Infield theory, theprimitive element theorem states that everyfiniteseparablefield extension issimple, i.e. generated by a single element. This theorem implies in particular that allalgebraic number fields over the rational numbers, and all extensions in which both fields are finite, are simple.

Let${\displaystyle E/F}$ be afield extension. An element${\displaystyle \alpha \in E}$ is aprimitive element for${\displaystyle E/F}$ if${\displaystyle E=F(\alpha ),}$ i.e. if every element of${\displaystyle E}$ can be written as arational function in${\displaystyle \alpha }$ with coefficients in${\displaystyle F}$. If there exists such a primitive element, then${\displaystyle E/F}$ is referred to as asimple extension.

If the field extension${\displaystyle E/F}$ has primitive element${\displaystyle \alpha }$ and is of finitedegree${\displaystyle n=[E:F]}$, then every element${\displaystyle \gamma \in E}$ can be written in the form

${\displaystyle \gamma =a_0+a_1\alpha +\cdots +a_{n-1}{\alpha }^{n-1},}$

for unique coefficients${\displaystyle a_0,a_1,\dots ,a_{n-1}\in F}$. That is, the set

${\displaystyle \{1,\alpha ,\dots ,{\alpha }^{n-1}\}}$

is abasis forE as avector space overF. The degreen is equal to the degree of theirreducible polynomial ofα overF, the unique monic${\displaystyle f(X)\in F[X]}$ of minimal degree withα as a root (a linear dependency of${\displaystyle \{1,\alpha ,\dots ,{\alpha }^{n-1},{\alpha }^n\}}$).

IfL is asplitting field of${\displaystyle f(X)}$ containing itsn distinct roots${\displaystyle {\alpha }_1,\dots ,{\alpha }_n}$, then there arenfield embeddings${\displaystyle {\sigma }_i:F(\alpha )\hookrightarrow L}$ defined by${\displaystyle {\sigma }_i(\alpha )={\alpha }_i}$ and${\displaystyle \sigma (a)=a}$ for${\displaystyle a\in F}$, and these extend to automorphisms ofL in theGalois group,${\displaystyle {\sigma }_1,\dots ,{\sigma }_n\in \mathrm{G}\mathrm{a}\mathrm{l}(L/F)}$. Indeed, for an extension field with${\displaystyle [E:F]=n}$, an element${\displaystyle \alpha }$ is a primitive element if and only if${\displaystyle \alpha }$ hasn distinct conjugates${\displaystyle {\sigma }_1(\alpha ),\dots ,{\sigma }_n(\alpha )}$ in some splitting field${\displaystyle L\supseteq E}$.

If one adjoins to therational 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 aslinear combinations of 1,${\displaystyle \sqrt{2}}$,${\displaystyle \sqrt{3}}$,${\displaystyle \sqrt{6}}$ withinteger coefficients. One can solve thissystem of linear equations for${\displaystyle \sqrt{2}}$ and${\displaystyle \sqrt{3}}$ over${\displaystyle \mathbb{Q}(\alpha )}$, to obtain${\displaystyle \sqrt{2}=\frac{1}{2}({\alpha }^3-9\alpha )}$ and${\displaystyle \sqrt{3}=-\frac{1}{2}({\alpha }^3-11\alpha )}$. This shows thatα is indeed a primitive element:

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

One may also use the following more general argument.^[1] The field${\displaystyle E=\mathbb{Q}(\sqrt{2},\sqrt{3})}$ clearly has four field automorphisms${\displaystyle {\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_4:E\to E}$ defined by${\displaystyle {\sigma }_i(\sqrt{2})=\pm \sqrt{2}}$ and${\displaystyle {\sigma }_i(\sqrt{3})=\pm \sqrt{3}}$ for each choice of signs. The minimal polynomial${\displaystyle f(X)\in \mathbb{Q}[X]}$ of${\displaystyle \alpha =\sqrt{2}+\sqrt{3}}$ must have${\displaystyle f({\sigma }_i(\alpha ))={\sigma }_i(f(\alpha ))=0}$, so${\displaystyle f(X)}$ must have at least four distinct roots${\displaystyle {\sigma }_i(\alpha )=\pm \sqrt{2}\pm \sqrt{3}}$. Thus${\displaystyle f(X)}$ has degree at least four, and${\displaystyle [\mathbb{Q}(\alpha ):\mathbb{Q}]\ge 4}$, but this is the degree of the entire field,${\displaystyle [E:\mathbb{Q}]=4}$, so${\displaystyle E=\mathbb{Q}(\alpha )}$.

The primitive element theorem states:

Everyseparable field extension of finite degree is simple.

