Primitive element field theory

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In mathematics, a primitive element for an extension of fields L/K is an element ζ of L such that

L = K(ζ),

or in other words such that L is generated by ζ over K. This means that every element of L can be written as a quotient of two polynomials in ζ with coefficients from K.

If the extension L/K admits a primitive element, then L is either a finite extension of K, in case ζ is an algebraic element of L over K, or L is isomorphic to the field of rational functions over K in one indeterminate, if ζ is a transcendental element of L over K.

The primitive element theorem of field theory answers the question of which finite field extensions have primitive elements. It is not, for example, immediately obvious that if one adjoins to the field Q of rational numbers roots of both polynomials

X2 − 2


X2 − 3,

say α and β respectively, to get a field K = Q(α, β) of degree 4 over Q, that K is Q(γ) for a primitive element γ. One can in fact check that with

γ = α + β

the powers γi for 0 ≤ i ≤ 3 can be written out as linear combinations of 1, α, β and αβ with integer coefficients. Taking these as a system of linear equations, one can solve for α and β over Q, which implies that this choice of γ is indeed a primitive element in this example.

The general primitive element theorem states:

The field extension L/K is finite and has a primitive element if and only if there are only finitely many intermediate fields F with KFL.

In this form, the theorem is somewhat unwieldy and rarely used. An important corollary states

Every finite separable extension L/K has a primitive element.

This corollary applies to the example considered above (and to many others like it), since Q has characteristic 0 and therefore every extension over Q is separable.

For non-separable extensions, one can at least state the following:

If the degree [L:K] is a prime number, then L/K has a primitive element.

If the degree is not a prime number and the extension is not separable, one can give counterexamples. For example if K is Fp(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, then there is no primitive element for L over K. In fact one can see that for any α in L, the element αp lies in K. Therefore we have [L:K] = p2 but there is no elements of L with degree p2 over K, as a primitive element must have.

See also