Factor of the product of rings cannot be free module? I need to prove the following statement:

Consider the product ring $R = R_1 \times R_2$. Prove that $R_1$ is
  never a free $R$-module.

From my basic knowledge,if $R_1$ was a free $R$-module then it would be of the form $R_1 \cong R^{(I)} = \oplus_{I} (R_1 \times R_2)$. Now, I ignore if this isomorphism is possible.
Do you have any suggestions? Is this the right way to go?
 A: The other answer has roughly the right idea, but there are some edge cases to be careful of.  That answer relies on the assumption that $(\langle 0,1\rangle)\neq (0)$ and that a free $R$-module cannot of nonzero annihilator.  But this is not correct: the ideal $(\langle 0,1\rangle)$ could be $(0)$, if $1=0$ in $R_2$ (in other words, if $R_2$ is the zero ring).  And a free $R$-module can have nonzero annihilator, if the module is the zero module.  That happens in this case if $R_1$ is the zero ring.
So, the argument does not work if $R_1$ or $R_2$ is the zero ring.  Indeed, the statement you are trying to prove is false in those cases: if $R_1=0$ then it is a free $R$-module on zero generators, and if $R_2=0$ then $R_1\cong R$ is a free $R$-module on one generator.
(Note also that it is absolutely crucial here that we are considering $R_1$ as an $R$-module via the projection map $R_1\times R_2\to R_1$.  If you have some different $R$-module structure on $R_1$, then pretty much anything could happen.)
A: (Updated due to rschwieb's comment)
$$R=R_1\times R_2\rightarrow R_1$$
$\text{Ann}_R(R_1)=\{0\}\times R_2\neq {0}$, so $R_1$ is not a free $R$-module since the annihilator would be $0$ (unless $R_1=R_2=0$).
