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When [''E'' : ''F''] = ''p''<sup>2</sup>, there may not be a primitive element (in which case there are infinitely many intermediate fields). The simplest example is <math>E=\mathbb{F}_p(T,U)</math>, the field of rational functions in two indeterminates ''T'' and ''U'' over the [[finite field]] with ''p'' elements, and <math>F=\mathbb{F}_p(T^p,U^p)</math>. In fact, for any α = ''g''(T,U) in <math>E \setminus F</math>, the [[Frobenius endomorphism]] shows that the element ''α''<sup>''p''</sup> lies in ''F'' , so α is a root of <math>f(X)=X^p-\alpha^p\in F[X]</math>, and α cannot be a primitive element (of degree ''p''<sup>2</sup> over ''F''), but instead ''F''(α) is a non-trivial intermediate field.
==Proof==▼
Starting with a simple finite extension ''E'' = ''F''(α), let ''f'' be the [[minimal polynomial (field theory)|minimal polynomial]] of α over ''F''. If ''K'' is an intermediate subfield, then let ''g'' be the minimal polynomial of α over ''K'', and let ''L'' be the field generated over ''F'' by the coefficients of ''g''. Then since ''L'' ⊆ ''K'', the minimal polynomial of α over ''L'' must be a multiple of ''g'', so it is ''g''; this implies that the degree of ''E'' over ''L'' is the same as that over ''K'', but since ''L'' ⊆ ''K'', this means that ''L'' = ''K''. Since ''g'' is a factor of ''f'', this means that there can be no more intermediate fields than factors of ''f'', so there are only finitely many.▼
Going in the other direction, if ''F'' is finite, then any finite extension ''E'' of ''F'' is automatically simple, so assume that ''F'' is infinite. Then ''E'' is generated over ''F'' by a finite number of elements, so it's enough to prove that ''F''(α, β) is simple for any two elements α and β in ''E''. But, considering all fields ''F''(α + ''x'' β), where ''x'' is an element of ''F'', there are only finitely many, so there must be distinct ''x<sub>0</sub>'' and ''x<sub>1</sub>'' in ''F'' for which ''F''(α + ''x<sub>0</sub>'' β) = ''F''(α + ''x<sub>1</sub>'' β). Then simple algebra shows that ''F''(α + ''x<sub>0</sub>'' β) = ''F''(α, β). For an alternative proof, observe that each of the finite number of intermediate fields is a proper linear subspace of ''E'' over ''F'', and that a finite union of proper linear subspaces of a vector space over an infinite field cannot equal the entire space. Then, taking any element in ''E'' that is not in any intermediate field, it must generate the whole of ''E'' over ''F''.<ref>Theorem 26, ''Galois Theory'', Emil Artin and Arthur N. Milgram, University of Notre Dame Press, 2nd ed., 1944.</ref><ref>[https://stacks.math.columbia.edu/tag/030N Lemma 9.19.1 (Primitive element)], [[Stacks Project|The Stacks project]]. Accessed on line July 19, 2023.</ref>▼
==Constructive results==
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This is almost immediate as a way of showing how Steinitz' result implies the classical result, and a bound for the number of exceptional ''c'' in terms of the number of intermediate fields results (this number being something that can be bounded itself by Galois theory and ''a priori''). Therefore, in this case trial-and-error is a possible practical method to find primitive elements.
▲==Proof==
▲Starting with a simple finite extension ''E'' = ''F''(α), let ''f'' be the [[minimal polynomial (field theory)|minimal polynomial]] of α over ''F''. If ''K'' is an intermediate subfield, then let ''g'' be the minimal polynomial of α over ''K'', and let ''L'' be the field generated over ''F'' by the coefficients of ''g''. Then since ''L'' ⊆ ''K'', the minimal polynomial of α over ''L'' must be a multiple of ''g'', so it is ''g''; this implies that the degree of ''E'' over ''L'' is the same as that over ''K'', but since ''L'' ⊆ ''K'', this means that ''L'' = ''K''. Since ''g'' is a factor of ''f'', this means that there can be no more intermediate fields than factors of ''f'', so there are only finitely many.
▲Going in the other direction, if ''F'' is finite, then any finite extension ''E'' of ''F'' is automatically simple, so assume that ''F'' is infinite. Then ''E'' is generated over ''F'' by a finite number of elements, so it's enough to prove that ''F''(α, β) is simple for any two elements α and β in ''E''. But, considering all fields ''F''(α + ''x'' β), where ''x'' is an element of ''F'', there are only finitely many, so there must be distinct ''x<sub>0</sub>'' and ''x<sub>1</sub>'' in ''F'' for which ''F''(α + ''x<sub>0</sub>'' β) = ''F''(α + ''x<sub>1</sub>'' β). Then simple algebra shows that ''F''(α + ''x<sub>0</sub>'' β) = ''F''(α, β). For an alternative proof, observe that each of the finite number of intermediate fields is a proper linear subspace of ''E'' over ''F'', and that a finite union of proper linear subspaces of a vector space over an infinite field cannot equal the entire space. Then, taking any element in ''E'' that is not in any intermediate field, it must generate the whole of ''E'' over ''F''.<ref>Theorem 26, ''Galois Theory'', Emil Artin and Arthur N. Milgram, University of Notre Dame Press, 2nd ed., 1944.</ref><ref>[https://stacks.math.columbia.edu/tag/030N Lemma 9.19.1 (Primitive element)], [[Stacks Project|The Stacks project]]. Accessed on line July 19, 2023.</ref>
== History ==
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