11
$\begingroup$

It is known that $A[[X]]$ is flat if $A$ is noetherian (see for example Bourbaki, Algèbre commutative, Ch. III, §3, Cor. 3 p. 146).

What happens if A is not noetherian? Is there an easy counter-example to the flatness of $A[[X]]$?

$\endgroup$

1 Answer 1

15
$\begingroup$

As a module, $A[[X]]$ is the product of a countable family of copies of $A$. It is known that the product of flat $A$-modules is flat if and only if the ring $A$ is coherent, that is, every finitely generated ideal is finitely presented ( http://www.ams.org/journals/tran/1960-097-03/S0002-9947-1960-0120260-3/S0002-9947-1960-0120260-3.pdf ).

If you look at the proof of Theorem 2.1 in that paper, you can show that if $k$ is a field and $A$ is the quotient a polynomial ring $k[t_1, t_2, \dots]$ in countably many variables by the ideal generated by the products $t_it_j$, the product of countably many copies of $A$ is not flat over $A$.

$\endgroup$
3
  • $\begingroup$ Thanks a lot for this answer and for the reference, Angelo. $\endgroup$ Jan 31, 2013 at 19:04
  • $\begingroup$ Dear Baptiste, you are very welcome. $\endgroup$
    – Angelo
    Jan 31, 2013 at 19:31
  • $\begingroup$ As I understand it, this paper was what launched the definition (and hence the whole study) of coherent rings in the first place. $\endgroup$ Feb 1, 2013 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.