primitive element theorem

Let $E/F$ be a field extension. An element $\backslash alpha\backslash in\; E$ is a primitive element for $E/F$ if $E=F(\backslash alpha),$ i.e. if every element of $E$ can be written as a rational function in $\backslash alpha$ 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 $\backslash alpha$ and is of finite degree $n\; =\; [E:F]$, then every element $\backslash gamma\backslash in\; E$ can be written in the form

:$\backslash gamma\; =a\_0+a\_1\{\backslash alpha\}+\backslash cdots+a\_\{n-1\}\{\backslash alpha\}^\{n-1\},$

for unique coefficients $a\_0,a\_1,\backslash ldots,a\_\{n-1\}\backslash in\; F$. That is, the set

:$\backslash \{1,\backslash alpha,\backslash ldots,\{\backslash alpha\}^\{n-1\}\backslash \}$

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 $f(X)\backslash in\; F[X]$ of minimal degree with α as a root (a linear dependency of $\backslash \{1,\backslash alpha,\backslash ldots,\backslash alpha^\{n-1\},\backslash alpha^n\backslash \}$).

If L is a splitting field of $f(X)$ containing its n distinct roots $\backslash alpha\_1,\backslash ldots,\backslash alpha\_n$, then there are n field embeddings $\backslash sigma\_i\; :\; F(\backslash alpha)\backslash hookrightarrow\; L$ defined by $\backslash sigma\_i(\backslash alpha)=\backslash alpha\_i$ and $\backslash sigma(a)=a$ for $a\backslash in\; F$, and these extend to automorphisms of L in the Galois group, $\backslash sigma\_1,\backslash ldots,\backslash sigma\_n\backslash in\; \backslash mathrm\{Gal\}(L/F)$. Indeed, for an extension field with $[E:\; F]=n$, an element $\backslash alpha$ is a primitive element if and only if $\backslash alpha$ has n distinct conjugates $\backslash sigma\_1(\backslash alpha),\backslash ldots,\backslash sigma\_n(\backslash alpha)$ in some splitting field $L\; \backslash supseteq\; E$.

If one adjoins to the rational numbers $F\; =\; \backslash mathbb\{Q\}$ the two irrational numbers $\backslash sqrt\{2\}$ and $\backslash sqrt\{3\}$ to get the extension field $E=\backslash mathbb\{Q\}(\backslash sqrt\{2\},\backslash sqrt\{3\})$ of degree 4, one can show this extension is simple, meaning $E=\backslash mathbb\{Q\}(\backslash alpha)$ for a single $\backslash alpha\backslash in\; E$. Taking $\backslash alpha\; =\; \backslash sqrt\{2\}\; +\; \backslash sqrt\{3\}$, the powers 1, α, α², α³ can be expanded as linear combinations of 1, $\backslash sqrt\{2\}$, $\backslash sqrt\{3\}$, $\backslash sqrt\{6\}$ with integer coefficients. One can solve this system of linear equations for $\backslash sqrt\{2\}$ and $\backslash sqrt\{3\}$ over $\backslash mathbb\{Q\}(\backslash alpha)$, to obtain $\backslash sqrt\{2\}\; =\; \backslash tfrac12(\backslash alpha^3-9\backslash alpha)$ and $\backslash sqrt\{3\}\; =\; -\backslash tfrac12(\backslash alpha^3-11\backslash alpha)$. This shows that α is indeed a primitive element:

:$\backslash mathbb\{Q\}(\backslash sqrt\; 2,\; \backslash sqrt\; 3)=\backslash mathbb\{Q\}(\backslash sqrt2\; +\; \backslash sqrt3).$

One may also use the following more general argument.{{Cite book |last=Lang |first=Serge |url=http://link.springer.com/10.1007/978-1-4613-0041-0 |title=Algebra |date=2002 |publisher=Springer New York |isbn=978-1-4612-6551-1 |series=Graduate Texts in Mathematics |volume=211 |location=New York, NY |pages=243 |doi=10.1007/978-1-4613-0041-0}} The field $E=\backslash Q(\backslash sqrt\; 2,\backslash sqrt\; 3)$ clearly has four field automorphisms $\backslash sigma\_1,\backslash sigma\_2,\backslash sigma\_3,\backslash sigma\_4:\; E\backslash to\; E$ defined by $\backslash sigma\_i(\backslash sqrt\; 2)=\backslash pm\backslash sqrt\; 2$ and $\backslash sigma\_i(\backslash sqrt\; 3)=\backslash pm\backslash sqrt\; 3$ for each choice of signs. The minimal polynomial $f(X)\backslash in\backslash Q[X]$ of $\backslash alpha=\backslash sqrt\; 2+\backslash sqrt\; 3$ must have $f(\backslash sigma\_i(\backslash alpha))\; =\; \backslash sigma\_i(f(\backslash alpha))\; =\; 0$, so $f(X)$ must have at least four distinct roots $\backslash sigma\_i(\backslash alpha)=\backslash pm\backslash sqrt\; 2\; \backslash pm\backslash sqrt\; 3$. Thus $f(X)$ has degree at least four, and $[\backslash Q(\backslash alpha):\backslash Q]\backslash geq\; 4$, but this is the degree of the entire field, $[E:\backslash Q]=4$, so $E\; =\; \backslash Q(\backslash alpha\; )$.

