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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.

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    $\begingroup$ 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. $\endgroup$
    – Mohan
    Apr 14, 2018 at 0:28
  • $\begingroup$ I don't understand why B have depth 3. Please give me a more explanation. $\endgroup$
    – Kitamado
    Apr 14, 2018 at 14:40
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    $\begingroup$ Did you mean depth 2? $\endgroup$
    – Mohan
    Apr 14, 2018 at 14:41
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    $\begingroup$ 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. $\endgroup$
    – Mohan
    Apr 14, 2018 at 14:50
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    $\begingroup$ @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... ) $\endgroup$
    – Kitamado
    Apr 14, 2020 at 16:53

1 Answer 1

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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.

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    $\begingroup$ I turned Mohan's comments into an answer in order to remove this question from the unanswered queue. $\endgroup$
    – user26857
    Apr 6, 2019 at 21:47

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