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Let $k$ be an algebraically closed field of characteristic $0$.

For $\alpha :=(a_1,\dots,a_{n+1})\in \mathbb N^{n+1}_{\ge 0}$ , let $\bar x^{\alpha}:= x_1^{a_1} \dots x_{n+1}^{a_{n+1}} \in k[x_1,\dots, x_{n+1}]$.

Consider a rational map $f=(f_1: \dots : f_{n+1}): \mathbb P^n \to \mathbb P^n$ where each

$f_i =\bar x^{\alpha_i}$ for some $\alpha_i \in \mathbb N^{n+1}_{\ge 0}$ and each $f_i$ has same degree say $d \ge 1$.

I am looking for a moderately elementary proof of the fact that $f$ is birational if and only if $\det (\alpha_1\dots\alpha_{n+1} ) = \pm d$ . So basically I'm looking for an elementary proof of Proposition 2.1 of this paper (Simis-Villarreal, Proc AMS 2002) when $m=n$ .

I would very much appreciate it if I can at least get such a proof for $n=2$.

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There is no characteristic restriction needed.

Here's a direct proof for $n=2$ (below the general case is proved based on elementary linear algebra). I write coordinates $(x:y:z)$.

Since your monomials are coprime, the degree of $x$ in one of the coordinates vanishes. Similarly for $y$ and $z$. If they all vanish at the same coordinate, we have $d=0$ which is excluded since $f$ is not constant. Otherwise, we have two cases (A) they vanish at distinct coordinates, (B), two vanish at one coordinate and the third at another. Then, after permuting the coordinates (which only affects the given determinant by change of sign), we have:

Case (A): $$(x:y:z)\mapsto (y^{D-a}z^{D+a}:x^{D-b}z^{D+b},x^{D+c}z^{D-c}),$$ with degree $d=2D$, and $D\pm a,D\pm b,D\pm c$ non-negative integers;

Case (B): $$(x:y:z)\mapsto (x^d:y^az^{d-a}:x^by^cz^{d-b-c}),$$ with degree $d$, and $a,b,c$ non-negative integers with $a,b+c\le d$.

In both cases, I will use affine coordinates ($(x,y)\mapsto (x:y:1)$). A monomial map in affine coordinates $(x,y)\mapsto (x^sy^t,x^uy^v)$ with $M=\begin{pmatrix}s & t\\ u & v\end{pmatrix}\in\mathrm{Mat}_2(\mathbf{Z})$ is dominant if and only $\det M\neq 0$ (clear), and birational if and only if $\det M=\pm 1$ (indeed when $|\det(M)|\neq 0$ generic fibers have cardinal $|\det(M)|$ by an easy linear algebra argument).

In case (A), in affine coordinates ($(x,y)\mapsto (x:y:1)$) it writes as $$(x,y)\mapsto (x^{-D-c}y^{c-a},x^{b-c}y^{c-D}).$$ In this case the affine determinant is $\delta=\begin{vmatrix}-d-c & c-a\\ b-c & c-d\end{vmatrix}=D^2+ab-ac-bc$. While the original determinant (as in the question) is $$\Delta=\begin{vmatrix}0 & D+b & D+c\\ D-a & 0 & D-c \\ D+a & D-b & 0 & \end{vmatrix}=(D-a)(D-b)(D+c)+(D+a)(D+b)(D-c)=2D\delta=d\delta.$$

In case (B), in affine coordinates, it writes as $$(x,y)\mapsto (x^{d-b}y^{-c},x^{-b}y^{a-c}).$$ In this case the affine determinant is $\delta=\begin{vmatrix}d-b & -b\\ -c & a-c\end{vmatrix}=da-dc-ab$; the original determinant is $$\Delta=\begin{vmatrix}d & 0 & b\\ 0 & a & c\\ 0 & d-a & d-b-c\end{vmatrix}=d\begin{vmatrix}a & c\\ d-a & d-b-c\end{vmatrix}=d\delta.$$

In both cases we deduce $\Delta=d\delta=\pm d$, since $\delta=\pm 1$.


Now let's prove the general case.

