Multivariable Puiseux Theorem for Convergent Generalised Power - PowerPoint PPT Presentation

The ring of formal generalised power series R � x ∗ � . Let x = ( x 1 , . . . , x m ) . α c α x α such that α ∈ [ 0 , ∞ ) m , c α ∈ R and F ( x ) = � Supp ( F ) := { α : c α � = 0 }⊆ S 1 × . . . × S m , where S i ⊆ [ 0 , ∞ ) well ordered. ( F has finitely many minimal monomials) α | c α | r | α | < ∞ (sup of all finite subsums). F convergent if ∃ r ∈ ( 0 , ∞ ) m � F induces a C 0 function (the sum of the series) on [ 0 , r ) m (analytic on ( 0 , r ) m ). R { x ∗ } = � r R { x ∗ } r is the ring of convergent generalised power series . Examples. • f ∈ O m , α i ∈ [ 0 , ∞ ); F ( x ) = f ( x α 1 1 , . . . , x α m m ) . Supp ( F ) ⊆ α 1 N × . . . × α m N .

The ring of formal generalised power series R � x ∗ � . Let x = ( x 1 , . . . , x m ) . α c α x α such that α ∈ [ 0 , ∞ ) m , c α ∈ R and F ( x ) = � Supp ( F ) := { α : c α � = 0 }⊆ S 1 × . . . × S m , where S i ⊆ [ 0 , ∞ ) well ordered. ( F has finitely many minimal monomials) α | c α | r | α | < ∞ (sup of all finite subsums). F convergent if ∃ r ∈ ( 0 , ∞ ) m � F induces a C 0 function (the sum of the series) on [ 0 , r ) m (analytic on ( 0 , r ) m ). R { x ∗ } = � r R { x ∗ } r is the ring of convergent generalised power series . Examples. • f ∈ O m , α i ∈ [ 0 , ∞ ); F ( x ) = f ( x α 1 1 , . . . , x α m m ) . Supp ( F ) ⊆ α 1 N × . . . × α m N . n x log n (Riemann’s ζ ). • F ( x ) = ζ ( − log x )= �

The ring of formal generalised power series R � x ∗ � . Let x = ( x 1 , . . . , x m ) . α c α x α such that α ∈ [ 0 , ∞ ) m , c α ∈ R and F ( x ) = � Supp ( F ) := { α : c α � = 0 }⊆ S 1 × . . . × S m , where S i ⊆ [ 0 , ∞ ) well ordered. ( F has finitely many minimal monomials) α | c α | r | α | < ∞ (sup of all finite subsums). F convergent if ∃ r ∈ ( 0 , ∞ ) m � F induces a C 0 function (the sum of the series) on [ 0 , r ) m (analytic on ( 0 , r ) m ). R { x ∗ } = � r R { x ∗ } r is the ring of convergent generalised power series . Examples. • f ∈ O m , α i ∈ [ 0 , ∞ ); F ( x ) = f ( x α 1 1 , . . . , x α m m ) . Supp ( F ) ⊆ α 1 N × . . . × α m N . n x log n (Riemann’s ζ ). Supp ( F ) = { log n } n ∈ N ր + ∞ . • F ( x ) = ζ ( − log x )= �

The ring of formal generalised power series R � x ∗ � . Let x = ( x 1 , . . . , x m ) . α c α x α such that α ∈ [ 0 , ∞ ) m , c α ∈ R and F ( x ) = � Supp ( F ) := { α : c α � = 0 }⊆ S 1 × . . . × S m , where S i ⊆ [ 0 , ∞ ) well ordered. ( F has finitely many minimal monomials) α | c α | r | α | < ∞ (sup of all finite subsums). F convergent if ∃ r ∈ ( 0 , ∞ ) m � F induces a C 0 function (the sum of the series) on [ 0 , r ) m (analytic on ( 0 , r ) m ). R { x ∗ } = � r R { x ∗ } r is the ring of convergent generalised power series . Examples. • f ∈ O m , α i ∈ [ 0 , ∞ ); F ( x ) = f ( x α 1 1 , . . . , x α m m ) . Supp ( F ) ⊆ α 1 N × . . . × α m N . n x log n (Riemann’s ζ ). Supp ( F ) = { log n } n ∈ N ր + ∞ . • F ( x ) = ζ ( − log x )= � ∞ 2 + n − 1 � 1 • F ( x ) = 2 i x 2 i n , i = 0

