THE FUNDAMENTAL THEOREM OF ALGEBRA
MARCO ADAMO SEVESO
Contents
1. The fundamental theorem of algebra
References
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1. The fundamental theorem of algebra
Lemma 1.1. Suppose that C/R is a finite extension with R a perfect field with the following properties.
• Every polynomial f (X) ∈ R [X] of odd degree has a root in R.
• Every degree two polynomial Q (X) ∈ C [X] has a root in C.
Then C is algebraically closed.
Proof. Suppose that we have given f (X) ∈ R [X] and let Rf ⊃ R be a splitting field of f over R. Then
Rf /R is Galois and it follows that Rf C/C is Galois and that we have
∼
GRf C/C → GRf /Rf ∩C ⊂ GRf /R .
(1)
Let P2 ⊂ GRf /R be a 2-Sylow subgroup, so that
h
i RfP2 : R = GRf /R : P2 ≡ 1 mod 2
and RfP2 /R is an extension of odd degree. We claim that, if E/R is an extension of odd degree, then E = R.
Indeed, since R is perfect, E/R is separable and, by the Primitive Element Theorem, we know that we
with fξ (X) ∈ R [X], the minimal polynomial of ξ, irreducible in R [X]. Since
have E = R (ξ) ' (fR[X]
ξ (X))
deg (fξ ) = [E : R] is odd, it follows that fξ (X) has a root in R and the only possibility is that deg (fξ ) = 1,
GR
/R
implying E = R. Hence RfP2 = R and RfP2 = R = Rf f gives GRf /R = P2 .
It follows from the equality GRf /R = P2 and (1) that GRf C/C is a group of order 2n a power of 2. If n ≥ 1
there is a subgroup H ⊂ GRf C/C such that GRf C/C : H = 2 (see, for example, [La, Ch. I, §6, Corollary
6.6]) and, consequently,
h
i H
(Rf C) : C = GRf C/C : H = 2.
We claim that there is no extension E/C such that [E : C] = 2. Indeed, since R is perfect and C/R
is finite, C is perfect; it follows that E/C is separable and, by the Primitive Element Theorem, that we
with fξ (X) ∈ C [X], the minimal polynomial of ξ, irreducible in C [X]. Since
have E = C (ξ) ' (fC[X]
ξ (X))
deg (fξ ) = [E : C] = 2, the polynomial fξ (X) has a root in C and we know that this is not posssible. This
contradition shows that n = 0 and then that Rf C = C. Hence Rf ⊂ Rf C = C and f (X) ∈ R [X] splits in
C [X].
Since C/R is algebraic and we have shown that every polynomial f (X) ∈ R [X] splits in C [X], it follows
from [Mi, Proposition 1.44] that C is algebraically closed.
Lemma 1.2. The extension C/R has the properties stated in Lemma 1.1.
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Proof. If f (X) ∈ R [X] has odd degree we have f (±∞) = ±∞ and consequently, the Intermediate Value
theorem of Calculus shows that there is α ∈ R such that f (α) = 0. Suppose now that Q (X) = aX 2 +bX +c ∈
C [X] is a degree 2 polynomial, so that a 6= 0. If we show that
2
(·) : C → C mapping x 7→ x2
will be a root of Q where ∆ ∈ C is such that ∆2 = b2 − 4ac.
is surjective then α := −b±∆
2a
Given x + yi ∈ C with x, y ∈ R, we have
2
(x + iy) = x2 + 2xyi − y 2 = x2 − y 2 + 2xyi.
Hence, given a + bi ∈ C with a, b ∈ R, we have to solve the equation
2
x − y2 = a
2xy = b.
(2)
We will use the fact that
2
(·) : R≥0 R≥0 mapping x 7→ x2
(3)
is surjective, which is again a consequence of the Intermediate Value theorem of Calculus (suffices to show
that R R≥0 is surjective because r2 = s2 if and only if r = ±s). For every y such that a + y 2 ≥ 0 there is
xa,y ∈ R such that x2a,y = a + y 2 , thanks to (3). Then we have
2xa,y y = ±b ⇔ 4 a + y 2 y 2 = 4x2a,y y 2 = b2 .
(4)
Expanding 4 a + y 2 y 2 we find that this means:
4y 4 + 4ay 2 − b2 = 0.
(5)
We now reverse the arguments and show that (5) and a+y 2 ≥ 0 can be achieved. This will give our claim (2)
because, as remarked, a + y 2 ≥ 0 implies that we may find xa,y such that x2a,y = a + y 2 ; then (5) is equivalent
to (4) and, since xa,y can be changed of sign without affecting x2a,y = a + y 2 , we see that 2xa,y y = ±b gives
2xa,y y = b.
Setting z := y 2 we have that 4z 2 + 4az − b2 = 0 has roots
p
1p 2
1
±
a + b2 , where a2 + b2 ≥ 0.
za,b
=− a±
2
2
±
Here we remark that, since a2 + b2 ≥ 0, we have indeed that ∃za,b
∈ R, thanks to (3). Furthermore, we claim
√
√
+
that za,b ≥ 0: indeed this is true if and only if a2 + b2 ≥ a; we may have a ≤ 0 ≤ a2 + b2 and then the
inequality is satisfied, or 0 ≤ a and then the inequality is satisfied if and only if a2 ≤ a2 + b2 , which is true.
+
+
2
Given our za,b
≥ 0 we may find, thanks to (3), ya,b ∈ R such that ya,b
= za,b
and (5) is satisfied by this ya,b .
We also have
1
1p 2
+
2
a + ya,b
= a + za,b
= a+
a + b2
2
2
When a ≥ 0 √
it is clear that this quantity if ≥ 0 and, when a ≤ 0 we have that this quantity is ≥ 0 if and
only if −a ≤ a2 + b2 ; since −a ≥ 0 this inequality is satisfied if and only if a2 ≤ a2 + b2 , which is true. As
explained this gives the existence of (x, y) satisfying (2).
The Fundamental Theorem of Algebra is now a consequence of Lemmas 1.2 and 1.1.
Theorem 1.3. The field C is algebraically closed.
Proof. Here is another proof using Liouville’s theorem in complex analysis. Suppose that f (X) ∈ C [X] is a
−1
polynomial which has no roots in C. Then f (X) defines a holomorphic function on C which is bounded.
By Liouville’s theorem it has to be constrant so that f (X) = c ∈ C× . Hence, if f (X) is non-constant it
has a root in C, proving that C is algebraically closed (see [Mi, Proposition 1.42] for the various equivalent
definitions of being algebraically closed).
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References
[La] S. Lang, ”Algebra”, Springer.
[Mi] J. S. Milne, ”Fields and Galois theory”, www.jmilne.org/math/CourseNotes/FT.pdf. Chapter 5, ”The fundamental theorem of algebra”.
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