# Irreducible components of a projective variety

I would like to understand the irreducible components of a projective algebraic set. Given an irreducible and homogeneous polynomial $$H(w,x,y)\in \mathbb{C}[w,x,y]$$ we define $$H_i(w,x_0,x_i):=H(w,x_0,x_i)\in \mathbb{C}[w,x_0,x_1,\dotsc,x_n]$$ and the projective algebraic set $$Z(H_1,\dotsc,H_n)\subseteq \mathbb{P}^{n+1}$$.

How many irreducible components of dimension one does this set have? Are all of them isomorphic? Does $$H$$ give some informations about the function field of those curves?

What I suppose is that there should be some symmetries between the these curves, but I don't know how to attack this problem. You may know some references dealing with the same kind of questions (maybe some intersection theory).

This problem arises from the following: I'm interested in finding explicitly non trivial embeddings of curves in a higher dimensional projective space. (By trivial embedding I mean $$Z(H(w,x_0,x_1),x_2−x_1,\dotsc,x_n−x_1)$$, I would like the curve to "spread" among all coordinates.)

• Consider the polynomial $H(w,x,y) = (x+w)y+xw$. One of the irreducible components of your algebraic set is $\text{Zero}(H(w,x_0,x_1),x_2-x_1,\dots,x_n-x_1)$, i.e., a smooth plane conic. Another irreducible component is $\text{Zero}(w,x_0)$, i.e., a projective linear subspace of codimension $2$. These irreducible components have different dimensions (as soon as $n+1 > 3$). Nov 28, 2021 at 22:21
• Thank you for the example. I'm interested to the components of dimension 1 i.e. the curves. Nov 29, 2021 at 10:43
• Please rewrite your question to clarify precisely what you are asking. Nov 29, 2021 at 12:14

We have either $$H(0,0,1)=0$$ or $$H(0,0,1) \neq 0$$. In the first case, the locus $$w=x_0=0$$ is contained in $$Z(H_1,\dots, H_n)$$ and is either an irreducible component of dimension $$n-1$$ or contained in an irreducible component of dimension $$n$$, and in the second case the locus $$w=x_0=$$ does not intersect $$Z(H_1,\dots, H_n)$$, so in either case we can ignore this locus, at least as long as $$n \geq 3$$. (If $$n=2$$, it might provide an irreducible component of dimension 1).
After removing this locus, $$Z(H_1,\dots, H_n)$$ maps to $$\mathbb P^1$$ by the projection with coordinates $$(w:x_0)$$. This projection is the fiber product of $$n$$ copies of the map $$Z(H) \to \mathbb P^1$$ with coordinates $$(w:x)$$. So we are looking at an $$n$$-fold fiber product of a map of curves.
If any fiber of the map $$Z(H) \to \mathbb P^1$$ has positive dimension, i.e. if $$Z(H)$$ contains a line of the form $$aw+bx=0$$, then the fiber product will contain the $$n$$'th power of this line, an irreducible component of dimension $$n$$.
Ignoring these components, we are taking the $$n$$-fold fiber product of a flat morphism of curves, which is a scheme of dimension $$1$$, so every irreducible component has dimension $$1$$. To calculate the irreducible components, we may restrict to an open set where the morphism is finite étale. Then we may calculate components by Grothendieck's Galois theory - the étale morphism $$Z(H) \to \mathbb P^1$$ corresponds to a finite set with an action of the fundamental group of this open subset, and each irreducible component corresponds to an orbit of this group action on the $$n$$'th power of the finite set.
There certainly are symmetries among the components, and it depends entirely on the combinatorics of this monodromy group action. For example, if the degree is $$d$$ and the monodromy group is the full symmetric group $$S_d$$, then components are governed by $$S_d$$-orbits on $$n$$-tuples of choices from a set of $$d$$ letters, which are the same thing as partitions of $$\{1,\dots,n\}$$ into at most $$d$$ parts, and two components are birational if their partitions have the same numbers of parts.