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I have seen the following property in my class note but I don´t know how to prove, could someone help me?

If we consider the product ring $R=R_1timesR_2$, then $R_1$ can not be a free right $R$-module

I think that it must have a simple proof, but I don't get it.

Thanks

Let $F$ be a nonzero free left $R$-module. Then $operatornameAnn_R(F)=rin R:rF=0=0$.

This is essentially obvious, because $F$ is a direct sum of copies of $R$.

Now note that, when $R=R_1times R_2$, $operatornameAnn_R(R_1times0)=0times R_2$. If neither $R_1$ nor $R_2$ is the trivial ring, we're done.

Let's assume that things are non-trivial, that is, both $R_1$ and $R_2$ are not the zero ring.

Now, multiplication in the product ring $R_1 times R_2$ is coordinate-wise, that is, for an element $(a_1, a_2) in R_1 times R_2$ with $a_1 in R_1$ and $a_2 in R_2$, and another element $(b_1, b_2) in R_1 times R_2$ with $b_1 in R_1$ and $b_2 in R_2$, we have
$$
(a_1, a_2) cdot (b_1, b_2) = (a_1 cdot a_2, b_1 cdot b_2).
$$

Moreover, we can identify the ring $R_1$ with the subring $R_1 times 0_R_2 subseteq R_1 times R_2$ where $0_R_2$ denotes the zero element of $R_2$. The identification map is
$$
iota_1 : R_1 rightarrow R_1 times R_2,quad a mapsto (a, 0_R_2).
$$
Now, $R_1 times 0_R_2$, being a subring of $R_1 times R_2$, is a right $(R_1 times R_2)$-module, where the module operation is given by ring multiplication. That is, if we have a "module element" $(a_1, 0_R_2) in R_1 times 0_R_2$ and a "coefficiont ring element" $(a_2, b_2) in R_1 times R_2$, then the product is
$$
(a_1, 0_R_2) cdot (a_2, b_2) = (a_1 cdot a_2, 0_R_2).
$$
Now, since both $R_1$ and $R_2$ are not the zero ring, we can pick $a_3 in R_1 setminus 0_R_1$ and $b_3 in R_2 setminus 0_R_2$. With these, we construct a nonzero "module element" $(a_3, 0_R_2) in R_1 times 0_R_2$ and a nonzero "coefficient ring element" $(0_R_1, b_3) in in R_1 times R_2$. Their product is then
$$
(a_3, 0_R_2) cdot (0_R_1, b_3) = (0_R_1, 0_R_2).
$$
That is, the product of a nonzero module element and a nonzero coefficient ring element is the zero module element.
This does not happen in a free module.

So we have shown that $R_1 times 0_R_2$ is not a free right $(R_1 times R_2)$-module.

Suppose $R_1$ and $R_2$ are nontrivial, and consider $R_1 = M$ as a right $R = R_1times R_2$-module via the natural action
beginalign*
Mtimes R&to M\
(m,(r,s))&mapsto mr.
endalign*

To show that $M$ is not free over $R,$ we must show that there is no basis of $M.$ Recall that a basis of $M$ is a subset $B = e_imid iin Isubseteq M$ such that the following two conditions hold:

1. For any $min M,$ there exists a finite subset $Jsubseteq I$ and collection of elements $r_jmid jin Jsubseteq R$ such that $$m = sum_jin J e_j r_j,$$ and


2. If $r_jmid jin Jsubseteq Isubseteq R$ is a finite collection of elements of $R$ such that $$sum_jin J e_j r_j = 0,$$ then $r_j = 0$ for all $j.$
   

To that end, suppose such a basis $B = e_i_iin Isubseteq M$ exists (as $R_1$ is nontrivial, the set $B$ is necessarily nonempty). Let $r = (0,1),$ and choose any $ein B.$ Then we have
$$
er = e(0,1) = ecdot 0 = 0,
$$
but $(0,1)in R$ is not zero (as $R_2$ is not the trivial ring). This contradicts our assumption that $B$ was a basis. Therefore, assuming that both $R_1$ and $R_2$ are nontrivial, we have shown $R_1$ cannot be a free right $R_1times R_2$-module via the natural action.