Let $f \in \mathbb{C}[x_1, ..., x_n]$ be a homogeneous polynomial of degree $d$. I was trying to understand the definition of degree of hypersurfaces.

It says on Wikipedia https://en.wikipedia.org/wiki/Degree_of_an_algebraic_variety that "The degree of a hypersurface F = 0 is the same as the total degree of the homogeneous polynomial F defining it (granted, in case F has repeated factors, that intersection theory is used to count intersections with multiplicity, as in Bézout's theorem)."

I was a bit confused with this. So is the degree of a hypersurface $f = 0$ always the total degree of $f$ or is it different when $f$ has a repeated factor?

Thank you!

  • $\begingroup$ When working with varieties, you assume that all generating polynomials are irreducible. This is remedied in the theory of schemes, where repeated factors are incorporated into the structure sheaf. $\endgroup$ – Arthur Jun 27 '15 at 18:28
  • $\begingroup$ @Arthur I see, thank you. So this definition of degree of hypersurface is only for when $f$ is irreducible? $\endgroup$ – Johnny T. Jun 27 '15 at 18:34

1) A nonzero homogeneous polynomial $f(x_1,\cdots, x_n)\in \mathbb C[x_1,\cdots, x_n]$ has a unique degree $d$, which it is unnecessary to call "total" degree.
For example $6x_1^3x_2^2-(1+i\sqrt{19})x_2x_3^4+\frac {22} {7}x_3^5 \in \mathbb C[x_1,x_2, x_3]$ is homogeneous of degree $d=5$.

2) A hypersurface $V(f)\subset \mathbb P^{n-1}(\mathbb C)$ of degree $d$ is determined by a homogeneous polynomial $f(x_1,\cdots, x_n)\in \mathbb C[x_1,\cdots, x_n]$ of degree $d$ and two polynomials $f,g\in \mathbb C[x_1,\cdots, x_n]$ determine the same hypersurface $V(f)=V(g)$ if and only $f=\lambda g$ for some $\lambda\in \mathbb C$.

3) And strangely the definition of "hypersurface" is not very important (which is why I didn't give it to you!) : there have been several definitions since more than a century, the latest (dating from the late 1950's) being through the notion of scheme.
But the brilliant Italian, German, French, English, ... algebraic geometers of the nineteenth century knew very well that the line $x_1=0$ is completely different from the conic $x_1^2=0$ !
To sum up: if $f=f_1^{e_1} \cdots f_r^{e_r}$ is the factorization of $f$ into irreducibles, you may consider the polynomial (of lower degree) $f_{red}=f_1 \cdots f_r$ but you must be aware that $V(f)$ and $V(f_{red})$ are different hypersurfaces.

  • $\begingroup$ Thank you very much for this very clear answer! I just have a quick question. Is it correct then that $V(f)$ and $V(f_{red})$ given in your answer are the same as set of points, but they are different hypersurfaces and of different degree, degree $(\sum e_i \deg f_i)$ and degree $\sum \deg f_i$, respectively? $\endgroup$ – Johnny T. Jun 29 '15 at 1:00
  • $\begingroup$ Dear Johnny, yes: everything you write is correct. $\endgroup$ – Georges Elencwajg Jun 29 '15 at 5:30

You have to consider the hypersurface as a scheme (which remembers multiplicity), not just as a set of points. For example, consider the "double point" $\newcommand{\Spec}{\operatorname{Spec}}\Spec \newcommand{\C}{\mathbb{C}}\C[x]/(x^2) \to \Spec \C[x] = \newcommand{\aff}{\mathbb{A}}\aff^1$ in the affine line. As sets, this is just the inclusion $\{0\} \subset \C$. However, $\C[x]/(x^2)$ has length $2$ as a $\C$-module, so the scheme structure can see that the point is "doubled". The case of higher-dimensional hypersurfaces is similar, and that's what's meant by counting with multiplicity.

As long as you count with multiplicity, the degree of the hypersurface is exactly the degree of the defining polynomial, whether or not there are repeated factors. Forgetting multiplicities is equivalent to forgetting how many times each factor occurs.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.