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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 $\Bbb{R}$?

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  • $\begingroup$ Are you familiar with the proof of the unique factorization theorem for the integers? The proof for polynomials is similar (and too long to write out in detail here), starting with the Division Theorem, developing the Euclidean Algorithm, etc. What happens in $\bf R$ is that there are polynomials like $x^2+1$ that don't split into linear factors. There is still unique factorization into irreducible factors, but now those factors could be quadratic. $\endgroup$ – Gerry Myerson Aug 25 '14 at 7:30
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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.

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