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Let $M$ be a flat $A$-module, and $N$ a $A$-module isomorphic to $M$, what can we say about the flatness of $N$?

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$N$ is flat? Or what do you want to hear? –  martini Mar 27 '12 at 18:56
why is $N$ flat? –  Jr. Mar 27 '12 at 19:09
If $M\cong N$ as $A$-modules, then for any $A$-module $P$, $P\otimes_A M\cong P\otimes_A N$. –  InvisiblePanda Mar 27 '12 at 19:24
Dear Jr., here is a meta-rule for you. Whenever mathematicians define a property P that some objects in a category may or may not have, you can be sure that if an object has property P, then any isomorphic object also has property P. –  Georges Elencwajg Mar 27 '12 at 21:39
Dear Georges , is there a formal proof of your statement? I mean, only using abstract category theory, does one could reach that conclusion? –  Jr. Mar 27 '12 at 23:12

1 Answer 1

Let $f:M\to N$ be an isomorphism of $A$-modules, and let $P$ be an arbitrary $A$-module. Then $P\times M\to P\otimes_A N$, $(p,m)\mapsto p\otimes f(m)$ is $A$-bilinear, hence we get an induced well-defined homomorphism $$\operatorname{id}\otimes f:P\otimes_A M\to P\otimes_A N, p\otimes m\mapsto p\otimes f(m).$$ In the same way, we have an inverse homomorphism $\operatorname{id}\otimes f^{-1}$, such that $P\otimes_A M\cong P\otimes_A N$.

Now $M$ being flat means that if $g:P\to P'$ is an injective morphism of $A$-modules, $g\otimes\operatorname{id}:P\otimes_A M\to P'\otimes_A M$ is, too. But then the map $$P\otimes_AN\xrightarrow{\sim}P\otimes_AM\xrightarrow{g\otimes\operatorname{id}}P'\otimes_AM\xrightarrow{\sim}P'\otimes_A N$$ is an injective $A$-homomorphism, which proves the flatness of $N$.

As for your question regarding Georges' answer, I'm not exactly sure what you mean. In general, one defines these properties such that they stay invariant under isomorphism in the respective category. But looking for a proof of this, as Georges calls it, meta-rule, wouldn't really make sense to me.

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