Theorem of the primitive element

The primitive element theorem is a mathematical theorem from algebra that specifies sufficient conditions for a body expansion to be a simple body expansion . If there are bodies, then the body expansion is called simply if it can be produced by the adjunction of a single element. Such, in general, not uniquely determined element with is primitive element called. The theorem of the primitive element was fully proven by Galois and can be found in a publication by Abel from 1829. ${\displaystyle K\subseteq L}$ ${\displaystyle a\in L}$ ${\displaystyle L=K(a)}$

sentence

There are two propositions called the primitive element proposition, the second proposition being a consequence of the first.

• A body extension is easy if it is of the form with an over algebraic element and over separable elements . ${\displaystyle L/K}$ ${\displaystyle L}$ ${\displaystyle L=K(a,c_{1},c_{2},\ldots ,c_{n})}$ ${\displaystyle K}$ ${\displaystyle a}$ ${\displaystyle K}$ ${\displaystyle c_{1},c_{2},\ldots ,c_{n}}$
• Every finite separable field extension is simple.

meaning

In particular, finite Galois extensions are of this form and therefore simple. If such an extension is, then an element of the Galois group , that is, an -automorphism of , is already uniquely determined by the value . With the help of a primitive element, the Galois group of a polynomial or a body extension can be determined. Hence the meaning of this sentence results in Galois theory. ${\displaystyle L=K(a)}$ ${\displaystyle K}$ ${\displaystyle \sigma }$ ${\displaystyle L}$ ${\displaystyle \sigma (a)}$

Examples

• ${\displaystyle \textstyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})}$is a body extension over . One possible primitive element is ${\displaystyle \textstyle \mathbb {Q} }$ ${\displaystyle \textstyle t\in \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})}$

${\displaystyle t={\sqrt {2}}+{\sqrt {3}}}$,

because with

${\displaystyle t^{2}=5+2{\sqrt {6}}}$, and ${\displaystyle \quad t^{3}=11{\sqrt {2}}+9{\sqrt {3}}\quad }$ ${\displaystyle \quad t^{4}=49+20{\sqrt {6}}}$

it follows that t is the zero of the polynomial and thus algebraically over . ${\displaystyle \textstyle x^{4}-10x^{2}+1}$ ${\displaystyle \textstyle \mathbb {Q} }$

You also get the equations:

${\displaystyle t^{3}-9t=2{\sqrt {2}}\quad }$ and . ${\displaystyle \quad t^{3}-11t=-2{\sqrt {3}}}$

This means that and can be replaced by polynomials with the variable t: ${\displaystyle \textstyle {\sqrt {2}}}$ ${\displaystyle \textstyle {\sqrt {3}}}$

${\displaystyle {\sqrt {2}}={\tfrac {1}{2}}(t^{3}-9t)}$and . ${\displaystyle {\sqrt {3}}=-{\tfrac {1}{2}}(t^{3}-11t)}$

So is

${\displaystyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})=\mathbb {Q} ({\sqrt {2}}+{\sqrt {3}})}$

and {1, t, t ² , t ³ } be a basis of a vector space over . Another possible basis is { }, i.e. H. ${\displaystyle \textstyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})}$ ${\displaystyle \textstyle \mathbb {Q} }$ ${\displaystyle \textstyle 1,{\sqrt {2}},{\sqrt {3}},{\sqrt {6}}}$

${\displaystyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})=\{a+b{\sqrt {2}}+c{\sqrt {3}}+d{\sqrt {6}}\mid a,b,c,d\in \mathbb {Q} \}}$ .

So it is an algebraic field expansion of degree four.

