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[parent] a condition of algebraic extension (Theorem)

Theorem. A field extension $ L/K$ is algebraic if and only if any subring of the extension field $ L$ containing the base field $ K$ is a field.

Proof. Assume first that $ L/K$ is algebraic. Let $ R$ be a subring of $ L$ containing $ K$. For any non-zero element $ r$ of $ R$, naturally $ K[r] \subseteq R$, and since $ r$ is an algebraic element over $ K$, the ring $ K[r]$ coincides with the field $ K(r)$. Therefore we have $ r^{-1} \in K[r] \subseteq R$, and $ R$ must be a field.

Assume then that each subring of $ L$ which contains $ K$ is a field. Let $ a$ be any non-zero element of $ L$. Accordingly, the subring $ K[a]$ of $ L$ contains $ K$ and is a field. So we have $ a^{-1} \in K[a]$. This means that there is a polynomial $ f(x)$ in the polynomial ring $ K[x]$ such that $ a^{-1} = f(a)$. Because $ af(a)-1 = 0$, the element $ a$ is a zero of the polynomial $ xf(x)-1$ of $ K[x]$, i.e. is algebraic over $ K$. Thus every element of $ L$ is algebraic over $ K$.

Bibliography

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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See Also: ring adjunction, field adjunction, overring


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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
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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.

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AMS MSC12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)

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