The equivalent definitions of linearly disjoint field extensions I am trying to proove that the following charactrerization of linearly disjontness holds.
We have the definition:

Given the following extensions, $K\subset L, M \subset N$, $M$ is said to be linearly disjoint from $L$ over $K$ in $N$ if $\forall S\subset M$ linearly independent over $K$, the subset $S\subset N$ is linearly independent over $L$.

I am trying to prove that this is equivalent to the fact that the map $t:L\otimes_kM\rightarrow N$, given by $t(l\otimes_k m)=lm$ is injective.
Let us make the following notation:
$$\theta:L\times M\rightarrow N,\ \theta(l,m)=lm$$
$L$ and $N$ are $K$-algebras. Thus, $L\otimes_K M$ is a ring.
$\theta$ is $K$-bilinear so $t$ is well-defined as a linear map. 
Furthemore, $t((l_1\otimes_K m_1)(l_2 \otimes_K m_2))=t(l_1l_2\otimes_k m_1m_2)=l_1l_2m_1m_2=l_1m_1l_2m_2=t(l_1\otimes m_1)t(l_2\otimes m_2)$, 
which makes $t$ a ring homomorphism.
"$\Leftarrow$" We assume that $t$ is injective. Let $S\subset M$ be linearly independent over $K$. Let $n\in\mathbb{N}$, $s_1,s_2,\dots,s_n\in S$ and $l_1,l_2,\dots l_n\in L$ such that
$l_1s_1+l_2s_2+\dots+l_ns_n=0.$
We want to prove that $S\subset N$ is linearly independent over $L$. Therefore, we would have to prove that $l_i=0,\ \forall i=\overline{1,n}$.
We have that $t(l_1\otimes_K s_1+l_2\otimes_K s_2+\dots+l_n\otimes_K s_n)=l_1s_1+l_2s_2+\dots+l_ns_n=0$. Hence $l_1\otimes_K s_1+l_2\otimes_K s_2+\dots+l_n\otimes_K s_n\in \ker(t)$, so we get from the injectivity of $t$ that:
$$l_1\otimes_K s_1+l_2\otimes_K s_2+\dots+l_n\otimes_K s_n=0$$
How do I go from here?
I realise that intuitively, the linear independence of $s_i$ forces $l_i$ to be 0, but how?
 A: $M$ can be seen as a free $K$-module. Since the $\{s_i|i=\overline{1,n}\}$ is an independent set, it can be completed up to a basis: $\mathcal{S}=\{s_j\vert j\in J,\ s_j\in M\}\supset S$. Every element in the tensor product $L\otimes_K M$ can be written in a unique way as: $\sum_{j\in J} l_j\otimes_K s_j$, where finitely many $l_j$ are different from $0$. Thus, the $l_i$ in the initial equation are all 0, making $S$ linearly independent over $L$. 
$"\Rightarrow$"
Let $\sum_{i=1}^m l_i\otimes_K s_i\in \ker(t)$. It follows that:
$\sum_{i=1}^m l_is_i=0$
If all the $l_i$ are zero, it follows that $\sum_{i=1}^n l_i\otimes_K s_i=0$. Otherwise, remove all the $l_i=0$, we get the equation, after renumbering
\begin{equation}\label{lin}
\sum_{i=1}^n l_is_i=0
\end{equation}
the set $S=\{s_1,\dots,s_n\}$ is not linearly independent over $L\Rightarrow S$ is not linearly independent over $K\Rightarrow \exists k_i\in K,\ i=\overline{1,n-1}$ such that $\sum_{i=1}^{n-1} k_is_i=s_n$. Replacing in the previous sum and reordering after $s_i$, we get a sum $\sum_{i=1}^{n-1} (l_i+k_i)s_i=0 $. In this sum, all the coefficients of $s_i$ are not $0$. (if they are, we can multiply the initial equation by a certain value different from $0$). Repeating the procedure, we reach in the end that $s_1=0$, making all the $s_i=0$, $i=\overline{1,n}$. We get that $\sum_{i=1}^m l_i\otimes_K s_i=0$ since in each term one of the factors is $0$, making $t$ injective.
