Proof the Existence and Uniqueness of Factorization Form of Polynomial with Complex Coefficient

Question

If $p\in P(\Bbb{C})$ is a nonconstant polynomial, then $p$ has a unique factorization (except for the order of the factors) of the form

$$p(z)=c(z-\lambda_1)....(z-\lambda_m)$$

where $c,\lambda_1,....\lambda_m \in \Bbb{C}$

I know how to prove the existence of the form. However, I don't know how to do it with the uniqueness of the form.

I know that $c$ must be unique, because $c$ equals the coefficient of $z^m$ in $p$. But how to prove that each $\lambda_j$ is unique?

And I also don't understand why the coefficient must be $\Bbb{C}$, but why doesn't this apply to $\Bbb{R}$?

Answer 1

You can show that in $\mathbb{C}$, all linear functions are prime, and there do not exist primes which are not linear.

Lemma 1: If $p$ is prime and $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$. Let $p$ not divide $a$. Then as $p$ is relatively prime to $a$, $gcd(p,a) = 1$ and so $\exists x,y \in F$ such that $xp + ya = 1$. Multiplying by $b$, we have that $bxp + yab = b$. And we know that $p$ divides $ab$ so $ab = mp$. Thus we have that $bxp + ymp = b$ and so $p(bx + ym) = b$ so $p$ divides b.

The Result: Now, let $f(x) = p_1 \cdot p_2 \cdot p_3 ... \cdot p_n$ where $p_1,.... p_n$ are prime factors. Suppose $f(x) = q_1 \cdot q_2 \cdot ... q_k$ where all $q_i$ are prime factors [We are trying a different factorisation]. Then, each $q_i$ must divide $f(x) = p_1 \cdot p_2 \cdot p_3 ... \cdot p_n$ and by Lemma 1, they must divide some $p_i$. But a prime is such that if $q_i$ divides $p_i$ then $q_i$ is an associate of $p_j$. If we ignore the difference between a function and its associate, we can see that $q_i = p_j$ [effectively] and so we must have that each $q_i$ corresponds to a different $p_j$ and so we have that $n = k$ and the factorisation is unique.

Note: In $\mathbb{R}$, there can be quadratic equations which are prime as pointed out in the comment (They have complex factors only). Therein lies a fundamental difference.

Answer 2

Lemma 1 Let $p$ be a complex polynomial such that $\deg(p) > 0$. Then $\lim_{z\rightarrow\infty}|p(z)| = \infty$.

Proof Let $p(z) = c_nz^n + c_{n-1}z^{n-1} + \cdots + c_1z + c_0$. Then $$ \begin{align*} |p(z)| &\geq |c_nz^n|-|c_{n-1}z^{n-1} + \cdots + c_1z + c_0|\\ &\geq |c_nz^n|-|c_{n-1}z^{n-1}| - \cdots - |c_1z| - |c_0|\\ &= |c_n||z|^n - |c_{n-1}||z|^{n-1} - \cdots - |c_1||z| - |c_0|\\ &\underset{z\rightarrow\infty}{\longrightarrow}\infty. \end{align*} $$ $\square$

Corollary Let $p, q$ be complex polynomials. Then $p$ and $q$ are identical as functions iff they have the same degree and the same respective coefficients.

Lemma 2 Let $p, q_1, q_2, r_1, r_2$ be complex polynomials such that $pq_1+r_1 = pq_2+r_2$ and such that $\deg(r_1), \deg(r_2) < \deg(p)$. Then $q_1=q_2$. (It is also true that $r_1=r_2$, but we won't need this fact here.)

Proof Define $d$ to be the polynomial $d := p(q_2-q_1)+(r_2-r_1)$. Then $d = 0$. Suppose to the contrary that $q_1\neq q_2$. Then the polynomial $q_2-q_1 \neq 0$. Then $\deg(d) \geq \deg(p) > 0$. Then by lemma 1 $\lim_{z\rightarrow\infty}|d(z)| = \infty$, a contradiction. $\square$

Theorem Let $m, n \in \{1,2,\dots\}$, let $p_1, \dots, p_m \in \mathbb{C}$ be distinct, let $q_1, \dots, q_n \in \mathbb{C}$ be distinct, let $a_1, \dots, a_m \in \{1,2,\dots\}$, and let $b_1, \dots, b_n \in \{1,2,\dots\}$. Suppose that $(z-p_1)^{a_1}\cdots(z-p_m)^{a_m}=(z-q_1)^{b_1}\cdots(z-q_n)^{b_n}$. Then $\{p_1, \dots, p_m\} = \{q_1, \dots, q_n\}$ (in particular, $m = n$), and $p_i = q_j \implies a_i = b_j$.

Proof To see that $\{p_1,\dots,p_m\}=\{q_1,\dots,q_n\}$, we will show that $\{p_1,\dots,p_m\}\subseteq\{q_1,\dots,q_n\}$ (the converse containment can be shown analogously). Suppose to the contrary that $p_i \notin\{q_1,\dots,q_n\}$ for some $i \in \{1,2,\dots,m\}$. Write $\alpha(z) := (z-p_1)^{a_1}\cdots(z-p_m)^{a_m}$ and $\beta(z) := (z-q_1)^{b_1}\cdots(z-q_n)^{b_n}$. Then, by assumption, $\alpha=\beta$, but $\alpha(p_i) = 0 \neq \beta(p_i)$, a contradiction.

Now suppose, to the contrary, that for some $i,j \in \{1,2,\dots,m\}$ $p_i = q_j$ but $a_i \neq b_j$. Suppose w.l.g. that $i = j = 1$ and that $a_1 < b_1$. Then $$ \begin{multline*} (z-p_1)^{a_1}\underbrace{(z-p_2)^{a_2}\cdots(z-p_m)^{a_m}}_{\gamma(z)} =\\ (z-q_1)^{a_1}\underbrace{(z-q_1)^{b_1-a_1}(z-q_2)^{b_2}\cdots(z-q_m)^{b_m}}_{\delta(z)}. \end{multline*} $$ Then by lemma 2, $\gamma=\delta$, but $\gamma(p_1) \neq 0 = \delta(p_1)$, a contradiction. $\square$