fibre with only one point is isomorphic to the spec of a field

Question

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

"commutative diagrams with unity" — did you mean commutative rings? — lisyarus, Aug 19 '20 at 12:48
Maybe I am misunderstanding, but what happens if $R$ is the zero ring? — Alex Youcis, Aug 19 '20 at 12:51
What about $R=k[X]$ with $Q=(X)$ and $T=k[X,Y]/(Y^2-X)$. Then $R_Q/QR_Q=k[X]/(X)=k$ and $T\otimes_R R_Q/QR_Q=k[X,Y]/(Y^2-X)\otimes_{k[X]}k=k[Y]/(Y^2)$ which is not a field. — Roland, Aug 19 '20 at 15:12
@AlexYoucis By assuming $R$ has a prime ideal $Q$, I'm avoiding $R$ is the zero ring — Johnny T., Aug 20 '20 at 08:01

score 2 · Accepted Answer · answered Aug 20 '20 at 10:25

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.

Thank you very much for this! This shows that $S^{-1}A/ p S^{-1}A$ is indeed a field. Is there a way to see that this is isomorphic to $B_p/p B_p$? — Johnny T., Aug 20 '20 at 15:20
This need not be the case always. Simply look at $\operatorname{Spec}L \rightarrow \operatorname{Spec k}$ where $L|_k$ has positive transcendence degree. However if the residue field is agebrically closed and the morphism is of finite type, then you can use the weak nullstellensatz to conclude the isomorphism — user6, Aug 20 '20 at 16:09

fibre with only one point is isomorphic to the spec of a field

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