FUNDAMENTAL THEOREM OF ALGEBRA THROUGH ALGEBRAIC
TOPOLOGY
SANATH K. DEVALAPURKAR
The fundamental theorem of algebra is a basic and very important fact, which is easy to state,
but not very easy to prove without some creativity. Its importance stems from the fact that it
“linearizes” complex polynomials.
P
Theorem 1. [The fundamental theorem of algebra] Let p(z) = k=0 ak z k be a nonconstant polynomial with coefficients in the complex numbers, i.e., ak ∈ C for all k ≥ 0. Then p(z) has at least
one complex root.
This is a rather incredible theorem. Consider the polynomial x2 + 1, whose coefficients are in
R. This clearly has no solution in R itself, but its solution in C exists and is given by ±i. There
are multiple proofs of this assertion (enough to warrant a MathOverflow question1!), but the goal
of this talk is not to go through all of them. Instead, I’ll give two completely different proofs of this
result, the second of which might be surprising if you have not studied algebraic topology.
The first proof is rather complex-analytic. I’ll introduce one piece of terminology, and state an
important result, whose proof I’ll give (by citing another theorem, whose proof I won’t give).
Definition 2. A complex function f (z) is entire if it is “differentiable” on C. (The appropriate
analogue of differentiability in the complex context is called holomorphicity.) Say that f (z) is
bounded if there is some
√ constant N such that |f (z)| ≤ N for all z ∈ C, where you should recall
that |z| is defined as z × z.
Theorem 3 (Liouville’s theorem). A bounded and entire function f (z) defined on C is constant.
P∞
Proof. Let f (z) = k=0 ak z k be the Taylor series of f (z). The Cauchy integral formula says that:
I
1
f (z)
ak =
dz
2πi C z k+1
where C is the circle of radius r around 0. Since we assume that f (z) is bounded (i.e. |f (z)| ≤ N ),
we observe that:
I
I
1
|f (z)|
1
N
ak ≤
|dz|
≤
|dz|
k+1
k+1
2πi C |z
|
2πi C |r
|
The last inequality is true because on C, we must have |z| = r. This therefore becomes:
ak ≤
1 N
N
(2πr) = k
2πi rk+1
r
We chose an arbitrary circle of radius r, so let r → ∞. Then ak = 0, so f (z) = a0 , the constant
term.
This is obviously not true for a bounded and everywhere differentiable function on R. For
example, let f (x) = sin x. This is bounded, but clearly isn’t constant!
To prove Theorem 1, we need something called the “reverse triangle inequality”.
Lemma 4. Let z and s be complex numbers. Then |z| − |s| ≤ |z − s|.
1http://mathoverflow.net/questions/10535/ways-to-prove-the-fundamental-theorem-of-algebra
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Proof. Let z = (z − s) + s. Then from the ordinary triangle inequality, we see that:
|z| = |(z − s) + s| ≤ |z − s| + |s|
Which gives us the required result.
We now have all the preliminary results required to prove Theorem 1. I learnt this proof from
https://www.math.uh.edu/~sanders/Math3364/fta.pdf.
Let p(z) be a polynomial of degree n ≥ 1. Let’s write p(z) = an z n + an−1 z n−1 + · · · + a0 . We
assume that an 6= 0 because otherwise p(z) wouldn’t be a polynomial of degree n. Let’s write
p(z) = an z n − −(an−1 z n−1 + · · · + a0 ) . From the reverse triangle inequality, we see that:
|p(z)| ≥ |an z n | − |an−1 z n−1 + · · · + a0 | ≥ |an ||z n | − |an−1 ||z n−1 | + · · · + |a0 |
Let z → ∞; this shows that p(z) → ∞. In particular, there is some R such that for all |z| > R, we
have |p(z)| ≥ 1.
But this is only for |z| strictly greater than R; what about the closed disk DR = {|z| ≤ R}? Our
polynomial p(z) can attain its minimum there. (If it doesn’t, then its minimum is 1 by the above
argument.) Using the extreme value theorem, we see that |p(z)| attains its minimum value p(zmin )
on the disk at some zmin ∈ DR .
If |p(zmin )| 6= 0, then p(z) never has a zero. Let’s assume this is true. Then, we can consider
the function f (z) = 1/p(z). Our arguments above show that |f (z)| ≤ min(p(z1min ),1) < ∞, so that
f (z) is bounded. At the same time, since p(z) is never zero, the function f (z) is defined on all of
C. In other words, f (z) is bounded and entire, so by Liouville’s theorem, it must be constant. But
this means that p(z) is constant, which is not true! Therefore, our assumption that p(z) never has
a zero must be false.
Now that I have convinced you that this is true, I want to give a very different proof of this
result, that is inherently geometric, unlike the analytic proof presented above. First, we need some
topological preliminaries.
