5.2. Fundamental Theorem of Algebra.

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We finally prove the misnamed4 fundamental theorem of algebra.

Theorem 5.5. The field C of complex numbers is algebraically closed.

Proof. Define C to be the splitting field of X2 +1 ∈ R[X], and let i be a root of X2 +1

in C;th us C = R[i]. We have to show (see 2.10) that every f(X) ∈ R[X] has a root in C.

The two facts we need to assume about R are:

• Positive real numbers have square roots.

• Every polynomial of odd degree with real coefficients has a real root.

Both are immediate consequences of the Intermediate Value Theorem, which says that a

continuous function on a closed interval takes every value between its maximum and minimum

values (inclusive). (Intuitively, this says that, unlike the rationals, the real line has no

“holes”.)

We first show that every element of C has a square root. Write α = a + bi, with a, b ∈ R,

and choose c, d to be real numbers such that

c2 =

(a +

√

a2 + b2)

2

, d2 =

(−a +

√

a2 + b2)

2

.

Then c2 − d2 = a and (2cd)2 = b2. If we choose the signs of c and d so that cd has the same

sign as b, then (c + di)2 = α.

Let f(X) ∈ R[X], and let E be a splitting field for f(X)(X2 + 1)—we have to show that

E = C. Since R has characteristic zero, the polynomial is separable, and so E is Galois over

R. Let G be its Galois group, and let H be a Sylow 2-subgroup of G.

Let M = EH. Then M is of odd degree over R, and M = R[α] some α (Theorem 5.1).

The minimum polynomial of α over R has odd degree, and so has a root in R. It therefore

has degree 1, and so M = R and G = H.

We now have that Gal(E/C) is a 2-group. If it is _= 1, then it has a subgroup N of index

2. The field EN has degree 2 over C, and can therefore be obtained by extracting the square

root of an element of C (see 3.23), but we have seen that all such elements already lie in C.

Hence EN = C, which is a contradiction. Thus E = C.

Corollary 5.6. (a) The field C is the algebraic closure of R.

(b) The set of all algebraic numbers is an algebraic closure of Q.

Proof. Part (a) is obvious from the definition of “algebraic closure”, and (b) follows from

the discussion on p15.

4Because it is not strictly a theorem in algebra: it is a statement about R whose construction is part of

analysis. In fact, I prefer the proof based on Liouville’s theorem in complex analysis to the more algebraic

proof given in the text: if f(z) is a polynomial without a root in C, then f(z)−1 will be bounded and

holomorphic on the whole complex plane, and hence (by Liouville) constant. The Fundamental Theorem

was quite a difficult theorem to prove. Gauss gave a proof in his doctoral dissertation in 1798 in which he

used some geometric arguments which he didn’t justify. He gave the first rigorous proof in 1816. The elegant

argument given here is a simplification by Emil Artin of earlier proofs.

FIELDS AND GALOIS THEORY 39

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