Namespaces
Variants
Actions

Extension of a field

From Encyclopedia of Mathematics
Revision as of 21:12, 23 December 2015 by Richard Pinch (talk | contribs) (Comments: distinguished classes)
Jump to: navigation, search

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


A field extension $K$ is a field containing a given field $k$ as a subfield. The notation $K/k$ means that $K$ is an extension of the field $k$. In this case, $K$ is sometimes called an overfield of the field $k$.

Let $K/k$ and $L/k$ be two extensions of a field $k$. An isomorphism of fields $\def\phi{\varphi}\phi:K\to L$ is called an isomorphism of extensions (or a $k$-isomorphism of fields) if $\phi$ is the identity on $k$. If an isomorphism of extensions exists, then the extensions are said to be isomorphic. If $K=L$, $\phi$ is called an automorphism of the extension $K/k$. The set of all automorphisms of an extension forms a group, $\textrm{Aut}(K/k)$. If $K/k$ is a Galois extension, this group is denoted by $\textrm{Gal}(K/k)$ and is called the Galois group of the field $K$ over $k$, or the Galois group of the extension $K/k$. An extension is called Abelian if its Galois group is Abelian.

An element $\def\a{\alpha}\a$ of the field $K$ is called algebraic over $k$ if it satisfies some algebraic equation with coefficients in $k$, and transcendental otherwise. For every algebraic element $\a$ there is a unique polynomial $f_\a(x)$ with coefficients in $k$ which is monic (with leading coefficient equal to 1), irreducible in the polynomial ring $k[x]$ and satisfying $f_\a(\a) = 0$; any polynomial over $k$ having $\a$ as a root is divisible by $f_\a(x)$. This polynomial is called the minimal polynomial of $\a$ over $k$. An extension $K/k$ is called algebraic if every element of $K$ is algebraic over $k$. An extension that is not algebraic is called transcendental. An extension is called normal if it is algebraic and if every irreducible polynomial in $k[x]$ having a root in $K$ factorizes into linear factors in $K[x]$. The subfield $k$ is said to be algebraically closed in $K$ if every element of $K$ that is algebraic over $k$ actually lies in $k$, that is, every element of $K\setminus k$ is transcendental over $k$. A field that is algebraically closed in all its extensions is called an algebraically closed field.

An extension $K/k$ is said to be finitely generated (or an extension of finite type) if there is a finite subset $S$ of $k$ such that $K$ coincides with the smallest subfield containing $S$ and $k$. In this case one says that $K$ is generated by $S$ over $k$. If $K$ is generated over $k$ by one element $\a$, then the extension is called simple or primitive and one writes $K=k(\a)$: the generator $a$ is termed a primitive element of the extension $K/k$. A simple algebraic extension $k(\a)$ is completely determined by the minimal polynomial $f_\a$ of $\a$. More precisely, if $\def\b{\beta}k(\b)$ is another simple algebraic extension and $f_\a = f_\b$, then there is an isomorphism of extensions $k(\a)\to k(\b)$ sending $\a$ to $\b$. Furthermore, for any irreducible polynomial $f\in k[x]$ there is a simple algebraic extension $k(\a)$ with minimal polynomial $f_\a = f$. It can be constructed as the quotient ring $k[x]/fk[x]$. On the other hand, for any simple transcendental extension $k(\a)$ there is an isomorphism of extensions $k(\a) \to k(x)$, where $k(x)$ is the field of rational functions in $x$ over $k$. Any extension of finite type can be obtained by performing a finite sequence of simple extensions.

An extension $K/k$ is called finite if $K$ is finite-dimensional as a vector space over $k$, and infinite otherwise. The dimension of this vector space is called the degree of $K/k$ and is denoted by $[K:k]$. Every finite extension is algebraic and every algebraic extension of finite type is finite. The degree of a simple algebraic extension coincides with the degree of the corresponding minimal polynomial. On the other hand, a simple transcendental extension is infinite.

Suppose one is given a sequence of extensions $K\subset L\subset M$. Then $M/K$ is algebraic if and only if both $L/K$ and $M/L$ are. Further, $M/K$ is finite if and only if $L/K$ and $M/K$ are, and then $$[M:K]=[M:L][L:K].$$ If $P/k$ and $Q/k$ are two algebraic extensions and $PQ$ is the compositum of the fields $P$ and $Q$ in a common overfield, then $PQ/k$ is also algebraic.

See also Separable extension; Transcendental extension.

References

[Bo] N. Bourbaki, "Eléments de mathématique. Algèbre", Masson (1981) pp. Chapt. 4–7 MR1994218 Zbl 1139.12001
[La] S. Lang, "Algebra", Addison-Wesley (1984) MR0783636 Zbl 0712.00001
[Wa] B.L. van der Waerden, "Algebra", 1–2, Springer (1967–1971) (Translated from German) MR0263582 MR0263583 Zbl 0724.12001 Zbl 0724.12002
[ZaSa] O. Zariski, P. Samuel, "Commutative algebra", 1, Springer (1975) MR0384768 Zbl 0313.13001

Comments

A distinguished class of extensions is a family $\mathfrak{E}$ with the properties (i) for $M / L / K$ we have $M/L,\,L/K \in \mathfrak{E} \Leftrightarrow M/K \in \mathfrak{E}$; (ii) $M / K,\,L/K \in \mathfrak{E} \Rightarrow ML/L \in \mathfrak{E}$. Examples of distinguished classes are: algebraic extensions; finite degree extensions; finitely generated extensions; separable extensions; purely inseparable extensions; solvable extensions.

References

[Ro] Steven Roman, Field Theory, Graduate Texts in Mathematics 158 (2nd edition) Springer (2007) ISBN 0-387-27678-5 Zbl 1172.12001
How to Cite This Entry:
Extension of a field. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Extension_of_a_field&oldid=36928
This article was adapted from an original article by O.A. Ivanova (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article