This theorem applies toalgebraic number fields, i.e. finite extensions of the rational numbersQ, sinceQ hascharacteristic 0 and therefore every finite extension overQ is separable.

Using thefundamental theorem of Galois theory, the former theorem immediately follows fromSteinitz's theorem.

For a non-separable extension${\displaystyle E/F}$ ofcharacteristic p, there is nevertheless a primitive element provided the degree [E : F] isp: indeed, there can be no non-trivial intermediate subfields since their degrees would be factors of the primep.

When [E : F] =p², there may not be a primitive element (in which case there are infinitely many intermediate fields bySteinitz's theorem). The simplest example is${\displaystyle E={\mathbb{F}}_p(T,U)}$, the field of rational functions in two indeterminatesT andU over thefinite field withp elements, and${\displaystyle F={\mathbb{F}}_p(T^p,U^p)}$. In fact, for any${\displaystyle \alpha =g(T,U)}$ in${\displaystyle E\setminus F}$, theFrobenius endomorphism shows that the element${\displaystyle {\alpha }^p}$ lies inF , soα is a root of${\displaystyle f(X)=X^p-{\alpha }^p\in F[X]}$, andα cannot be a primitive element (of degreep² overF), but insteadF(α) is a non-trivial intermediate field.

Suppose first that${\displaystyle F}$ is infinite. By induction, it suffices to prove that any finite extension${\displaystyle E=F(\beta ,\gamma )}$ is simple. For${\displaystyle c\in F}$, suppose${\displaystyle \alpha =\beta +c\gamma }$ fails to be a primitive element,${\displaystyle F(\alpha )⊊F(\beta ,\gamma )}$. Then${\displaystyle \gamma \notin F(\alpha )}$, since otherwise${\displaystyle \beta =\alpha -c\gamma \in F(\alpha )=F(\beta ,\gamma )}$. Consider the minimal polynomials of${\displaystyle \beta ,\gamma }$ over${\displaystyle F(\alpha )}$, respectively${\displaystyle f(X),g(X)\in F(\alpha )[X]}$, and take a splitting field${\displaystyle L}$ containing all roots${\displaystyle \beta ,{\beta }^{\prime },\dots }$ of${\displaystyle f(X)}$ and${\displaystyle \gamma ,{\gamma }^{\prime },\dots }$ of${\displaystyle g(X)}$. Since${\displaystyle \gamma \notin F(\alpha )}$, there is another root${\displaystyle {\gamma }^{\prime }\ne \gamma }$, and a field automorphism${\displaystyle \sigma :L\to L}$ which fixes${\displaystyle F(\alpha )}$ and takes${\displaystyle \sigma (\gamma )={\gamma }^{\prime }}$. We then have${\displaystyle \sigma (\alpha )=\alpha }$, and:

${\displaystyle \beta +c\gamma =\sigma (\beta +c\gamma )=\sigma (\beta )+c\sigma (\gamma )}$, and therefore${\displaystyle c=\frac{\sigma (\beta )-\beta }{\gamma -\sigma (\gamma )}}$.

Since there are only finitely many possibilities for${\displaystyle \sigma (\beta )={\beta }^{\prime }}$ and${\displaystyle \sigma (\gamma )={\gamma }^{\prime }}$, only finitely many${\displaystyle c\in F}$ fail to give a primitive element${\displaystyle \alpha =\beta +c\gamma }$. All other values give${\displaystyle F(\alpha )=F(\beta ,\gamma )}$.

For the case where${\displaystyle F}$ is finite, we simply take${\displaystyle \alpha }$ to be aprimitive root of the finite extension field${\displaystyle E}$.

In his First Memoir of 1831, published in 1846,^[2]Évariste Galois sketched a proof of the classical primitive element theorem in the case of asplitting field of a polynomial over the rational numbers. The gaps in his sketch could easily be filled^[3] (as remarked by the refereePoisson) by exploiting a theorem^[4][5] ofLagrange from 1771, which Galois certainly knew. It is likely that Lagrange had already been aware of the primitive element theorem for splitting fields.^[5] Galois then used this theorem heavily in his development of theGalois group. Since then it has been used in the development ofGalois theory and thefundamental theorem of Galois theory.

The primitive element theorem was proved in its modern form byErnst Steinitz, in an influential article onfield theory in 1910, which also containsSteinitz's theorem;^[6] Steinitz called the "classical" resultTheorem of the primitive elements and his modern versionTheorem of the intermediate fields.

Emil Artin reformulated Galois theory in the 1930s without relying on primitive elements.^[7][8]

• 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)

• Field (mathematics)
• Theorems in abstract algebra

• CS1 German-language sources (de)
• Articles with short description
• Short description matches Wikidata