Unformatted text preview: 322 7. FIELD EXTENSIONS – FIRST LOOK 7.3. Adjoining Algebraic Elements to a Field
The ﬁelds in this section are general, not necessarily subﬁelds of the complex numbers.
A ﬁeld extension L of a ﬁeld K is, in particular, a vector space over
K . You are asked to check this in the Exercises. It will be helpful to review
the deﬁnition of a vector space in Section 3.3
Since a ﬁeld extension L of a ﬁeld K is a K -vector space, in particular,
L has a dimension over K , possibly inﬁnite. The dimension of L as a K vector space is denoted dimK .L/ (or, sometimes, ŒL W K.) Dimensions of
ﬁeld extensions have the following multiplicative property:
Proposition 7.3.1. If K Â L Â M are ﬁelds, then
dimK .M / D dimK .L/ dimL .M /: Proof. Suppose that f 1 ; : : : ; r g is a subset of L that is linearly independent over K , and that f 1 ; : : : ; s g is a subset of M that is linearly
independent over L. I claim that f i PW 1 Ä i Ä r; 1 ÄP Ä s g is linearly
independent over K . In fact, if 0 D i;j kij i j D j . i kij i / j ,
with kij 2 K , then linear independence of f j g over L implies that
i kij i D 0 for all j , and then linear independence of f i g over K
implies that kij D 0 for all i; j , which proves the claim.
In particular, if either dimK .L/ or dimL .M / is inﬁnite, then there are
arbitrarily large subsets of M that are linearly independent over K , so
dimK .M / is also inﬁnite.
Suppose now that dimK .L/ and dimL .M / are ﬁnite, that the set
f 1 ; : : : ; r g is a basis of L over K , and that the set f 1 ; : : : ; s g is a
basis of M over L. The fact that f j g spans M over L and that f i g
spans L over K implies that the set of products f i j g spans M over K
(exercise). Hence, f i j g is a basis of M over K .
I Deﬁnition 7.3.2. A ﬁeld extension K Â L is called ﬁnite if L is a ﬁnite–
dimensional vector space over K .
Now, consider a ﬁeld extension K Â L. According to Proposition
6.2.7, for any element ˛ 2 L there is a ring homomorphism (“evaluation”
at ˛ ) from KŒx into L given by '˛ .f .x// D f .˛/. That is,
'˛ .k0 C k1 x C C kn x n / D k0 C k1 ˛ C C kn ˛ n : (7.3.1) ...
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