View source for Linearly-disjoint extensions


{{MSC|12Fxx}}
{{TEX|done}}


Two subextensions $A$ and $B$ of an extension $\def\O{\Omega}\O$ of $k$ are called linearly disjoint  if the
subalgebra generated by $A$ and $B$ in $\O$ is (isomorphic to) the
[[Tensor product|tensor product]] $A\otimes B$ over $k$ (cf.
[[Extension of a field|Extension of a field]]). Let $A$ and $B$ be
arbitrary subrings of an extension $\O$ of $k$, containing $k$, and let
$C$ be the subring of $\O$ generated by $A$ and $B$. There is always a
ring homomorphism $\phi:A\otimes B \to C$ that associates with an element $x\otimes y\in A\otimes B$, $x\in A$, $y\in B$,
the product $xy$ in $C$. The algebras $A$ and $B$ are said to be
linearly disjoint over $k$ if $\phi$ is an isomorphism of $A\otimes B$ onto
$C$. In this case, $A\cap B = k$. For $A$ and $B$ to be linearly disjoint over
$k$ it is sufficient that there is a basis of $B$ over $k$ that is
independent over $A$. If $A$ is a finite extension of $k$, then the
degree of the extension $[B(A):B]$ does not exceed the degree of extension
$A:k$ and equality holds if and only if $A/k$ and $B/k$ are linearly
disjoint.

====References====
{|
|-
|valign="top"|{{Ref|Bo}}||valign="top"| N. Bourbaki, "Algebra", ''Elements of mathematics'', '''1''', Springer (1988) pp. Chapts. 4–7 (Translated from French) {{MR|1994218}} {{ZBL|1139.12001}}
|-
|valign="top"|{{Ref|ZaSa}}||valign="top"| O. Zariski, P. Samuel, "Commutative algebra", '''1''', Springer (1975) {{MR|0384768}} {{ZBL|0313.13001}}
|-
|}