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I am reading "Algebraic Geometry, a first course", then I can't solve the following question that is an application of Bertini's theorem:

Exercise $17.17$: Use Bertini's theorem to show that (a) the general hypersurface of degree $d$ in $\mathbb{P}^{n}$ is smooth and more generally (b) for $k<n$ if $F_{1},...,F_{k}$ are general homogeneous polynomials of degree $d_{1},..,d_{k}$ in $n+1$ variables the corresponding hyperurfaces in $\mathbb{P}^{n}$ intersect transversely in a smooth $(n-k)$-dimensional variety.

In the book, Bertini's theorem is stated as: "If $X$ is any quasi-projective variety, $f: X \to \mathbb{P}^{n}$ a regular map, $H\subset \mathbb{P}^{n}$ a general hyperplane, and $Y = f^{-1} (H)$, then $Y_{sing}=X_{sing}\cap Y$.

What is a general hypersurfaces? Because I know that there is a hypersurface that is not smooth, then general does not mean any hypersurface.

I really don't have idea how to solve this question.

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Let's first discuss the term "general". There is a natural bijection between the set of hyperplanes $H \subset \mathbb P^n$ and $\mathbb P^n$ itself. To spell it out, this bijection maps a hyperplane $V(a_0 x_0 + \dots + a_n x_n) \subset \mathbb P^n$ to its dual point $[a_0 : \dots : a_n ]$ in the dual $\mathbb P^n$. We now endow the set of hyperplanes $H \subset \mathbb P^n$ with a topology: the Zariski topology on the dual $\mathbb P^n$. And when we say that a statement is true "for a general hyperplane", we mean that the statement is true on an open subset of the space of all hyperplanes, where openness is defined with respect to this topology. (Or intuitively, the statement is true for all hyperplanes except "special" ones.)

Now, let's work out some special cases of the statement of Bertini's theorem that you've written down. First, let start with:

If $X$ is a smooth projective variety in $\mathbb P^n$ and $H \subset \mathbb P^n$ a general hyperplane, then $Y := X \cap H$ is smooth.

This follows from the statement you wrote down, with $f$ being the natural embedding of $X$ into $\mathbb P^n$. (Note that $X$ is assumed smooth, hence $X_{\rm sing} = \emptyset$, hence $Y_{\rm sing} = X_{\rm sing} \cap Y = \emptyset$, hence $Y$ is smooth.)

If $X$ is a smooth projective variety in $\mathbb P^n$ and $V(f) \subset \mathbb P^n$ is a general hypersurface of degree $d$, then $X \cap V(f) $ is smooth.

Here, the trick is to embed $X$ into $\mathbb P^{{{n+d}\choose n} - 1}$ using the Veronese embedding. The degree-$d$ hypersurface in $\mathbb P^n$, when viewed within this $\mathbb P^{{{n+d}\choose n} - 1}$, is a hyperplane. When we talk of general degree-$d$ hypersurfaces in $\mathbb P^n$, we're talking about general hyperplanes in this $\mathbb P^{{{n+d}\choose n} - 1}$. Once you appreciate this Veronese embedding trick, the new version of the statement is follows in the same way as the previous one.


Finally, let's address the task at hand. We prove the result by induction.

  • For a general $f_1$, $V(f_1) = \mathbb P^n \cap V(f_1)$ is smooth (applying Bertini to $X = \mathbb P^n$ itself).
  • If we then pick a general $f_2$, then $V(f_1, f_2) = V(f_1) \cap V(f_2) $ is smooth (applying Bertini to $X = V(f_1)$, remembering that $V(f_1)$ is smooth from the previous step).
  • If we then pick a general $f_3$, then $V(f_1, f_2, f_3) = V(f_1, f_2) \cap V(f_3)$ is smooth (applying Bertini with $X = V(f_1, f_2)$ this time).

And so on...

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  • $\begingroup$ I have a question how this relates to the hypersurface intersection in $\mathbb{P}^n$. Would this be irreducible as the preimage of the irreducible variety in the bigger space? $\endgroup$
    – vasar
    Nov 17, 2020 at 12:54
  • $\begingroup$ also I wonder how this argument works with different degree hyperplanes.. does it work with the same embedding? or would we embed into a much larger space? sorry been thinking about this for a while and I'm confused $\endgroup$
    – vasar
    Nov 17, 2020 at 15:45

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