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 <b>subfields of a larger field</b>, say $K$.  (I am happy with the equivalence of the various definitions I've seen in this case.)

For example, see [Wikipedia][1] 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.  Equivalently, 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 <b>NOT subfields of a larger one</b>, even when the field extensions are not algebraic (so there is no tacit assumption that they live in the algebraic closure).  

Examples I can think of right now:

1) Eisenbud, <i>Commutative Algebra</i>, 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 $K$."

2) Liu, <i>Algebraic Geometry and Arithmetic Curves</i>, 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 $k$.

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

I'm hesitant here to start making things up ad-hoc if there might be some standard interpretation I should know about...

<B>Edit:</b> My first guess was (and still is) to say that the tensor product is a domain...

  [1]: http://en.wikipedia.org/wiki/Tensor_product_of_fields#Compositum_of_fields