# Primitive element theorem

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Primitive element theorem

In mathematics, more specifically in field theory, the primitive element theorem provides a characterization of the finite field extensions which are simple and thus can be generated by the adjunction of a single primitive element.

Primitive element theorem

:A field extension "L"/"K" is finite and has a primitive element if and only if there are only finitely many intermediate fields "F" with "K" &sube; "F" &sube; "L".

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.In more concrete language, every separable extension "L"/"K" of finite degree n is generated by a single element x satisfying a polynomial equation of degree n, xn +c1xn-1+..+cn=0, with coefficients in "K". The primitive element x provides a basis [1,x,x2,...,xn-1] for "L" over "K".

This corollary applies to algebraic number fields, which are finite extensions of the rational numbers Q, 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 α"p" lies in "K". Therefore we have ["L":"K"] = "p"2 but there is no element of "L" with degree "p"2 over "K", as a primitive element must have.

Example

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

:"X"2 − 2

and

:"X"2 − 3,

say α and &beta; respectively, to get a field "K" = Q(α, &beta;) of degree 4 over Q, that the extension is simple and there exists a primitive element &gamma; in "K" so that "K" = Q(&gamma;). One can in fact check that with

:&gamma; = α + &beta;

the powers &gamma;"i" for 0 &le; "i" &le; 3 can be written out as linear combinations of 1, α, &beta; and α&beta; with integer coefficients. Taking these as a system of linear equations, or by factoring, one can solve for α and &beta; over Q(&gamma;), which implies that this choice of &gamma; is indeed a primitive element in this example.

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

* Primitive element (finite field)

References

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