Definition 5. Let X be a space (any collection of points that has a notion of what an open and
closed set is, much like the open and closed intervals from calculus in R), and let x ∈ X. The
fundamental group, denoted π1 (X, x), of loops based at x is a collection of homotopy classes of
loops in X based at x, where a homotopy class of loops consists of all loops homotopic to one
another. Here, homotopic means that two maps can be deformed into one another.
π1 (R2 ) at any point x ∈ R2 is just 0 because every loop can be contracted to the trivial loop
at a point. A more interesting and nontrivial fundamental group is that of the circle S 1 . Fix a
basepoint x ∈ S 1 , so that we don’t have to continue writing down extra symbols. Any loop based at
x must either go clockwise or anticlockwise around the circle, and must also loop around the circle
an integer number of times (if not, then it’s not a loop).
We should think of S 1 as lying in the complex plane, as the unit circle centered at 0. Since
2πi
e
= 1, and this traces out a loop, directed anticlockwise, from the point 1 to itself, we can
precisely write that a loop going anticlockwise that loops around n times is defined by e2πin , whose
homotopy class is traditionally denoted [ωn ]. Likewise, if the loop is directed clockwise, and loops
around n times, it can be defined as e−2πin . Observe, however, that ever such loop depends on some
integer n, and is completely specified by the choice of this integer n. Therefore, π1 (S 1 ) is just the
integers Z! There’s additional structure, though, namely that concatenating loops is equivalent to
adding integers (left to the reader). In particular, we seeQthat π1 (S 1 ) is “generated” by e2πi because
n 2πi
every element of π1 (S 1 ) can be obtained from e2πi via
e , i.e., [ωn ] = n[ω1 ].
All this is inherently geometric, and at first sight it seems impossible that this discussion is
relevant to our proof of the fundamental theorem of algebra. But as we shall see, the fundamental
theorem of algebra is a simple consequence of the fact that π1 (S 1 ) is Z. I learnt this proof from
Hatcher’s Algebraic Topology.
Pn
0 k
Let’s consider aPpolynomial f (z) =
k=0 ak z . We may divide by an everywhere to get a
n
k
0
polynomial p(z) = k=0 ak z where ak = ak /an , so that the coefficient of z n is 1. Let’s assume for
2
the sake of contradiction that p(z) has no roots. Define a loop for a fixed r ∈ C and s ∈ [0, 1] as:
fr (s) =
p(re2πis )/p(r)
|p(re2πis )/p(r)|
p(r)/p(r)
Why is this a loop? When s = 0 and when s = 1, we have fr (s) = |p(r)/p(r)|
= 1, and for all
2πis
s ∈ [0, 1], this is continuous. Because we scale by the “size” of p(re
)/p(r), this is a loop on the
unit circle. Note that this never jumps to ∞ because we assumed that p(z) has no zeroes.
As r varies, this loop contracts into a point. More precisely, when r = 0, this is just 1. Therefore,
for any fixed value of r ∈ C, the loop fr (s) is homotopic to the trivial loop. This does not in itself
imply that the polynomial p(z) is constant.
For a fixed r such that r > |a0 | + · · · + |an−1 | and r > 1, and z such that |z| = r, we see that:
|z n | = r × rn−1 > (|a0 | + · · · + |an−1 |)|z n−1 | ≥ |a0 + · · · + an−1 z n−1 |
Define a polynomial pt (z) for t ∈ [0, 1] via:
pt (z) = z n + t(a0 + · · · + an−1 z n−1 )
This polynomial has no roots when |z| = r and t ∈ [0, 1] because as we observed above, |z n | >
|a0 + · · · + an−1 z n−1 |. Now, when t = 1, we get p(z), and when t = 0, we get z n . If we replace p(z)
in our definition of fr (s) with pt (z), we get a homotopy from fr (s) to the loop that goes around the
circle n times.
Let’s be more imprecise. When t = 1, since we get p(z), we start at fr (s). But at t = 0,
(re2πis )n /r n
Since
we get z n , so the function fr (s) has now been deformed into the function |(re
2πis )n /r n | .
2πis n n
2πins
2πis n n
2πins
(re
) /r = e
, and |(re
) /r | = |e
| = 1, we observe that the function fr (s) has
been deformed into the function ωn defined as e2πins = ωn (s), which lies in the homotopy class [ωn ].
Summarizing what we have done: we found that fr is homotopic to the trivial loop for any fixed
r. We also found that fr is homotopic to ωn for a fixed r. Writing this in terms of homotopy classes,
we see that [ωn ] = [fr ], and that [fr ] = 0; but this means that [ωn ] = 0. We said above that π1 (S 1 )
is generated by [ω1 ], i.e., that [ωn ] = n[ω1 ]. This must mean that n = 0; but n is the degree of p!
So deg p = 0, and therefore p(z) is constant.
These proofs are incredibly different, but prove the same thing. The algebro-topological proof is
particularly beautiful, and displays how different branches of math can converge.
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