In field theory, the primitive element theorem states that every finite separable field extension is simple, i.e. generated by a single element. This theorem implies in particular that all algebraic number fields over the rational numbers, and all extensions in which both fields are finite, are simple.

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

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

for unique coefficients ${\displaystyle a_{0},a_{1},\ldots ,a_{n-1}\in F}$. That is, the set

is a basis for E as a vector space over F. The degree n is equal to the degree of the irreducible polynomial of α over F, the unique monic ${\displaystyle f(X)\in F[X]}$ of minimal degree with α as a root (a linear dependency of ${\displaystyle \{1,\alpha ,\ldots ,\alpha ^{n-1},\alpha ^{n}\}}$).

If L is a splitting field of ${\displaystyle f(X)}$ containing its n distinct roots ${\displaystyle \alpha _{1},\ldots ,\alpha _{n}}$, then there are n field 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 of L in the Galois group, ${\displaystyle \sigma _{1},\ldots ,\sigma _{n}\in \mathrm {Gal} (L/F)}$. Indeed, for an extension field with ${\displaystyle [E:F]=n}$, an element ${\displaystyle \alpha }$ is a primitive element if and only if ${\displaystyle \alpha }$ has n distinct conjugates ${\displaystyle \sigma _{1}(\alpha ),\ldots ,\sigma _{n}(\alpha )}$ in some splitting field ${\displaystyle L\supseteq E}$.

If one adjoins to the rational 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 as linear combinations of 1, ${\displaystyle {\sqrt {2}}}$, ${\displaystyle {\sqrt {3}}}$, ${\displaystyle {\sqrt {6}}}$ with integer coefficients. One can solve this system of linear equations for ${\displaystyle {\sqrt {2}}}$ and ${\displaystyle {\sqrt {3}}}$ over ${\displaystyle \mathbb {Q} (\alpha )}$, to obtain ${\displaystyle {\sqrt {2}}={\tfrac {1}{2}}(\alpha ^{3}-9\alpha )}$ and ${\displaystyle {\sqrt {3}}=-{\tfrac {1}{2}}(\alpha ^{3}-11\alpha )}$. This shows that α is indeed a primitive element:

One may also use the following more general argument. 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 ${\displaystyle [\mathbb {Q} (\alpha ):\mathbb {Q} ]\geq 4}$, but this is the degree of the entire field, ${\displaystyle [E:\mathbb {Q} ]=4}$, so ${\displaystyle E=\mathbb {Q} (\alpha )}$.