Why this happens for homogeneous polynomials?

Question

I read this:

Definition 1.1. The complex projective line $\mathbb{CP}^1$ (or just $\mathbb{P}^1$) is the set of all ordered pairs of complex numbers $\{(x, y)\in\mathbb{C}^2\mid(x, y)\neq(0, 0)\}$ where we identify pairs $(x, y)$ and $(x', y')$ if one is a scalar multiple of the other: $(x, y)=\lambda(x', y')$ for some $\lambda\in\mathbb{C}^*$, where $\mathbb{C}^*$ is the set of nonzero complex numbers.

$\quad$ So for example $(1, 2)$, $(3, 6)$, and $(2+3i, 4+6i)$ all represent the same point of $\mathbb{P}^1$.

$\quad$ This construction is an example of the quotient of a set by an equivalence relation. See Exercise 2.

$\quad$ The idea is that $\mathbb{P}^1$ can be thought of as the union of the set of complex numbers $\mathbb{C}$ and a single point "at infinity". To see this, consider the following subset $U_0 \subset \mathbb{P}^1$:

$$ U_0 = \{(x_0, x_1) \in\mathbb{P}^1\mid x_0\neq 0 \}.$$

Then $U_0$ is in one-to-one correspondence with $\mathbb{C}$ via the map

$$\tag{1} \phi_0 : U_0 \to \mathbb{C} : \qquad (x_0, x_1) \mapsto \frac{x_1}{x_0}. $$

Note that $\phi_0$ is well defined on $U_0$. First of all, $x_0$ is not $0$ so the division makes sense. Secondly, if $(x_0, x_1)$ represent the same point of $\mathbb{P}^1$ as $(x_0', x_1')$ then $x_0 = \lambda x_0'$ and $x_1 = \lambda x_1'$ for some nonzero $\lambda \in \mathbb{C}$. Thus, $\phi_0((x_0, x_1)) = x_1 / x_0 = (\lambda x_1')/(\lambda x_0')$ $= x_1' / x_0' = \phi_0((x_0', x_1')) $ and $\phi_0$ is well defined as claimed. The inverse map is given by

$$ \psi_0 : \mathbb{C}\to U_0 : \qquad z \mapsto (1, z). $$

$\quad$ The complement of $U_0$ is the set of all points of $\mathbb{P}^1$ of the form $(0, x_1)$. But since $(0, x_1) = x_1(0, 1)$, all of these points coincide with $(0,1)$ as a point of $\mathbb{P}^1$. So $\mathbb{P}^1$ is obtained from a copy of $\mathbb{C}$ by adding a single point.

$\quad$ This point can be thought of as the point of infinity. To see this, consider a complex number $t$, and identify it with a point of $U_0$ using $\psi_0$; i.e., we identify it with $\psi_0(t) = (1, t)$. Now let $t \to \infty$. The beautiful feature is that the limit now exists in $\mathbb{P}^1$! To see this, rewrite $(1, t)$ as $(1/t, 1)$ using scalar multiplication by $1/t$. This clearly approaches $(0,1)$ as $t \to \infty$, so $(0, 1)$ really should be thought of as the point at infinity!

$\quad$ We have been deliberately vague about the precise meaning of limits in $\mathbb{P}^1$. This is a notion from topology, which we will deal with later in Chapter 4. The property that limits exist in a topological space is a consequence of the compactness of the space, and the process of enlarging $\mathbb{C}$ to the compact space $\mathbb{P}^1$ is our first example of the important process of compactification. This makes the solutions to enumerative problems well-defined, by preventing solutions from going off to infinity. A precise definition of compactness will be given in Chapter 4.

$\quad$ We now have to modify our description of complex polynomials by associating to them polynomials $F(x_0, x_1)$ on $\mathbb{P}^1$. Before turning to their definition, note that the equation $F(x_0, x_1) = 0$ need not make sense as a well-defined equation of $\mathbb{P}^1$, since it is conceivable that a point could have different representations $(x_0, x_1)$ and $(x_0', x_1')$ such that $F(x_0, x_1) = 0$ while $F(x_0', x_1') \neq 0$. We avoid this problem by requiring that $F(x_0, x_1)$ be a homogenous polynomial; i.e., all terms in $F$ have the same total degree, which is called the degree of $F$. So

$$\tag{2} F(x_0, x_1) = \sum_{i=0}^d a_i x_0^i x_1^{d-i} $$

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In the last paragraph, I tried to take one non-homogeneous polynomial and checked that indeed, that problem happens but when I took an homogeneous polynomial, the problem vanishes! I am baffled: it looks like sorcery! I tried to explain myself why that happens but couldn't so, why does that happen?

Answer 1

The coordinates of different representatives of the same point differ by a scalar multiple; if $(x_0,x_1)$ and $(x_0',x_1')$ represent the same point then there exists some nonzero scalar $\lambda$ such that $x_0'=\lambda x_0$ and $x_1'=\lambda x_1$. Then for a homogeneous polynomial $F$ of degree $d$ you have $$F(x_0',x_1')=F(\lambda x_0,\lambda x_1)=\lambda^dF(x_0,x_1),\tag{1}$$ which means that $F(x_0',x_1')=0$ if and only if $F(x_0,x_1)=0$. Of course $(1)$ fails for polynomials in general, so we only consider homogeneous polynomials as functions.

Answer 2

What about $F(x) =x+1$? Is it mystifying that $F(-1)=0$ but $F(2\cdot - 1)\neq 0$? The property homogenous polynomials have is that scaling the coordinates scales all coefficients of the polynomial uniformly, which is what you need. That doesn't happen for any polynomial that isn't homogeneous.

Answer 3

Suppose $F$ is homogeneous of degree $d$, so that $$F(x_0, x_1) = \sum_{i = 0}^d a_i x_0^i x_1^{d - i} .$$ Any other nonzero point $(x_0', x_1')$ on the line spanned by $(x_0, x_1) \neq (0, 0)$ can be written as $(\lambda x_0, \lambda x_1)$ for some $\lambda \neq 0$, and substituting into the formula for $F$ gives $$F(\lambda x_0, \lambda x_1) = \sum_{i = 0}^d a_i (\lambda x_0)^i (\lambda x_1)^{d - i} = \sum_{i = 0}^d a_i \lambda^i x_0^i \lambda^{d - i} x_1^{d - i} \stackrel{(\ast)}{=} \lambda^d \sum_{i = 0}^d a_i x_0^i x_1^{d - i} = \lambda^d F(x_0, x_1) .$$ In particular, $F(x_0, x_1) = 0$ if and only if $F(\lambda x_0, \lambda x_1) = \lambda^d (0) = 0$. Put another way, for homogeneous $F$ the equation $F(x_0, x_1) = 0$ is well-defined on $\Bbb P^1$.

On the other hand, if $F$ is not homogeneous, we cannot factor all of the occurrences of $\lambda$ out of the sum as we did in the equality marked $(\ast)$ above, and so the conclusion does not follow.