The ring of formal generalised power series R � x ∗ � . Let x = ( x 1 , . . . , x m ) . α c α x α such that α ∈ [ 0 , ∞ ) m , c α ∈ R and F ( x ) = � Supp ( F ) := { α : c α � = 0 }⊆ S 1 × . . . × S m , where S i ⊆ [ 0 , ∞ ) well ordered. ( F has finitely many minimal monomials) α | c α | r | α | < ∞ (sup of all finite subsums). F convergent if ∃ r ∈ ( 0 , ∞ ) m � F induces a C 0 function (the sum of the series) on [ 0 , r ) m (analytic on ( 0 , r ) m ). R { x ∗ } = � r R { x ∗ } r is the ring of convergent generalised power series . Examples. • f ∈ O m , α i ∈ [ 0 , ∞ ); F ( x ) = f ( x α 1 1 , . . . , x α m m ) . Supp ( F ) ⊆ α 1 N × . . . × α m N . n x log n (Riemann’s ζ ). Supp ( F ) = { log n } n ∈ N ր + ∞ . • F ( x ) = ζ ( − log x )= � ∞ 1 − √ x � √ x 2 + n − 1 � 2 i , solution of ( 1 − x ) F ( x ) = x + 1 1 � � � • F ( x ) = 2 i x 2 x F . n , i = 0

The ring of formal generalised power series R � x ∗ � . Let x = ( x 1 , . . . , x m ) . α c α x α such that α ∈ [ 0 , ∞ ) m , c α ∈ R and F ( x ) = � Supp ( F ) := { α : c α � = 0 }⊆ S 1 × . . . × S m , where S i ⊆ [ 0 , ∞ ) well ordered. ( F has finitely many minimal monomials) α | c α | r | α | < ∞ (sup of all finite subsums). F convergent if ∃ r ∈ ( 0 , ∞ ) m � F induces a C 0 function (the sum of the series) on [ 0 , r ) m (analytic on ( 0 , r ) m ). R { x ∗ } = � r R { x ∗ } r is the ring of convergent generalised power series . Examples. • f ∈ O m , α i ∈ [ 0 , ∞ ); F ( x ) = f ( x α 1 1 , . . . , x α m m ) . Supp ( F ) ⊆ α 1 N × . . . × α m N . n x log n (Riemann’s ζ ). Supp ( F ) = { log n } n ∈ N ր + ∞ . • F ( x ) = ζ ( − log x )= � ∞ 1 − √ x � √ x 2 + n − 1 � 2 i , solution of ( 1 − x ) F ( x ) = x + 1 1 � � � • F ( x ) = 2 i x 2 x F . n , i = 0 1 2 3 4 5 6

The ring of formal generalised power series R � x ∗ � . Let x = ( x 1 , . . . , x m ) . α c α x α such that α ∈ [ 0 , ∞ ) m , c α ∈ R and F ( x ) = � Supp ( F ) := { α : c α � = 0 }⊆ S 1 × . . . × S m , where S i ⊆ [ 0 , ∞ ) well ordered. ( F has finitely many minimal monomials) α | c α | r | α | < ∞ (sup of all finite subsums). F convergent if ∃ r ∈ ( 0 , ∞ ) m � F induces a C 0 function (the sum of the series) on [ 0 , r ) m (analytic on ( 0 , r ) m ). R { x ∗ } = � r R { x ∗ } r is the ring of convergent generalised power series . Examples. • f ∈ O m , α i ∈ [ 0 , ∞ ); F ( x ) = f ( x α 1 1 , . . . , x α m m ) . Supp ( F ) ⊆ α 1 N × . . . × α m N . n x log n (Riemann’s ζ ). Supp ( F ) = { log n } n ∈ N ր + ∞ . • F ( x ) = ζ ( − log x )= � ∞ 1 − √ x � √ x 2 + n − 1 � 2 i , solution of ( 1 − x ) F ( x ) = x + 1 1 � � � • F ( x ) = 2 i x 2 x F . n , i = 0 1 2 3 4 5 6 Theorem (vdDries-Speissegger, ’98). R { x ∗ } generates a polynomially bounded o-minimal expansion R an ∗ of R an .

Our main result

Our main result � R { x ∗ 1 , . . . , x ∗ x �→ 1 � � Definition. A = m } ∪ x m ∈ N

Our main result � R { x ∗ 1 , . . . , x ∗ x �→ 1 � � Definition. A = m } ∪ x m ∈ N An A -cell is a cell such that the defining functions are A -terms.

