# When an integral extension of integral domains is flat?

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?

• In general this is false. For example, take $A=k[x^4, y^4,z]\subset k[x^4, x^3y, xy^3, y^4, z]=B$. 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 3, while $B$ has depth 2. Apr 14, 2018 at 0:28
• I don't understand why B have depth 3. Please give me a more explanation. Apr 14, 2018 at 14:40
• Did you mean depth 2? Apr 14, 2018 at 14:41
• Since $z$ is a non-zero divisor, suffices to check $B'=k[x^4,x^3y,xy^3,y^4]$ has depth one. 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, $x^2y^2\not\in B'$ and integral, so $B'$ is not normal. Of course, this can also done by simple calculation without resorting to Serre. Apr 14, 2018 at 14:50
• @Fawzy Hegab | $A$ may not be a PID. An element of $A$ is a sum of $a_n \otimes x_n$ with $a_n \in A$, $x_n \in B$, which is not always written the single term form $a \otimes x$. (Excuse me, my English is terrible... ) Apr 14, 2020 at 16:53

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.