"up to associates" phrase meaning in proof of irreducible zeroes

Question

This is what is trying to be proved:

Let $f(x)$ be an irreducible polynomial over a field $F$. If $F$ has characteristic $0$, then $f(x)$ has no multiple $0$'s.

The proof starts off with: If $f(x)$ has a multiple zero, then, by a previous theorem, $f(x)$ and $f '(x)$ have a common divisor of positive degree in $F[x]$. Since the only divisor of positive degree of $f(x)$ in $F[x]$ is $f(x)$ itself (up to associates), we see that $f(x)$ divides $f'(x)$.

What does the phrase "up to associates" mean here? I know that two elements $(a,b)$ of an integral domain are associates iff $a = ub$, where $u$ is a unit of the integral domain. But I don't understand what up to associates means in this context.

Answer 1

In this context it means that the only divisors of $f$ are (non-zero) constants and polynomials of the form $\lambda f$, with $\lambda\in\Bbb F^\times$.

In general, up to X means "if we pretend that two things differing only by X are the same", and X is generally some kind of equivalence relation. Examples :

There is only one group with two elements, up to isomorphism.

Here "up to isomorphism" means that we don't make a distinction between two isomorphic groups, so that $(\{1,-1\},\cdot )$ is the same as $\mathbb{Z}/2\mathbb{Z}$.

Every natural number greater that $1$ can be factorised as a product of prime numbers, and this factorisation is unique up to the order of the factors.

Here "up to the order of the factors" means that we don't make a distinction between two factorisations which have the same primes but in a different order, so for example $6=2\times 3$ and $6=3\times 2$ are considered to be the same factorisation.