Our main result � R { x ∗ 1 , . . . , x ∗ x �→ 1 � � Definition. A = m } ∪ x m ∈ N An A -cell is a cell such that the defining functions are A -terms. THEOREM. x = ( x 1 , . . . , x m ) , r ∈ ( 0 , ∞ ) , f ∈ R { x ∗ , y ∗ } r with f ( 0 , 0 ) = 0.

Our main result � R { x ∗ 1 , . . . , x ∗ x �→ 1 � � Definition. A = m } ∪ x m ∈ N An A -cell is a cell such that the defining functions are A -terms. THEOREM. x = ( x 1 , . . . , x m ) , r ∈ ( 0 , ∞ ) , f ∈ R { x ∗ , y ∗ } r with f ( 0 , 0 ) = 0. Then, ∃ W ⊆ R m + 1 nbd of 0 and ∃ an A -cell decomposition of W ∩ [ 0 , r ) m + 1

Our main result � R { x ∗ 1 , . . . , x ∗ x �→ 1 � � Definition. A = m } ∪ x m ∈ N An A -cell is a cell such that the defining functions are A -terms. THEOREM. x = ( x 1 , . . . , x m ) , r ∈ ( 0 , ∞ ) , f ∈ R { x ∗ , y ∗ } r with f ( 0 , 0 ) = 0. Then, ∃ W ⊆ R m + 1 nbd of 0 and ∃ an A -cell decomposition of W ∩ [ 0 , r ) m + 1 ( x , y ) ∈ [ 0 , r ) m + 1 : f ( x , y ) = 0 � � which is compatible with .

Our main result � R { x ∗ 1 , . . . , x ∗ x �→ 1 � � Definition. A = m } ∪ x m ∈ N An A -cell is a cell such that the defining functions are A -terms. THEOREM. x = ( x 1 , . . . , x m ) , r ∈ ( 0 , ∞ ) , f ∈ R { x ∗ , y ∗ } r with f ( 0 , 0 ) = 0. Then, ∃ W ⊆ R m + 1 nbd of 0 and ∃ an A -cell decomposition of W ∩ [ 0 , r ) m + 1 ( x , y ) ∈ [ 0 , r ) m + 1 : f ( x , y ) = 0 � � which is compatible with . y f = 0 W f = 0 f < 0 f < 0 f < 0 f > 0 x

Our main result � R { x ∗ 1 , . . . , x ∗ x �→ 1 � � Definition. A = m } ∪ x m ∈ N An A -cell is a cell such that the defining functions are A -terms. THEOREM. x = ( x 1 , . . . , x m ) , r ∈ ( 0 , ∞ ) , f ∈ R { x ∗ , y ∗ } r with f ( 0 , 0 ) = 0. Then, ∃ W ⊆ R m + 1 nbd of 0 and ∃ an A -cell decomposition of W ∩ [ 0 , r ) m + 1 ( x , y ) ∈ [ 0 , r ) m + 1 : f ( x , y ) = 0 � � which is compatible with . y f = 0 W f = 0 f < 0 f < 0 f < 0 f > 0 x In particular, the solutions of f = 0 are of the form y = ϕ ( x ) , where ϕ : C → R is an A -term and C ⊆ R m is an A -cell.

Strategy of proof

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that:

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ;

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ;

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form )

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form ), so f ◦ ρ = 0 has only trivial solutions;

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form ), so f ◦ ρ = 0 has only trivial solutions; 4) ρ respects y:

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form ), so f ◦ ρ = 0 has only trivial solutions; 4) ρ respects y: • ρ 0 does not depend on Y , so ρ 0 : [ 0 , ε ) m ∋ X �→ x ∈ [ 0 , r ) m ;

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form ), so f ◦ ρ = 0 has only trivial solutions; 4) ρ respects y: • ρ 0 does not depend on Y , so ρ 0 : [ 0 , ε ) m ∋ X �→ x ∈ [ 0 , r ) m ; • ρ 0 is bijective outside a small set and the components of ρ − 1 are A -terms; 0

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form ), so f ◦ ρ = 0 has only trivial solutions; 4) ρ respects y: • ρ 0 does not depend on Y , so ρ 0 : [ 0 , ε ) m ∋ X �→ x ∈ [ 0 , r ) m ; • ρ 0 is bijective outside a small set and the components of ρ − 1 are A -terms; 0 • ρ 1 ( X , · ) : Y �→ y is monotonic for almost all X .