• The polynomial has the zeros and therefore has a decay field . As shown above, is a primitive element and the four zeros can thus be represented as polynomials p ₁ , p ₂ , p ₃ , p ₄ with the variable t ₁ : ${\displaystyle \textstyle x^{4}-5x^{2}+6=(x^{2}-2)(x^{2}-3)}$ ${\displaystyle \textstyle x_{1}={\sqrt {2}},\quad x_{2}=-{\sqrt {2}},\quad x_{3}={\sqrt {3}},\quad x_{4}=-{\sqrt {3}}}$ ${\displaystyle \textstyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})}$ ${\displaystyle \textstyle t_{1}={\sqrt {2}}+{\sqrt {3}}}$

${\displaystyle x_{1}=p_{1}(t_{1})={\tfrac {1}{2}}(t_{1}^{3}-9t_{1})}$,

${\displaystyle x_{2}=p_{2}(t_{1})=-{\tfrac {1}{2}}(t_{1}^{3}-9t_{1})}$,

${\displaystyle x_{3}=p_{3}(t_{1})=-{\tfrac {1}{2}}(t_{1}^{3}-11t_{1})}$,

${\displaystyle x_{4}=p_{4}(t_{1})={\tfrac {1}{2}}(t_{1}^{3}-11t_{1})}$,

The primitive element t ₁ is - as calculated above - the zero of the over irreducible polynomial . The other zeros of this polynomial are obtained by pulling the roots twice - together with the relationship : ${\displaystyle \textstyle \mathbb {Q} }$ ${\displaystyle \textstyle x^{4}-10x^{2}+1=\left(x^{2}-5\right)^{2}-24}$ ${\displaystyle 5\pm 2{\sqrt {6}}=({\sqrt {2}}\pm {\sqrt {3}})^{2}}$

${\displaystyle t_{2}=-{\sqrt {2}}+{\sqrt {3}},\quad t_{3}={\sqrt {2}}-{\sqrt {3}},\quad t_{4}=-{\sqrt {2}}-{\sqrt {3}}}$.

If the variable t ₁ in the polynomials p ₁ , ... p _{4 is} replaced by t ₂ , t ₃ or t ₄ , the zeros x ₁ , x ₂ , x ₃ , x _{4 of} the output polynomial result , however, in a different order. These permutations of the zeros each correspond to an operation of an element of the Galois group on these zeros.

Inserting t ₁ provides the identity, the other relationships result from recalculation:

${\displaystyle p_{1}(t_{1})=x_{1},\quad p_{2}(t_{1})=x_{2},\quad p_{3}(t_{1})=x_{3},\quad p_{4}(t_{1})=x_{4},\quad \Longrightarrow \quad \sigma _{1}:\left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{1},x_{2},x_{3},x_{4}\right)}$,

${\displaystyle p_{1}(t_{2})=x_{2},\quad p_{2}(t_{2})=x_{1},\quad p_{3}(t_{2})=x_{3},\quad p_{4}(t_{2})=x_{4},\quad \Longrightarrow \quad \sigma _{2}:\left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{2},x_{1},x_{3},x_{4}\right)}$,

${\displaystyle p_{1}(t_{3})=x_{1},\quad p_{2}(t_{3})=x_{2},\quad p_{3}(t_{3})=x_{4},\quad p_{4}(t_{3})=x_{3},\quad \Longrightarrow \quad \sigma _{3}:\left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{1},x_{2},x_{4},x_{3}\right)}$,

${\displaystyle p_{1}(t_{4})=x_{2},\quad p_{2}(t_{4})=x_{1},\quad p_{3}(t_{4})=x_{4},\quad p_{4}(t_{4})=x_{3},\quad \Longrightarrow \quad \sigma _{4}:\left(x_{1},x_{2},x_{3},x_{4}\right)\mapsto \left(x_{2},x_{1},x_{4},x_{3}\right)}$.

{ } is the Galois group as a permutation group of zeros, as a group of body automorphisms it results as follows: ${\displaystyle \textstyle \sigma _{1},\sigma _{2},\sigma _{3},\sigma _{4}}$

Under are and are swapped, the same applies to for and . The sign changes for both roots below . The elements of the Galois group as body automorphisms are thus: ${\displaystyle \textstyle \sigma _{2}}$ ${\displaystyle \textstyle {\sqrt {2}}}$ ${\displaystyle -{\sqrt {2}}}$ ${\displaystyle \textstyle \sigma _{3}}$ ${\displaystyle \textstyle {\sqrt {3}}}$ ${\displaystyle \textstyle -{\sqrt {3}}}$ ${\displaystyle \textstyle \sigma _{4}}$

