a condition of algebraic extension
Proof. Assume first that is algebraic. Let be a subring of containing . For any non-zero element of , naturally , and since is an algebraic element over , the ring coincides with the field . Therefore we have , and must be a field.
Assume then that each subring of which contains is a field. Let be any non-zero element of . Accordingly, the subring of contains and is a field. So we have . This means that there is a polynomial in the polynomial ring such that . Because , the element is a zero of the polynomial of , i.e. is algebraic over . Thus every element of is algebraic over .
- 1 David M. Burton: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).
|Title||a condition of algebraic extension|
|Date of creation||2013-03-22 17:53:34|
|Last modified on||2013-03-22 17:53:34|
|Last modified by||pahio (2872)|