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form ), so f ◦ ρ = 0 has only trivial solutions; 4) ρ respects y: • ρ 0 does not depend on Y , so ρ 0 : [ 0 , ε ) m ∋ X �→ x ∈ [ 0 , r ) m ; • ρ 0 is bijective outside a small set and the components of ρ − 1 are A -terms; 0 • ρ 1 ( X , · ) : Y �→ y is monotonic for almost all X . Remark. The existence of a family F satisfying 1,2,3 is well known (see [Rol.-Sanz-Vill.; Rol.-S.’13], inspired by [vdDr.-Speis.’98; Rol.-Speis.-Wil.’03]).

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form ), so f ◦ ρ = 0 has only trivial solutions; 4) ρ respects y: • ρ 0 does not depend on Y , so ρ 0 : [ 0 , ε ) m ∋ X �→ x ∈ [ 0 , r ) m ; • ρ 0 is bijective outside a small set and the components of ρ − 1 are A -terms; 0 • ρ 1 ( X , · ) : Y �→ y is monotonic for almost all X . Remark. The existence of a family F satisfying 1,2,3 is well known (see [Rol.-Sanz-Vill.; Rol.-S.’13], inspired by [vdDr.-Speis.’98; Rol.-Speis.-Wil.’03]). The monomialising tools ( admissible transformations ) are essentially blow-ups with real exponents and translations by elements of R { x ∗ } .

Strategy of proof Find a finite family F of vertical admissible transformations [ 0 , ε ) m + 1 [ 0 , r ) m + 1 ρ : − → ( X , Y ) �− → ( x , y ) = ( ρ 0 ( X , Y ) , ρ 1 ( X , Y )) such that: 1) f ◦ ρ ∈ R { X ∗ , Y ∗ } ε ; 2) ∃ W ⊆ R m + 1 nbd of 0 such that W ∩ [ 0 , r ) m + 1 = � [ 0 , ε ) m + 1 � � ρ ∈F ρ ; 3) f ◦ ρ = X α Y β U ( X , Y ) for some ( α, β ) ∈ [ 0 , ∞ ) m + 1 , U ∈ R { X ∗ , Y ∗ } × ( monomialised form ), so f ◦ ρ = 0 has only trivial solutions; 4) ρ respects y: • ρ 0 does not depend on Y , so ρ 0 : [ 0 , ε ) m ∋ X �→ x ∈ [ 0 , r ) m ; • ρ 0 is bijective outside a small set and the components of ρ − 1 are A -terms; 0 • ρ 1 ( X , · ) : Y �→ y is monotonic for almost all X . Remark. The existence of a family F satisfying 1,2,3 is well known (see [Rol.-Sanz-Vill.; Rol.-S.’13], inspired by [vdDr.-Speis.’98; Rol.-Speis.-Wil.’03]). The monomialising tools ( admissible transformations ) are essentially blow-ups with real exponents and translations by elements of R { x ∗ } . The novelty here lies in 4, which allows to solve f = 0 ( verticality ).

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) .

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q .

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) .

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells.

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells. By verticality, ρ 0 is invertible and ρ 1 ( X , · ) is monotonic, outside a small-dimensional set

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells. By verticality, ρ 0 is invertible and ρ 1 ( X , · ) is monotonic, outside a small-dimensional set (wlog, an A -cell, by induction on the dimension).

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells. By verticality, ρ 0 is invertible and ρ 1 ( X , · ) is monotonic, outside a small-dimensional set (wlog, an A -cell, by induction on the dimension). Hence, it is enough to prove: A ⊆ R m + 1 A -cell and ρ ↾ A as above ⇒ ρ ( A ) is a fin. disj. union of A -cells.

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells. By verticality, ρ 0 is invertible and ρ 1 ( X , · ) is monotonic, outside a small-dimensional set (wlog, an A -cell, by induction on the dimension). Hence, it is enough to prove: A ⊆ R m + 1 A -cell and ρ ↾ A as above ⇒ ρ ( A ) is a fin. disj. union of A -cells. Wlog, A = { ( X , Y ) : X ∈ C , Y ∗ t ( X ) } , with C ⊆ R m A -cell, ∗ ∈ { = , < } and t : C → R A -term.

