# How does bilinearity induce a module homomorphism?

In many writings, it's often said that bilinearity induces a module homomorphism.

For example, just looking at the first link, say you have a bilinear homomorphism $\varphi\colon M\times N\to S$. What exactly is the induced homomorphism $M\otimes N\to S$? As in how is it explicitly defined based on the behavior of $\varphi$?

-
The homomorphism sends each elementary tensor $m \otimes n$ to the value $\varphi(m,n)$. From the construction of the tensor product module one can check that this really does extend additively to an honest linear map (the subtlety is that an element of $M \otimes N$ can be a sum of elementary tensors in many different ways). – KCd Jan 24 '12 at 0:51

Suppose $M$ and $N$ are $R$-modules. The tensor product $M\otimes N$ is constructed so that the following statement is true: there is a $R$-bilinear map $\psi:M\times N\to M\otimes N$ such that for any $R$-module $P$ and $R$-bilinear map $\varphi:M\times N\to P$, there is a $R$-module homomorphism $\overline{\varphi}:M\otimes N\to P$ such that $\varphi=\overline{\varphi}\circ \psi$. I will describe the construction of $M\otimes N$ and the map $\psi$, which will demonstrate what $\overline{\varphi}$, the induced homomorphism, must be.
The tensor product $M\otimes N$ is constructed as follows. For each $(m,n)\in M\times N$, let $x_{(m,n)}$ be an indeterminate. Let $F$ be the free $R$-module generated by the objects $x_{(m,n)}$. Let $T$ be the submodule of $F$ generated by elements of the form $$x_{(rm,n)}-r\cdot x_{(m,n)}$$ $$x_{(m,rn)}-r\cdot x_{(m,n)}$$ $$x_{(m_1+m_2,n)}- x_{(m_1,n)}-x_{(m_2,n)}$$ $$x_{(m,n_1+n_2)}- x_{(m,n_1)}-x_{(m,n_2)}$$ Then $M\otimes N$ is defined to be $F/T$. Let $m\otimes n$ denote the equivalence class of $x_{(m,n)}$ in $F/T$. Since the free $R$-module generated by a collection of objects consists precisely of finite $R$-linear combinations of those objects, the quotient module $M\otimes N$ consists of finite $R$-linear combinations of elements of the form $m\otimes n$. However, while in $F$ the elements $x_{(m,n)}$ satisfied no relations with each other (being indeterminates), in $M\otimes N$, the following relations hold (by design): $$r\cdot(m\otimes n)=(rm)\otimes n=m\otimes(rn)$$ $$(m_1+m_2)\otimes n=(m_1\otimes n)+(m_2\otimes n)$$ $$m\otimes(n_1+n_2)=(m\otimes n_1)+(m\otimes n_2)$$ The $R$-bilinear map $\psi:M\times N\to M\otimes N$ described above is simply the map sending $(m,n)$ to $m\otimes n$. Thus, if $\varphi=\overline{\varphi}\circ\psi$ is to be true, the $R$-module homomorphism $\overline{\varphi}:M\otimes N\to P$ induced by the $R$-bilinear map $\varphi: M\times N\to P$ must act as follows: $$\overline{\varphi}(m\otimes n)=\varphi(m,n)$$ and then extending by linearity, $$\overline{\varphi}\left(\sum_i r_i(m_i\otimes n_i)\right)=r_i\sum_i\phi(m,n)$$