Why do we care about the reducibility and factorization of polynomials?

Question

This may be a naive question, but why do we care about the reducibility and factorization of polynomials aside from finding roots and/or extending fields?

What are some other uses (mathematical or otherwise) of knowing whether a polynomial is reducible and what its factorization is?

Answer 1

A matrix is diagonalizable if its characteristic polynomial can be factored into unique linear factors. A matrix is similar to triangular matrix if it can be factorized into linear factors up to multiplicity (all complex matrices are similar to triangular matrices).

Answer 2

It matters in a geometric sense.

Take for instance the polynomial $f=x^2 + y^2$. Over $\mathbb{Q}$ (or $\mathbb{R}$), this has a single point of $(0,0)$ as the zero set, but over $\mathbb{C}$, this factors as $f=x^2+y^2=(x-iy)(x+iy)$ giving the zero set a structure as the union of two (complex) lines $x-iy=0$ and $x+iy=0$.

If we know the polynomial $f$ is irreducible over an algebraically closed field $k$, then the zero set of f, $X=Z(f)$ forms a variety (not a union of other polynomial zero sets) with the coordinate ring $A(X)=k[x,y]/(f)$ a domain. You can see quickly that $A(X)$ wont be a domain if $f$ is reducible.

Answer 3

Factoring in number fields often amounts to factoring polynomials over a finite field due to a result of Dedekind. (Maybe this still falls into the "extending fields" category...)

Suppose $K/\mathbb{Q}$ is a finite extension with ring of integers $\DeclareMathOperator{\O}{\mathcal{O}} \O_K$, and further assume that $\O_K = \mathbb{Z}[\alpha]$ for some (integral) $\alpha \in K$. Given a prime $p \in \mathbb{Z}$, even though the ideal $p\mathbb{Z}$ is prime, the ideal $p \O_K$ might not be. However, since $\O_K$ is a Dedekind domain then $p \O_K$ can be written uniquely as a product of prime ideals. To determine how $p \O_K$ factors, we first factor the minimal polynomial $m_\alpha$ of $\alpha$ mod $p$. If $\overline{m_\alpha(x)} = \pi_1(x)^{e_1} \cdots \pi_g(x)^{e_g}$ is this factorization into primes, then $\DeclareMathOperator{\p}{\mathfrak{p}} p \O_K = \p_1^{e_1} \cdots \p_g^{e_g}$, where $\p_i = (p, \pi_i(\alpha))$ is a prime ideal of $\O_K$.

You can find more on this result in this note by Keith Conrad and this post.