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells. By verticality, ρ 0 is invertible and ρ 1 ( X , · ) is monotonic, outside a small-dimensional set (wlog, an A -cell, by induction on the dimension). Hence, it is enough to prove: A ⊆ R m + 1 A -cell and ρ ↾ A as above ⇒ ρ ( A ) is a fin. disj. union of A -cells. Wlog, A = { ( X , Y ) : X ∈ C , Y ∗ t ( X ) } , with C ⊆ R m A -cell, ∗ ∈ { = , < } and t : C → R A -term. ( x , y ) : x ∈ ρ 0 ( C ) , y ∗ ′ ρ 1 ρ − 1 ρ − 1 � � � ��� Then, ρ ( A ) = ( x ) , t ( x ) . 0 0

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells. By verticality, ρ 0 is invertible and ρ 1 ( X , · ) is monotonic, outside a small-dimensional set (wlog, an A -cell, by induction on the dimension). Hence, it is enough to prove: A ⊆ R m + 1 A -cell and ρ ↾ A as above ⇒ ρ ( A ) is a fin. disj. union of A -cells. Wlog, A = { ( X , Y ) : X ∈ C , Y ∗ t ( X ) } , with C ⊆ R m A -cell, ∗ ∈ { = , < } and t : C → R A -term. ( x , y ) : x ∈ ρ 0 ( C ) , y ∗ ′ ρ 1 ρ − 1 ρ − 1 � � � ��� Then, ρ ( A ) = ( x ) , t ( x ) . 0 0 By induction on the dimension, ρ 0 ( C ) is a fin. disj. union of A -cells.

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells. By verticality, ρ 0 is invertible and ρ 1 ( X , · ) is monotonic, outside a small-dimensional set (wlog, an A -cell, by induction on the dimension). Hence, it is enough to prove: A ⊆ R m + 1 A -cell and ρ ↾ A as above ⇒ ρ ( A ) is a fin. disj. union of A -cells. Wlog, A = { ( X , Y ) : X ∈ C , Y ∗ t ( X ) } , with C ⊆ R m A -cell, ∗ ∈ { = , < } and t : C → R A -term. ( x , y ) : x ∈ ρ 0 ( C ) , y ∗ ′ ρ 1 ρ − 1 ρ − 1 � � � ��� Then, ρ ( A ) = ( x ) , t ( x ) . 0 0 By induction on the dimension, ρ 0 ( C ) is a fin. disj. union of A -cells. ρ − 1 ρ − 1 � � �� By verticality, ρ 1 ( x ) , t ( x ) is an A -term. 0 0

Why is this enough? Recall: W ∩ [ 0 , r ) m + 1 = � � [ 0 , ε ) m + 1 � and f ◦ ρ = X α Y β U ( X , Y ) ; ρ ∈F ρ x = ρ 0 ( X ) ; y = ρ 1 ( X , Y ) . Partition [ 0 , ε ) m + 1 into finitely many subquadrants Q of dim ≤ m + 1 s.t. f ◦ ρ has constant sign on Q . Then so does f on ρ ( Q ) . So, it is enough to show that ρ ( Q ) is a finite disjoint union of A -cells. By verticality, ρ 0 is invertible and ρ 1 ( X , · ) is monotonic, outside a small-dimensional set (wlog, an A -cell, by induction on the dimension). Hence, it is enough to prove: A ⊆ R m + 1 A -cell and ρ ↾ A as above ⇒ ρ ( A ) is a fin. disj. union of A -cells. Wlog, A = { ( X , Y ) : X ∈ C , Y ∗ t ( X ) } , with C ⊆ R m A -cell, ∗ ∈ { = , < } and t : C → R A -term. ( x , y ) : x ∈ ρ 0 ( C ) , y ∗ ′ ρ 1 ρ − 1 ρ − 1 � � � ��� Then, ρ ( A ) = ( x ) , t ( x ) . 0 0 By induction on the dimension, ρ 0 ( C ) is a fin. disj. union of A -cells. ρ − 1 ρ − 1 � � �� By verticality, ρ 1 ( x ) , t ( x ) is an A -term. 0 0

Examples of vertical blow-ups (m=2)

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1 ( λ + x 2 ) β = � � β λ β − n x n � 2 n ∈ N n

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1 ( λ + x 2 ) β = � � β λ β − n x n � ( x 2 becomes an analytic variable) 2 n ∈ N n

