7
$\begingroup$

Let $F: \mathbb{P}^2 --\to \mathbb{P}^2$ be given by the partial derivatives of $x^2y + x^2z + y^3$ $$ [x:y:z] \mapsto [2x(y+z):x^2 + 3y^2:x^2] $$ How can I find the ideal of the subscheme of $\mathbb{P}^2\times\mathbb{P}^2$ representing the closure of the graph of $F$ where it is well defined?

$\endgroup$
6
  • $\begingroup$ I would guess that the equations cutting out the graph of any rational function $[f_0: \ldots : f_m] : P^n \to P^m$ are $y_i = f_i$ for $i = 0, \ldots, m$... is there something missing from this? They certainly vanish on the graph, and they cut out a subscheme of the correct dimension. I'm not sure how to show that $V(y_i = f_i)$ is integral though... and perhaps that can fail? $\endgroup$ – Lorenzo Najt Feb 13 '17 at 20:10
  • $\begingroup$ @AreaMan I would expect the polynomial to be bihomogeneous. If you look at a closed subscheme of $\mathbb{P}^n\times\mathbb{P}^m$ then you are looking at $Proj$ of a bigraded ring $\mathbb{C}[x_0,\ldots,x_n,y_0,\ldots,y_m]$ modulo some ideal $I$ which is homogeneous. So, $y_0 - x_0^2$ is not bihomogenous, while, $y_0x_1^2 - x_0^2y_1 \in \Gamma(\mathcal{O}(2,1))$ is. $\endgroup$ – 54321user Feb 13 '17 at 20:39
  • $\begingroup$ Yes, you're right. The zero locus of $y_i = f_i$ isn't well defined on pairs of lines... thanks. $\endgroup$ – Lorenzo Najt Feb 13 '17 at 22:36
  • $\begingroup$ What about $y_i f_j = y_j f_i$ ? They will be bihomogeneous, as each $f_i$ is of the same degree, and they will vanish on the image. Then if you work in the afine patch $y_i \not = 0$ (hence $f_i \not = 0$ on the domain, which is affine), this should agree with $y_{k/i} = f_k / f_i$, which are the equations we get for the graph of the corresponding rational map from $A^n \to A^m$ (though I think we have to cover the domain some more to say precisely this). (By the way, this looks like the equations describing a blow up, right...) $\endgroup$ – Lorenzo Najt Feb 14 '17 at 1:29
  • $\begingroup$ And because they are the blow up equations, we know that the locus they cut out is integral and of the correct dimension. I think that is a proof... $\endgroup$ – Lorenzo Najt Feb 15 '17 at 15:11
1
$\begingroup$

Edit: It seems that this is wrong. Here is the Macaulay2 code verifying Richard's counter example:

R = QQ[x_1, x_2, x_3, y_1, y_2, y_3]

I = ideal(y_2*x_2*x_3 - y_1*( x_1*x_3 + x_2^2), y_3*x_2*x_3 - y_1* x_3^2, y_ _3 * ( x_1*x_3 + x_2^2) - y_2 *x_3^2)

i11 : radI = radical I

i12 : radI == I

o12 = false

(This is the first hint that something is wrong -- the ideal should definately be radical if it is supposed to cut out the reduced induced subscheme!)

J = ideal(x_2,x_3)

W = radI:J

I think the error is in the argument trying to show that these equations cut out the blow-up (Claim 1).


Let $\phi [f_0 : \ldots: f_m] : P^n \to P^m$ be a rational function, with $f_i \in O(k)$ for some $k$. Let $x_i$ be the homogeneous coordinates on $P^n$, and $y_i$ the homogeneous coordinates on $P^m$.

Define the graph of $\phi$, $\Gamma_{\phi}$, to be the closure in $P^n \times P^m$ of the set $\{ (u, \phi(u)) : u \in U \} \subset P^n \times P^m $, where $U$ is a nonempty open set on which $\phi$ is regular. As a scheme $\Gamma_{\phi}$ is given the reduced induced subscheme structure.

Let $Y$ be the scheme cut out by $g_{ij} = y_i f_j - y_j f_i= 0$, for all pairs of $i,j = 0 \ldots m$.

Claim 1: The projection onto the first coordinate map $\pi : Y \to P^n$ coincides with the blow-up map $\rho: Bl_X P^n$, where $X$ is the subscheme $V(f_0, \ldots, f_k)$ in $P^n$.

Claim 2: The equations $g_{ij} = y_i f_j - y_j f_i= 0$, for all pairs of $i,j = 0 \ldots m$, cut out $\Gamma_\phi$ in $P^n \times P^m$.

Corollary: The graph (with projection) is the blow up (with its projection). Thus, blow ups resolve rational maps, in the sense that the projetion from the graph onto the second factor is regular, and agreed with the rational map on its domain of definition.

Proof of Claim 1:

We cover $P^n$ with affine patches $Spec B$ (say $D(x_i)$), where $X \cap Spec B = X' = V(g_0,\ldots, g_m)$ - I mean in particular that $f_i$ pulls back to $g_i$.

Next we prove / recall that on these patches, $Bl_{X'} Spec B = Spec B \times Proj[Y_0, \ldots, Y_m] / (Y_i g_j = Y_j g_i)$:

In 22.3.1 in Ravi Vakil's notes, it is shown that $Bl_{V(x_1, \ldots, x_n)} spec \mathbb{Z} [x_1,\ldots, x_n]$ is the locus in $spec \mathbb{Z} [x_1, \ldots, x_n] \times Proj \mathbb{Z} [X_1, \ldots, X_n]$ cut out by $x_i X_j - x_j X_i = 0$.

