A question on a presumably erroneous proof

Question

I'm working through an exercise in an algebra book that asks one to prove that given the field $F$, considering the ring of polynomials over $F$ given by $F[x]$, that if $f$ is a degree 1 polynomial, then $(f)$, the principal ideal generated by $f$, is maximal.

I understand this is special case of the fact that the ideals generated by irreducible elements are maximal. I've attempted to prove this and came to something that looks like a solution, but it appears I am either somewhere implicitly using that $\text{deg} f = 1$, or otherwise unable to see my error.

My proof: Assume for the sake of contradiction that $(f)$ is not maximal, i.e. there is some ideal $I$ such that: $$(f) \subsetneq I \subsetneq R$$ Obviously then, $I \setminus (f)$ is nonempty then. By the well-ordering principle, we can consider then a polynomial of minimal degree in $I \setminus (f)$. \ \ By definition, $f \nmid g$, as this would have $g \in (f)$. Appealing to the division algorithm for polynomials then, we have a unique decomposition of $g$ given by the following: \begin{equation} \begin{aligned} g = fq + r \\ q, r \in F[x], \; \text{deg}(r) < \text{deg}(g) \end{aligned} \end{equation} As $f \nmid g$ then, it must be $r \neq 0$. Reorganizing, we get $$r = g - fq,$$ where $r \in I$ as $g \in I$ and $-fq = f(-q) \in (f) \subsetneq I$, appealing to fact that $I$ as an ideal is additive subgroup. There are two possibilities then, that $r \in (f)$ or $r \in I \setminus (f)$. If $r \in (f)$, then using the first line in (1) and the properties of ideals, it is clear then that we must have $g \in (f)$, a contradiction. If $r \in I \setminus (f)$ instead, then as $\text{deg}(r) < \text{deg}(g)$, the minimality of the degree of $g$ is contradicted. Thus both cases yield a contradiction, and so it must be $(f)$ is maximal.

Any help in ascertaining the error or providing insight would be appreciated.

Answer 1

You are essentially inlining the ubiquitous proof that ideals in a Euclidean domain are principal, generated by any minimal element. This implies $\,I = (g)\,$ so $\,(g)\supsetneq (f)\,\Rightarrow\, g\mid f\,$ properly, hence $f$ irreducible $\Rightarrow \,g\,$ is a unit, so $I=(g)=(1)$.

Note $ $ Viewed constructively the descent step yields Gauss's algorithm for computing inverses mod primes: if prime $\,p\nmid a\,$ then we can invert $\,a\,$ modulo $\,p\,$ by using the descent $\,a \to p\bmod a\,$ to get a decreasing chain of multiples of $a$, eventually terminating with $1$, so $(p,a)=(1).\,$

In the OP special case where $f$ has degree $1$ we can instead use the general fact that in a Euclidean domain if $p\neq 0$ is a nonunit of minimal Euclidean size (here degree) then remainders mod $p\,$ are units $\,u\,$ (or $\,0),\,$ so if $\,p\nmid a\, $ then $\,a = q\,p + u\,$ so unit $\,u = a-q\,p\in (a,p)\Rightarrow (a,p)=(1),\,$ thus $(p)$ is max (said structurally: $(p)$ is max by $R/p\,$ is a field, by mod $p\,$ nonzero elements are units).

Applied in $F[x]\!:\:\!$ the minimal degree nonunits $\,p\,$ are precisely the polynomials of degree $1,\,$ so by above $(p)$ is max (so $p$ is prime). Applied in $\Bbb Z$ this merely proves that $\color{#c00}{2,3}$ are prime, but applied in quadratic number rings it can be much more useful, e.g. allowing us easily to deduce that certain quadratic number rings are not Euclidean. For example if $\pm1$ are the only units of an integral domain $R$ then $R$ is not Euclidean if $\,a^3−a\,$ is coprime to $\,b^3-b\,$ for some $\,a,b\,$ in $R\,$ or, said equivalently, if neither $\,\Bbb Z/\color{#c00}2\,$ nor $\,\Bbb Z/\color{#c00}3\,$ is an image of $R,\,$ i.e. $R$ has neither parity nor triality (which yields a slick proof by Lenstra that the PID $\,\Bbb Z[1 + \sqrt{-19})/2]\,$ is not Euclidean).