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1 ( λ + x 2 ) β = � � β λ β − n x n � ( x 2 becomes an analytic variable) 2 n ∈ N n  π 0 � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, 0 ≤ x 2 < ε x δ � � = 1    π λ � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, λ x δ 1 ≤ x 2 < ( λ + ε ) x δ � � = • 1 � � π ∞ � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε x 1 /δ  [ 0 , ε ) 3 � = , 0 ≤ x 2 < ε   2

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1 ( λ + x 2 ) β = � � β λ β − n x n � ( x 2 becomes an analytic variable) 2 n ∈ N n  π 0 � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, 0 ≤ x 2 < ε x δ � � = 1    π λ � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, λ x δ 1 ≤ x 2 < ( λ + ε ) x δ � � = • , so 1 � � π ∞ � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε x 1 /δ  [ 0 , ε ) 3 � = , 0 ≤ x 2 < ε   2 [ 0 , r ) 3 ∩ W = � π λ � [ 0 , ε ) 3 � λ ∈ R ∪{∞}

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1 ( λ + x 2 ) β = � � β λ β − n x n � ( x 2 becomes an analytic variable) 2 n ∈ N n  π 0 � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, 0 ≤ x 2 < ε x δ � � = 1    π λ � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, λ x δ 1 ≤ x 2 < ( λ + ε ) x δ � � = • , so 1 � � π ∞ � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε x 1 /δ  [ 0 , ε ) 3 � = , 0 ≤ x 2 < ε   2 [ 0 , r ) 3 ∩ W = � π λ � [ 0 , ε ) 3 � (need only finitely many λ , by compactness) λ ∈ R ∪{∞}

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1 ( λ + x 2 ) β = � � β λ β − n x n � ( x 2 becomes an analytic variable) 2 n ∈ N n  π 0 � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, 0 ≤ x 2 < ε x δ � � = 1    π λ � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, λ x δ 1 ≤ x 2 < ( λ + ε ) x δ � � = • , so 1 � � π ∞ � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε x 1 /δ  [ 0 , ε ) 3 � = , 0 ≤ x 2 < ε   2 [ 0 , r ) 3 ∩ W = � π λ � [ 0 , ε ) 3 � (need only finitely many λ , by compactness) λ ∈ R ∪{∞} • π λ = π λ 0 , π λ π λ � � respects y , for λ ∈ R ∪ {∞} : 1 = id (so, monotonic), 1

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1 ( λ + x 2 ) β = � � β λ β − n x n � ( x 2 becomes an analytic variable) 2 n ∈ N n  π 0 � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, 0 ≤ x 2 < ε x δ � � = 1    π λ � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, λ x δ 1 ≤ x 2 < ( λ + ε ) x δ � � = • , so 1 � � π ∞ � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε x 1 /δ  [ 0 , ε ) 3 � = , 0 ≤ x 2 < ε   2 [ 0 , r ) 3 ∩ W = � π λ � [ 0 , ε ) 3 � (need only finitely many λ , by compactness) λ ∈ R ∪{∞} • π λ = π λ 0 , π λ π λ � � respects y , for λ ∈ R ∪ {∞} : 1 = id (so, monotonic), 1 π λ 0 does not depend on y and is bijective outside { x 1 = 0 } , { x 2 = 0 }

Examples of vertical blow-ups (m=2) Fix δ ∈ ( 0 , ∞ ) .  π 0 : ( x 1 , x 2 , y ) �→ x 1 , x δ � � 1 x 2 , y ( chart at 0 )   π λ : ( x 1 , x 2 , y ) �→  x 1 , x δ � � 1 ( λ + x 2 ) , y ( λ ∈ R \ { 0 } ) ( regular charts) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 x 1 /δ  , x 2 , y ( chart at ∞ )   2 2 y γ and λ ∈ R ∪ {∞} , then f ◦ π λ ∈ R { x ∗ � 1 x β c αβγ x α 1 , x ∗ 2 , y ∗ } : • If f = α,β,γ ( λ + x 2 ) β y γ x 1 , x δ α,β,γ c αβγ x α + δβ � � = � f 1 ( λ + x 2 ) , y 1 ( λ + x 2 ) β = � � β λ β − n x n � ( x 2 becomes an analytic variable) 2 n ∈ N n  π 0 � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, 0 ≤ x 2 < ε x δ � � = 1    π λ � [ 0 , ε ) 3 � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε, λ x δ 1 ≤ x 2 < ( λ + ε ) x δ � � = • , so 1 � � π ∞ � ( x 1 , x 2 , y ) : 0 ≤ x 1 < ε x 1 /δ  [ 0 , ε ) 3 � = , 0 ≤ x 2 < ε   2 [ 0 , r ) 3 ∩ W = � π λ � [ 0 , ε ) 3 � (need only finitely many λ , by compactness) λ ∈ R ∪{∞} • π λ = π λ 0 , π λ π λ � � respects y , for λ ∈ R ∪ {∞} : 1 = id (so, monotonic), 1 π λ 0 does not depend on y and is bijective outside { x 1 = 0 } , { x 2 = 0 } � − 1 : ( x 1 , x 2 ) �→ � � x 1 x − 1 /δ π λ x 1 , x 2 x − δ x 1 , x 2 x − δ � � � � � ; − λ ; , x 2 A -terms 0 1 1 2

