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…
Best Answer
You are absolutely right: those definitions in the book are rather sloppy!
If you have a homogeneous polynomial $F(X_0,\dots,X_n)$ of degree $d$ you should decompose it into irreducible factors as $F=F_1^{m_1}\dots F_r^{m_r}$ and associate to this decomposition the so-called divisor $$V(F)=m_1V(F_1)+\dots+m_rV(F_r)$$
The sets $V(F_i)$ are called the irreducible components of the divisor $V(F)$ and the set $V_{red}(F)=V(F_1)\cup\dots \cup V(F_r)$ is called the support of the divisor $V(F)$.
The degree of the divisor $V(F)$ is of course the degree of $F$ and $$\operatorname {deg} F=\sum m_i \operatorname {deg} F_i$$
Contrary to what one might naïvely think it is not a good idea to replace $F$ by the product without multiplicities $F_{red}=F_1\dots F_r$.
The reason is that if you consider a family of irreducible polynomials like, say, $$F_t(X_0,\dots,X_n)=X_0^2+t(X^2_1\dots+X_n^2)$$ you want the limit of the $V(F_t)$'s for $t$ tending to zero (strangely, this makes sense!) to be $V(F_0(X_0,\dots,X_n))=V(X^2_0)$ and not $V(X_0)$, in order to obtain a flat pencil of divisors , flatness being a very useful more advanced concept.
Conclusion
As you very aptly say we must consider hypersurfaces which know the multiplicities in the polynomials which define them: the divisor concept is the key to that knowledge.
In his much more elementary, freely available, online book Algebraic Curves Fulton defines, at the very beginning of chapter 3, a curve exactly in this way: as a divisor.
[As to canonical bundles, I would advise you to only consider them in the case of an irreducible smooth hypersurface $V(F)$ ($F$ irreducible ) ]