2.2 Splitting Fields and Algebraic Closure

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Let F1, F2 be fields, and assume that   : F1 ! F2 is a field homomorphism.

Define the homomorphism ˆ   : F1[x] ! F2[x] simply by applying   to the

coefficients of polynomials in F1[x]. We have the following two results.

Proposition 2.2.1 Let F1 be a field and let K1 = F1(_1), where _1 is

algebraic over F1, with minimal polynomial f1(x) 2 F1[x]. Suppose we have

  : F1

_=

−! F2,

ˆ   : F1[x] _= −! F2[x],

where ˆ   is defined as above. Let K2 = F2(_2) _ F2, where _2 is a root of

f2(x) = ˆ  (f1(x)). Then there exists an isomorphism

¯   : K1

_=

−! K2,

such that ¯  (_1) = _2, and ¯  |F1 =  .

Proposition 2.2.2 Let F1 be a field, let f1(x) 2 F1[x], and let K1 be a

splitting field over F1 for f1(x). Let

  : F1

_=

−! F2,

let f2(x) = ˆ  (f1(x)) 2 F2[x], and let K2 be a splitting field over F2 for f2(x).

Then there is a commutative diagram

K1

¯   - K2

F1

6

  - K2

6

where the vertical maps are inclusions, and where ¯   is an isomorphism.

Let F be a field and let F _ F[x]. By a splitting field for F we mean a

field extension K _ F such that every polynomial in F splits completely in

K, and K is minimal in this respect.

48 CHAPTER 2. FIELD AND GALOIS THEORY

Proposition 2.2.3 Let F1 be a field, let F1 _ F1[x], and let K1 be a

splitting field over F1 for F1. Let

  : F1

_=

−! F2,

let F2 = ˆ  (F1) _ F2[x], and let K2 be a splitting field a over F2 for F2.

Then there is commutative diagram

K1

¯   - K2

F1

6

  - F2

6

where the vertical maps are the obvious inclusions, and where ¯   is an isomorphism.

Corollary 2.2.3.1 Let F be a field and let F _ F[x]. Then any splitting

field for F over F is unique, up to an isomorphism fixing F element-wise.

If F = F[x], then a splitting field for F over F is called an algebraic

closure of F. Furthermore, if every polynomial f(x) 2 F[x] splits completely

in F[x], we call F algebraically closed .

Lemma 2.2.4 Let ¯F _ F be an algebraic closure. Then ¯F is algebraically

closed.

Theorem 2.2.5 Let F be a field. Then there exists an algebraic closure of

F.

The idea of the proof of the above is first to construct a “very large”

algebraically closed field E _ F and then let ¯F be the subfield of F generated

by the roots of all polynomials f(x) 2 F.

Note that the algebraic closure of the field F, whose existence is guaranteed

by the above theorem, is essentially unique (in the sense of Corollary

10, above).

2.2. SPLITTING FIELDS AND ALGEBRAIC CLOSURE 49

Exercises 2.2

1. Let f(x) = xn − 1 2 Q[x]. In each case below, construct a splitting

field K over Q for f(x), and compute [K : Q].

(i) n = p, a prime.

(ii) n = 6.

(iii) n = 12.

Any conjectures? We’ll discuss this problem in Section 8.

2. Let f(x) = xn − 2 2 Q[x]. Construct a splitting field for f(x) over Q.

(Compare with Exercise 5 of Section 2.1.)

3. Let f(x) = x3 + x2 − 2x − 1 2 Q[x].

(a) Prove that f(x) is irreducible.

(b) Prove that if _ 2 C is a root of f(x), so is _2 − 2.

(c) Let K _ Q be a splitting field over Q for f(x). Using part (b),

compute [K : Q].

4. Let _ = e2_i/7 2 C, and let _ = _ + _−1. Show that m_(x) = x3 +

x2 − 2x − 1 (as in Exercise 3 above), and that _2 − 2 = _2 + _−2.

5. If _ = e2_i/11 and _ = _ + _−1, compute m_(x) 2 Q[x].

6. Let K _ F be a splitting field for some set F of polynomials in F[x].

Prove that K is algebraic over F.

50 CHAPTER 2. FIELD AND GALOIS THEORY

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