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.

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] = p², there may not be a primitive element (in which case there are infinitely many intermediate fields by Steinitz's theorem). The simplest example is $E=\backslash mathbb\{F\}\_p(T,U)$, the field of rational functions in two indeterminates T and U over the finite field with p elements, and $F=\backslash mathbb\{F\}\_p(T^p,U^p)$. In fact, for any $\backslash alpha=g(T,U)$ in $E\; \backslash setminus\; F$, the Frobenius endomorphism shows that the element $\backslash alpha^p$ lies in F , so α is a root of $f(X)=X^p-\backslash alpha^p\backslash in\; F[X]$, and α cannot be a primitive element (of degree p² over F), but instead F(α) is a non-trivial intermediate field.

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

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

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

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

In his First Memoir of 1831, published in 1846,{{Cite book|last=Neumann|first=Peter M.|url=https://www.worldcat.org/oclc/757486602|title=The mathematical writings of Évariste Galois|date=2011|publisher=European Mathematical Society|isbn=978-3-03719-104-0|location=Zürich|oclc=757486602}} 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{{Cite book|last=Tignol|first=Jean-Pierre | author-link=Jean-Pierre Tignol |url=https://www.worldscientific.com/worldscibooks/10.1142/9719|title=Galois' Theory of Algebraic Equations|date=February 2016|publisher=WORLD SCIENTIFIC|isbn=978-981-4704-69-4|edition=2|location=|pages=231|language=en|doi=10.1142/9719|oclc=1020698655}} (as remarked by the referee Poisson) by exploiting a theorem{{Cite book|last=Tignol|first=Jean-Pierre|url=https://www.worldscientific.com/worldscibooks/10.1142/9719|title=Galois' Theory of Algebraic Equations|date=February 2016|publisher=WORLD SCIENTIFIC|isbn=978-981-4704-69-4|edition=2|pages=135|language=en|doi=10.1142/9719|oclc=1020698655}}{{Cite book|last=Cox|first=David A.|url=https://www.worldcat.org/oclc/784952441|title=Galois theory|date=2012|publisher=John Wiley & Sons|isbn=978-1-118-21845-7|edition=2nd|location=Hoboken, NJ|pages=322|oclc=784952441}} of Lagrange from 1771, which Galois certainly knew. It is likely that Lagrange had already been aware of the primitive element theorem for splitting fields. Galois then used this theorem heavily in his development of the Galois group. Since then it has been used in the development of Galois theory and the fundamental theorem of Galois theory.

The primitive element theorem was proved in its modern form by Ernst Steinitz, in an influential article on field theory in 1910, which also contains Steinitz's theorem;{{Cite journal|last=Steinitz|first=Ernst|date=1910|title=Algebraische Theorie der Körper.|url=https://gdz.sub.uni-goettingen.de/id/PPN243919689_0137?tify=%7B%22view%22:%22info%22,%22pages%22:%5B171%5D%7D|journal=Journal für die reine und angewandte Mathematik|language=de|volume=1910|issue=137 |pages=167–309|doi=10.1515/crll.1910.137.167|s2cid=120807300 |issn=1435-5345|url-access=subscription}} Steinitz called the "classical" result Theorem of the primitive elements and his modern version Theorem of the intermediate fields.

Emil Artin reformulated Galois theory in the 1930s without relying on primitive elements.{{cite book|last=Kleiner|first=Israel|title=A History of Abstract Algebra|date=2007|publisher=Springer|isbn=978-0-8176-4685-1|pages=64|chapter=§4.1 Galois theory|chapter-url=https://books.google.com/books?id=udj-1UuaOiIC&pg=PA64}}{{Cite book|last=Artin|first=Emil|url=https://www.worldcat.org/oclc/38144376|title=Galois theory|date=1998|publisher=Dover Publications|others=Arthur N. Milgram|isbn=0-486-62342-4|edition=Republication of the 1944 revised edition of the 1942 first publication by The University Notre Dame Press|location=Mineola, N.Y.|oclc=38144376}}

{{Reflist}}

• [http://www.jmilne.org/math/CourseNotes/ft.html J. Milne's course notes on fields and Galois theory]
• [http://www.mathreference.com/fld-sep,pet.html The primitive element theorem at mathreference.com]
• [http://planetmath.org/ProofOfPrimitiveElementTheorem The primitive element theorem at planetmath.org]
• [http://www.math.cornell.edu/~kbrown/6310/primitive.pdf The primitive element theorem on Ken Brown's website (pdf file)]

Category:Field (mathematics)

Category:Theorems in abstract algebra