[proofplan]
This is a direct application of the universal property of the tensor product. The forward direction uses the linearisation: if the tensor sum is zero, the induced linear map sends it to zero. The reverse direction is a contrapositive argument: if the tensor sum is nonzero, we exhibit a specific bilinear map (the canonical one into $M \otimes_R N$ itself) that detects it.
[/proofplan]
[step:Prove the forward direction: if $\sum_i m_i \otimes n_i = 0$ then $\sum_i f(m_i, n_i) = 0$ for every bilinear $f$]
Let $L$ be an $R$-module and let $f: M \times N \to L$ be an $R$-bilinear map. By the [Universal Property of Tensor Product](/theorems/2908), there exists a unique $R$-module homomorphism $h: M \otimes_R N \to L$ satisfying $h(m \otimes n) = f(m, n)$ for all $m \in M$, $n \in N$. Since $h$ is $R$-linear:
\begin{align*}
\sum_{i=1}^\ell f(m_i, n_i) = \sum_{i=1}^\ell h(m_i \otimes n_i) = h\Bigl(\sum_{i=1}^\ell m_i \otimes n_i\Bigr) = h(0) = 0.
\end{align*}
[guided]
The universal property tells us that every $R$-bilinear map $f: M \times N \to L$ factors through $M \otimes_R N$: there exists a unique $R$-linear map $h: M \otimes_R N \to L$ with $f = h \circ i_{M \otimes N}$, i.e., $h(m \otimes n) = f(m, n)$.
Now suppose $\sum_{i=1}^\ell m_i \otimes n_i = 0$ in $M \otimes_R N$. Applying the $R$-linear map $h$:
\begin{align*}
\sum_{i=1}^\ell f(m_i, n_i) = \sum_{i=1}^\ell h(m_i \otimes n_i) = h\Bigl(\sum_{i=1}^\ell m_i \otimes n_i\Bigr) = h(0) = 0.
\end{align*}
The second equality uses the $R$-linearity of $h$ (which allows us to pull the finite sum outside), and the third uses the hypothesis that the tensor sum is zero. This holds for every choice of $L$ and $f$.
[/guided]
[/step]
[step:Prove the reverse direction: if $\sum_i m_i \otimes n_i \neq 0$ then some bilinear map detects it]
We prove the contrapositive. Suppose $\sum_{i=1}^\ell m_i \otimes n_i \neq 0$ in $M \otimes_R N$. We exhibit a specific $R$-module $L$ and $R$-bilinear map $f: M \times N \to L$ such that $\sum_{i=1}^\ell f(m_i, n_i) \neq 0$.
Take $L = M \otimes_R N$ and $f = i_{M \otimes N}: M \times N \to M \otimes_R N$, the canonical $R$-bilinear map $(m, n) \mapsto m \otimes n$. Then:
\begin{align*}
\sum_{i=1}^\ell f(m_i, n_i) = \sum_{i=1}^\ell i_{M \otimes N}(m_i, n_i) = \sum_{i=1}^\ell m_i \otimes n_i \neq 0.
\end{align*}
Hence the condition "$\sum_i f(m_i, n_i) = 0$ for every $R$-module $L$ and every $R$-bilinear map $f$" fails. By contrapositive, the reverse implication holds.
[guided]
For the reverse direction, we need: if $\sum_i f(m_i, n_i) = 0$ for **every** bilinear $f$, then $\sum_i m_i \otimes n_i = 0$. The contrapositive is: if $\sum_i m_i \otimes n_i \neq 0$, then **some** bilinear map detects this.
Where do we find such a bilinear map? The tensor product itself provides one. The canonical bilinear map $i_{M \otimes N}: M \times N \to M \otimes_R N$ defined by $(m, n) \mapsto m \otimes n$ is $R$-bilinear by construction. Taking $L = M \otimes_R N$ and $f = i_{M \otimes N}$:
\begin{align*}
\sum_{i=1}^\ell f(m_i, n_i) = \sum_{i=1}^\ell m_i \otimes n_i \neq 0.
\end{align*}
This is the natural "test": the tensor product is itself the universal recipient of bilinear maps, so if an element is nonzero there, the canonical bilinear map will witness it. The argument is a one-line observation once you realize that $i_{M \otimes N}$ is a valid choice of $f$.
Combining both directions completes the proof: $\sum_i m_i \otimes n_i = 0$ if and only if $\sum_i f(m_i, n_i) = 0$ for every $R$-bilinear $f: M \times N \to L$ and every $R$-module $L$.
[/guided]
[/step]