First start with the necessary purely linear-algebraic elementary background. let $K$ be a field (or even ring) of coefficients. Define $\Psi:\mathrm{M}_n(K)\to\mathrm{M}_{n+1}(K)$ by $$\Psi((a_{ij})_{1\le i,j\le n})=(b_{ij})_{1\le i,j\le n+1} ,\quad b_{ij}=a_{ij},\;b_{i,n+1}=0,\;b_{n+1,j}=-\sum_{\ell=1}^na_{\ell j}.$$ That is, $\Psi(A)$ is obtained from $A$ by adding a zero last column and filling the last line so as to make all columns have sum zero.

Define $\mathrm{M}_{n+1}^\sharp(K)$ be the subalgebra of matrices $A$ all of whose columns have the same sum $d(A)$. If $v$ is the line-matrix $(1,\dots,1)$, then $\mathrm{M}_{n+1}^\sharp(K)$ is the set of $A\in\mathrm{M}_{n+1}(K)$ such that $Av$ is collinear to $v$, and then $Av=d(A)v$. In particular, $d:\mathrm{M}_{n+1}^\sharp(K)\to K$ is a $K$-algebra homomorphism.

Let $Q(K)$ be the set of matrices in $\mathrm{M}_{n+1}(K)$ all of whose lines are constant. Clearly $Q(K)\subset\mathrm{M}_{n+1}^\sharp(K)$, and, restricted to $Q(K)$, $d$ equals the trace.

Then $\det(\Psi(A)+q)=\mathrm{Trace}(q)\det(A)$ for all $A\in\mathrm{M}_n(K)$ and all $q\in Q(K)$.

[Indeed, after a change of basis, this amounts to the same formula, when instead $(\Psi,Q(K))$ is replaced with $(\Psi',Q'(K))$, where $\Psi'$ is defined as the standard embedding (adding only zeros on both the last column and last line) and $q'(K)$ is the set of matrices that is zero outside the last column, and in this case this the formula is obvious.]

Now assume that there is a total ordering on $K$, invariant under addition, and define $q(A)$ by $q_{ij}=-\min(b_{i1},\dots,b_{in})$, where $B=\Psi(A)$. Define $\Phi(A)=\Psi(A)+q(A)$. By construction, $\Phi(A)\in\mathrm{M}_{n+1}^\sharp(K)$, it has entries $\ge 0$, and the min of each line is zero. From the last formula, we have

For every $A\in\mathrm{M}_n(K)$, one has $\det(\Phi(A))=\mathrm{Trace}(q(A))\det(A)$.

Now let us explain why this is exactly what we need, namely that $\Phi$ encodes the map that passes a monomial transformation from its affine to its projective form, and that $\mathrm{Trace}\circ q$ is precisely the degree.

Precisely, start from $A=(a_{ij})_{1\le i,j\le n}\in\mathrm{M}_n(\mathbf{Z})$. It defines the monomial transformation $f_A$ in affine coordinates $(z_1,\dots,z_n)\mapsto (w_1,\dots,w_n)$ with $w_i=\prod_{j=1}^n z_j^{a_{ij}}$ (which is defined at least when all $z_i$ are nonzero). In projective coordinates this is $(z_1:\dots:z_n:1)\mapsto (w_1:\dots:w_n:1)$. Hence $f_A$ maps $$(z_1:\dots :z_{n+1})\mapsto (w'_1:\dots,w'_n:1),\;w'_i=\prod_{j=1}^n (z_j/z_{n+1})^{a_{ij}}$$ (defined at least when all $z_i$ are nonzero). Hence, $w'_i=\prod_{j=1}^{n+1}z_j^{b_{ij}}$ for all $1\le i\le i+1$, where $B=(b_{ij})_{1\le i,j\le n+1}=\Psi(A)$.

By multiplying by a suitable monomial, we can clear denominators and common factors in the numerators, and get $f_A$ in the form. $$(z_1:\dots :z_{n+1})\mapsto (\xi_1:\dots,\xi_n:\xi_{n+1}),\;\xi_i=\prod_{j=1}^{n+1} z_j^{c_{ij}}.$$ By definition $\Phi(A)$ is the matrix $(c_{ij})_{1\le i,j\le n+1}$. This is a matrix with entries in $\mathbf{N}$ and with the property that all its columns have the same sum $d=d(\Phi(A))=\deg(f_A)$, which is the degree of the above monomial map, and all its lines have minimum zero. Hence by the previous formula (applied to $K=\mathbf{Z}$), we have

For every $A\in\mathrm{M}_n(\mathbf{Z})$ we have $\det(\Phi(A))=\deg(f_A)\det(A)$.

