Further explanation for the submodule we quotient out by in defining the tensor products of two $R$-modules.

Question

Let $R$ be a commutative ring and let $F^R(M\times N)=R^{\oplus (M\times N)}$ the free $R$-module over $M \times N$.

The tensor product of $M$ and $N$ over $R$ is defined as $$M \otimes_R N := F^R(M \times N)/K$$ where $K$ the submodule generated by elements of the form $$(m, r_1n_1+r_2n_2)-r_1(m,n_1)-r_2(m,n_2) \quad \text{ and }\\ (r_1m_1+r_2m_2,n)-r_1(m_1, n)-r_2(m_2,n).$$

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My question is: What is $K$?

I know that it will contain the relations needed so that whenever we have an $R$-bilinear map $f: M\times N \to P$ between $R$-modules, we get a unique $R$-module homomorphism $\bar f: M \otimes_R N \to P$.

But, beyond the fact that it contains these needed relations, what is $K$? What do the elements look like?

If $x,y \in K$, then $rx+y \in K$. Is there something special about $rx+y$?

Answer 1

I'm not sure this fully answers your question, but using the notation $m\otimes n$ for the congruence class of $(m,n)$ modulo $K$, $K$ is the smallest submodule of $R^{(M\times N)}$ such that the relations \begin{align} m\otimes(r_1n_1+r_2n_2) = r_1(m\otimes n_1) + r_2(m\otimes n_2) \\ (r_1m_1+r_2m_2)\otimes n = r_1(m_1\otimes n) + r_2(m_2\otimes n) \end{align} are satisfied for all $m, m_1,m_2\in M$, $\;n,n_1, n_2\in N$, $\;r_1,r_2\in R$ – in other words, such that the canonical map: \begin{align} M\times N&\longrightarrow R^{(M\times N)}\Big/ K \\ (m,n)&\longmapsto m\otimes n \end{align} is bilinear.