PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Owner confidence rating: Very high Entry average rating: No information on entry rating
[parent] field adjunction (Definition)

Let $ K$ be a field and $ E$ its extension field. If $ \alpha \in E$, then the smallest subfield of $ E$, that contains $ K$ and $ \alpha$, is denoted by $ K(\alpha)$. We say that $ K(\alpha)$ is obtained from the field $ K$ by adjoining the element $ \alpha$ to $ K$ via field adjunction.

Theorem 1   $ K(\alpha)$ is identical with the quotient field $ Q$ of $ K[\alpha]$.

Proof. (1) Because $ K[\alpha]$ is an integral domain (as a subring of the field $ E$), all possible quotients of the elements of $ K[\alpha]$ belong to $ E$. So we have

$\displaystyle K\cup\{\alpha\} \subseteq K[\alpha] \subseteq Q \subseteq E,$
and because $ K(\alpha)$ was the smallest, then $ K(\alpha) \subseteq Q.$

(2) $ K(\alpha)$ is a subring of $ E$ containing $ K$ and $ \alpha$, therefore also the whole ring $ K[\alpha]$, that is, $ K[\alpha] \subseteq K(\alpha)$. And because $ K(\alpha)$ is a field, it must contain all possible quotients of the elements of $ K[\alpha]$, i.e., $ Q \subseteq K(\alpha)$.

In addition to the adjunction of one single element, we can adjoin to $ K$ an arbitrary subset $ S$ of $ E$: the resulting field $ K(S)$ is the smallest of such subfields of $ E$, i.e. the intersection of such subfields of $ E$, that contain both $ K$ and $ S$. We say that $ K(S)$ is obtained from $ K$ by adjoining the set $ S$ to it. Naturally,

$\displaystyle K \subseteq K(S) \subseteq E.$
The field $ K(S)$ contains all elements of $ K$ and $ S$, and being a field, also all such elements that can be formed via addition, subtraction, multiplication and division from the elements of $ K$ and $ S$. But such elements constitute a field, which therefore must be equal with $ K(S)$. Accordingly, we have the
Theorem 2   $ K(S)$ constitutes of all rational expressions formed of the elements of the field $ K$ with the elements of the set $ S$.

Notes.
1. $ K(S)$ is the union of all fields $ K(T)$ where $ T$ is a finite subset of $ S$.
2. $ K(S_1 \cup S_2) = K(S_1)(S_2)$.
3. If, especially, $ S$ also is a subfield of $ E$, then one may denote $ K(S) = KS$.

Bibliography

1
B. L. VAN DER WAERDEN: Algebra. Erster Teil. Siebte Auflage der Modernen Algebra. Springer-Verlag; Berlin, Heidelberg, New York (1966).



"field adjunction" is owned by pahio.
(view preamble)

View style:

See Also: ground fields and rings, forcing, polynomial ring over a field, a condition of algebraic extension

Other names:  simple extension
Keywords:  adjunction

This object's parent.

Attachments:
simple field extension (Definition) by pahio
Log in to rate this entry.
(view current ratings)

Cross-references: union, rational expressions, division, multiplication, subtraction, addition, intersection, subset, adjunction, ring, quotients, subring, integral domain, quotient field, contains, subfield, extension field, field
There are 6 references to this entry.

This is version 12 of field adjunction, born on 2004-05-30, modified 2008-01-11.
Object id is 5875, canonical name is FieldAdjunction.
Accessed 2812 times total.

Classification:
AMS MSC12F99 (Field theory and polynomials :: Field extensions :: Miscellaneous)

Pending Errata and Addenda
None.
[ View all 2 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add derivation | add example | add (any)