I don’t understand this argument. Just because functional extensionality is undecidable for arbitrary functions doesn’t mean that it is undecidable for every class of functions.
In the specific situation, let’s say that by an array we mean a finite, ordered list whose entries are indexed by the numbers 0, 1, …, n - 1 for some natural number n. Let’s also say that two arrays are equal if they have the same length and the same value at each position (in other words, they have “the same elements in the same order”).
If we now want to represent a function f as an array arr such that f(i) = arr[i] for every possible input i of f, then this will only be possible for some very specific functions: those whose domain are the set {0, 1, …, n - 1} for some natural number n. But for any two such functions f, g : {0, 1, …, n - 1} → t, their extensional equality is logically equivalent to the equality of the corresponding arrays: you really can check that f and g are extensionally equal by checking that they are represented by equal arrays.
It’s also the standard notation for zero objects, i.e., for terminal objects that are also initial. This entails all abelian/additive/preadditive categories, such as categories of modules, vector spaces, or abelian groups. (But there are also counterexamples, such as the categories of groups and of pointed sets.)
But I’d agree that it’s not standard notation to use 0 for a terminal object in an arbitrary category. I’d guess that most people use 1 instead, so that for example 1 × X ≅ X. (The post talks about group objects in the category of algebraic varieties (over some field), in which case 1 seems to be more appropriate than 0.)
I think these are two different questions:
- Why care about radicals?
- Why try to solve polynomial equations in terms of radicals?
For the first question:
Taking Nth powers is a fairly basic operation, which occurs all the time in mathematics.
Taking Nth roots is simply the inverse operation, so it is fairly natural to be interested in it/having to deal with it.
For the second question:
Let’s pretend for a moment that we didn’t know how the quadratic formula looked like.
Could we nevertheless say anything about it?
The quadratic formula is supposed to give us the solutions to the equation a x^2 + b x + c = 0.
A special case of this general quadratic equation is x^2 - p = 0.
There are two ways of solving this specialized equation:
either by taking a square root, giving us the two solutions ±√p, or by using the general quadratic formula (with a = 1, b = 0, c = -p).
Both of these approaches need to give us the same results, since they are both correct.
This tells us that if we simplify the quadratic formula with a = 1, b = 0, c = -p, then a square root needs to appear.
How can this happen?
Well, the most basic guess is that the quadratic formula contained at least one square root to begin with.
Looking at the actual quadratic formula tells us that this guess is correct:
the formula uses the four basic arithmetic operations (addition, subtraction, multiplication, division) and a square root.
We can repeat the same thought experiment for cubic equations, and we find that the cubic formula should probably contain third roots.
Looking up the formula confirms this suspicion.
However, it should be noted that the cubic equation does not only contain third roots, but also square roots.
The situation for the quartic equation is similar:
we suspect that the quartic formula contains fourth roots.
And thanks to our experience with the cubic formula, we may also suspect that the quartic formula contains third roots and square roots.
Looking up the formula, we see that it contains both third roots and square roots, but not (directly) any fourth roots.
(Our original idea breaks down a bit because fourth roots can be expressed as iterated square roots.
This makes it possible that the general quartic formula does not contain fourth roots, even though its simplified version will contain them.)
So what about a general polynomial equations of degree N >= 5?
Our original observation tells us that a solution formula needs to contain some sort of operation(s) that, when the formula is applied to certain special cases, gives us Nth roots.
Just as before, the most basic guess is that the formula will contain Kth roots, and the previous examples suggest that one should expect K = 2, ..., N to occur.
Summary:
To find a formula for polynomials equations of degree N >= 2, we are forced to use additional operations apart from the four basic arithmetic operations.
In certain special cases, these additional operations need to simplify to roots.
This suggests using roots in the formula, and the cases N = 2, 3, 4 support this idea.
Heuristically speaking, we are not trying to use roots because we want to, but because they seem to be the bare minimum required to even hope of finding a formula.
In the specific situation, let’s say that by an array we mean a finite, ordered list whose entries are indexed by the numbers 0, 1, …, n - 1 for some natural number n. Let’s also say that two arrays are equal if they have the same length and the same value at each position (in other words, they have “the same elements in the same order”).
If we now want to represent a function f as an array arr such that f(i) = arr[i] for every possible input i of f, then this will only be possible for some very specific functions: those whose domain are the set {0, 1, …, n - 1} for some natural number n. But for any two such functions f, g : {0, 1, …, n - 1} → t, their extensional equality is logically equivalent to the equality of the corresponding arrays: you really can check that f and g are extensionally equal by checking that they are represented by equal arrays.