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 $k[x_1,\dots,x_r]$.
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Transcendental extension. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Transcendental_extension&oldid=36929