Primitive element theorem

In field theory, the primitive element theorem or Artin's theorem on primitive elements is a result characterizing the finite degree field extensions that possess a primitive element, or simple extensions. It says that a finite extension is simple if and only if there are only finitely many intermediate fields. In particular, finite separable extensions are simple.

Terminology

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

If there exists such a primitive element, then $E/F$ is referred to as a simple extension. If the field extension is of finite degree $n = [E:F]$, then every element x of E can be written in the form

$x=f_{n-1}{\alpha}^{n-1}+\cdots+f_1{\alpha}+f_0,$

where $f_i\in F$ for all i, and $\alpha\in E$ is fixed. That is, if $E/F$ is a separable extension of degree n, there exists $\alpha\in E$ such that the set

$\{1,\alpha,\cdots,{\alpha}^{n-1}\}$

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

For example, the extension $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$ is a simple extension and, as shown below, so is $\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q}.$

Classical Primitive Element Theorem

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

Existence statement

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

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

Corollary

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

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

Counterexamples

For non-separable extensions, necessarily in characteristic p with p a prime number, then at least when the degree [L : K] is p, L / K has a primitive element, because there are no intermediate subfields. When [L : K] = p², there may not be a primitive element (and therefore there are infinitely many intermediate fields). This happens, for example if K is

F_p(T, U),

the field of rational functions in two indeterminates T and U over the finite field with p elements, and L is obtained from K by adjoining a p-th root of T, and of U. In fact one can see that for any α in L, the element α^p lies in K, but a primitive element must have degree p² over K.

Constructive results

Generally, the set of all primitive elements for a finite separable extension L / K is the complement of a finite collection of proper K-subspaces of L, 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 K 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

$\gamma = \alpha + c \beta\$

with c in K, that fail to generate the subfield containing both elements. 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. See the Example.

Example

It is not, for example, immediately obvious that if one adjoins to the field $\mathbb{Q}$ of rational numbers roots of both polynomials

$x^2 - 2\$

and

$x^2 - 3,\$

say $\sqrt{2}$ and $\sqrt{3}$ respectively, to get a field K = $\mathbb{Q}(\sqrt{2},\sqrt{3})$ of degree 4 over $\mathbb{Q}$, that the extension is simple and there exists a primitive element γ in K so that K = $\mathbb{Q}(\gamma)$. One can in fact check that with

$\gamma = \sqrt{2} + \sqrt{3}$

the powers γ ⁱ for 0 ≤ i ≤ 3 can be written out as linear combinations of 1, $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{2}\sqrt{3}=\sqrt{6}$ with integer coefficients. Taking these as a system of linear equations, or by factoring, one can solve for $\sqrt{2}$ and $\sqrt{3}$ over $\mathbb{Q}(\gamma)$ (one gets, for instance, Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \sqrt{2} = \scriptstyle\frac{\gamma^3-9\gamma}2 ), which implies that this choice of γ is indeed a primitive element in this example. A simpler argument, assuming the knowledge of all the subfields as given by Galois theory, is to note the independence of 1, $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{2}\sqrt{3}$ over the rationals; this shows that the subfield generated by γ cannot be that generated by $\sqrt{2}$ or $\sqrt{3}$ or $\sqrt{2}\sqrt{3}$, exhausting all the subfields of degree 2. Therefore it must be the whole field.

See also

• Primitive element (finite field)

External links

• 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)