All definitions I've seen for the statement "$E,F$ are linearly disjoint extensions of $k$" are only meaningful when $E,F$ are given as subfields of a larger field, say $K$. I am happy with the equivalence of the various definitions I've seen in this case. Lang's Algebra VIII.3-4 and (thanks to Pete) Zariski & Samuel's Commutative Algebra 1 II.15-16 have good coverage of this.

"Ambient" definitions of linear disjointness:

Wikipedia says it means the map $E\otimes_k F\to E.F$ is injective, where $E.F$ denotes their compositum in $K$, the smallest subfield of $K$ containing them both.

An equivalent (and asymmetric) condition is that any subset of $E$ which is linearly independent over $k$ is also linearly independent over $F$ (hence the name); this all happens inside $K$.

However, I often see the term used for field extensions which are NOT subfields of a larger one, even when the field extensions are not algebraic (so there is no tacit assumption that they live in the algebraic closure). Some examples of these situations are given below.

Question: What is the definition of "linearly disjoint" for field extensions which are not specified inside a larger field?

ANSWER: (After reading the helpful responses of Pete L. Clark, Hagen Knaf, Greg Kuperberg, and JS Milne -- thanks guys! -- I now have a satisfying and fairly exhaustive analysis of the situation.)

There are two possible notions of abstract linear disjointness for two field extensions $E,F$ of $k$ (proofs below):

(1) "Somewhere linearly disjoint", meaning
"There exists an extension $K$ with maps $E,F\to K$, the images of $E,F$ are linearly disjoint in $K$."
This is equivalent to the tensor product $E\otimes_k F$ being a domain.

(2) "Everywhere linearly disjoint", meaning
"For any extension $K$ with maps $E,F\to K$, the images of $E,F$ are linearly disjoint in $K$."
This is equivalent to the tensor product $E\otimes_k F$ being a field.

Results:

(A) If either $E$ or $F$ is algebraic, then (1) and (2) are equivalent.

(B) If neither $E$ nor $F$ is algebraic, then (2) is impossible.

Depending on when theorems would read correctly, I'm not sure which of these should be the "right" definition... (1) applies in more situations, but (2) is a good hypothesis for implicitly ruling out pairs of transcendental extensions. So I'm just going to remember both of them :)

PROOFS: (for future frustratees of linear disjointness!)

(1) Linear disjointness in some field $K$, by the Wikipedia defintion above, means the tensor product injects to $K$, making it a domain. Conversely, if the tensor product is a domain, then $E,F$ are linearly disjoint in its field of fractions.

(2) If the tensor product $E\otimes_k F$ is a field, since any map from a field is injective, by the Wikipedia definition above, $E,F$ are linearly disjoint in any $K$. Conversely, if $E \otimes_k F$ is not a field, then it has a non-trivial maximal ideal $m$, with quotient field say $K$, and then since $E\otimes_k F\to K$ has non-trivial kernel $m$, by definition $E,F$ are not linearly disjoint in $K$.

(A) Any two field extensions have some common extension (take a quotient of their tensor product by any maximal ideal), so (2) always implies (1).

Now let us first show that (1) implies (2) supposing $E/k$ is a finite extension. By hypothesis the tensor product $E\otimes_k F$ is a domain, and finite-dimensional as a $F$-vector space, and a finite dimensional domain over a field is automatically a field: multiplication by an element is injective, hence surjective by finite dimensionality over $F$, so it has an inverse map, and the image of $1$ under this map is an inverse for the element. Hence (1) implies (2) when $E/k$ is finite.

Finally, supposing (1) and only that $E/k$ is algebraic, we can write $E$ as a union of its finite sub-extensions $E_\lambda/k$. Since tensoring with fields is exact, $E_\lambda\otimes_k F$ naturally includes in
$E\otimes_k F$, making it a domain and hence a field by the previous argument. Then $E\otimes_k F$ is a union of fields, making it a field, proving (1) implies (2).

(B) Now this is easy. Let $t_1\in E$, $t_2\in F$ be transcendental elements. Identify $k(t)=k(t_1)=k(t_2)$ by $t\mapsto t_1 \mapsto t_2$, making $E$,$F$ extensions of $k(t)$. Let $K$ be a common extension of $E,F$ over $k(s)$ (any quotient of $E\otimes_{k(s)} F$ by a maximal ideal will do). Then $E,F$ are not linearly disjoint in $K$ because their intersection is not $k$: for example the set { $1,t$ }$\subseteq E$ is linearly independent over $k$ but not over $F$, so they are not linearly disjoint by the equivalent definition at the top.

Examples in literature of linear disjointness referring to abstract field extensions:

• Eisenbud, Commutative Algebra, Theorem A.13 (p.564 in my edition) says, in characteristic $p$,

"$K$ is separable over $k$ iff $k^{1/p^{\infty}}$ is linearly disjoint from $K$."

• Liu, Algebraic Geometry and Arithmetic Curves, Corollary 2.3 (c) (p. 91) says, for an integral algebraic variety $X$ over a field $k$ with function field $K(X)$,

"$X$ is geometrically integral iff $K(X)$ and $\overline{k}$ are linearly disjoint over $k$.

(Follow-up: Since in both these situations, one extension is algebraic, the two definitions summarized in the answer above are equivalent, so everything is fantastic.)

Old edit: My first guess was (and still is) to say that the tensor product is a domain...