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.)

  • 2
    $\begingroup$ 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 $n+1 > 3$). $\endgroup$ Nov 28, 2021 at 22:21
  • $\begingroup$ Thank you for the example. I'm interested to the components of dimension 1 i.e. the curves. $\endgroup$
    – Vanja
    Nov 29, 2021 at 10:43
  • 1
    $\begingroup$ Please rewrite your question to clarify precisely what you are asking. $\endgroup$ Nov 29, 2021 at 12:14

1 Answer 1


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.

So the exact number of components depends on the degree of this covering and its monodromy group (inside the symmetric group of that degree).

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.


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