Finite extension is Algebraic extension (Proof) + “Converse”

These two are useful lemmas in Galois/Field Theory.

Finite extension is Algebraic extension (Proof)
Let L/K be a finite field extension. Then L/K is an algebraic extension.
Proof:
Let L/K be a finite extension, where [L:K]=n. Let \alpha\in L. Consider \{1,\alpha,\alpha^2,\dots,\alpha^n\} which has to be linearly dependent over K since there are n+1 elements. Thus, there exists c_i\in K (not all zero) such that \sum_{i=0}^n c_i\alpha^i=0, so \alpha is algebraic over K.

Finitely Generated Algebraic Extension is Finite (Proof)
Let L/K be a finitely generated algebraic extension. Then L/K is a finite extension.

Proof:
Since L/K is finitely generated, L=K(\alpha_1,\dots,\alpha_n) for some \alpha_1,\dots,\alpha_n\in K. Since L/K is algebraic, each \alpha_i is algebraic over K. Denote L_i:=K(\alpha_1,\dots,\alpha_i) for 1\leq i\leq n. Then L_i=L_{i-1}(\alpha_i) for each i. Since \alpha_i is algebraic over K, it is also algebraic over L_{i-1}, so there exists a polynomial g_i with coefficients in L_{i-1} such that g_i(\alpha_i)=0. Thus [L_i:L_{i-1}]\leq\deg g_i<\infty. Similarly [L_1:K]<\infty. By Tower Law, [L:K]=[L_n:L_{n-1}][L_{n-1}:L_{n-2}]\dots[L_1:K]<\infty.

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One Response to Finite extension is Algebraic extension (Proof) + “Converse”

  1. learningcubed says:

    Great post with a lot of detail.

    Liked by 1 person

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