I’m reading a construction/proof of existence of the tensor product $T$ of $A$-modules $M$ and $N$, and I’d like to ask some clarifying questions. Here’s the construction.
Let $C$ denote the free $A$-module $A^{(M \times N)} .$ The elements of $C$ are formal linear combinations of elements of $M \times N$ with coefficients in $A$, i.e. they are expressions of the form $\sum_{i=1}^{n} a_{i} \cdot\left(x_{i}, y_{i}\right)\left(a_{i} \in A, x_{i} \in M, y_{i} \in N\right)$. Let $D$ be the submodule of $C$ generated by all elements of $C$ of the following types: $$ \begin{gathered} \left(x+x^{\prime}, y\right)-(x, y)-\left(x^{\prime}, y\right) \\ \left(x, y+y^{\prime}\right)-(x, y)-\left(x, y^{\prime}\right) \\ (a x, y)-a \cdot(x, y) \\ (x, a y)-a \cdot(x, y) \end{gathered} $$
Let $T=C / D$. For each basis element $(x, y)$ of $C$, let $x \otimes y$ denote its image in $T$. Then $T$ is generated by the elements of the form $x \otimes y$, and from our definitions we have $$ \begin{gathered} \left(x+x^{\prime}\right) \otimes y=x \otimes y+x^{\prime} \otimes y, \quad x \otimes\left(y+y^{\prime}\right)=x \otimes y+x \otimes y^{\prime}, \\ (a x) \otimes y=x \otimes(a y)=a(x \otimes y) \end{gathered} $$ Equivalently, the mapping $g: M \times N \rightarrow T$ defined by $g(x, y)=x \otimes y$ is $A$-bilinear.
Any map $f$ of $M \times N$ into an $A$-module $P$ extends by linearity to an $A$ module homomorphism $\bar{f}: C \rightarrow P$. Suppose in particular that $f$ is $A$-bilinear. Then, from the definitions, $\bar{f}$ vanishes on all the generators of $D$, hence on the whole of $D$, and therefore induces a well-defined $A$-homomorphism $f^{\prime}$ of $T=C / D$ into $P$ such that $f^{\prime}(x \otimes y)=f(x, y)$. The mapping $f^{\prime}$ is uniquely defined by this condition, and therefore the pair $(T, g)$ satisfy the conditions of the proposition.
My questions are:
- I’m trying to verify one of the properties of $\otimes$ from above, namely $\left(x+x^{\prime}\right) \otimes y=x \otimes y+x^{\prime} \otimes y$. Since $T$ is a quotient module, I suppose I can write (and verify) the above as
$$[(x+x’,y)] = [(x,y)] + [(x’,y)],\ \text{i.e.},$$
$$(x + x’,y) + D = (x,y) + (x’,y) + D$$
Now suppose $p \in (x + x’,y) + D$, so $p = (x+x’,y) + d$, for $d \in D$. Looking at the generators of $D$, I think I can write $p$ as, for $k,k’,l,l’ \in A$, \begin{align} p &= (kx + k’x’,ly + l’y)\\ &= (x,y)+(x’,y)+(k_1x + k_1’x,l_1y+l_1’y) \end{align} and so $p \in (x,y) + (x’,y) + D$. Is this the right approach, and which elements of $C$ then would not be in the module generated by $D$? - I’m trying to make sense of the last quoted paragraph diagrammatically. When the text says “$f$ $\dots$ extends by linearity to an $A$-module homomorphism $\bar{f}: C \to P$”, would it be correct to interpret it as $M \times N$ being embedded in $C$ by inclusion, and so $(x,y) \in M \times N$ would be $1 \cdot (x,y) \in C$? The picture I’m thinking of is as follows:
