primitive element theorem
Let and be arbitrary fields, and let be an extension of of finite degree. Then there exists an element such that if and only if there are finitely many fields with .
Let be an indeterminate. Then is not generated over by a single element (and there are infinitely many intermediate fields ). To see this, suppose it is generated by an element . Then clearly must be transcendental, or it would generate an extension of finite degree. But if is transcendental, we know it is isomorphic to , and this field is not isomorphic to : for example, the polynomial has no roots in the first but it has two roots in the second. It is also clear that it is not sufficient for every element of to be algebraic over : we know that the algebraic closure of has infinite degree over , but if is algebraic over then will be finite.
This theorem has the corollary:
Let be a field, and let be finite and separable. Then there exists such that . In fact, we can always take to be an -linear combination (http://planetmath.org/LinearCombination) of and .
To see this (in the case of characteristic ), we need only show that there are finitely many intermediate fields. But any intermediate field is contained in the splitting field of the minimal polynomials of and , which is Galois with finite Galois group. The explicit form of comes from the proof of the theorem.
For more detail on this theorem and its proof see, for example, Field and Galois Theory, by Patrick Morandi (Springer Graduate Texts in Mathematics 167, 1996).
|Title||primitive element theorem|
|Date of creation||2013-03-22 11:45:48|
|Last modified on||2013-03-22 11:45:48|
|Last modified by||alozano (2414)|