ago
0 like 0 dislike
0 like 0 dislike
When V is an affine algebraic set, we say that the quotient K\[x₁, ..., xₙ\]/I(V) is the affine algebra of V. I've always had trouble grasping the notion of quotients in group/ring theory and I really don't understand what these are. According to Wikipedia, the elements of a quotient R / I are equivalence classes with respect to the equivalence relation a\~b iff a-b ∈ I. By this logic, the elements of K\[x₁, ..., xₙ\]/I(V) should be equivalence classes of n-variable polynomials in which the equivalence class of a polynomial f is the set of polynomials g such that f(v) = g(v) for all v in I(V). My problem with this is that the lecture notes seem to treat K\[x₁, ..., xₙ\]/I(V) as a set of polynomials, not as a set of equivalence classes of polynomials. If we are to see the quotient as a set of polynomials, then **which** polynomials belong to K\[x₁, ..., xₙ\]/I(V)?
ago
0 like 0 dislike
0 like 0 dislike
You can represent each equivalent class of polynomial with a polynomial, and this suffices in practice. Because the intuition behind the quotient is that 2 polynomials in the same equivalent class would evaluate to the same value on the affine variety (since we are quotienting out by all the polynomials that evaluate to identically 0 on the variety). So in the appropriate context (e.g. you're thinking about the variety itself, not its embedding in an affine space), then it's fine to conflate the polynomial with its equivalence class.
ago
0 like 0 dislike
0 like 0 dislike
It's probably easiest to see this by example. Let's take a quadric curve in C^2 defined by
  
x^2 + y^2 - 1 = 0

which we cill call V.
Polynomials on C^2 are given by the ring C[x,y]. What about polynomials on V? Clearly, any polynomial on C^2 can be restricted to a polynomial on V, but this restriction isn't injective. For example, the polynomial x^2 + y^2 and the polynomial 1 both take the same value on V (hence are the same polynomial on V) but are clearly different when considering all of C^2 .

More generally, two polynomials p(x,y) and q(x,y) are the same on V if their difference is zero on V. This means (by some basic AG theorem) that

p(x,y)-q(x,y) =  f(x,y)(x^2 + y^2 - 1)

i.e. p(x,y)-q(x,y) lies in the ideal I(V).

Conversely, take a polynomial p(x,y) defined only on V. We want to write down a formula for this polynomial in terms of x and y, so we want to express it as a polynomial on all of C^2. As we have seen, we have many inequivalent choices of expressions for p(x,y), but all of them reduce to the same polynomial when restricted to V.

Hence, if we're talking about p(x,y) as a function on V, it is precise to instead refer to the set of all polynomials in two variables that restrict to the same function on V. If we want a formula, we may pick one from this set (the "representative"), but all choices are equivalent on V.

(This is why some people write p(x,y) for the polynomial on C^2 , and p(x,y) + I(V) for the polynomial on V, since any expression of the form p(x,y)+ i(x,y) for i in I(V) restricts to the same polynomial on V)
ago

No related questions found

33.4k questions

135k answers

0 comments

33.7k users

OhhAskMe is a math solving hub where high school and university students ask and answer loads of math questions, discuss the latest in math, and share their knowledge. It’s 100% free!