- Primitive element theorem
In

mathematics , more specifically infield theory , the**primitive element theorem**provides a characterization of thefinite field extension s 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" ⊆ "F" ⊆ "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, x^{n}+c_{1}x^{n-1}+..+c_{n}=0, with coefficients in "K". The primitive element x provides a basis [1,x,x^{2},...,x^{n-1}] for "L" over "K".This corollary applies to

algebraic number field s, 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 "F

_{p}"("T","U"), the field of rational functions in two indeterminates "T" and "U" over thefinite 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**ofrational number s roots of bothpolynomial s:"X"

^{2}− 2and

:"X"

^{2}− 3,say α and β respectively, to get a field "K" =

**Q**(α, β) of degree 4 over**Q**, that the extension is simple and there exists a primitive element γ in "K" so that "K" =**Q**(γ). One can in fact check that with:γ = α + β

the powers γ

^{"i"}for 0 ≤ "i" ≤ 3 can be written out aslinear combination s of 1, α, β and αβ with integer coefficients. Taking these as asystem of linear equations , or by factoring, one can solve for α and β over**Q**(γ), which implies that this choice of γ 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.

**See also***

Primitive element (finite field) **References**

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