primitive element theorem PrimitiveElementTheorem 2001-10-15 20:25:57 2007-06-13 11:57:06 Theorem number theory % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem{theorem}{Theorem} \newtheorem{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} \begin{theorem} 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 $F\subseteq L\subseteq K$. \end{theorem} Note that this implies that every finite separable extension is not only finitely generated, it is generated by a single element. 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: \begin{cor} 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 $F$-\PMlinkname{linear combination}{LinearCombination} of $\beta$ and $\gamma$. \end{cor} To see this (in the case of characteristic $0$), we need only show that there are finitely many intermediate fields. But any intermediate field is contained in the splitting field of the minimal polynomials of $\beta$ and $\gamma$, which is Galois with finite Galois group. The explicit form of $\alpha$ comes from the proof of the theorem. For more detail on this theorem and its proof see, for example, \emph{Field and Galois Theory}, by Patrick Morandi (Springer Graduate Texts in Mathematics 167, 1996).