# primitive element theorem

###### Theorem 1.

Let $F$ and $K$ be arbitrary fields, and let $K$ be an extension  of $F$ of finite degree. Then there exists an element $\alpha\in K$ such that $K=F(\alpha)$ if and only if there are finitely many fields $L$ with $F\subseteq L\subseteq K$.

Let $X$ be an indeterminate  . Then $\mathbb{Q}(X,i)$ is not generated over $\mathbb{Q}$ by a single element (and there are infinitely many intermediate fields $\mathbb{Q}(X,i)/L/\mathbb{Q}$). To see this, suppose it is generated by an element $\alpha$. Then clearly $\alpha$ must be transcendental, or it would generate an extension of finite degree. But if $\alpha$ is transcendental, we know it is isomorphic   to $\mathbb{Q}(X)$, and this field is not isomorphic to $\mathbb{Q}(X,i)$: for example, the polynomial    $Y^{2}+1$ has no roots in the first but it has two roots in the second. It is also clear that it is not sufficient for every element of $K$ to be algebraic over $F$: we know that the algebraic closure  of $\mathbb{Q}$ has infinite degree over $\mathbb{Q}$, but if $\alpha$ is algebraic over $\mathbb{Q}$ then $[\mathbb{Q}(\alpha):\mathbb{Q}]$ will be finite.

This theorem has the corollary:

###### Corollary 1.

Let $F$ be a field, and let $[F(\beta,\gamma):F]$ be finite and separable. Then there exists $\alpha\in F(\beta,\gamma)$ such that $F(\beta,\gamma)=F(\alpha)$. In fact, we can always take $\alpha$ to be an $F$-linear combination  (http://planetmath.org/LinearCombination) of $\beta$ and $\gamma$.

For more detail on this theorem and its proof see, for example, Field and Galois Theory  , by Patrick Morandi (Springer Graduate Texts in Mathematics 167, 1996).

Title primitive element theorem PrimitiveElementTheorem 2013-03-22 11:45:48 2013-03-22 11:45:48 alozano (2414) alozano (2414) 18 alozano (2414) Theorem msc 12F05 msc 65-01 SimpleFieldExtension PrimitiveElementOfBiquadraticField2 PrimitiveElementOfBiquadraticField