How can I find the scheme representing the graph of the closure of this rational map?

Question

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?

Answer 1

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).

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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.