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � x 1 , x α/β x 2 , y  1    f ◦ π λ = f � � x 1 , x α/β ( λ + x 2 ) , y 1  f ◦ π ∞ = f � � x 1 x 1 /δ  , x 2 , y   2

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β x 2 , y  1 1    f ◦ π λ = f � � x 1 , x α/β ( λ + x 2 ) , y 1  f ◦ π ∞ = f � � x 1 x 1 /δ  , x 2 , y   2

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � x 1 , x α/β ( λ + x 2 ) , y 1  f ◦ π ∞ = f � � x 1 x 1 /δ  , x 2 , y   2

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α ( λ + x 2 ) , y 1 1  f ◦ π ∞ = f � � x 1 x 1 /δ  , x 2 , y   2

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � x 1 x 1 /δ  , x 2 , y   2

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. g ◦ ˜

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜ π λ : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ b) regular charts for λ ∈ R \ { 0 } ˜ ( λ + y ) 1

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜ π λ : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ b) regular charts for λ ∈ R \ { 0 } ˜ ( λ + y ) 1 π λ = x α 1 ( 1 − ( λ + y ) γ ( 1 + η 0 ( x 1 , x 2 , y ))) g ◦ ˜

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜ π λ : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ b) regular charts for λ ∈ R \ { 0 } ˜ ( λ + y ) 1 π λ = x α 1 ( 1 − ( λ + y ) γ ( 1 + η 0 ( x 1 , x 2 , y ))) g ◦ ˜ 1 ( 1 − λ γ + η ( x 1 , x 2 , y )) , with η ( 0 , 0 , 0 ) = 0 = x α

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜ π λ : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ b) regular charts for λ ∈ R \ { 0 } ˜ ( λ + y ) 1 π λ = x α 1 ( 1 − ( λ + y ) γ ( 1 + η 0 ( x 1 , x 2 , y ))) g ◦ ˜ 1 ( 1 − λ γ + η ( x 1 , x 2 , y )) , with η ( 0 , 0 , 0 ) = 0 = x α (y analytic)

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜ π λ : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ b) regular charts for λ ∈ R \ { 0 } ˜ ( λ + y ) 1 π λ = x α 1 ( 1 − ( λ + y ) γ ( 1 + η 0 ( x 1 , x 2 , y ))) g ◦ ˜ 1 ( 1 − λ γ + η ( x 1 , x 2 , y )) , with η ( 0 , 0 , 0 ) = 0 = x α (y analytic) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 y γ/α , x 2 , y c) chart at ∞ ˜

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜ π λ : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ b) regular charts for λ ∈ R \ { 0 } ˜ ( λ + y ) 1 π λ = x α 1 ( 1 − ( λ + y ) γ ( 1 + η 0 ( x 1 , x 2 , y ))) g ◦ ˜ 1 ( 1 − λ γ + η ( x 1 , x 2 , y )) , with η ( 0 , 0 , 0 ) = 0 = x α (y analytic) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 y γ/α , x 2 , y c) chart at ∞ ˜ ! NOT VERTICAL !

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜ π λ : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ b) regular charts for λ ∈ R \ { 0 } ˜ ( λ + y ) 1 π λ = x α 1 ( 1 − ( λ + y ) γ ( 1 + η 0 ( x 1 , x 2 , y ))) g ◦ ˜ 1 ( 1 − λ γ + η ( x 1 , x 2 , y )) , with η ( 0 , 0 , 0 ) = 0 = x α (y analytic) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 y γ/α , x 2 , y c) chart at ∞ ˜ ! NOT VERTICAL ! However, y γ > > x α π ∞ � [ 0 , ε ) 3 � 1 on ˜ , so g cannot vanish there.

