https://www.zhihu.com/question/264128942?from=profile_question_card
$f:X\to Y,\textrm{ where }$$X=\operatorname{Spec}B,Y=\operatorname{Spec}A.$ $f^{\sharp}(Y):A\to B.$
Let $K=A^{(I)},L=A^{(J)}$.
$\alpha:A^{(I)}\to A^{(J)}$
$\alpha_B:A^{(I)}\otimes_AB\to A^{(J)}\otimes_AB$ \begin{array}[c]{ccc} \widetilde{A^{(I)}\otimes_AB}&\stackrel{\textrm{associated to $\alpha_B$}}\rightarrow&\widetilde{A^{(J)}\otimes_AB}\\ \downarrow\scriptstyle{\varphi}&&{\downarrow}\scriptstyle{\psi}\\ f^{-1}\widetilde{A^{(I)}}\otimes_{f^{-1}\mathcal O_Y}\mathcal O_X &\stackrel{\beta}{\rightarrow}&f^{-1}\widetilde{A^{(J)}}\otimes_{f^{-1}\mathcal O_Y}\mathcal O_X \end{array} $\forall \mathfrak p\in \operatorname{Spec}B, \mathfrak q=(f^{\sharp}(Y))^{-1}(\mathfrak p).$ $\textrm{According to Lemma 1.2.10 and Corollary 1.1.11 in Liu Qing,}$ $(A^{(I)}\otimes_AB)_{\mathfrak p}\simeq (A^{(I)}\otimes_AB)\otimes_BB_{\mathfrak p}\simeq A^{(I)}\otimes_AB_{\mathfrak p}\simeq (A^{(I)}\otimes_AA_{\mathfrak q})\otimes_{A_{\mathfrak q}}B_{\mathfrak p}\simeq (A^{(I)})_{\mathfrak q}\otimes_{A_{\mathfrak q}}B_{\mathfrak p}$ $\forall g\in B, \textrm{we have}$ $(A^{(I)}\otimes_AB)_g\simeq (A^{(I)}\otimes_AB)\otimes_BB_g\simeq A^{(I)}\otimes_AB_g $
$\varphi(D(g))\textrm{ is defined by}$ $(A^{(I)}\otimes_AB)_g\simeq A^{(I)}\otimes_AB_g\to (f^{-1}\widetilde{A^{(I)}})(D(g))\otimes_{(f^{-1}\mathcal O_Y)(D(g))}B_g\to (f^{-1}\widetilde{A^{(I)}}\otimes_{f^{-1}\mathcal O_Y}\mathcal O_X)(D(g))$ $\psi(D(g))\textrm{ is defined the same as above.}$
It is easy to verify $\varphi,\psi$ are respectively isomorphic and the above diagram is commutative.
Is my argument correct?