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Suppose $R_1$ and $R_2$ are unital rings. Consider $R_1 \oplus \{0\}$ an $R_1 \oplus R_2$-module. Is this a free module?

I am thinking it's not, since there are relations. How can I take this idea further?

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Certainly not, since it has a non-zero annihilator: $\{0\}\times R_2$.

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  • $\begingroup$ Thanks! I actually just found out that result. I am pretty new to module theory; is there a way to show nontrivial annihilator $\implies$ not free using only the definition of a free module as one with a basis? $\endgroup$ – TorsionSquid May 5 '15 at 0:20
  • $\begingroup$ Here it is simple: as it's a cyclic module; i. e. a quotient of $R_1\times R_2$ by a non-zero ideal, generated by $\overline{(1,0)}$, which is killed by any $(0,r_2)$. $\endgroup$ – Bernard May 5 '15 at 0:34
  • $\begingroup$ Unless $R_2 = 0$. :) $\endgroup$ – darij grinberg May 5 '15 at 3:00
  • $\begingroup$ Yes. Impilcitly, I consider rings with $1$. $\endgroup$ – Bernard May 5 '15 at 7:26
  • $\begingroup$ @Bernard $\{0\}$ is a ring with identity! ;-). By the way, over rings without an identity, the notion of free module doesn't make much sense: even the ring itself has no basis. $\endgroup$ – egreg May 5 '15 at 13:21
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Your example is actually an ideal of a ring: in the commutative case it is a free module iff it is principal and the generating element is not a zero-divisor. In your case it is principal being generated by $(1,0)$ but is n a zero divisor as pointed out by Bernard, and hence not free.

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