Let consider the birational Cremona transformation $\sigma: \mathbb{P}^2 \dashrightarrow \mathbb{P}^2 $ defined by $$(X:Y:Z) \mapsto (X^{-1}:X^{-1}: Z^{-1})=(YZ:XZ:XY)$$. 


In language of birational geometry the sections $YZ,XZ,XY \in H^0(\mathbb{P}^2, O\_{\mathbb{P}^2}(2))$ induce it as 


$$\sigma: \mathbb{P}^2 \backslash B = \mathbb{P}^2\_{YZ} \cup \mathbb{P}^2\_{XZ} \cup \mathbb{P}^2\_{XY} \to \mathbb{P}^2$$.


with the *base locus* $B= \operatorname{Supp}V(YZ) \cap \operatorname{Supp}V(XZ) \cap \operatorname{Supp}V(XY) $ where $\sigma$ is not defined and $\mathbb{P}^2\_{YZ}:= \{p \in \mathbb{P}^2 \ \vert \ (YZ)\_p \neq 0 \}$ the non vanishing locus. Similar for $XZ$ and $XY$.


By construction $\mathbb{P}^2\_{YZ}= \sigma^{-1}(D\_+ (X)), \mathbb{P}^2\_{XZ}= \sigma^{-1}(D\_+ (Y)), \mathbb{P}^2\_{XY}= \sigma^{-1}(D\_+ (Z))$. Obviously $\sigma$ is not defined at points $(0:0:1), (0:1:0),(1:0:0) \in B$.


*Up to now we introduced some notations; now the essential part of my question*:


Let $V(F)=C \subset \mathbb{P}^2 $ be a curve defined by homogeoneous polynomial $F(X,Y,Z)= \sum\_{i+j+k=n}a\_{ijk}X^iY^jZ^k$ with $deg(F)=n$.


We assume that $C$ has at point $(0:0:1)$ multiplicity $d$, $(0:1:0)$ multiplicity $e$ and $(1:0:0)$ multiplicity $f$.


Recall, how multiplicity is defined: $(0:0:1) \in D\_+(Z)$ and when we dehomogenize $F$ with respect $Z$ we get a new polynomial $f^Z$ in variables $x=X/Z,y=Y/Z$ that has structure 


$$f^Z(x,y)= \sum\_{i+j+k=n}a\_{ijk}x^iy^j= \\ \sum\_{i+j+k=n, i+j=d}a\_{ijk}x^iy^j + \sum\_{i+j+k=n, i+j >d}a\_{ijk}x^iy^j =: f^Z\_d +f^Z\_{>d} $$


Recall again that by definition of multiplicity $d$ is minimal natural number with this property: for all $a\_{ijk} \neq 0$ we have $i+j \ge d$. This is equivalent to say that $d$ is maximal with the property that the $d$-th power $(x,y)^d$ of the maxiaml ideal $(x,y) \subset k[x,y]$ contains $f^Z(x,y)$.


Analogously for multiplities $e$ and $f$ with respect to $(0:1:0)$ and $(1:0:0)$.


**QUESTION:** Why the pushforward of $C$ by $\sigma$ is the vanishing set


$$ \sigma\_\*C = (X^{-f}Y^{-e}Z^{-d}F(YZ,XZ, XY) =0) \ \ \ ?$$


By symmetry reasons it is clear that it suffice to show locally that this holds for restriction to affine open subscheme $D\_+(Z)= \operatorname{Spec} \ k[X/Z,Y/Z] \subset \mathbb{P}^2$.


We know from construction that 


$$\sigma\_\*C(D\_+(Z))=C(\sigma^{-1}D\_+(Z))= C(\mathbb{P}^2\_{XY})= \\ C(D\_+(XY))= C \vert \_{D\_+(X)}(D(y)) = k[y,z]\_y/f^X(y,z).$$


Recall, $k[y,z]\_y= k[y^{\pm 1},z]$ is the localization 
by $y$. If we now denote 


$H:= X^{-f}Y^{-e}Z^{-d}F(YZ,XZ, XY)$ and $S:= V(H)$, then $H$ localized 
at $Z$ is $h^Z(x,y):=x^{-f}y^{-e} F(y,x, xy)$ and thus


$$S(D\_+(Z))= k[x,y]/h^Z(x,y).$$


And we have to verify $k[y,z]\_y/f^X(y,z)=k[x,y]/h^Z(x,y)$. 


Although up to now I only used definitions (in correct way?) the equation I have to verify seems not to make any sense. I think that somewhere I have already done a hard error but I can't find it. Can anybody help? Is there an easier way to verify the claim in the question?


Thanks for your help.



Cremona transformation $\sigma: \mathbb{P}^2 \dashrightarrow \mathbb{P}^2 $ and pushforward of divisor
I think that you need to assume that $F$ is not divisible by $X$, $Y$ or $Z$. This being done, the equation of $\sigma\_\*(F)$ on the open subset where $XYZ\not=0$ is given by $F(YZ,XZ,XY)$. Hence, $\sigma\_\*(F)$ is equal to 
$$F(YZ,XZ,XY)X^aY^bZ^c$$
for some integers $a,b,c\in \mathbb{Z}$. As you want to remove $X=0$, $Y=0$ and $Z=0$ from the polynomial $F(YZ,XZ,XY)$, you can choose $a,b,c$ to be $\le 0$ and then define them as $-\alpha,-\beta,-\gamma$ where $\alpha,\beta,\gamma$ are the multiplicities of $F(YZ,XZ,XY)$ along $X=0,Y=0,Z=0$ respectively.


Let us do the case of $X$, the rest being symmetrical. Write $$F=\sum\_{i\ge f} X^{n-i} a\_i(Y,Z)$$ where each $a\_i$ is homogeneous of degree $i$ and $a\_f\not=0$ (as $f$ is for you the multiplicity of $F$ at $[1:0:0]$). Then $$F(YZ,XZ,XY)=\sum\_{i\ge f} (YZ)^{n-i} a\_i(XZ,XY)=X^f(\sum\_{i\ge f} (YZ)^{n-i} X^{i-f}a\_i(Z,Y))$$
which proves that $\alpha=f$. Similarly, you find $\beta=e$ and $\gamma=d$, giving the desired result.


