# Why is $M \otimes N$ free with a basis $\{x_i \otimes y_j \}_{i,j}$?

Theorem $:$

Suppose $M,N$ are $R$-modules with $M=\oplus \sum_{i=1}^n M_i$ and $N=\oplus \sum_{j=1}^m N_j$ (where $M_i$ and $N_j$ are submodules of $M$ and $N$ respectively). Then $$M \otimes N \simeq \oplus \sum_{i,j} (M_i \otimes N_j)$$

I have understood the proof of the above theorem from the lecture note given by our instructor. After the completion of the proof our instructor has left a corollary for us as an easy exercise which is as follows $:$

Corollary $:$

Suppose $M$ and $N$ are free $R$-modules having bases $\{x_1,x_2, \cdots , x_n \}$ and $\{y_1,y_2, \cdots , y_m \}$ respectively. Then $M \otimes N$ is free with a basis $\{x_i \otimes y_j \}_{i,j}$.

I am trying to prove the above corollary as follows $:$

I first note that $M=\oplus \sum_{i=1}^n Rx_i$ and $N = \oplus \sum_{j=1}^m Ry_j$. So by the above theorem we have $$M \otimes N \simeq \oplus \sum_{i,j} (Rx_i \otimes Ry_j).$$ Now I observe that $Rx_i \otimes Ry_j = R(x_i \otimes y_j)$. So from this observation we conclude that $$M \otimes N \simeq \oplus \sum_{i,j} R (x_i \otimes y_j).$$ From here can I say that $M \otimes N$ is free with a basis $\{x_i \otimes y_j \}_{i,j}$? But that requires the fact that $\oplus \sum_{i,j} R (x_i \otimes y_j)$ is free which is possible if $R(x_i \otimes y_j)$ is a free cyclic $R$-module for all $i,j$ i.e. if $\mathrm {Ann}\ (x_i \otimes y_j) = \{0 \}$ for all $i,j$.

EDIT $:$

Let us fix some $i,j$. We want to show that $\mathrm {Ann}\ (x_i \otimes y_j) = \{0 \}$. Assume in contrary that $\mathrm {Ann}\ (x_i \otimes y_j) \neq \{0 \}$. Then $\exists$ $r \in R \setminus \{0 \}$ such that $r(x_i \otimes y_j)=0$. Let us now construct a map $f : M \times N \longrightarrow R$ defined by $f(p,q) = c_i d_j$ where $p=\sum_{k=1}^n c_k x_k$ and $q = \sum_{l=1}^m d_l y_l$, $p \in M$, $q \in N$. Then $f$ is a well-defined $R$-bilinear map. So by the univesal property of tensor product we have a unique $R$-linear map $g : M \otimes N \longrightarrow R$ defined by $g(p \otimes q) = c_i d_j$. Now if we apply $g$ to the equation $r (x_i \otimes y_j) = 0$ we get $r g(x_i \otimes y_j)=0$ $\implies r=0$ since $g(x_i \otimes y_j) = 1$ by the definition of $g$ which contradicts the fact that $r \in R \setminus \{0 \}$. Thus we are through.

Is my above reasoning correct at all? Please check it @Max.

• Well assume $r x_i\otimes y_j =0$, then create a bilinear map $M\times N \to$ something, that shows that this is impossible – Max Apr 28 '18 at 7:15
• Well some interesting module, $R$, $M$, $N$ for instance – Max Apr 28 '18 at 7:48
• Try with $R$, and sending $(x_i, y_j)$ to $1$ – Max Apr 28 '18 at 7:58