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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 $A[[X]]$?

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up vote 12 down vote accepted

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 $A$.

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Thanks a lot for this answer and for the reference, Angelo. –  Baptiste Calmès Jan 31 '13 at 19:04
    
Dear Baptiste, you are very welcome. –  Angelo Jan 31 '13 at 19:31
    
As I understand it, this paper was what launched the definition (and hence the whole study) of coherent rings in the first place. –  Neil Epstein Feb 1 '13 at 15:07

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