M953 Homework 6 Solutions

Due Friday, March 15, 2002

[1] Let R = k[x₁, ..., x_n] and let R_d denote the subset of all polynomials in R of degree d or less, including 0. Now let S = k[x₀, ..., x_n] and let S_d denote the subset of all homogeneous polynomials in S of degree d, including 0. Note that both R_d and S_d are k vector spaces. (Part (b) is optional; it's really just for those people who haven't seen "stars and bars" before.)

(a) Show that dim_kR_d = dim_kS_d.
(b) Show that dim_kS_d is equal to the binomial coefficient "n+d choose d". [Hint: count the number of monomials of degree d by writing each monomial as a product, not using exponents. For example, say n = 3, and consider x₀²x₂x₃³ to be x₀x₀x₂x₃x₃x₃. But you can indicate the same information by replacing the variables by *'s, and separating the different variables by |'s. So x₀x₀x₂x₃x₃x₃ becomes **||*|***, meaning two x₀'s, no x₁'s, one x₂, and three x₃'s.]

Solution:

(a) A basis for R_d is given by the set A = {x₁^m₁...x_n^m_n : m₁ + ... + m₁ <= d} of monomials in x₁, ..., x_n of degree at most d. A basis for S_d is given by the set B = {x₀^m₀...x_n^m_n : m₀ + ... + m₁ = d} of monomials in x₀, ..., x_n of degree exactly d. We get a map f : B -> A by plugging 1 into x₀. This map is bijective, since given f(b) it is easy to write down b (if d = 7 and f(b) = x₁²x₂, then obviously b = x₀⁴x₁²x₂). Thus f is injective. And for any monomial M in A, deg(M) <= d, so x₀^d-deg(M)M is in B and f(x₀^d-deg(M)M) = M, so f is surjective. Thus R_d and S_d have the same dimension.
(b) A monomial of degree d is specified by saying how many factors of each variable x_i occur in the monomial. This can be done by writing d stars, and inserting n bars to indicate which variable each star corresponds to. (We only need n bars, because just as one cut splits a sandwich into two pieces, n bars split the stars into n+1 subsets, one subset per variable.) Thus there is a bijective correspondence between elements of S and arrangements of d stars and n bars. But there are n+d choose d ways to arrange a set of n+d objects, split into two subsets of n identical objects, and d identical objects different from the other n.

[2] In this problem assume, F is in k[x,y], where k is the complex numbers.

(a) Find each singular point p of F = x^2*y - y^2 + 2*y - 1, and for each one, find its multiplicity m_p(F) and all tangent lines at p. Check your answer using Macaulay2.
(b) Use Macaulay2 to find the singular points of F = x^4 + 2*x^2*y^2 + 2*x^3 + 2*x*y^2 + x^2 + y^2. For each singular point p, find its multiplicity m_p(F) and all tangent lines at p.
Notes: Recall that a point p is singular if and only if p is in V(F, F_x, F_y), where F_x and F_y represent partial derivatives with respect to x and y, resp. In Macaulay2, for any ideal I, we can find V(I) via the command decompose I. You can also use Macaulay2 to simplify substitutions: given F(x,y) = x^2 - y^3, to find F(x+1,y-3), simply input (x+1)^2 - (y-3)^3, as in the following example:
```
     i11 : (x+1)^2 - (y-3)^3

              3    2     2
     o11 = - y  + x  + 9y  + 2x - 27y + 28
```

Solution:

(a) The set of singular points is V(F, F_x, F_y). Here F = x^2*y - y^2 + 2*y - 1, F_x = 2*x*y, and F_y = x^2 - 2*y + 2. But F_x = 0 means either x = 0 or y = 0. If x = 0, then F = 0 gives - y^2 + 2*y - 1 = 0, so y = 1, and since (0,1) is in the 0-locus of F_y, we see that (0,1) is a singular point. If y = 0, then F = -1, so (0,1) is the only singular point. To find m_p(F), we simplify F(x,y+1). We find F(x,y+1) = x^2*y + x^2 - y^2, so m_p(F) = 2, and the tangents at p (in the p-centered coordinates) are the factors of x^2 - y^2; i.e., x-y and x+y. In the original coordinates, the tangents are x-(y-1) and x+(y-1). Here is how to do it in Macaulay2:

i1 : R = QQ[x,y]

o1 = R

o1 : PolynomialRing

i2 : I = ideal(x^2*y - y^2 + 2*y - 1, 2*x*y, x^2-2*y+2)

