Primitive element theorem

Let ${\displaystyle E/F}$ be a field extension. An element ${\displaystyle \alpha \in E}$ is a primitive element for ${\displaystyle E/F}$ when ${\displaystyle E=F(\alpha ).}$

If there exists such a primitive element, then ${\displaystyle E/F}$ is referred to as a simple extension. If the field extension is of finite degree ${\displaystyle n=[E:F]}$, then every element x of E can be written 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, and ${\displaystyle \alpha \in E}$ is fixed. That is, if ${\displaystyle E/F}$ is a simple extension of degree n, there exists ${\displaystyle \alpha \in E}$ such that 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 ${\displaystyle \mathbb {Q} (\alpha )}$, for example ${\displaystyle {\sqrt {2}}={\tfrac {1}{2}}(\alpha ^{3}-9\alpha )}$. This shows α is indeed a primitive element:

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

Another argument is to note the independence of 1, ${\displaystyle {\sqrt {2}}}$, ${\displaystyle {\sqrt {3}}}$, ${\displaystyle {\sqrt {6}}}$ over the rationals; this shows that the subfield generated by α cannot be that generated by ${\displaystyle {\sqrt {2}}}$ or ${\displaystyle {\sqrt {3}}}$ or ${\displaystyle {\sqrt {6}}}$, exhausting all the subfields of degree 2 as given by Galois theory. Therefore, ${\displaystyle \mathbb {Q} (\alpha )}$ must be the whole field.

Let ${\displaystyle E/F}$ be a separable extension of finite degree. Then ${\displaystyle E=F(\alpha )}$ for some ${\displaystyle \alpha \in E}$; that is, the extension is simple and ${\displaystyle \alpha }$ is a primitive element.

The interpretation of the theorem changed with the formulation of the theory of Emil Artin, around 1930. From the time of Galois, the role of primitive elements had been to represent a splitting field as generated by a single element. This (arbitrary) choice of such an element was bypassed in Artin's treatment.[1] At the same time, considerations of construction of such an element receded: the theorem becomes an existence theorem.

The following theorem of Artin then takes the place of the classical primitive element theorem.

Theorem

Let ${\displaystyle E/F}$ be a finite degree field extension. Then ${\displaystyle E=F(\alpha )}$ for some element ${\displaystyle \alpha \in E}$ if and only if there exist only finitely many intermediate fields K with ${\displaystyle E\supseteq K\supseteq F}$.

A corollary to the theorem is then the primitive element theorem in the more traditional sense (where separability was usually tacitly assumed):

Corollary

Let ${\displaystyle E/F}$ be a finite degree separable extension. Then ${\displaystyle E=F(\alpha )}$ for some ${\displaystyle \alpha \in E}$.

The corollary 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.

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 E, 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.

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 for 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. 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 Artin's 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.

• Primitive element (finite field)

• 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)