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by sykh 2989 days ago
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.

1 comments

Technically, people have polynomials with uncountably many variables, so you can't order them.
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.