Blow-ups in action 2 y γ ( 1 + x 1 y ) Example. f ( x 1 , x 2 , y ) = x α 1 − x β • π blow-up of ( x 1 , x 2 ) with exponent δ = α β  f ◦ π 0 = f � � � 2 y γ ( 1 + x 1 y ) � x 1 , x α/β = x α 1 − x β = x α x 2 , y 1 · unit  1 1    f ◦ π λ = f � � � 1 − y γ ( 1 + x 1 y ) ( λ + x 2 ) β � x 1 , x α/β = x α = x α ( λ + x 2 ) , y 1 · unit 1 1  f ◦ π ∞ = f � � � 1 − y γ � �� x 1 x 1 /δ = x β 1 + x 1 x β/α  x α , x 2 , y y   2 2 2 1 − y γ � � 1 + x 1 x β/α Now we look at g ( x 1 , x 2 , y ) = x α y 2 π = blow-up of ( x 1 , y ) with exponent α • ˜ γ π 0 : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ a) chart at 0 ˜ y 1 π 0 = x α 1 ( 1 − y γ ( 1 + η 0 ( x 1 , x 2 , y ))) , with η 0 ( 0 , 0 , 0 ) = 0. (trivial solution) g ◦ ˜ π λ : ( x 1 , x 2 , y ) �→ � � x 1 , x 2 , x α/γ b) regular charts for λ ∈ R \ { 0 } ˜ ( λ + y ) 1 π λ = x α 1 ( 1 − ( λ + y ) γ ( 1 + η 0 ( x 1 , x 2 , y ))) g ◦ ˜ 1 ( 1 − λ γ + η ( x 1 , x 2 , y )) , with η ( 0 , 0 , 0 ) = 0 = x α (y analytic) π ∞ : ( x 1 , x 2 , y ) �→ � � x 1 y γ/α , x 2 , y c) chart at ∞ ˜ ! NOT VERTICAL ! However, y γ > > x α π ∞ � [ 0 , ε ) 3 � 1 on ˜ , so g cannot vanish there. (trivial solution)

i = 1 x α i 1 x β i General strategy. f ( x 1 , x 2 , y ) = � d 2 y γ i U i ( x 1 , x 2 , y ) U i units.

i = 1 x α i 1 x β i General strategy. f ( x 1 , x 2 , y ) = � d 2 y γ i U i ( x 1 , x 2 , y ) U i units. � d � x α i 1 x β i 2 y γ i Monomialisation algorithm : the set of minimal monomials i = 1 determines the choice of pairs of variables and exponents in the blow-ups.

i = 1 x α i 1 x β i General strategy. f ( x 1 , x 2 , y ) = � d 2 y γ i U i ( x 1 , x 2 , y ) U i units. � d � x α i 1 x β i 2 y γ i Monomialisation algorithm : the set of minimal monomials i = 1 determines the choice of pairs of variables and exponents in the blow-ups. ———

i = 1 x α i 1 x β i General strategy. f ( x 1 , x 2 , y ) = � d 2 y γ i U i ( x 1 , x 2 , y ) U i units. � d � x α i 1 x β i 2 y γ i Monomialisation algorithm : the set of minimal monomials i = 1 determines the choice of pairs of variables and exponents in the blow-ups. ——— Problem: solving systems of equations (joint work with J.-P. Rolin). � f ( x 1 , x 2 , y 1 , y 2 ) = 0 Given f , g ∈ R { x ∗ 1 , x ∗ 2 , y ∗ 1 , y ∗ 2 } , find the solutions of g ( x 1 , x 2 , y 1 , y 2 ) = 0.

i = 1 x α i 1 x β i General strategy. f ( x 1 , x 2 , y ) = � d 2 y γ i U i ( x 1 , x 2 , y ) U i units. � d � x α i 1 x β i 2 y γ i Monomialisation algorithm : the set of minimal monomials i = 1 determines the choice of pairs of variables and exponents in the blow-ups. ——— Problem: solving systems of equations (joint work with J.-P. Rolin). � f ( x 1 , x 2 , y 1 , y 2 ) = 0 Given f , g ∈ R { x ∗ 1 , x ∗ 2 , y ∗ 1 , y ∗ 2 } , find the solutions of g ( x 1 , x 2 , y 1 , y 2 ) = 0. Possible strategy. Find a solution y 2 = t ( x 1 , x 2 , y 1 ) ( t A -term) of f = 0 and replace in g .