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I am trying to show that if $X$ is an irreducible projective variety, its graph under a regular map into projective space is an irreducible subvariety [Harris Ex 2.24]. I tried to modify a proof I gave before for the affine case.

If $X\subset P^n$ is a projective variety, it is the vanishing set of some homogenous polynomials $G_j$ in $n+1$ variables. The regular map $\phi: X \rightarrow P^m$ is of the form $[\phi_0([X_0:\ldots:X_n]):\ldots: \phi_m([X_0:\ldots:X_n])]$, where each $\phi_i$ is regular on $X$. Call the graph of regular map $\phi$, $\Gamma= \{[x:\phi(x)]:x\in X\} \subset P^n\times P^m$. To show that this graph is a variety, we can exhibit a set of polynomials who zero-locus is $\Gamma$. Consider the set of polynomials $S = \{G_j\} \cup \{ x_{n+i}-\phi_i\}$ where $i=0,\ldots, m$ and its vanishing set $V(S)$. First we show $\Gamma \subset V(S)$. Take some element $[x:\phi(x)]$ in $\Gamma$, by definition $x\in X$ so we know first $n+1$ coordinates of $x$ satisfy all the $G_j$ and the last $m+1$ coordinates satisfy them trivially as they do not appear. We need to see that $x$ satisfy the other $m$ polynomial. The polynomials $x_{n+i}-\phi_i$ are trivially satisfied by the definition of the graph thus $\Gamma \subset V(S)$. So we check $\Gamma \supset V(S)$. Satisfying the $x_{n+i}-\phi_i$ polynomials guarantees we are on the bigger graph $\Gamma' = \{(x,f(x)):x\in P^n\}$ and the other polynomials $G_i$ intersect the first $n$ coordinates with $X$ as a subspace, thus giving $\Gamma$. To see that $\Gamma$ is irreducible if $X$ is, we can look at the ideals of the polynomials. We know the ideal generated by $\{G_j\}$ is prime, and the ideal generated by $\{-\phi_i+x_{n+i}\}$ are all prime because of the $x_{n+i}$ term being irreducible and degree 1. Since the ideal generators do not intersect, the ideal generated by their union is still prime and therefore the graph is irreducible.

Two issues I have: the set $S$ needs to be all homogenous polynomials, and the $-\phi_i+x_{n+i}$ are clearly not since $x_{n+i}$ is degree 1 and $\phi_i$ can be any degree. I was thinking of raising $x_{n+1}$ to the degree of $\phi_i$ but I am not sure that would fix anything. Also, then my argument about irreducibility would not work. To be frank, I am not full convinced it works, but I tried several examples and it seems to be the case that if you have generates $S_i$ for a prime ideal and generators $T_i$ for a prime ideal and if $\{S_i\}\cap \{T_i\}=\emptyset$ then $\{S_i\}\cup \{T_i\}$ generates a prime ideal.

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Not all $X \to \mathbb{P}^k$ are of the form $[\phi_0([X_0: \ldots : X_n]), \ldots, \phi_k([X_0: \ldots : X_n])]$ for regular $\phi_i$. Even the identity map $\mathbb{P}^k \to \mathbb{P}^k$ is not of this form. The reason is that on projective varieties, the only globally defined regular maps are constant. –  xyzzyz Apr 5 '13 at 22:03
How is that identity not of that form? Isn't the identity just that with $\phi_i([X_0:\ldots:X_k])=X_i$? If I were to assume all regular maps were of this form where each $\phi_i$ is homogenous, do you have any advice for the question? I agree with you that this form of a regular map does not cover all of them, as I've read in Harris's book. –  Steven-Owen Apr 5 '13 at 22:10
$\phi_i([X_0 : \ldots : X_k]) = X_i$ is not even a map. What's the value of $\phi_0$ on $[1 : 0] = [2 : 0] \in \mathbb{C}\mathbb{P}^1$? –  xyzzyz Apr 5 '13 at 22:11
If $\phi_i$ are homogeneous of the same degree, and have no common zero, then indeed they define a morphism $X \to \mathbb{P}^k$. Still, not all morphisms between projective varieties are of this form. To talk about line bundles, you don't have to talk about schemes. Topological definition of a vector bundle works in algebraic context just fine: you just demand that all maps involved are algebraic morphisms. –  xyzzyz Apr 5 '13 at 22:37
Subvarieties of $\mathbb P^n \times\mathbb P^m$ are given by bihomogeneous polynomials, not homogeneous ones. Also, irreducibility is a purely topological concept: the image of an irrreducible topological space under a continuous map is irreducible. –  Georges Elencwajg Apr 6 '13 at 9:42
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1 Answer

