This is probably a simple-minded question, but I haven't been able to prove it or find a counterexample. This [old question](https://mathoverflow.net/questions/26723/does-smooth-target-space-and-smooth-fibers-imply-smooth-total-space?rq=1) seems to dance around my question, but I don't think any of the answers address exactly my situation. (Please tell me if I am wrong!)


Suppose $\varphi: X\to Y$ is a surjective morphism of algebraic varieties (reduced, irreducible, separated schemes, finite type over an algebraically closed field), and furthermore assume that $Y$ is smooth. Also, I can assume that $\varphi$ is finitely presented.


If the fiber $X\_y$ is smooth and equidimensional of dim $n$ for any $y\in Y$, is the morphism flat? I know that equidimensionality alone does not mean flat, but I wonder if the smoothness assumptions are enough. Obviously, I want to conclude that $X$ is smooth and this is enough.


The equidimensionality assumption rules out the blow-up examples in the above cite problem, and the normalization of a node on a curve is ruled out by smoothness of $Y$. I would be happy with a counterexample though.



smooth equidimensional fibers over a smooth base
The flatness statement you would like follows from Theorem 3.3.27 of Schoutens's [book](https://www.springer.com/us/book/9783642133671) on ultraproducts. 


