslash p$. Then we have the following 1-1 correspondences $$f^{-1}(p)\leftrightarrow\{P\in \operatorname{Spec}A : P\cap B=p \}\leftrightarrow \{P \in \operatorname {Spec}A : P\supseteq pA \ , \ P \cap {B\backslash p}=\phi \} $$ $$\leftrightarrow \operatorname {Spec} \frac{S^{-1}A}{pS^{-1}A} \cong \operatorname {Spec} A\otimes_ B \frac{B_p }{pB_p}=X\times_Y \operatorname {Spec}\kappa(p) $$
Now you use a patching argument to complete the proof.
So you are asking when $\frac{A_p }{pA_p}$ is a field assuming $\operatorname {Spec} \frac{A_p}{pA_p}$ is a singleton. Let $P\in \operatorname {Spec} {A}$ be the unique prime ideal such that $P\cap B\backslash p =\phi $ and $P\supset pA$. Then $\frac{A_p }{pA_p}$ is a field iff $pA_p =PA_p$, namely the maximal ideal of $\mathcal O_{Y,p}$ generates the maximal ideal of $\mathcal O_{X,P}$ which is precisely what's given in the question you linked.