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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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    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.2012-02-12

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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!2012-02-12