Consider a map $\mathbb{Z} [e_1, \ldots, e_m] \to B$, sending $e_i$ to $g_i$, so that $X' = V(g_i)$ is the scheme theoretic pullback of the origin. Thus we know that $Bl_{X'} Spec B = Spec B \times Proj[Y_0, \ldots, Y_m] / (Y_i g_j = Y_j g_i)$, by what is called the blow-up closure lemma in Ravi Vakil's notes.

Because of naturality of the blow up, since we showed that $Bl_X P^n \to P^n$ pulls back $Spec B \times Proj[Y_0, \ldots, Y_m] / (Y_i g_j = Y_j g_i) \to Spec B$ on the patch $Spec B \to P^n$, and since we know that $P^n \times Proj[Y_0, \ldots, Y_m] / (Y_i f_j = Y_j f_i) \to P^n$ pulls back to the same (literally the same scheme over the same base, with the same map, so descent condition on overlaps is automatic I hope / think), we can patch together an isomorphism from the blow up to $V(f_i Y_j - f_j Y_i | i,j = 0 \ldots m) \subset P^n \times P^m$.

Proof Claim 2:

Let $Y = V(g_{ij})$.

a) First note that the $g_{ij}$ are bihomogeneous of bi-degree $(k,1)$ and hence in particular are global sections of $O(k) \boxtimes O(1)$. Thus they define a subscheme $V(g_{ij}) = Y$.

We pass to the affine patch of the form $D(f_i) \times A^m$ defined by setting $y_i \not = 0$. On this patch, our function is regular and takes the form $\phi_i = (f_0 / f_i, \ldots , f_i / f_i = 1, \ldots, f_m / f_i)$. The equations $g_{ij}= y_i f_j - y_j f_i = 0$ can be rewritten $\frac{y_j}{y_i} = \frac{f_j}{f_i}$. Thus they cut out the image of $\phi_i$ in $D(f_i) \times A^m$.

Note that $\cup D(f_i) = U \subset P^n$, hence we have shown that $im(\phi) = \Gamma_{\phi} \cap (U \times P^m = V(g_{ij}) \cap (U \times P^m)$. This implies that $Y \supset \Gamma_{\phi}$, since the latter is the closure of this set.

b) Since we know that $Y = Bl_X P^n$, we know that $Y$ is dimension n and integral (if we blow up an integral scheme, it's still integral - except for a very stupid case, which is when we blow up $Z$ at $Z$... the empty set isn't irreducible right??). Thus $Y = X$, since if we have a containment of two integral schemes of the same dimension, they are equal.

$\endgroup$
9
  • $\begingroup$ There's something to question here, whether the answer is right or wrong. I had an example where the indeterminate $X$ was a point, yet $J = <f_0, f_1, f_2>$ was not prime. I ended up with the scheme Y both non-reduced and reducible. The problem was only resolved by taking $Y$ to be the scheme given by the ideal $rad<g_{ij}>:J'$. where $J' = rad J$ which is the ideal of the point. $\endgroup$ – Richard Birkett Oct 2 '19 at 19:33
  • $\begingroup$ @RichardBirkett It could definitely be wrong -- I'm not an expert on this now, and I certainly wasn't a few years ago when I wrote this answer while studying Ravi Vakil's notes. Do you have the example? $\endgroup$ – Lorenzo Najt Oct 2 '19 at 19:47
  • $\begingroup$ The example is given by expanding $(x_1, x_2) \mapsto (x_2, x_1 + x_2^2)$ to $\mathbb{P}^2$. We get that the indeterminate point is $[1:0:0]$ and when you take the ideal $I = \langle g_{ij}\rangle$ (or its radical) you get that every term contains $x_2$ or $x_3$, hence the (reduced) scheme cut out here has a full $\mathbb{P}^2$ in the fibre over $[1:0:0]$. However this is wrong. When you cut away the fiber and close up, by taking $P = \sqrt{I}: \langle x_2, x_3 \rangle $ we find $P$ is prime and then of $V(P)\cap V(\langle x_2, x_3 \rangle) $ is merely $V(\langle x_2, x_3, y_3\rangle)$. $\endgroup$ – Richard Birkett Oct 2 '19 at 20:50
  • $\begingroup$ @RichardBirkett I agree with you. When you extend that function to a rational map, you get $[(X_2/X_3) , (X_1/X_3) + (X_2/X_3)^2: 1 ] = [X_2 X_3 : X_1 X_3 + X^2_2 : X_3^2 ]$. There is a full $P^2$ in the fiber above $[1:0:0]$, when using the given equations. I also checked with Macaulay2 that the ideal of $g_{ij}$ is not radical, and it's colon ideal is as you as computed. So I agree with you that something is wrong. It does seem to be the case that the closure of the graph is the blow-up. I would guess that my mistake was in claiming these equations cut out the blow -up. $\endgroup$ – Lorenzo Najt Oct 3 '19 at 1:24
  • $\begingroup$ Even worse, its not a blow up! In fact the scheme I wrote is singular at ([1:0:0],[0:1:0]). If it was a blow up of a point on a smooth surface it should be smooth. You can graph it effectively and see it has corners. Technically this can be verified by checking the partial derivatives of the generators of the ideal $P$ and seeing that the rank is $1$ when it should be $4-2=2$. $\endgroup$ – Richard Birkett Oct 3 '19 at 20:50

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.