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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. http://en.wikipedia.org/wiki/Linearly_disjoint2012-09-12
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    You are right, I read this question http://math.stackexchange.com/questions/57414/linearly-disjoint-field-extensions and skimmed through wikipedia and thought that was the same definition.2012-09-12
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    Nonetheless, this definition http://www.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?2012-09-12
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    The compositum is the sub-$k$-algebra generated by $A\cup B$, it is not a field in general.2012-09-12
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    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.2012-09-13

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