Posted by: holdenlee | July 2, 2010

## Fundamental Theorem of Algebra

### Fundamental Theorem of Algebra

Any complex, nonconstant polynomial has a zero in $\mathbb{C}$. Hence by induction it splits into linear factors.

We give three standard proofs. By scaling, it suffices to show this for $P(x)=x^n+a_{n-1}x^{n-1}+\ldots +a_0$.

#### Proof 1: Elementary methods (with a bit of analysis)

Main idea: If $P$ does not have a zero, take $x$ such that it $|P(x)|$ attains minimum. We can move a small amount in one direction to make $|P(x)|$ smaller, a contradiction.

Step 1: $|P|$ attains a minimum.

Note that if $x$ has absolute value at least $S$, then $|P(x)|$ is at least $S^n-|a_{n-1}|S^{n-1}-\cdots$. This approaches infinity as $S$ approaches infinity. Hence the infimum of $|P|$ is the same as if we restrict it to some sufficiently large ball around 0 ($|x|\leq S$ for some $S$). But this is a compact set (as it is closed and bounded in $\mathbb{R}^2$), so $|P|$ attains a minimum here, say at $x_0$. Suppose that $x_0\neq 0$.

Step 2: Adjusting $x$.

For convenience, set $Q(x)=\frac{P(x_0+x)}{Q(x_0)}$; that is, $Q$ is a shifted and scaled version of $P$ so that $Q(0)=1$ is the minimum of $|Q|$. Write $Q(x)=1+b_kx^k+b_{k+1}x^{k+1}\cdots$ where $b_k\neq 0$. Then $Q(re^{i\theta})=1+b_k(r^ke^{i\theta k})+O(r^{k+1})$ as $r$ approaches 0. However, we can choose the direction $\theta$ so that the second term is negative, and choosing $r>0$ small enough we get $Q(re^{i\theta})<1$, contradiction.

Source: Real Analysis, Walter Rudin

### Proof 2: Algebraic Topology

Main idea: Let $x$ range over the circles $g(\theta)=re^{2\pi i\theta}, \theta\in[0,1]$ for different $r$. For $r=0$, $P(x)$ is just the constant path $a_0$ and does not go around 0; for large $r$, the $x^n$ term dominates, and the path resembles $g(\theta)^n$ which goes around 0 $n$ times. Thus for some intermediate $r$, $P(x)$ passes through 0. We need to make precise what it means for $P(x)$ to “wind” $n$ times around 0, and prove that it is invariant under “deforming” a path (without having it pass through 0). (This part seems intuitively obvious, but it is the bulk of the proof. Our proof will be longer than is necessary, so as to show an important idea in algebraic topology.)

Step 1: Smoothly deforming curves.

Suppose that $P$ has no zero. Obviously, $a_0\neq 0$ as else $x=0$ is a zero.

Let $P_r(\theta)=P(g_r(\theta))$. We think of this as a family of functions (or paths) $[0,1]\to \mathbb{C}$. For any $r$, we have that

(1.1) $P_0\equiv a_0$ deforms continuously to $P_r$,

as $P_r(\theta)$ is continuous in $r$ and $\theta$

Now, $x^n$ winds $n$ times around 0 as $x$ goes around the circle $g_r(\theta)$, and $|x^n|> |a_{n-1}x^{n-1}+\ldots|$ for sufficiently large $r$. Thus for such a $r=R$, $P_R(\theta)$ sticks close to $x^n$ as $x$ varies around the circle, following $x^n$ around 0 $n$ times. Indeed,

(1.2) $P_R(\theta)$ deforms continuously into $r^ne^{2\pi in}$.

To see this, define $f_t(\theta)=(g_R(\theta))^n+t (a_{n-1}g_R(\theta)^{n-1}+\ldots)$. Then $f_t(x)$ for $0\leq t\leq 1$ deforms from $f_0(\theta)=r^ne^{2\pi n i\theta}$ to $f_1(\theta)=P_R(\theta)$ in $\mathbb{C}-\{0\}$ (since $|x^n|> |a_{n-1}x^{n-1}+\ldots|$, none of the $f_t$ pass through 0).

