primitive element theorem

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.

Terminology

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

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

:$\gamma =a_0+a_1+\cdots+a_^,$

for unique coefficients $a_0,a_1,\ldots,a_\in F$. 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 f(X)\in F of minimal degree with ''α'' as a root (a linear dependency of $\$).

If ''L'' is a splitting field of $f(X)$ containing its ''n'' distinct roots $\alpha_1,\ldots,\alpha_n$, then there are ''n'' field embeddings $\sigma_i : F(\alpha)\hookrightarrow L$ defined by $\sigma_i(\alpha)=\alpha_i$ and $\sigma(a)=a$ for $a\in F$, and these extend to automorphisms of ''L'' in the Galois group, $\sigma_1,\ldots,\sigma_n\in \mathrm(L/F)$. Indeed, for an extension field with : Fn , an element $\alpha$ is a primitive element if and only if $\alpha$ has ''n'' distinct conjugates $\sigma_1(\alpha),\ldots,\sigma_n(\alpha)$ in some splitting field $L \supseteq E$.

Example

If one adjoins to the rational numbers $F = \mathbb$ the two irrational numbers $\sqrt$ and $\sqrt$ to get the extension field $E=\mathbb(\sqrt,\sqrt)$ of degree 4, one can show this extension is simple, meaning $E=\mathbb(\alpha)$ for a single $\alpha\in E$. Taking $\alpha = \sqrt + \sqrt$, the powers 1, ''α'', ''α''², ''α''³ can be expanded as linear combinations of 1, $\sqrt$, $\sqrt$, $\sqrt$ with integer coefficients. One can solve this system of linear equations for $\sqrt$ and $\sqrt$ over $\mathbb(\alpha)$, to obtain $\sqrt = \tfrac12(\alpha^3-9\alpha)$ and $\sqrt = -\tfrac12(\alpha^3-11\alpha)$. This shows that ''α'' is indeed a primitive element:
:$\mathbb(\sqrt 2, \sqrt 3)=\mathbb(\sqrt2 + \sqrt3).$
One may also use the following more general argument. The field $E=\Q(\sqrt 2,\sqrt 3)$ clearly has four field automorphisms $\sigma_1,\sigma_2,\sigma_3,\sigma_4: E\to E$ defined by $\sigma_i(\sqrt 2)=\pm\sqrt 2$ and $\sigma_i(\sqrt 3)=\pm\sqrt 3$ for each choice of signs. The minimal polynomial f(X)\in\Q of $\alpha=\sqrt 2+\sqrt 3$ must have $f(\sigma_i(\alpha)) = \sigma_i(f(\alpha)) = 0$, so $f(X)$ must have at least four distinct roots $\sigma_i(\alpha)=\pm\sqrt 2 \pm\sqrt 3$. Thus $f(X)$ has degree at least four, and Q(\alpha):\Qgeq 4 , but this is the degree of the entire field, :\Q4 , so $E = \Q(\alpha )$.

Theorem statement

The primitive element theorem states:
:Every separable field extension of finite degree is simple.

This theorem applies to algebraic number fields, i.e. finite extensions of the rational numbers Q, since Q has characteristic 0 and therefore every finite extension over Q is separable.

Using the fundamental theorem of Galois theory, the former theorem immediately follows from Steinitz's theorem.

Characteristic ''p''

For a non-separable extension $E/F$ of characteristic p, there is nevertheless a primitive element provided the degree 'E'' : ''F''is ''p:'' indeed, there can be no non-trivial intermediate subfields since their degrees would be factors of the prime ''p''.

When 'E'' : ''F''= ''p''², there may not be a primitive element (in which case there are infinitely many intermediate fields by Steinitz's theorem). The simplest example is $E=\mathbb_p(T,U)$, the field of rational functions in two indeterminates ''T'' and ''U'' over the finite field with ''p'' elements, and $F=\mathbb_p(T^p,U^p)$. In fact, for any $\alpha=g(T,U)$ in $E \setminus F$, the Frobenius endomorphism shows that the element $\alpha^p$ lies in ''F'' , so ''α'' is a root of f(X)=X^p-\alpha^p\in F /math>, and ''α'' cannot be a primitive element (of degree ''p''² over ''F''), but instead ''F''(''α'') is a non-trivial intermediate field.

Proof

Suppose first that $F$ is infinite. By induction, it suffices to prove that any finite extension $E=F(\beta,\gamma)$ is simple. For $c\in F$, suppose $\alpha = \beta+ c\gamma$ fails to be a primitive element, $F(\alpha)\subsetneq F(\beta,\gamma)$. Then $\gamma\notin F(\alpha)$, since otherwise $\beta = \alpha-c\gamma\in F(\alpha)=F(\beta,\gamma)$. Consider the minimal polynomials of $\beta,\gamma$ over $F(\alpha)$, respectively f(X), g(X) \in F(\alpha) /math>, and take a splitting field $L$ containing all roots $\beta,\beta',\ldots$ of $f(X)$ and $\gamma,\gamma',\ldots$ of $g(X)$. Since $\gamma\notin F(\alpha)$, there is another root $\gamma'\neq \gamma$, and a field automorphism $\sigma:L\to L$ which fixes $F(\alpha)$ and takes $\sigma(\gamma)=\gamma'$. We then have $\sigma(\alpha) =\alpha$, and:
:$\beta + c \gamma = \sigma(\beta + c \gamma) = \sigma(\beta) + c \, \sigma(\gamma)$, and therefore $c = \frac$.
Since there are only finitely many possibilities for $\sigma(\beta)=\beta'$ and $\sigma(\gamma)=\gamma'$, only finitely many $c\in F$ fail to give a primitive element $\alpha=\beta+c\gamma$. All other values give $F(\alpha)=F(\beta,\gamma)$.

For the case where $F$ is finite, we simply take $\alpha$ to be a primitive root of the finite extension field $E$.

History

In his First Memoir of 1831, published in 1846, Évariste Galois sketched a proof of the classical primitive element theorem in the case of a splitting field of a polynomial over the rational numbers. The gaps in his sketch could easily be filled (as remarked by the referee Poisson) by exploiting a theorem of Lagrange

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