Base of transcendence

Transcendence base is an algebraic term from the theory of field extensions , which can be seen in analogy to the concept of the vector space basis of linear algebra . The power of such a transcendent base, the so-called degree of transcendence , represents a measure of the size of a transcendent body expansion.

Concept formation

Let it be a body extension, that is, it is a part of the body . A -elementige amount is algebraically independent over if it except the zero polynomial no polynomial with there. Any subset is said to be algebraically independent over if every finite subset of es is. ${\displaystyle L/K}$ ${\displaystyle K}$ ${\displaystyle L}$ ${\displaystyle n}$ ${\displaystyle \{a_{1},\ldots ,a_{n}\}\subset L}$ ${\displaystyle K}$ ${\displaystyle f\in K[t_{1},\ldots ,t_{n}]}$ ${\displaystyle f(a_{1},\ldots ,a_{n})=0}$ ${\displaystyle A\subset L}$ ${\displaystyle K}$ ${\displaystyle A}$

A maximum algebraically independent set in , which cannot be expanded by any further element to form an algebraically independent set, is called a transcendence basis of the field expansion . ${\displaystyle L}$ ${\displaystyle K}$ ${\displaystyle L/K}$

Note the analogy to linear algebra, in which a vector space basis can be characterized as a maximally linearly independent set.

Existence and properties of bases of transcendence

Just as the existence of a Hamel basis is proved in linear algebra , the existence of a transcendence basis is obtained by showing that every union of increasing sets of algebraically independent sets is again algebraically independent and then applying the lemma of Zorn .

There are other ways to characterize bases of transcendence. For example, the following statements are equivalent for a field extension and an algebraically independent set : ${\displaystyle L/K}$ ${\displaystyle B\subset L}$

• ${\displaystyle B}$is a transcendence base of . ${\displaystyle L/K}$
• ${\displaystyle L/K(B)}$is algebraic, where the smallest field in is that contains and (see body adjunct ). ${\displaystyle K(B)}$ ${\displaystyle L}$ ${\displaystyle K}$ ${\displaystyle B}$

Degree of transcendence

In analogy to Steinitz's exchange lemma of linear algebra, one shows that two bases of transcendence of a body extension are equally powerful. Therefore, the mightiness of a transcendent base is an invariant of the body expansion , which is called its degree of transcendence and is designated with . The spelling is based on the English term transcendence degree . From it follows that is infinite, because the integer powers of a transcendent element are linearly independent over , whereby a body extension by a transcendent element already has infinite degrees; the degree of transcendence does not correspond to the degree of a body expansion. ${\displaystyle L/K}$ ${\displaystyle \mathrm {Trg} (L:K)}$ ${\displaystyle \mathrm {trdeg} (L:K)}$ ${\displaystyle \mathrm {trdeg} (L:K)>0}$ ${\displaystyle \mathrm {deg} (L:K)}$ ${\displaystyle t}$ ${\displaystyle K}$ ${\displaystyle K(t):K}$

Furthermore one has

• The following applies to bodies . ${\displaystyle K\subset M\subset L}$ ${\displaystyle \mathrm {Trg} (L:K)\,=\,\mathrm {Trg} (L:M)+\mathrm {Trg} (M:K)}$

Purely transcendent body extensions

A body expansion is called purely transcendent if there is a transcendent base with . It follows that every element from transcendent is over . Every body expansion can be split into an algebraic and a purely transcendent body expansion, as the following sentence shows: ${\displaystyle L/K}$ ${\displaystyle A}$ ${\displaystyle L=K(A)}$ ${\displaystyle L\setminus K}$ ${\displaystyle K}$

If there is a body extension, there is an intermediate body, so the following applies ${\displaystyle L/K}$ ${\displaystyle M}$

• ${\displaystyle M/K}$ is purely transcendent.
• ${\displaystyle L/M}$ is algebraic.

To prove this we take for transcendence degree over . ${\displaystyle M=K(A)}$ ${\displaystyle A\subset L}$ ${\displaystyle K}$

Examples

• A field extension is algebraic if and only if the empty set is a basis of transcendence. Again, this is equivalent to having . ${\displaystyle L/K}$ ${\displaystyle \mathrm {Trg} (L:K)=0}$
• If the body of the rational functions is over , then the body expansion has the transcendence basis and it is therefore valid ${\displaystyle K(t)}$ ${\displaystyle K}$ ${\displaystyle K(t)/K}$ ${\displaystyle \{t\}}$ ${\displaystyle \mathrm {Trg} (K(t):K)=1}$
• If the body of the rational functions is indefinite , then it holds . This results from the above formula for calculating the degree of transcendence with the help of intermediate bodies from the last example. ${\displaystyle K(t_{1},\ldots ,t_{n})}$ ${\displaystyle n}$ ${\displaystyle K}$ ${\displaystyle \mathrm {Trg} (K(t_{1},\ldots ,t_{n}):K)=n}$
• For reasons of power, the following applies (read "beth one", see Beth function ). ${\displaystyle \mathrm {Trg} (\mathbb {C} :\mathbb {Q} )=\beth _{1}}$
• The body extensions and are purely transcendent, with the non-trivial fact of the transcendence of Euler's number above being used for the latter . ${\displaystyle K(t)/K}$ ${\displaystyle \mathbb {Q} (e)/\mathbb {Q} }$ ${\displaystyle e}$ ${\displaystyle \mathbb {Q} }$
• The body expansion is non-algebraic, but also not purely transcendent, since algebraic is over . ${\displaystyle \mathbb {Q} ({\sqrt {2}},e)/\mathbb {Q} }$ ${\displaystyle {\sqrt {2}}}$ ${\displaystyle \mathbb {Q} }$

Individual evidence

1. ↑ Gerd Fischer , Reinhard Sacher: Introduction to Algebra. 2nd, revised edition. Teubner, Stuttgart 1978, ISBN 3-519-12053-4 , Appendix 4.
2. ^ Kurt Meyberg: Algebra. Volume 2. Carl Hanser, Munich et al. 1976, ISBN 3-446-12172-2 , sentence 6.10.10.
3. ^ Kurt Meyberg: Algebra. Volume 2. Carl Hanser, Munich et al. 1976, ISBN 3-446-12172-2 , sentence 6.10.11.
4. ↑ Gerd Fischer, Reinhard Sacher: Introduction to Algebra. 2nd, revised edition. Teubner, Stuttgart 1978, ISBN 3-519-12053-4 , Appendix 4, sentence 2.