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Difference between revisions of "Transcendental extension"

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A transcendental extension of a field $k$ is
 
A transcendental extension of a field $k$ is
a field extension that is not algebraic (cf.
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a [[field extension]] that is not an [[algebraic extension]]. An extension $K/k$ is
[[Extension of a field|Extension of a field]]). An extension $K/k$ is
 
 
transcendental if and only if the field $K$ contains elements that are
 
transcendental if and only if the field $K$ contains elements that are
 
transcendental over $k$, that is, elements that are not roots of any
 
transcendental over $k$, that is, elements that are not roots of any
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and only if the extension $K/k$ is algebraic.
 
and only if the extension $K/k$ is algebraic.
  
====References====
 
{|
 
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|valign="top"|{{Ref|Bo}}||valign="top"| N. Bourbaki, "Algebra", ''Elements of mathematics'', '''1''', Springer (1988) pp. Chapt. 4–6 (Translated from French)
 
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|valign="top"|{{Ref|ZaSa}}||valign="top"| O. Zariski, P. Samuel, "Commutative algebra", '''1''', Springer (1975)
 
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====Comments====
 
 
The Noether normalization lemma says that if $A$ is
 
The Noether normalization lemma says that if $A$ is
 
an integral domain that is finitely generated as a ring over a field
 
an integral domain that is finitely generated as a ring over a field
 
$k$, then there exist $x_1,\dots,x_r \in A$ that are algebraically independent over $k$
 
$k$, then there exist $x_1,\dots,x_r \in A$ that are algebraically independent over $k$
 
such that $A$ is integral over $k[x_1,\dots,x_r]$.
 
such that $A$ is integral over $k[x_1,\dots,x_r]$.
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====References====
 
====References====
 
{|
 
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|valign="top"|{{Ref|Co}}||valign="top"| P.M. Cohn, "Algebra", '''1–2''', Wiley (1989) pp. Vol. 2, 350; Vol. 3, 168ff
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|valign="top"|{{Ref|Bo}}||valign="top"| N. Bourbaki, "Algebra", ''Elements of mathematics'', '''1''', Springer (1988) pp. Chapt. 4–7 (Translated from French) {{MR|1994218}} {{ZBL|1139.12001}} 
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|-
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|valign="top"|{{Ref|Co}}||valign="top"| P.M. Cohn, "Algebra", '''1–2''', Wiley (1989) pp. Vol. 2, 350; Vol. 3, 168ff {{MR|1006872}} {{MR|1002977}}  {{ZBL|0674.10023}} {{ZBL|0703.00002}}
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|valign="top"|{{Ref|ZaSa}}||valign="top"| O. Zariski, P. Samuel, "Commutative algebra", '''1''', Springer (1975) {{MR|0384768}} {{ZBL|0313.13001}} 
 
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Latest revision as of 19:41, 13 December 2015

2020 Mathematics Subject Classification: Primary: 12Fxx [MSN][ZBL]

A transcendental extension of a field $k$ is a field extension that is not an algebraic extension. An extension $K/k$ is transcendental if and only if the field $K$ contains elements that are transcendental over $k$, that is, elements that are not roots of any non-zero polynomial with coefficients in $k$.

The elements of a set $X\subset K$ are called algebraically independent over $k$ if for any finite set $x_1,\dots,x_m \in X$ and any non-zero polynomial $F(X_1,\dots,X_m)$ with coefficients in $k$, $$F(x_1,\dots,x_m)\ne 0.$$ The elements of $X$ are transcendental over $k$. If $X\subset K$ is a maximal set of algebraically independent elements over $k$, then $X$ is called a transcendence basis of $K$ over $k$. The cardinality of $X$ is called the transcendence degree of $K$ over $k$ and is an invariant of the extension $K/k$. For a tower of fields $L\supset K\supset k$, the transcendence degree of $L/k$ is equal to the sum of the transcendence degrees of $L/K$ and $K/k$.

If all elements of a set $X$ are algebraically independent over $k$, then the extension $k(X)$ is called purely transcendental. In this case the field $k(X)$ is isomorphic to the field of rational functions in the set of variables $X$ over $k$. Any field extension $L/k$ can be represented as a tower of extensions $L\supset K\supset k$, where $L/K$ is an algebraic and $K/k$ is a purely transcendental extension. If $K$ can be chosen so that $L/K$ is a separable extension, then the extension $K/k$ is called separably generated, and the transcendence basis of $K$ over $k$ is called a separating basis. If $L$ is separably generated over $k$, then $L$ is separable over $k$. In the case when the extension $L/k$ is finitely generated, the converse holds as well. By definition, an extension $K/k$ is separable if and only if any derivation (cf. Derivation in a ring) of $k$ extends to $K$. Such an extension is uniquely determined for any derivation if and only if the extension $K/k$ is algebraic.

The Noether normalization lemma says that if $A$ is an integral domain that is finitely generated as a ring over a field $k$, then there exist $x_1,\dots,x_r \in A$ that are algebraically independent over $k$ such that $A$ is integral over $k[x_1,\dots,x_r]$.


References

[Bo] N. Bourbaki, "Algebra", Elements of mathematics, 1, Springer (1988) pp. Chapt. 4–7 (Translated from French) MR1994218 Zbl 1139.12001
[Co] P.M. Cohn, "Algebra", 1–2, Wiley (1989) pp. Vol. 2, 350; Vol. 3, 168ff MR1006872 MR1002977 Zbl 0674.10023 Zbl 0703.00002
[ZaSa] O. Zariski, P. Samuel, "Commutative algebra", 1, Springer (1975) MR0384768 Zbl 0313.13001
How to Cite This Entry:
Transcendental extension. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Transcendental_extension&oldid=21155
This article was adapted from an original article by L.V. Kuz'min (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article