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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?

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  • $\begingroup$ It's a consequence that these will be also in $C$. $\endgroup$
    – Berci
    May 29 '20 at 20:01
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    $\begingroup$ 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? $\endgroup$
    – alf262
    May 29 '20 at 20:10
  • $\begingroup$ Yes, see my answer. $\endgroup$
    – Berci
    May 29 '20 at 20:25
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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 $\lambda=-1$.

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  • $\begingroup$ 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. $\endgroup$
    – alf262
    May 29 '20 at 21:00
  • $\begingroup$ You're just there: $(e_{xn+x,\,y}-e_{xn, y} - e_{x,y})\, +\, (e_{xn, y} - ne_{x,y})\ \in C$. $\endgroup$
    – Berci
    May 29 '20 at 21:07
  • $\begingroup$ 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? $\endgroup$
    – alf262
    May 29 '20 at 22:25
  • $\begingroup$ Hmm.. You're right, it's my mistake. In this form they are present in the commutative setting only.. $\endgroup$
    – Berci
    May 29 '20 at 23:19
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    $\begingroup$ 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 $e_{-nx,y}+ne_{x,y} \in C$ $\endgroup$
    – Berci
    May 30 '20 at 11:00

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