From (1.1) and (1.2), the point $a_0$ deforms continuously into the function $re^{2\pi in}$ without passing through 0. A family of functions not passing though 0 connects them. It is more convenient to look at the projections of these paths on the unit circle, rotated so they have the same starting and ending point. Replacing each function $f$ by $\frac{f/f(0)}{|f/f(0)|}$, we have that the constant path at 1 is homotopic to a path going around the circle $n$ times, i.e. one can be continuously deformed into another keeping the endpoints the same (there exists $g_t(\theta)$ so that $g_t(\theta)$ is continuous in $t,\theta$; this $g_t$ is just the projections of the $P_t$ for $0\leq t\leq R$ followed by $f_t$ for $1\geq t\geq 0$ with suitable reparametrization). We show that this implies $n=0$, i.e. $P$ is constant; then we are done. To do this, we need to classify the homotopy classes (equivalence classes under homotopy) of paths (loops) on the circle starting at ending at the same point.

Step 2: The fundamental group of the circle

The homotopy classes with the same basepoint have a group structure (which is irrelevant here, but is nevertheless an important idea) where the operation is just concatenation of paths. For simplicity represent the points on the circle with real numbers modulo 1 (i.e. that tell the number of revolutions). We show the fundamental group simply consists of the classes of paths that go around the circle $n$ times, for $n\in \mathbb{Z}$ (the direction depending on the sign on $n$). Composing a path that goes around $n$ times and a path that goes around $m$ times gives a path going around $m+n$ paths. We prove the following.

Theorem: The fundamental group $\pi_1(S_1)$ of the circle $S_1$ is isomorphic to $\mathbb{Z}$, with the isomorphism $\Phi:\mathbb{Z}\to \pi_1(S_1)$, where $\Phi_n$ is the path $[0,1]\to S_1$ given by $\Phi_n(t)=nt\bmod{1}$, i.e. it sends $n$ to the path going around the circle $n$ times.

To prove this, we find a space $R$ which maps to $S_1$, such that every path $f$ in $S_1$ lifts to a unique path $g$ in $R$. That is, if $\phi:R\to S_1$ is the map, and $r,t$ are given so that $\phi(r)= f(t)$, then there is a unique $g$ such that $f=\phi g$ and $g(t)=r$. We want $R$ to have the property that the lift of each distinct homotopy class with $g(0)=0$ is the class of a path with distinct endpoint in $R$ (so it’s easy to distinguish homotopy classes). Think of $R$ as an infinite helix above the circle, and $\Phi$ as the projection onto the plane of the circle. $R$ has the same structure as $\mathbb{R}$, so we just consider the elements of $R$ as real numbers, and $\Phi$ as mapping a number to its fractional part.

Step 2.1. Cover $S_1$ with the open sets $O_1=(0,.6)$ and $O_2=(.5,1.1)$. Now the inverse image under $\Phi$ of each is an union of disconnected open intervals in $R$,–simply the numbers with fractional parts in $(0,.6)$ and in $(.5,1)\cup [0,.1)$, respectively. Each of these open intervals is homeomorphic to its image via $\Phi$. $R$ is said to cover $S_1$.

Step 2.2. Construct a lift $g$ locally. For each path $f$ and $t\in [0,1]$, $f(t)$ must be contained in one of the open sets in (2.1), say $O$. By continuity of $f$, we can find a neighborhood $N_t$ around $t$ so that $f(t)\in O$ for all $t\in N_t$. Since the preimage of $O$ under $\phi$ is disconnected, if we fix a value for $g(t)$ then $g$ must stay in the same component $O'$ of $\phi^{-1}(O)$. Since $O'$ maps homeomorphically to $O$, there is a unique and natural way to define $g$; just let $g(t)=\phi^{-1}[f(t)]$.

