Graphics Reference
In-Depth Information
Definition.
Let k be a subfield of a field K. An element a ΠK is said to be
transcen-
dental over k
if k(a) ª k(X). It is said to be
algebraic over k
if it is the root of a poly-
nomial f(X) Πk[X]. If it is the root of an irreducible polynomial of degree n, then it
is said to be
algebraic over k of degree n
.
This definition of transcendental and algebraic agrees with the definition given for
rings in Section B.7.
B.11.1. Theorem.
Let k be a subfield of a field K. If an element a ΠK is algebraic
over k, then a is the root of a unique irreducible polynomial with leading coefficient
equal to 1 called the
minimum polynomial
of a over k.
Proof.
See [Jaco64].
Definition.
An extension K of a field k is called an
algebraic
extension of k if every
element of K is algebraic over k. If K is not an algebraic extension of k, then it is called
a
transcendental
extension.
Definition.
An extension K of a field k is called a
finite extension of k of degree n
if K
is a finite dimensional vector space over k of dimension n. Otherwise, K is called an
infinite extension
of k of degree •. The degree of the extension is denoted by [K : k].
B.11.2. Theorem.
If K is a finite extension of a field k, then K is an algebraic exten-
sion of k.
Proof.
See [Mill58].
B.11.3. Theorem.
If K is a finite extension of a field k, then K = k(q
1
,q
2
,...,q
n
),
where the q
i
are algebraic over k.
Proof.
See [Mill58].
B.11.4. Theorem.
(Theorem of the Primitive Element) If k is a field of characteris-
tic 0 and if q
1
, q
2
,..., and q
n
belong to some extension field of k and are algebraic
over k, then
(
)
=
()
k
qq
,
,...,
q
k
q
12
n
for some element q (which is algebraic over k). In other words, every finite extension
of a field of characteristic 0 is simple.
Proof.
See [Mill58].
B.11.5. Theorem.
Let q be a transcendental element over a field k and let K be a
field such that k à K Õ k(q) with k π K. Then there exists an element s in K which is
transcendental over k and K = k(s).
Proof.
See [Walk50].
