TITLE: How can hypersurfaces "know" the degree of their defining polynomials?
QUESTION [11 upvotes]: I'm currently trying to learn some complex and projective geometry.
There is one issue bugging me again and again, from different perspectives, and I just can't get my head around it.
One incarnation of my problem:
In the book by Griffiths and Harris, they define the degree of a variety in chapter 1.3.
One of their definitions is:

In case [the variety] $V \subset \mathbb P^n$ is a hypersurface, we have seen that it may be given in terms of homogeneous coordinates $X_0 \dots X_n$ as the locus $V = \big( F(X_0 \dots X_n) = 0 \big)\;$ of a homogeneous polynomial $F$. If $F$ has degree $d$, then [...] $V$ has degree $d$.

A consequence would be for example that the canonical bundle of $V$ is $\mathcal O(d - n - 1) \big|_V$.
My problem is:
As far as I understand it, the object $V$ here is just the hypersurface as a geometrical object (an algebraic variety was defined as the locus of a collection of polynomials and nothing more).
Hence $F$ and $F^2$ would define the same object $V$.
But using $F^2$ instead of $F$ gives us a different degree / a different canonical bundle, which doesn't make sense...

REPLY [2 votes]: Here is a topological way of recovering the degree. If $V \subset \mathbb{CP}^n$ is a (smooth) hypersurface, then by the Lefschetz hyperplane theorem the map
$$H^2(\mathbb{CP}^n, \mathbb{Z}) \to H^2(V, \mathbb{Z})$$
is injective. The LHS has a distinguished element $\omega$ given by the Chern class of $\mathcal{O}(1)$, and hence so does the RHS. 
Now, the normal bundle to $V$ in $\mathbb{CP}^n$ is a complex line bundle on $V$, and so it also has a Chern class. If $V$ is a hypersurface of degree $d$ then in fact the normal bundle to $V$ is $\mathcal{O}(d) \mid_V$, and so its Chern class is $d \omega$. 
So we find that we can recover the degree using only 1) the Chern class of $\mathcal{O}(1)$ in $H^2(\mathbb{CP}^n, \mathbb{Z})$ and 2) the almost complex structure on the normal bundle to $V$.