# Linearly disjoint vs. free field extensions

Consider two field extensions $K$ and $L$ of a common subfield $k$ and suppose $K$ and $L$ are both subfields of a field $\Omega$, algebraically closed.

Lang defines $K$ and $L$ to be 'linearly disjoint over $k$' if any finite set of elements of $K$ that are linearly independent over $k$ stays linearly independent over $L$ (it is, in fact, a symmetric condition). Similarly, he defines $K$ and $L$ to be 'free over $k$' if any finite set of elements of $K$ that are algebraically independent over $k$ stays algebraically independent over $L$.

He shows right after that if $K$ and $L$ are linearly disjoint over $k$, then they are free over $k$.

Anyway, Wikipedia gives a different definition for linearly disjointness, namely $K$ and $L$ are linearly disjoint over $k$ iff $K \otimes_k L$ is a field, so I was wondering:

do we have a similar description of 'free over $k$' in terms of the tensor product $K \otimes_k L$?

It should be a weaker condition than $K \otimes_k L$ being a field, perhaps it needs to be a integral domain?

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The wikipedia's definition for linear disjointness is different from yours. en.wikipedia.org/wiki/Linearly_disjoint – Makoto Kato Sep 12 '12 at 20:41
You are right, I read this question math.stackexchange.com/questions/57414/… and skimmed through wikipedia and thought that was the same definition. – Niccolò Sep 12 '12 at 21:39
Nonetheless, this definition encyclopediaofmath.org/index.php/Linearly-disjoint_extensions means that when $A$ and $B$ are field extensions and not just $k$-algebras, their tensor products is isomorphic to their compositum, which is a field. And by the way, in the definition you linked, isn't the map $A \otimes_k B \rightarrow AB$, mapping $a \otimes b$ to $ab$, always surjective? – Niccolò Sep 12 '12 at 21:46
The compositum is the sub-$k$-algebra generated by $A\cup B$, it is not a field in general. – user18119 Sep 12 '12 at 21:52
Linear disjointness and freeness are broadly discussed in Zariski-Samuel, Commutative Algebra I, Chapter 3. In particular the point, that QiL mentions in his answer, is emphasized. – Hagen Knaf Sep 13 '12 at 6:55

The condition of being linearly disjoint or free depends much on the "positions" of $K, L$ inside $\Omega$, while the isomorphism class of the $k$-algebra $K\otimes_k L$ doesn't. For instance, consider $\Omega=\mathbb C(X,Y)$, $K=\mathbb C(X)$, $L_1=\mathbb C(Y)$ and $L_2=K$. Then $$K\otimes_\mathbb C L_1\simeq K\otimes_{\mathbb C} L_2$$ as $\mathbb C$-algebras. But $K, L_1$ are linearly disjoint (so free) in $\Omega$, not $K, L_2$. This example shows that in general, the linear disjointness nor the freeness can be determined by intrinsic properties of $K\otimes_k L$.
If $K$ or $L$ is algebraic over $k$, then it is true that linear disjointness is equivalent to $K\otimes_k L$ is a field. But in this situation the freeness is automatic whenever the tensor product is a field or not (can even be non-reduced).