${\displaystyle \sigma _{1}':a+b{\sqrt {2}}+c{\sqrt {3}}+d{\sqrt {6}}\mapsto a+b{\sqrt {2}}+c{\sqrt {3}}+d{\sqrt {6}}}$,

${\displaystyle \sigma _{2}':a+b{\sqrt {2}}+c{\sqrt {3}}+d{\sqrt {6}}\mapsto a-b{\sqrt {2}}+c{\sqrt {3}}-d{\sqrt {6}}}$,

${\displaystyle \sigma _{3}':a+b{\sqrt {2}}+c{\sqrt {3}}+d{\sqrt {6}}\mapsto a+b{\sqrt {2}}-c{\sqrt {3}}-d{\sqrt {6}}}$,

${\displaystyle \sigma _{4}':a+b{\sqrt {2}}+c{\sqrt {3}}+d{\sqrt {6}}\mapsto a-b{\sqrt {2}}-c{\sqrt {3}}+d{\sqrt {6}}}$.

You can see that the body remains fixed element by element next to the base body . At and are the fixed bodies and . ${\displaystyle \textstyle \sigma _{2}'}$ ${\displaystyle \textstyle \mathbb {Q} }$ ${\displaystyle \textstyle \mathbb {Q} ({\sqrt {3}})}$ ${\displaystyle \textstyle \sigma _{3}'}$ ${\displaystyle \textstyle \sigma _{4}'}$ ${\displaystyle \textstyle \mathbb {Q} ({\sqrt {2}})}$ ${\displaystyle \textstyle \mathbb {Q} ({\sqrt {6}})}$

Because the initial polynomial is not irreducible over , the Galois group does not operate transitively on the set of zeros of this polynomial: for example, there is no element of the Galois group that maps the zero to the zero . ${\displaystyle \textstyle x^{4}-5x^{2}+6=(x^{2}-2)(x^{2}-3)}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle {\sqrt {2}}}$ ${\displaystyle {\sqrt {3}}}$

• The algebraic conjugates of the primitive element , i.e. the zeros ${\displaystyle t_{1}={\sqrt {2}}+{\sqrt {3}}}$

${\displaystyle t_{2}=-{\sqrt {2}}+{\sqrt {3}}}$, and , ${\displaystyle t_{3}={\sqrt {2}}-{\sqrt {3}}}$ ${\displaystyle t_{4}=-{\sqrt {2}}-{\sqrt {3}}}$

are also primitive elements; H. the following applies:

${\displaystyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})=\mathbb {Q} ({\sqrt {2}}+{\sqrt {3}})=\mathbb {Q} (-{\sqrt {2}}+{\sqrt {3}})=\mathbb {Q} ({\sqrt {2}}-{\sqrt {3}})=\mathbb {Q} (-{\sqrt {2}}-{\sqrt {3}})}$.

Web links

Wikiversity: A Proof of the Theorem  - Course Materials

Individual evidence

1. ↑ Niels Henrik Abel: Mémoire sur une classe particulière d'equations résolubles algébriquement , J. reine angew. Math. Volume 4 (1829), pages 131–156
2. ↑ Helmut Koch: Introduction to Classical Mathematics I , Springer-Verlag, ISBN 3-540-16665-3 , chap. 7.5: The theorem of the primitive element
3. ^ Christian Karpfinger, Kurt Meyberg: Algebra. Groups - rings - bodies. Spektrum Akademischer Verlag, Heidelberg 2009, ISBN 978-3-8274-2018-3 , pp. 259-260
4. ↑ Kurt Meyberg, Algebra II , Carl Hanser Verlag (1976), ISBN 3-446-12172-2 , sentence 6.9.17
5. ↑ Kurt Meyberg, Algebra II , Carl Hanser Verlag (1976), ISBN 3-446-12172-2 , Chapter 7.2: Determination of some Galois groups
6. ^ Nieper-Wißkirchen: Galois theory. University of Augsburg, p. 126, Proposition 4.8. ( PDF )
7. ^ Nieper-Wißkirchen: Galois theory. University of Augsburg 2013, p. 119, Proposition 4.4. ( PDF )