It's a little odd to say a polynomial "isn't cubic" just because the degree-3 coefficient is zero. Ordinarily what matters is that every coefficient of degree higher than three is zero.
Think of the result that you can fit a polynomial of degree n to go through n+1 points. It's easy to show that the polynomial indicated by that theorem which goes through a set of collinear points must be a line (degree-1 polynomial), no matter how many points are given[1]. But we don't say "as long as the points aren't collinear, this theorem holds"; we just accept that a line is a special case of cubic polynomial.
[1] Proof: the polynomial of degree n going through n+1 points is unique. The line going through n+1 collinear points goes through all of those points, and therefore must be the unique polynomial of degree n to do so.
I do see in the wikipedia article, though, that "cubic surface" appears to be defined to exclude polynomials which have any nonzero term of degree other than three. It's a weird world. :/
Well, it's defining it in projective space; in projective space, the polynomial has to be homogeneous to make sense. If you were working in affine space, you wouldn't have that requirement -- but then the theorem wouldn't be true without first passing to projective space, I imagine. Which would involve taking your polynomial and adding an extra variable so as to make it homogeneous.
> Proof: the polynomial of degree n going through n+1 points is unique.
Suppose you have nine points lined up in an exact 3x3 grid. What, according to your proposed definition, is the unique degree 8 polynomial going through them?
There are two obvious cubic polynomials I can think of: One, cover the nine points by three horizontal lines and multiply the equations of the lines. Two, do the same with the vertical lines.
I can't really think of any reasonable way to distinguish any one of these (or another) polynomial over other candidates.
In contrast, with the definition that a cubic polynomial requires its coefficients to be of degree 3, you get things like Bezout's theorem:
I was bothered as I was writing my comment by an issue related to this one. A curve covering a horizontal/vertical 3x3 grid of points cannot be a polynomial, as, if it is interpreted as a function, it has the range {true,false}, but some other domain, probably ℝ^2. The result I mention should be applicable to a set of points which can be represented as a function from ℝ to ℝ; no curve covering two different points which are vertically aligned can be represented that way (well, you could transpose the axes, but for a 3x3 grid that won't work either).
I do have some questions, because I don't fully understand what you're saying:
Suppose the 9 points are {-1,0,1} × {-1,0,1}. The horizontal line equations are y = -1, y = 0, and y = 1. If I multiply those together, I get y^3 = 0, which is a single horizontal line. I feel I must have done something wrong there.
If I instead do (y + 1)(y - 0)(y - 1) = 0 I get y^3 - y = 0, which at least has y = -1, y = 0, and y = 1 as solutions. Since a cubic equation can't have four roots, this graph consists exactly of three horizontal lines. It's not a polynomial, since it has the " = 0" constraint embedded. (Using the definition on wikipedia, "[i]n mathematics, a polynomial is an expression consisting of variables (or indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents".) Is this what you meant?
I know I've seen a diagram of exactly the example you're talking about somewhere, as something in the spirit of "math fun facts". But I can't find it; do you know of a writeup you could point me toward?
I don't know of a good writeup, but y^3 - y = 0, or alternatively x^3 - x = 0, among other polynomials, is what you want.
In "ordinary" algebra, typically we consider polynomials of one variable, which we write f(x). The graph of the polynomial represents the set of solutions to y = f(x), or alternatively y - f(x) = 0.
In algebraic geometry, the solution set to y - f(x) = 0 is indeed a plane curve, but we think of y - f(x) as a special case of a polynomial in the two variables x and y. All such polynomials (except constant polynomials) also define plane curves. This is the concept that generalizes.
In particular: y^3 - y = 0 and x^3 - x = 0 are both plane curves, they are both polynomials in x and y (i.e. we can think of y^3 - y = 0 as a polynomial in not only y but also x, even though x doesn't appear), and the intersection of these two curves is the set of nine points in question.
Think of the result that you can fit a polynomial of degree n to go through n+1 points. It's easy to show that the polynomial indicated by that theorem which goes through a set of collinear points must be a line (degree-1 polynomial), no matter how many points are given[1]. But we don't say "as long as the points aren't collinear, this theorem holds"; we just accept that a line is a special case of cubic polynomial.
[1] Proof: the polynomial of degree n going through n+1 points is unique. The line going through n+1 collinear points goes through all of those points, and therefore must be the unique polynomial of degree n to do so.
I do see in the wikipedia article, though, that "cubic surface" appears to be defined to exclude polynomials which have any nonzero term of degree other than three. It's a weird world. :/