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.


(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...

  • $\begingroup$ @Andrew and everybody: the top three answers are all correct and helpful. Is there more to be desired? If not, vote 'em up! $\endgroup$ Dec 9 '09 at 18:46
  • $\begingroup$ Indeed, all four answers below were helpful to me. Please vote them up! $\endgroup$ Dec 9 '09 at 20:47
  • $\begingroup$ @AndrewCritch Indeed ! and (+1) to your question too, thanks a lot (I'm on the track of algebraic disjunction, but must understand the subtleties of linear disjunction first). $\endgroup$ Jul 12 '20 at 13:05

The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.

I am not (yet) getting voter support, but I stand my ground! :-)

First, clearly if $E \otimes_k F$ is a field, then it is isomorphic to every compositum.

Second, if $E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly disjoint, namely the fraction field of $E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly disjoint in any compositum.

If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly disjoint, when $E \otimes_k F$ is an integral domain, and strongly linearly disjoint, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.

(I needed to think about this issue in this paper.)

Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$. $E'$ and $F'$ are only weakly linearly disjoint, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly disjoint extensions have different behavior.

  • $\begingroup$ I don't think your last sentence is correct for extensions of infinite degree, because then I don't think an injective map to a compositum $E \otimes_k F \to E.F$ is necessarily surjective, in which case I don't see how to infer the tensor product is a field. Am I missing something? $\endgroup$ Dec 9 '09 at 7:42
  • $\begingroup$ I now support Greg's answer. It makes clear though that the "abstract linear disjointness" is not a straightforward generalization of "ambient linear disjointness" for transcendental extensions. $\endgroup$ Dec 9 '09 at 18:40
  • $\begingroup$ I was missing something, and this clears up everything, thank-you! $\endgroup$ Dec 9 '09 at 19:00

I have only seen the concept of "abstract linear disjointness of $K_1, K_2$ over $k$" used when at least one of the k-algebras can be embedded in the algebraic closure of the other. (I.e., only in this case can you omit the ambient field.) Both of your examples above are of this form.

In complete generality I don't think the concept of "abstract linear disjointness" makes much sense. For instance, consider two purely transcendental field extensions of $k$, say $k(s)$ and $k(t)$ with s and t algebraically independent indeterminates. (Even this statement seems to be implicitly speaking of some overfield!) The tensor product construction does not see the difference between $k(s) \otimes_k k(t)$ and
$k(t) \otimes_k k(t)$, but inside k(s,t), k(s) and k(t) are linearly disjoint and k(t) and k(t) are not!

There is also a little voice inside my head that says that, when applicable, the right criterion for linear disjointess is that the tensor product be a domain, not a field (as Greg Kuperberg says). Indeed, isn't that what happens in my example $k(s) \otimes_k k(t)$ inside k(s,t) above? But I am going to ignore this little voice and go to bed. We'll see what tomorrow brings.

  • 2
    $\begingroup$ k(t) ⊗ k(t) is indeed not a field, the kernel I of the multiplication map to k(t) is non-trivial (I/I² is Kahler differentials). $\endgroup$
    – user2035
    Dec 9 '09 at 8:06

Regard all the fields as subfields of the algebraic closure of K (or K(X)). More precisely, choose an algebraic closure of K and form $k^{1/p^{\infty}}$ inside of it.

Added: Linear disjointness is only defined for subfields of some big field. If you choose a different algebraic closure of K, then an isomorphism from it to the first will carry $k^{1/p^{\infty}}$ onto $k^{1/p^{\infty}}$ (with my definition), so you get an isomorphic situation. A similar remark applies to the second example (take $\bar k$ to be the algebraic closure of k in an algebraic closure of K(X)). This is why some authors don't bother to make this explicit (judging by this discussion, they should).

There is no ambiguity.

  • $\begingroup$ Do you know if this is equivalent to the tensor product being a domain? (I'm now appending this conjecture to the question.) I ask because I want to adopt a general definition if possible, not just an interpretation for these two theorems... $\endgroup$ Dec 9 '09 at 7:49
  • $\begingroup$ In the general case, there is more than way to regard $E$ and $F$ as subfields of the algebraic closure of $k$. The ambiguity is exactly the ambiguity in the choice of a compositum, so it doesn't change anything. $\endgroup$ Dec 9 '09 at 15:23
  • $\begingroup$ @GK: Are you perhaps overlooking the fact that E and F need not be algebraic extensions of k? They need not be realizable as subfields of any algebraic closure of k. (Everything you have said is correct in the case that E and F are finite extensions of k.) $\endgroup$ Dec 9 '09 at 15:49
  • $\begingroup$ Well I was overlooking that, you're right. See my extended answer. $\endgroup$ Dec 9 '09 at 17:09

Linear disjointness and its relation to tensor products is explained in detail in Zariski+Samuel, Commutative Algebra - I forgot which volume of the two.

There linear disjointness over $k$ of two $k$-algebras $A,B$ is defined only for algebras that are contained in some larger ring $C$.

The tensor product is introduced as follows: a product of $A$ and $B$ is a $k$-algebra $C$ and $k$-algebra morphisms $f:A\rightarrow C$, $g:B\rightarrow C$ such that the smallest subalgebra of $C$ containing $f(A),g(B)$ is $C$ itself.

A product $C$ of $A$ and $B$ is called tensor product, if $f(A)$ and $g(B)$ are linearly disjoint over $k$.

As for the case of two field extensions $E,F$ of $k$ one of which is algebraic over $k$ ($F$ say) one has a surjective map $E\otimes_k F\rightarrow E.F$, where $E.F$ denotes the smallest subring of the algebraic closure of $E$ that contains $E$ and $F$.

Since $F/k$ is algebraic, $E.F$ also is the smallest subfield that contains $E$ and $F$.

We get the equivalent statements:

(1) $E$ and $F$ are linearly disjoint over $k$ within the algebraic closure of $E$.

(2) the tensor product $E\otimes_k F$ is a field.

  • $\begingroup$ Thanks for the reference, it quite helped put things into perspective :) $\endgroup$ Dec 9 '09 at 21:02

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