Conceptual question about the strategy used in the following theorem: Every Ideal of $F[x]$ is Principal

To refresh everyone, the following picture from Pinter's "A Book of Abstract Algebra" details the proof for the theorem that Every Ideal of $F[x]$ is Principal:

This strategy is pretty common, so I have seen it plenty of times. However, my question arises from the fact that:

$\operatorname{deg}(0)$ is undefined

i.e. the degree of the $0$ polynomial is undefined. As such, it almost feels like this proof is comparing oranges to apples...in the sense that it feels like it is saying:

Well, if the $r(x)$ polynomial has this property (i.e. $\neq 0$), its degree is a number that would be contradictory...so, it must be this other thing, whose degree is 'undefined'.

Is something that is undefined still "a number"? Or is it something that is entirely "non-number"? How exactly does one logically evaluate this? Any clarification would be greatly appreciated!

Edit: Another way of reframing this question is:

How does one compare an assumption that describes a numerical property to an object that has an undefined numerical property? i.e. $b(x)$ is described as having some $n \in \mathbb Z$ degree that must be the smallest number...how am I supposed to compare an undefined number (i.e. $\operatorname{deg} (r(x)=0))$ to this $n$?

If I cannot make this comparison, how can I decide whether or not it is a contradiction?

Solution 1:

The proof is fine - it's meant (though not explicitly stated) that $b(x)$ has the lowest degree among non-zero polynomials, which avoids that issue at that point in the proof. The later part of the proof never references $\deg 0$ - note that it quotes the remainder theorem in the following sense:

[We can] write $$a(x)=b(x)q(x)+r(x)$$ where $r(x) = 0$ or $\deg r(x) < \deg b(x)$.

Observe that there are two alternatives: either $r(x)=0$ or we take its degree to be lesser than that of $b(x)$. It finds out that the second alternative is absurd, and thus concludes the first alternative. It's formally just "We know $A$ or $B$ is true and $B$ is not true. Therefore, $A$ is true."

Note that it's also somewhat common to say that $\deg 0 = -\infty$ to preserve various properties of degree and to make the remainder theorem unconditionally state $\deg r(x) < \deg b(x)$. Then we still want to choose $b(x)$ to have the lowest degree among non-zero polynomials, but when we find out that $\deg r(x) < \deg b(x)$, we immediately know that $r(x)$ is not a non-zero polynomial (i.e. $r(x) = 0$)

Solution 2:

The proof never uses $\deg 0.\,$ Below is an abstraction of the argument that may help clarify this, including a view as a prototypical least counterexample (minimal criminal) descent.

Say $\,f\in \bar J := J\backslash 0\,$ is $ $ good $ $ if $\,f\,$ divides every element of $\,\bar J\,$ (else say $\,f\,$ is $ $ bad, $ $ i.e. not good).

Division Algorithm $\Rightarrow$ bad $f$ are not minimal degree (by constructing a smaller degree $\,f'\in \bar J,\,$ i.e. the remainder $\,f' := g\bmod f,\,$ using $\,f\,$ bad $\,\Rightarrow\,f\nmid g\,$ for some $\,g\in\bar J).$

So a min degree $\,f\in \bar J$ must be good (if it were bad the above descent would yield a smaller deg $\,f'\in J,\,$ contra minimality). Note min degree $f$ exist by $\Bbb N$ is well-ordered (and $\bar J$ not empty).

Summary $ $ By $\Bbb N\,$ well-ordered, minimal degree elements exist; furthermore, by Euclidean remainder (mod) descent, bad elements are not minimal, thus a minimal element is good.

This method works generally - requiring only a descent on bads using a well-ordered set $N$ of "sizes", i.e. every nonempty subset of $N$ has a least element. $\,N = \Bbb N\,$ in the OP.

We can view the proof as a descent on bads (counterexamples), i.e. if all elements are bad, then choose $f\,$ to be a least degree bad (minimal criminal). As above, the division yields a smaller (degree) bad, contra minimality. So not all elements are bad, therefore a good element exists.

Or, equivalently, well-ordering implies that if we iterate our descent process to generate a descending chain of bads then the chain is finite, necessarily terminating at a good element, since, equivalently, well-ordering means there are no infinite descending chains.

This type of induction (descent) is ubiquitous in mathematics so it is well-worth the effort to master it early (in simple instances like this, since it will be more difficult to do so in more complex cases).

Remark $ $ The proof generalizes to any Euclidean domain, i.e. any domain enjoying division with smaller remainder. The key idea is that ideals are closed under remainder (mod), therefore the "least" $\,d\in I\,$ must divide every $\,i\in I,\,$ else $\,0\neq i\ {\rm mod}\ d\,$ is in $\,I\,$ and smaller than $\,d,\,$ contra minimality of $\,d.\,$ The descent in this proof can be interpreted constructively as computing a generator of $\,I\,$ by computing the gcd of its elements (by taking repeated remainders).

The idea extends to PIDs: (Dedekind-Hasse criterion) a domain $\rm\,D\,$ is a PID iff given $\rm\:0\neq a, b \in D,\:$ either $\rm\:a\:|\:b\:$ or some D-linear combination $\rm\:a\,d+b\,c\:$ is "smaller" than $\rm\,a.\,$ In a PID we can choose the number of prime factors as a measure of (Euclidean) size.