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If $R,R'$ are two rings, then $R\!\times\!0$ is a projective $R\!\times\!R'$-module, since it is a direct summand of a free module: $(R\!\times\!0)\oplus\!(0\!\times\!R')=R\!\times\!R'$.

What would be some sufficient conditions on $R$ and $R'$, so that $R\!\times\!0$ is not a free $R\!\times\!R'$-module? For example, if $R$ and $R'$ are unital, or commutative unital, does this suffice?

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Dear Leon: If $R$ and $R'$ are unital, and $R'$ is nonzero, then $R$ is not $(R\times R')$-free, because it is annihilated by $(0,1)\neq0$. –  Pierre-Yves Gaillard Feb 12 '12 at 9:41
    
Ah, of course, free modules have trivial annihilators. My line of thought was that if $f:R\!\times\!0\!\rightarrow\!(R\!\times\!R')^{(I)}$ is a $R\!\times\!R'$-linear bijection, then it is also a $0\!\times\!R'$-linear bijection. However, as $0\!\times\!R'$-modules, we don't have $R\!\times\!0\!\cong\!0\!\times\!0$, unless $R\!=\!0$, so I was worried. Thank you. You can write that as an answer, so I can accept. –  Leon Lampret Feb 12 '12 at 9:47
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1 Answer

up vote 5 down vote accepted

If $R$ and $R'$ are unital, and $R'$ is nonzero, then $R\times0$ is not $(R\times R')$-free, because it is annihilated by $(0,1)\neq0$.

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Surely the simplest, hence best, answer! –  Georges Elencwajg Feb 12 '12 at 11:50
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