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a condition of algebraic extension
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(Theorem)
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Theorem. A field extension is algebraic if and only if any subring of the extension field containing the base field is a field.
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).
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"a condition of algebraic extension" is owned by pahio.
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Cross-references: zero of the polynomial, polynomial ring, polynomial, contains, ring, algebraic, proof, field, base field, extension field, subring, field extension
There is 1 reference to this entry.
This is version 4 of a condition of algebraic extension, born on 2008-03-08, modified 2008-03-08.
Object id is 10379, canonical name is AConditionOfAlgebraicExtension.
Accessed 231 times total.
Classification:
| AMS MSC: | 12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions) |
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Pending Errata and Addenda
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