$\mathcal{O}_{X_y,x}=\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$ In a proof (proof of theorem 4.3.36 in Liu's book) I need the equality $\mathcal{O}_{X_y,x}=\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$. 
The hypothesis of the theorem are the following: $Y$ is a regular locally noetherian scheme and $f:X\to Y$ is a smooth morphism.
With affine scheme (all is local here) this mean (I guess) $(A\otimes_B B_\mathfrak{q}/\mathfrak{q}B_{\mathfrak{q}})_{\mathfrak{p}^e}=A_\mathfrak{p}/\mathfrak{q}A_\mathfrak{p}$.
I have a few ideas but this mixture of tensorial product, quotient and localization is difficult and I never know what is really "authorised" or not.
 A: I think I found but I'm not sure that all these rules are corrects: as we have $M\otimes_AS^{-1}A=S^{-1}M$ then with notation $A_\mathfrak{q}=(B\setminus\mathfrak{q})^{-1}A$,
$$ A\otimes_B B_\mathfrak{q}=A\otimes_B (B\setminus\mathfrak{q})^{-1} B=(B\setminus\mathfrak{q})^{-1}A=A_\mathfrak{q}$$
Then as we have $M\otimes_A A/I=M/IM$ then
$$A\otimes_B B_\mathfrak{q}/\mathfrak{q}B_\mathfrak{q}=A\otimes_B B_\mathfrak{q}\otimes_{B_\mathfrak{q}}B_\mathfrak{q}/\mathfrak{q}B_\mathfrak{q}=A_\mathfrak{q}\otimes_{B_\mathfrak{q}}B_\mathfrak{q}/\mathfrak{q}B_\mathfrak{q}=A_\mathfrak{q}/\mathfrak{q}A_\mathfrak{q}$$
Then as we have $(A/I)_\mathfrak{p}=A_\mathfrak{p}/I_\mathfrak{p}$ then
$$(A\otimes_B B_\mathfrak{q}/\mathfrak{q}B_\mathfrak{q})_{\mathfrak{p^e}}=(A_\mathfrak{q}/\mathfrak{q}A_\mathfrak{q})_{\mathfrak{p}_\mathfrak{q}}=(A_\mathfrak{q})_\mathfrak{p}/(\mathfrak{q}A_\mathfrak{q})_\mathfrak{p}=A_\mathfrak{p}/\mathfrak{q}A_\mathfrak{p}$$
One of my problems is that I used rules the are corrects for modules and not for rings
