View source for Transcendental extension

{{MSC|12Fxx}}
{{TEX|done}}

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|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|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|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====
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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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|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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