Flatness and normality I have just read: Direct proof of non-flatness and wondered what is exactly the claim that Alex Youcis is referring to: "...but are you aware of the fact that flatness preserves normality. In your case $A$ is non-normal and $B$ is normal, so $B/A$ can't be flat". Can one please explain what exactly "flatness preserves normality" means? Given $A \subseteq B$ (commutative rings) with $B$ flat over $A$, does "flatness preserves normality" mean: (1) If $B$ is normal, then $A$ is normal. (2) If $A$ is normal, then $B$ is normal. (3) Something else?

The simplest claim that would make Alex's point is that a flat integral extension of rings is faithfully flat, and that the only faithfully flat extensions of domains with the same field of fractions are identity maps. In other words if $B$ is integrally closed and integral and flat over $A$, then $B=A$, and in particular $A$ is also integrally closed.

That handles the situation of $k[x^2,x^3]\subset k[x]$. But the stronger claim (1) is false, as you can see by taking $B$ to be the field of fractions of any non-integrally closed $A$. (2) is also false, since if $A$ is a field then every $B$ is flat.

**EDIT** : The one other thing you might wonder is whether an integral extension of an integrally closed ring is integrally closed, and again the answer is no: $\mathbb{Z}[X^2,X^3]/\langle X^5+2\rangle$ is an integral extension of $\mathbb{Z}$ which is not integrally closed, since it doesn't contain $X$.