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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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    @Randal'Thor Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-13

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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.