# A variation in the construction of the tensor product of modules

Let $$A$$ be a ring, $$E$$ a right $$A$$-module and $$F$$ a left $$A$$-module. Consider the free $$\mathbf{Z}$$-module $$\mathbf{Z}^{(E\times F)}$$ which comes with the injective canonical mapping $$\phi:E\times F\rightarrow\mathbf{Z}^{(E\times F)},\,(x,y)\mapsto e_{x,y}$$, where $$e_{x,y}:=(\delta_{(x,y),z})_{z\in E\times F}$$ for $$(x,y)\in E\times F$$.

Bourbaki defines the tensor product of $$E$$ and $$F$$ as the quotient $$\mathbf{Z}$$-module $$(\mathbf{Z}^{(E\times F)})/C$$, where $$C$$ is the submodule of $$\mathbf{Z}^{(E\times F)}$$ generated by the elements of the form $$(e_{x_1+x_2,y}-e_{x_1,y}-e_{x_2,y})$$, $$(e_{x,y_1+y_2}-e_{x,y_1}-e_{x,y_2})$$ and $$e_{x\lambda,y}-e_{x,\lambda y}$$ for $$x,x_1,x_2\in E$$ and $$y,y_1,y_2\in F$$ and $$\lambda\in L$$.

Elsewhere, I have seen the element of the form $$ne_{x,y}-e_{xn,y}$$, with $$x\in E$$, $$y\in F$$ and $$n\in\mathbf{Z}$$, added to the list above. Is this necessary? Why does Bourbaki leave it out?

• It's a consequence that these will be also in $C$. May 29 '20 at 20:01
• I suppose I would have to show that $ne_{x,y}-e_{xn,y}$ is a linear combination of the family of elements $(e_{x_1+x_2,y}-e_{x_1,y}-e_{x_2,y})$, correct? May 29 '20 at 20:10
• Yes, see my answer. May 29 '20 at 20:25

It's indeed not necessary, as these elements will already be in $$C$$ even for Bourbaki's definition.
Specifically, for $$n\ge 1$$, use induction to see it (let $$x_1=nx$$ and $$x_2=x$$ in the induction step).
For $$n\le 0$$, use the rule $$e_{x\lambda, \, y} - e_{x,\, \lambda y} \in C$$ with $$\lambda=0$$ and $$\lambda=-1$$.
• Can you please clarify the induction step for me? I have $e_{x(n+1),y}-(n+1)e_{x,y}=e_{xn+x,y}-ne_{x,y}-e_{x,y}$. By induction hypothesis, $e_{xn,y}-ne_{x,y}\in C$. If I take $x_1=xn$ and $x_2=x$, then $e_{xn+x,y}-e_{xn,y}-e_{x,y}\in C$ by definition. But here I have $e_{xn,y}$ instead of $ne_{x,y}$. I am not sure how to utilize the induction hypothesis. May 29 '20 at 21:00
• You're just there: $(e_{xn+x,\,y}-e_{xn, y} - e_{x,y})\, +\, (e_{xn, y} - ne_{x,y})\ \in C$. May 29 '20 at 21:07
• Sorry to bother you again–I am not clear about where you got the rule "$e_{x\lambda,y}-\lambda e_{x,y}\in C$" from. I have $e_{x\lambda,y}-e_{x,\lambda y}\in C$ instead. Is that a typo? May 29 '20 at 22:25
• For $n>0$ use again the condition for addition: we have $e_{nx, y} - ne_{x,y}\in C$, and $e_{0, y}—e_{nx,y}-e_{-nx, y} \in C$, and also $e_{0,y}\in C$, hence also $e_{-nx,y}+ne_{x,y} \in C$ May 30 '20 at 11:00