# Module isomorphic to a flat module

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

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 $N$.