simple field extension
Let be obtained from the field via the of the element , which is called the primitive element of the field extension . We shall settle the of the field .
We consider the substitution homomorphism , where
According to the ring homomorphism theorem, the image ring is isomorphic with the residue class ring , where is the ideal of polynomials having as their zero. Because is, as subring of the field , an integral domain, then also has no zero divisors, and hence is a prime ideal. It must be principal, for is a principal ideal ring.
There are two possibilities:
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1.
, where is an irreducible polynomial with . Because every non-zero prime ideal of is maximal, the isomorphic image of is a field, and it must give the of . We say that is algebraic with respect to (or over ). In this case, we have a finite field extension .
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2.
. This means that the homomorphism is an isomorphism between and , i.e. all expressions behave as the polynomials . Now, is no field because is not such, but the isomorphy of the rings implies the isomorphy of the corresponding fields of fractions. Thus the simple extension field is isomorphic with the field of rational functions in one indeterminate . We say that is transcendental (http://planetmath.org/Algebraic) with respect to (or over ). This time we have a simple infinite field extension .
Title | simple field extension |
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Canonical name | SimpleFieldExtension |
Date of creation | 2013-03-22 14:23:06 |
Last modified on | 2013-03-22 14:23:06 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 25 |
Author | pahio (2872) |
Entry type | Definition |
Classification | msc 12F99 |
Related topic | PrimitiveElementTheorem |
Related topic | CanonicalFormOfElementOfNumberField |
Defines | primitive element |