I feel like that example counters your claim in two different ways.
First, (1, 2) only makes sense if you assume that x is first and y is second, or in other words that they correspond to X_1 and X_2 for some vector X. This is probably a reasonable assumption for x and y, but what if you have some other arbitrary choice of variables?
a ρ + 2 = 0
Then the only unambiguous way to write a solution is with explicit labels: a = 1, ρ = -2.
Second, another way to “solve” the equation “x y + 2 = 0” is: “y = -2/x, for any x ∈ ℝ \ {0}”. You could argue that this is also a different kind of “equation” than the one you started with: the reason it can be seen as a solution is not (just) that it’s simpler than the original equation, but that it provides an algorithm to enumerate the set of individual solutions, as well as the set of solutions given some proposed x value. (That is, a set with one or zero members depending on whether x = 0.) However, even if it is a different kind of object, there’s no way to represent it in standard notation that doesn’t conflate ‘questions’ with ‘answers’. If you really want to use tuples, you could go for “{(x, -2/x) | x ∈ ℝ \ {0}}”, which avoids the equals sign – but the ∈ is playing a similar role, an algorithm (enumerate all members of this set) disguised as a test (is this value a member of the set?).
edit: Upon further reflection, I might actually just be expressing violent agreement with the point you were trying to make. shrug
Polynomials in multiple variables always have the variables ordered. Sometimes the ordering is not important or explicitly stated but in reality they are supposed to be ordered.
If we start with x y + 2 = 0 we can apply the function
f(x) = x - 2
to both sides of the equation. This gives us the equation
x y = -2
This is a different equation than the one we started with but these two equations have the same solution set. Can go a step further and transform this equation to
y = -2/x
This equation has the same solution set as the first equation. All three equations are equivalent. Solutions are ordered pairs of numbers.
Commonly in basic courses like calculus we tell students that the last form is preferable and we write solutions as (x, -2/x). We call this set the graph of the equation but really it’s the solution set of the equation.
In algebraic geometry x y + 2 = 0 is preferable. The solution set is called an algebraic variety. The solutions are ordered pairs in affine space.
The rules of algebra, as taught in low level courses, are rules that allow one to transform a given equation into simpler equation. In one variable the goal is to end up with something like x = 3 because such an equation is easy to solve. The solution is the 1-tuple 3.
Not all sets are recursively enumerate so using enumerate as you did can cause problems.
Each polynomial has only finitely many variables in it. If you were to work with uncountably many variables I assume one would impose a well ordering on those variables. I’ve never worked with a polynomial ring with uncountably many variables. No one is going to solve an equation with uncountably many variables so I’m not sure the point you are trying to make.
First, (1, 2) only makes sense if you assume that x is first and y is second, or in other words that they correspond to X_1 and X_2 for some vector X. This is probably a reasonable assumption for x and y, but what if you have some other arbitrary choice of variables?
a ρ + 2 = 0
Then the only unambiguous way to write a solution is with explicit labels: a = 1, ρ = -2.
Second, another way to “solve” the equation “x y + 2 = 0” is: “y = -2/x, for any x ∈ ℝ \ {0}”. You could argue that this is also a different kind of “equation” than the one you started with: the reason it can be seen as a solution is not (just) that it’s simpler than the original equation, but that it provides an algorithm to enumerate the set of individual solutions, as well as the set of solutions given some proposed x value. (That is, a set with one or zero members depending on whether x = 0.) However, even if it is a different kind of object, there’s no way to represent it in standard notation that doesn’t conflate ‘questions’ with ‘answers’. If you really want to use tuples, you could go for “{(x, -2/x) | x ∈ ℝ \ {0}}”, which avoids the equals sign – but the ∈ is playing a similar role, an algorithm (enumerate all members of this set) disguised as a test (is this value a member of the set?).
edit: Upon further reflection, I might actually just be expressing violent agreement with the point you were trying to make. shrug