As I wrote in the case $n=2$, if $\det(A)=0$ then $f_A$ is not dominant. If $\det(A)\neq 0$ then $|\det(A)|$ is the cardinal of a generic fiber. Hence $f_A$ is birational iff $\det(A)=\pm 1$. This is therefore (for $n\ge 1$) equivalent to $\det(\Phi(A))=\pm\deg(f_A)\neq 0$.

(Not much to say about degree zero, i.e., $f_A$ constant: it is birational iff $n=0$.)

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  • $\begingroup$ In your proof of first part case (A) , how come the degree is only allowed to be $2D$ an even number ? Can't it very well be an odd number like can't the map be say $(xy^2 : yz^2 : zx^2)$ ? $\endgroup$
    – sdey
    Oct 14 '19 at 2:11
  • $\begingroup$ Or is it that your $D$ is not necessarily an integer ? $\endgroup$
    – sdey
    Oct 14 '19 at 2:15
  • $\begingroup$ @sde sorry for some reason I didn't see your message. No $D$ is not supposed to be an integer. $\endgroup$
    – YCor
    Oct 28 '19 at 22:41
  • $\begingroup$ Why do generic fibres have $|det(M)|$ cardinality when $|det(M)| \neq 0$ $\endgroup$
    – Ri-Li
    Apr 9 '20 at 4:54
  • $\begingroup$ @Ri-Li from the classification of subgroups of $\mathbf{Z}^n$ (i.e., standard linear algebra over $\mathbf{Z}$), this reduces to the case of a diagonal matrix, in which case it's clear. $\endgroup$
    – YCor
    Apr 9 '20 at 8:00
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As it is explained at this MathSE post, the condition is that the determinant of the matrix formed by $\alpha_{ij}$ is equal to $\pm d$.

Here is a simple proof (in any dimension and characteristic). If the map is given by monomials, it is toric. In particular, it restricts to an endomorphism of the torus $$ \{ (x_0:\dots:x_n) \in \mathbb{P}^n \mid x_i \ne 0\ \forall i \} \cong \mathbb{G}_{\mathrm{m}}^n. $$ Since the torus is dense open in the projective space, the map is birational, if and only if it is an automorphism of the torus. Since the map is compatible with the group law of the torus, the condition is that the kernel is trivial. Finally, note that the torus comes in an exact sequence $$ 1 \to \mathbb{G}_{\mathrm{m}} \to \mathbb{G}_{\mathrm{m}}^{n+1} \to \mathbb{G}_{\mathrm{m}}^n \to 1, $$ and the map is induced by the endomorphism of the torus in the middle that is given by the matrix of $\alpha_{ij}$, and the restriction of the map to the 1-dimensional subtorus is equal to $d$. Thus, the endomorphism of the quotient torus is an automorphism if and only if its determinant is $\pm1$, and by multiplicativity of determinant this holds true if and only if the determinant of $(\alpha_{ij})$ is $\pm d$.

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    $\begingroup$ I already know that ... I'm just looking for an elementary proof ... see my post carefully please $\endgroup$
    – sdey
    Oct 6 '19 at 6:48
  • $\begingroup$ Sorry, I was in haste when I was writing the answer, see now a proof. $\endgroup$
    – Sasha
    Oct 6 '19 at 18:33
  • $\begingroup$ Certainly if the map restricted on Torus has an inverse, it'll be Birational, but why is the converse true ? $\endgroup$
    – sdey
    Oct 6 '19 at 19:04
  • $\begingroup$ Since it takes the torus to the torus, it is birational if and only if it is birational on the torus. And on the torus it is a group homomorphism, so it is birational if and only if it is an isomorphism. $\endgroup$
    – Sasha
    Oct 6 '19 at 19:09
  • $\begingroup$ Okay, that makes sense, thanks ... one more question, how do we know that the Torus is mapped onto surjectively ? $\endgroup$
    – sdey
    Oct 6 '19 at 19:14

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