up vote 6 down vote accepted

1) The key to your problem is to remember that a subset $T\subset \mathbb P^n\times \mathbb P^m$ is closed (=is a subvariety) if it is the zero locus $F_i(z,w)=0 \; (i\in I)$ of a family $F_i(Z,W)$ of bihomogeneous polynomials in the variables $Z=(Z_0,\ldots,Z_n), W=(W_0,\ldots,W_m)$ where $Z$ and $W$ are the homogeneous coordinates of $\mathbb P^n$ and $ \mathbb P^m$.

2) Suppose now that $X\subset \mathbb P^n$ is given by the vanishing of the polynomials $G_j(Z)$.
There is a covering of $\mathbb P^n$ by open subsets $U\subset \mathbb P^n$ such that on $U\cap X$ the morphism $\phi$ is given by $\phi(x)=[\phi_0(x):\ldots:\phi_m(x))]$ where the $\phi_k(Z)$ are homogeneous polynomials of the same degree $r$. Then the graph graph $\Gamma$ is given in $U\times \mathbb P^m$ by the vanishing of the family consisting of the $G_j(Z)$ and of the $W_i\phi_k(Z)-W_k\phi_i(Z)$, these last polynomials being of bidegree $(r,1)$ in the variables$(Z,W)$.

3) So $\Gamma\cap (U \times \mathbb P^m)\subset U \times \mathbb P^m$ is closed and since these $U \times \mathbb P^m$ form an open covering of $\mathbb P^n\times \mathbb P^m,$ conclude that $\Gamma$ is closed in $\mathbb P^n\times \mathbb P^m$.

4) As usual in Harris's book the exercise is undoable by a beginner in algebraic geometry, due to the rarity or absence of preliminary explicit calculations .
Beginners can by all means consult Harris, a treasure trove of classical algebraic geometry, but should use it in conjunction with another book to learn the basics and find reasonable, encouraging exercises.

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Indeed bihomogeneity is key when dealing with things like $\Bbb{P}^n \times \Bbb{P}^m$ as I found out when doing a problem on the Segre embedding! –  fpqc Apr 6 '13 at 13:12
First of all, thank you so much for your help. Second, Which book do you recommend for an absolute beginner? Lots of books seem to offer pretty convoluted definitions for concepts that I imagine have rich intuitive meaning and a lot of the beauty seems to be getting lost in commutative algebra (which I also have never studied). –  Steven-Owen Apr 6 '13 at 14:44
Dear ricky: you are welcome. And here is a link which you might find useful in helping you to choose a suitable textbook. I am a great admirer of Perrin's and Shafarevich's books mentioned there. –  Georges Elencwajg Apr 6 '13 at 18:29
@GeorgesElencwajg Dear Georges, recently I posted a question here but the answer I received does not seem satisfactory. Would you like to post an answer to my question? Thanks. –  fpqc Jun 21 '13 at 23:44
Dear @Benja, unfortunately I have too little time now (and moreover I'm not sure at all that with more time I could give a good answer!). Thanks for asking, anyway. –  Georges Elencwajg Jun 23 '13 at 7:58
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