Namespaces
Variants
Actions

Linearly-disjoint extensions

From Encyclopedia of Mathematics
Revision as of 17:02, 7 February 2011 by 127.0.0.1 (talk) (Importing text file)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

of a field

Two subextensions and of an extension of such that the subalgebra generated by and in is (isomorphic to) the tensor product over (cf. Extension of a field). Let and be arbitrary subrings of an extension of , containing , and let be the subring of generated by and . There is always a ring homomorphism that associates with an element , , , the product in . The algebras and are said to be linearly disjoint over if is an isomorphism of onto . In this case, . For and to be linearly disjoint over it is sufficient that there is a basis of over that is independent over . If is a finite extension of , then the degree of the extension does not exceed the degree of extension and equality holds if and only if and are linearly disjoint.

References

[1] N. Bourbaki, "Algebra" , Elements of mathematics , 1 , Springer (1988) pp. Chapts. 4–7 (Translated from French)


Comments

References

[a1] O. Zariski, P. Samuel, "Commutative algebra" , 1 , Springer (1975)
How to Cite This Entry:
Linearly-disjoint extensions. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Linearly-disjoint_extensions&oldid=21153
This article was adapted from an original article by O.A. Ivanova (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article