For a flat morphism $f:X \rightarrow B$ and a sub scheme $Z$ of $B$ we know that the strict and total transforms of $X$ with respect to the blow up at $Z$ agree.


I want to know what happens when $f$ fails to be flat. Suppose that $f$ fails to be flat at some point $P$ and we blow up $P$, could the strict transform of $X$ agree with the total transform? It feels as if the non-flat locus of the total transform of $f$ should contain the entire exceptional divisor which in many cases would show that this is not the case. My first question is suppose we have $f:X \rightarrow B$ a morphism of varieties not flat at $P$ in $B$ and we blow up $P$ can the product $X \times\_B Bl\_P B$ be flat over $Bl\_P B$?


In general suppose that we have a non flat morphism of varieties $f:X \rightarrow Y$ and a proper surjective morphism $p:Z \rightarrow Y$ could the product $Z\times\_Y X$ ever be flat over $Z$?



Can pullbacks resolve non-flatness?
I am just posting my comments above as an answer. If $Y$ is allowed to be nonreduced, then there does exist a finitely presented, non-flat morphism $\pi:X\to Y$ and a proper, birational morphism $Z\to Y$ such that $Z\times\_Y X \to Z$ is flat. One example is when $Y=\text{Spec}\ k[u,v]/\langle u^2,uv \rangle,$ when $X$ is the closed subscheme $\text{Spec}\ k[u,v]/\langle u \rangle,$ and when $Z$ is the blowing up of $Y$ at the ideal $\mathfrak{m}= \langle \overline{u},\overline{v} \rangle.$


If you assume that $Y$ is reduced, then the positive result follows from the "valuative criterion of flatness", Théorème 11.8.1 of EGA $IV\_3.$ For simplicity, assume that $Y$ is integral. If $X\to Y$ is not flat, then there exists a point $y\in Y$ and a DVR $R$ with fraction field $\text{Frac}(Y)$ containing $\mathcal{O}\_{Y,y}$ such that the base change of $\pi$ by $i:\text{Spec}\ R \to Y$ is not flat over $\text{Spec}\ R.$ 


For every proper, birational morphism $f:Z\to Y,$ by the valuative criterion of properness, there exists a morphism $g:\text{Spec}\ R \to Z$ such that $i$ equals $f\circ g.$ If the pullback of $\pi$ by $f$ is flat over $Z,$ then the further pullback along $g$ is flat over $\text{Spec}\ R.$ This contradicts that the pullback of $\pi$ by $i$ is not flat over $\text{Spec}\ R.$


