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...