Let $A$ and $B$ be integral domains. $A$ is integrally closed and $A \rightarrow B$ is integral. ($\star$)
This is sufficient to show that the going-down theorem. If $A \rightarrow B$ is flat, then the extension satisfies the going-down property.
I'm wondering if ($\star$) is sufficient to show that $A \rightarrow B$ is flat. Is there a counter example?
Thank you for your directions.
In general this is false. For example, take $A=k[x^4,y^4]⊂k[x^4,x^3y,xy^3,y^4]=B$. This is an integral extension of domains, $A$ is regular, being a polynomial ring. If flat, $B$ would be free as an $A$-module and in particular will have depth $2$, while $B$ has depth $1$.
Easy to see that $B$ satisfies Serre condition $(R_1)$ and so if it had depth $2$, it would be normal by Serre criterion. But, $x^2y^2\notin B$ and integral, so $B$ is not normal. Of course, this can also be done by simple calculation without resorting to Serre.