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[parent] a finite extension of fields is an algebraic extension (Theorem)
Theorem 1   Let $L/K$ be a finite field extension. Then $L/K$ is an algebraic extension.
Proof. In order to prove that $L/K$ is an algebraic extension, we need to show that any element $\alpha\in L$ is algebraic, i.e., there exists a non-zero polynomial $p(x)\in K[x]$ such that $p(\alpha)=0$ .

Recall that $L/K$ is a finite extension of fields, by definition, it means that $L$ is a finite dimensional vector space over $K$ . Let the dimension be $$[L\colon K]=n$$ for some $n\in \Nats$ .

Consider the following set of ``vectors'' in $L$ : $$\mathcal{S}=\{ 1, \alpha, \alpha^2,\alpha^3,\ldots,\alpha^n\}$$ Note that the cardinality of $S$ is $n+1$ , one more than the dimension of the vector space. Therefore, the elements of $S$ must be linearly dependent over $K$ , otherwise the dimension of $S$ would be greater than $n$ . Hence, there exist $k_i\in K,\ 0\leq i \leq n$ , not all zero, such that $$k_0+k_1\alpha+k_2\alpha^2+k_3\alpha^3+\ldots+k_n\alpha^n=0$$ Thus, if we define $$p(X)=k_0+k_1X+k_2X^2+k_3X^3+\ldots+k_nX^n$$ then $p(X)\in K[X]$ and $p(\alpha)=0$ , as desired.

$ \qedsymbol$

NOTE: The converse is not true. See the entry ``algebraic extension'' for details.




"a finite extension of fields is an algebraic extension" is owned by alozano.
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See Also: algebraic, algebraic extension, proof of transcendental root theorem

Keywords:  algebraic, polynomial, finite

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the field extension $\mathbb{R}/\mathbb{Q}$ is not finite (Corollary) by alozano
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Cross-references: converse, linearly dependent, cardinality, dimension, vector space, finite dimensional, fields, polynomial, algebraic, element, order, algebraic extension, finite field extension
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This is version 3 of a finite extension of fields is an algebraic extension, born on 2003-09-11, modified 2003-09-17.
Object id is 4725, canonical name is AFiniteExtensionOfFieldsIsAnAlgebraicExtension.
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AMS MSC12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)

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