Step 2.3. Construct a lift globally. The $N_t$ in (2.2) for an open cover for $[0,1]$. Since this interval is compact, a finite number of them cover $[0,1]$. There is exactly one way to define a lift $g$ with $g(0)=0$: (2.2) determines the lift on the open set $N_t$ containing 0; then this determines a lift on the open sets overlapping $N_t$, and so on.

Step 2.3B. Do the same thing for homotopies. Each homotopy $f_s(t)$ can be thought of as a map from $[0,1]^2$ (instead of $[0,1]$) to $S_1$. Again we get a unique lift to $g_s(t)$. This shows that if two paths are homotopic in $R$ iff their images in $S$ are homotopic. (The forward assertion is obvious by applying $\phi$.)

Step 2.4. By (2.3B), the homotopy classes in $S_1$ are associated with homotopy classes in $R$ starting at 0 and ending in $\phi^{-1}(0)=\mathbb{Z}$. Such homotopies are classified by their endpoints, since it is easy to see any two paths with same starting and ending point in $R$ are homotopic (exercise). Thus the homotopy classes in $S_1$ are just those in the image of $\Phi$, and these are distinct.

Source: Algebraic Topology, Allen Hatcher

### Proof 3: Complex Analysis

Main Idea: If $P$ has no zeros, then $\frac{1}{P}$ is bounded and holomorphic (complex differentiable) on the whole complex plane. (It is bounded because $P(z)$ is either constant or approaches infinity as $|z|$ approaches infinity.) Then it must be constant by Liouville’s Theorem (see below).

Index: The index (or winding number) of $a$ with respect to the curve $\gamma$ is defined by

$n(\gamma, a)=\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z-a}.$

If $\gamma$ is a piecewise differentiable curve not passing through $a$, then the index is an integer. The index offers another way of describing what we mean by a curve going $n$ times around 0, without reference to the fundamental group.

Cauchy’s Integral Formula: Suppose that $f(z)$ is holomorphic in an open disk and let $\gamma$ be a closed curve in the disk. Then for any $a$ not on $\gamma$,

$n(\gamma, a)f(a)=\frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z-a}.$

Liouville’s Theorem: A function which is holomorphic and bounded in the whole plane is constant.

Proof: Apply Cauchy’s Integral Formula to a circle $C$ of radius $r$ around $a$ to get

$f'(a)=\frac{1}{2\pi i}\int_{C} \frac{f(z)dz}{(z-a)^2}.$

If $|f|\leq M$, then this gives $|f'(a)|\leq M/r$. By taking $r$ to infinity, we get that $f'(a)=0$, i.e. $f$ is constant.

Source: Complex Analysis, Lars Ahlfors

### Proof 4: Field Theory

We first show that all polynomials with real coefficients are reducible over the complex numbers, by induction on the highest power of 2 dividing the degree. For odd degree, the the statement follows since the polynomial has different signs near at $\pm \infty$. Now assuming the induction hypothesis, suppose $\deg(f)=2^km$ where $k$ is odd. Choose a splitting field $L$ of $f$, and write $P(x)=(x-r_1)\cdots (x-r_n)$. Consider the polynomial

$\displaystyle P_t(x)=\prod_{i\leq i

Its degree is $\frac{n(n-1)}{2}=2^{k-1}m(n-1)$. Since its coefficients are symmetric polynomials in the $r_i$, by hypothesis it has a complex zero, i.e. $r_i+r_j+tr_ir_j$ is real for some $i,j$. Since this is true for infinitely many values of $t$, we must have that $r_i+r_j+tr_ir_j$ is real for all $t$ some $i,j$. This means $r_i+r_j$ and $r_ir_j$ are both real. Then $r_i,r_j$ are roots of the quadratic $x^2-(r_i+r_j)x+r_ir_j$ so they are complex roots of $P(x)$. This concludes the induction.

Next for an arbitrary polynomial $P(x)$, consider the real polynomial $P(x) \overline{P(x)}$. (We take the conjugate of the coefficients, not $x$.) By the above, it factors entirely into linear factors. $P(x)$ divides $P(x) \overline{P(x)}$, so it splits as well.