Unramification stable under change base I want to show that if $f:X\to Y$ is an unramified scheme morphism (ie $m_y\mathcal{O}_{X,x}=m_x\mathcal{O}_{X,x}$ and $k(x)\leftarrow k(y)$ finite and separable) then any base change $X\times_Y Z\to Z$ stay unramified.
I think algebraic traduction is: if $\varphi:A\to B$ such that $\mathfrak{p}B_\mathfrak{q}=\mathfrak{q}B_\mathfrak{q}$ and $k(\mathfrak{p})\to k(\mathfrak{q})$ finite separable then $i:C\to B\otimes_A C$ verify $\mathfrak{r}(B\otimes_A C)_\mathfrak{s}=\mathfrak{s}(B\otimes_A C)_\mathfrak{s}$ and $k(\mathfrak{r})\to k(\mathfrak{s})$ finite separable if $i^{-1}(\mathfrak{s})=\mathfrak{r}$.
I don't care first with the field problem.
I had following idea for localisation:
$$ \mathfrak{s}(B\otimes_A C)_\mathfrak{s}=(\mathfrak{q}\otimes\mathfrak{r})(B_\mathfrak{q}\otimes_{A_\mathfrak{p}} C_\mathfrak{r})=\mathfrak{q}B_\mathfrak{q}\otimes_{A_\mathfrak{p}}\mathfrak{r}C_\mathfrak{r}=\mathfrak{p}B_\mathfrak{q}\otimes_{A_\mathfrak{p}}\mathfrak{r}C_\mathfrak{r}=B_\mathfrak{q}\otimes_{A_\mathfrak{p}}\mathfrak{r}C_\mathfrak{r}=\mathfrak{r}(B_\mathfrak{q}\otimes_{A_\mathfrak{p}}C_\mathfrak{r})=\mathfrak{r}(B\otimes_A C)_\mathfrak{s} $$
But I guess there is a few problems with these equalities:
First: $\mathfrak{s}=\mathfrak{q}\otimes\mathfrak{s}$ is not right in general
Second: $(B\otimes A)_\mathfrak{s}=B_\mathfrak{q}\otimes_{A_\mathfrak{p}}C_\mathfrak{r}$ is right for modules but not for algebra
And maybe others problems I don't see
So my question is: is my idea correct  (and in this case how to solve my problems) and if not how to prove the equality. 
 A: It's easier to think about unramified as finitely presented and $\Omega_{X/Y}^1$ vanishing. It encodes what you're trying to do into one neat package. So, now suppose that $f:X\to Y$ is unramified, and $g:Z\to Y$ is some morphism. Consider the diagram
$$\begin{matrix}X\times_Y Z & \xrightarrow{p} & X\\ \downarrow & & \downarrow^f \\ Z & \xrightarrow{g} & Y\end{matrix}$$
Then, it's easy to show that the pullback is finitely presented, and that $\Omega^1_{X\times_Y Z/Z}=p^\ast(\Omega_{X/Y}^1)$. But, it clearly follows then that if $\Omega^1_{X/Y}=0$ then so does $\Omega^1_{X\times_Y Z/Z}$.
Or, it's also easier in this context to think about unramified as having your diagional map be an open embedding. Then the claim is clear from the fact that open embeddings are invariant under base change and setting up the right fibered diagram to relate the diagionals of the original map and the base change's diagional.
A: You can definitely solve this without passing through Kähler differentials.
Let $f:X\longrightarrow Y$ be an unramified morphism, locally of finite type, $g:Y'\longrightarrow Y$ another morphism and $h:X\times_Y Y'\longrightarrow Y'$ the base change morphism.
You might know that a morphisms is unramified if and only if its fibers are unramified (this equivalence comes from the canonical isomorphism $\mathcal{O}_{{X_y},x}\simeq\mathcal{O}_{{Y},y}/\mathfrak{m}_x\mathcal{O}_{{Y},y}$ where $f:X\longrightarrow Y$ is a morphism and $x\in X$, $y=f(x)$).
Hence, you need to show that the fibers of $h$ are unramified, that is $(X\times_Y Y')_{y'}\longrightarrow \mathrm{Spec}(k(y'))$ is unramified, whenever $y'\in Y'$.
Furthermore, if you set $y=g(y')$, you have the following canonical isomorphisms:
$(X\times_Y Y')_{y'} = (X\times_Y Y')\times_{Y'}\mathrm{Spec}(k(y'))\simeq X\times_Y\mathrm{Spec}(k(y'))\simeq (X\times_Y\mathrm{Spec}(k(y)))\times_{\mathrm{Spec}(k(y))}\mathrm{Spec}(k(y'))=X_y\times_{\mathrm{Spec}(k(y))}\mathrm{Spec}(k(y'))$.
Since you know that the morphism $X_y\longrightarrow \mathrm{Spec}(k(y))$ is unramified, you are reduced to the case where $Y$ and $Y'$ are spectra of fields, say $Y=\mathrm{Spec}(k)$ and $Y'=\mathrm{Spec}(k')$.
Since ramification is a local problem, one may assume that $X$ is affine, id est $X=\mathrm{Spec}(A)$. Now, since $X$ is unramified over $k$, one may write $A$ as a finite product of finite separable field extensions of $k$ (this characterizes unramified morphisms locally of finite type over a field: see for example the proposition I.3.2. of Milnor's Étale cohomology).
Then $X\times_Y Y'=\mathrm{Spec}(A\otimes_k k')$ where $A\otimes_k k'$ is again a product of finite separable field extensions of $k'$. Since the latter characterizes unramified morphisms over fields, the base change morphism is itself unramified.
A: Here is a direct argument. Everything is local, so replace everything by local rings. Let $X\to Y$ be unramified, and let $Y'\to Y$ be arbitrary, and let $X'=X\times_YY'$. We want to show $X'\to Y'$ is also unramified:$\require{AMScd}$
\begin{CD}
X' @>{f'}>> Y'\\
@V{g'}VV @V{g}VV\\
X @>{f}>> Y
\end{CD}
For an arbitrary $z\in X'$, let $w=f'(z)\in Y'$, $x=g'(z)\in X$, and $y=f\circ g'(z)\in Y$. We have the following pullback square (where for example $y$ denotes a point with structure sheaf $\kappa(y)$, and $g^{-1}(y):=Y'\times_Yy$ is the scheme-theoretic preimage):
\begin{CD}
(f')^{-1}(w) @>{f'}>> w\\
@VVV @VVV\\
(g\circ f')^{-1}(x) @>{f'}>> g^{-1}(y)\\
@V{g'}VV @V{g}VV\\
x @>{f}>> y.
\end{CD}
Here, the top square is obtained from the diagram
\begin{CD}
(f')^{-1}(w) @>>> w\\
@VVV @VVV\\
X' @>{f'}>> Y'
\end{CD}
and the bottom square is the pullback of the first diagram along $y\to Y$ (here, $x=X\times_Yy$ since $f$ is unramified; this is why I replaced everything by local rings).
Now, this tells us $(f')^{-1}(w)=Spec(\kappa(x)\otimes_{\kappa(y)}\kappa(w))$, where $\kappa(x)\otimes_{\kappa(y)}\kappa(w)$ is a direct product of fields since $\kappa(x)/\kappa(y)$ is separable (which are moreover separable extensions of $\kappa(w)$). Thus, $\kappa(z)$ must be one of the direct summands, and so $\mathcal O_w/m_z\mathcal O_w=\kappa(w)$ and $\kappa(z)/\kappa(w)$ is separable. That is, $f'$ is unramified.
