7.3. Adjoining Algebraic Elements to a Field
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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 Kvector 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 L M are fields, then dimK.M / D dimK.L/ dimL.M /: Proof. Suppose that f1; : : : ; r g is a subset of L that is linearly independent over K, and that f1; : : : ; sg is a subset of M that is linearly independent over L. I claim that fi j W 1 i r; 1 j sg is linearly independent over K. In fact, if 0 D Pi;j kij i j D Pj .Pi kij i /j , with kij 2 K, then linear independence of fj g over L implies that
P i kij i D 0 for all j , and then linear independence of fi 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 f1; : : : ; r g is a basis of L over K, and that the set f1; : : : ; sg is a basis of M over L. The fact that fj g spans M over L and that fi g spans L over K implies that the set of products fi j g spans M over K (exercise). Hence, fi j g is a basis of M over K.
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Definition 7.3.2. A field extension K L is called finite if L is a finite– dimensional vector space over K.
Now, consider a field extension K 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 k1x C C knxn/ D k0 C k1˛ C C kn˛n:
(7.3.1)
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7.3. ADJOINING ALGEBRAIC ELEMENTS TO A FIELD
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Definition 7.3.3. An element ˛ in a field extension L of K is said to be algebraic over K if there is some polynomial p.x/ 2 KŒx such that p.˛/ D 0; equivalently, there are elements k0; k1; : : : kn 2 K such that k0 C k1˛ C C kn˛n D 0. Any element that is not algebraic is called transcendental. The field extension is called algebraic if every element of L is algebraic over K.
In particular, the set of complex numbers that are algebraic over Q are called algebraic numbers and those that are transcendental over Q are called transcendental numbers. It is not difficult to see that there are (only) countably many algebraic numbers, and that, therefore, there are uncountably many transcendental numbers. Here is the argument: The rational numbers are countable, so for each natural number n, there are only countably many distinct polynomials with rational coefficients with degree no more than n. Consequently, there are only countably many polynomials with rational coefficients altogether, and each of these has only finitely many roots in the complex numbers. Therefore, the set of algebraic numbers is a countable union of finite sets, so countable. On the other hand, the complex numbers are uncountable, so there are uncountably many transcendental numbers.
We show that any finite field extension is algebraic:
Proposition 7.3.4. If K L is a finite field extension, then (a) L is algebraic over K, and (b) there are finitely many (algebraic) elements a1; : : : ; an 2 L such that L D K.a1; : : : ; an/.
Proof. Consider any element ˛ 2 L. Since dimK.L/ is finite, the powers 1, ˛, ˛2, ˛3, . . . of ˛ cannot be linearly independent over K. Therefore, there exists a natural number N and there exist 0, 1, . . . N 2 K, not all zero, such that 0 C 1˛ C C n˛N D 0. That is, ˛ is algebraic over K. Since ˛ was an arbitrary element of L, this means that L is algebraic over K.
The second statement is proved by induction on dimK.L/. If dimK.L/ D 1, then K D L, and there is nothing to prove. So suppose that dimK.L/ > 1, and suppose that whenever K0 L0 are fields and dimK0.L0/