simple field extension
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:
. 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|
|Date of creation||2013-03-22 14:23:06|
|Last modified on||2013-03-22 14:23:06|
|Last modified by||pahio (2872)|