Factor of a product ring can not be free

Question

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 $R=R_1\times{}R_2$, 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

Answer 1

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.

Answer 2

Suppose $R_1$ and $R_2$ are nontrivial, and consider $R_1 = M$ as a right $R = R_1\times R_2$-module via the natural action \begin{align*} M\times R&\to M\\ (m,(r,s))&\mapsto mr. \end{align*}

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 $B = \{e_i\mid i\in I\}\subseteq M$ such that the following two conditions hold:

1. For any $m\in M,$ there exists a finite subset $J\subseteq I$ and collection of elements $\{r_j\mid j\in J\}\subseteq R$ such that $$m = \sum_{j\in J} e_j r_j,$$ and
2. If $\{r_j\mid j\in J\subseteq I\}\subseteq R$ is a finite collection of elements of $R$ such that $$\sum_{j\in J} e_j r_j = 0,$$ then $r_j = 0$ for all $j.$

To that end, suppose such a basis $B = \{e_i\}_{i\in I}\subseteq M$ exists (as $R_1$ is nontrivial, the set $B$ is necessarily nonempty). Let $r = (0,1),$ and choose any $e\in B.$ Then we have $$ er = e(0,1) = e\cdot 0 = 0, $$ 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 $R_1\times R_2$-module via the natural action.

Answer 3

Let $F$ be a nonzero free left $R$-module. Then $\operatorname{Ann}_R(F)=\{r\in R:rF=0\}=\{0\}$.

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

Now note that, when $R=R_1\times R_2$, $\operatorname{Ann}_R(R_1\times\{0\})=\{0\}\times R_2$. If neither $R_1$ nor $R_2$ is the trivial ring, we're done.