I have seen the following definition for the linearly disjoint field extensions $E$ and $F$:
"Two extension fields $E$ and $F$ of a field $k$ contained in a common field $L$, such that any finite set of elements in $E$ that is linearly independent when $E$ is regarded as a vector space over $k$ remains linearly independent when $E$ is regarded as a vector space over $F$."
- Does this imply that if $E$ is algebraic over $k$ and $E,F\supset k$ then $E$ and $F$ are linearly disjoint over $k?$
- By this can we say $E\bigotimes_K F$ is a field?
It's maybe helpful to know that if $E, F$ are finite Galois extensions of $k$, then $E$ and $F$ are linearly disjoint if and only if $E \cap F = k$. So linear disjointness requires a lot more than just algebraicity of the field extensions. For example, $E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, $F = \mathbb{Q}(\sqrt{2}, \sqrt{5})$ are not linearly disjoint over $\mathbb{Q}$; the $\mathbb{Q}$-linearly independent set $\{1, \sqrt{2}\} \subset E$ obviously does not remain linearly independent over $F$, since both elements lie in $F$ already.
If $E, F$ are linearly disjoint finite extensions over $k$, then the natural map $E \otimes_k F \to EF$ is an isomorphism, where $EF$ is the compositum inside $L$. You can check this explicitly by noting that a $k$-basis of the former gets sent to a $k$-basis of the latter. Letting $e_1, \dots, e_n$ be a $k$-basis for $E$ and $f_1, \dots, f_m$ be a $k$-basis for $F$, a $k$-basis for the tensor product is $e_i \otimes f_j$. The elements $e_if_j \in EF$ are linearly independent over $k$ by the linear disjointness condition: if $\sum_{i, j} a_{ij} e_i f_j = 0$, where $a_{ij} \in k$, then linear independence of the $e_i$ over $F$ implies $\sum_{j} a_{ij} f_j = 0$, for all $i$, hence all $a_{ij} = 0$. Since $\dim_k EF \leq \dim_k E \cdot \dim_k F$, we conclude that the homomorphism is an isomorphism.
However, linear disjointness does not guarantee that $E \otimes_k F$ is a field in general. For example, consider $\mathbb{C}(t), \mathbb{C}(s)$ as fields over $\mathbb{C}$, sitting inside the ambient field $\mathbb{C}(s, t)$. You can check that $\mathbb{C}(t) \otimes_{\mathbb{C}} \mathbb{C}(s)$ is not a field, although the map $\mathbb{C}(t) \otimes_{\mathbb{C}} \mathbb{C}(s) \to \mathbb{C}(s, t)$ is injective.