Primitive element theorem

Let ${\displaystyle E/F}$ be a field extension. An element ${\displaystyle \alpha \in E}$ is a primitive element for ${\displaystyle E/F}$ if ${\displaystyle E=F(\alpha ),}$ i.e. if every element of ${\displaystyle E}$ can be written as a rational function in ${\displaystyle \alpha }$ with coefficients in ${\displaystyle F}$. If there exists such a primitive element, then ${\displaystyle E/F}$ is referred to as a simple extension.

If the field extension ${\displaystyle E/F}$ has primitive element ${\displaystyle \alpha }$ and is of finite degree ${\displaystyle n=[E:F]}$, then every element x of E can be written uniquely in the form

${\displaystyle x=f_{n-1}{\alpha }^{n-1}+\cdots +f_{1}{\alpha }+f_{0},}$

where ${\displaystyle f_{i}\in F}$ for all i. That is, the set

${\displaystyle \{1,\alpha ,\ldots ,{\alpha }^{n-1}\}}$

is a basis for E as a vector space over F.

If one adjoins to the rational numbers ${\displaystyle F=\mathbb {Q} }$ the two irrational numbers ${\displaystyle {\sqrt {2}}}$ and ${\displaystyle {\sqrt {3}}}$ to get the extension field ${\displaystyle E=\mathbb {Q} ({\sqrt {2}},{\sqrt {3}})}$ of degree 4, one can show this extension is simple, meaning ${\displaystyle E=\mathbb {Q} (\alpha )}$ for a single ${\displaystyle \alpha \in E}$. Taking ${\displaystyle \alpha ={\sqrt {2}}+{\sqrt {3}}}$, the powers 1, α, α², α³ can be expanded as linear combinations of 1, ${\displaystyle {\sqrt {2}}}$, ${\displaystyle {\sqrt {3}}}$, ${\displaystyle {\sqrt {6}}}$ with integer coefficients. One can solve this system of linear equations for ${\displaystyle {\sqrt {2}}}$ and ${\displaystyle {\sqrt {3}}}$ over ${\displaystyle \mathbb {Q} (\alpha )}$, to obtain ${\displaystyle {\sqrt {2}}={\tfrac {1}{2}}(\alpha ^{3}-9\alpha )}$ and ${\displaystyle {\sqrt {3}}=-{\tfrac {1}{2}}(\alpha ^{3}-11\alpha )}$. This shows that α is indeed a primitive element:

${\displaystyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})=\mathbb {Q} ({\sqrt {2}}+{\sqrt {3}}).}$

The classical primitive element theorem states:

Every separable field extension of finite degree is simple.

This theorem applies to algebraic number fields, i.e. finite extensions of the rational numbers Q, since Q has characteristic 0 and therefore every finite extension over Q is separable.

The following primitive element theorem (Ernst Steinitz[1]) is more general:

A finite field extension ${\displaystyle E/F}$ is simple if and only if there exist only finitely many intermediate fields K with ${\displaystyle E\supseteq K\supseteq F}$.

Using the fundamental theorem of Galois theory, the former theorem immediately follows from the latter.

For a non-separable extension ${\displaystyle E/F}$ of characteristic p, there is nevertheless a primitive element provided the degree [E : F] is p: indeed, there can be no non-trivial intermediate subfields since their degrees would be factors of the prime p.

When [E : F] = p², there may not be a primitive element (in which case there are infinitely many intermediate fields). The simplest example is ${\displaystyle E=\mathbb {F} _{p}(T,U)}$, the field of rational functions in two indeterminates T and U over the finite field with p elements, and ${\displaystyle F=\mathbb {F} _{p}(T^{p},U^{p})}$. In fact, for any α = g(T,U) in ${\displaystyle E\setminus F}$, the Frobenius endomorphism shows that the element α^p lies in F , so α is a root of ${\displaystyle f(X)=X^{p}-\alpha ^{p}\in F[X]}$, and α cannot be a primitive element (of degree p² over F), but instead F(α) is a non-trivial intermediate field.

Starting with a simple finite extension E = F(α), let f be the 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 x₀ and x₁ in F for which F(α + x₀ β) = F(α + x₁ β). Then simple algebra shows that F(α + x₀ 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.[2][3]

Generally, the set of all primitive elements for a finite separable extension E / F is the complement of a finite collection of proper F-subspaces of E, namely the intermediate fields. This statement says nothing in the case of finite fields, for which there is a computational theory dedicated to finding a generator of the multiplicative group of the field (a cyclic group), which is a fortiori a primitive element (see primitive element (finite field)). Where F is infinite, a pigeonhole principle proof technique considers the linear subspace generated by two elements and proves that there are only finitely many linear combinations

${\displaystyle \gamma =\alpha +c\beta \ }$

with c in F, that fail to generate the subfield containing both elements:

as ${\displaystyle F(\alpha ,\beta )/F(\alpha +c\beta )}$ is a separable extension, if ${\displaystyle F(\alpha +c\beta )\subsetneq F(\alpha ,\beta )}$ there exists a non-trivial embedding ${\displaystyle \sigma :F(\alpha ,\beta )\to {\overline {F}}}$ whose restriction to ${\displaystyle F(\alpha +c\beta )}$ is the identity which means ${\displaystyle \sigma (\alpha )+c\sigma (\beta )=\alpha +c\beta }$ and ${\displaystyle \sigma (\beta )\neq \beta }$ so that ${\displaystyle c={\frac {\sigma (\alpha )-\alpha }{\beta -\sigma (\beta )}}}$. This expression for c can take only ${\displaystyle [F(\alpha ):F][F(\beta ):F]}$ different values. For all other value of ${\displaystyle c\in F}$ then ${\displaystyle F(\alpha ,\beta )=F(\alpha +c\beta )}$.

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.

In his First Memoir of 1831,[4] Évariste Galois sketched a proof of the classical primitive element theorem in the case of a splitting field of a polynomial over the rational numbers. The gaps in his sketch could easily be filled[5] (as remarked by the referee Siméon Denis Poisson; Galois' Memoir was not published until 1846) by exploiting a theorem[6][7] of Joseph-Louis Lagrange from 1771, which Galois certainly knew. It is likely that Lagrange had already been aware of the primitive element theorem for splitting fields.[7] Galois then used this theorem heavily in his development of the Galois group. Since then it has been used in the development of Galois theory and the fundamental theorem of Galois theory. The two primitive element theorems were proved in their modern form by Ernst Steinitz, in an influential article on field theory in 1910;[1] Steinitz called the "classical" one Theorem of the primitive elements and the other one Theorem of the intermediate fields. Emil Artin reformulated Galois theory in the 1930s without the use of the primitive element theorems.[8][9]

• J. Milne's course notes on fields and Galois theory
• The primitive element theorem at mathreference.com
• The primitive element theorem at planetmath.org
• The primitive element theorem on Ken Brown's website (pdf file)