             2     2                  2
o2 = ideal (x y - y  + 2y - 1, 2x*y, x  - 2y + 2)

o2 : Ideal of R

i3 : decompose I

o3 = {ideal (y - 1, x)}

o3 : List

 -- THIS SHOWED THAT THE ONLY SINGULAR POINT IS x = 0, y = 1; I.E. p = (0,1).

i4 : x^2*(y+1) - (y+1)^2 + 2*(y+1) - 1

      2     2    2
o4 = x y + x  - y 

 -- HERE WE SAW THAT m_p(F) = 2, AND
 -- THAT THE TANGENT LINES AT p ARE THE FACTORS 
 -- OF x^2 - y^2, OR x-y AND x+y.
 -- BUT THOSE COORDINATES ARE CENTERED AT p. 
 -- IN TERMS OF OUR ORIGINAL COORDINATES,
 -- THE TANGENT LINES ARE x - (y-1) AND x + (y-1).

(b) 
i5 : I = ideal(x^4 + 2*x^2*y^2 + 2*x^3 + 2*x*y^2 + x^2 + y^2, 4*x^3+4*y^2*x+6*x^2+2*y^2+2*x, 4*y*x^2+4*x*y+2*y)

             4     2 2     3       2    2    2    3       2     2     2         2
o5 = ideal (x  + 2x y  + 2x  + 2x*y  + x  + y , 4x  + 4x*y  + 6x  + 2y  + 2x, 4x y + 4x*y + 2y)

o5 : Ideal of R

i6 : decompose I

o6 = {ideal (y, x), ideal (y, x + 1)}

o6 : List

i7 : (x-1)^4 + 2*(x-1)^2*y^2 + 2*(x-1)^3 + 2*(x-1)*y^2 + (x-1)^2 + y^2

      4     2 2     3       2    2    2
o7 = x  + 2x y  - 2x  - 2x*y  + x  + y 

-- SO WE SEE THAT THE ONLY SINGULAR POINTS ARE 
-- (0,0) AND (-1,0), THAT FOR BOTH m_p(F) = 2 
-- AND THAT THE TANGENTS AT (0,0) ARE THE FACTORS
-- OF x^2 + y^2, OR x + iy and x - iy.
-- AT (-1,0), IN COORDINATES CENTERED AT (-1,0), 
-- THE TANGENTS ARE AGAIN THE FACTORS OF x^2 + y^2, 
-- OR x + iy and x - iy. IN TERMS OF OUR ORIGINAL 
-- COORDINATES, THEY ARE (x+1) - iy and (x+1) + iy.


[3] Calculate the intersection multiplicity 
I(p, F^G) [note: I'm using ^ to denote intersection]
for each F and G below. In each case, assume p = (0,0). Check your answers
with Macaulay2.

(a) F = y^2-x^5, G = y^2*x^2-x^5-y^5
(b) F = y^2-x^3, G = (y^2+x^2)^2+3*x^2*y-y^3
(c) F = y^4+x^4-x^3 G = y^3*x^2+y^5+x^2




Solution:

(a) Let H = G - F = y^2*x^2 - y^5 - y^2 = y^2*(x^2 - y^3 - 1). Then 
I(p, F^G) = I(p, F^H), but since (x^2 - y^3 - 1) doesn't vanish at (0,0),
we can factor it out of H and we have I(p, F^H) = I(p, F^(y^2))
= I(p, (y^2)^F) = I(p, (y^2)^(F-y^2)) = I(p, (y^2)^-(F-y^2))
= I(p, (y^2)^(x^5)), and since y^2 and x^5 have no tangents in common at p 
we can just multiply the multiplicities of the tangents to get I(p, (y^2)^(x^5)) = 2*5 = 10.
(b) Since F = y^2 - x^3, any place a y^2 appears in G we can replace it by x^3.
The new G will differ from the old one by a multiple of F, and the intersection multiplicities
will be unchanged. Thus I(p, F^G) = I(p, F^H), if H = (x^3+x^2)^2 + 3*x^2*y - y*x^3.
But H = x^2((x^2+x)^2 + 3*y - y*x), so I(p, F^H) = I(p, F^(x^2)) + 
I(p, F^((x^2+x)^2 + 3*y - y*x)) = 2*2 + I(p, F^((x^2+x)^2 + 3*y - y*x)).
Now I(p, F^((x^2+x)^2 + 3*y - y*x)) = 
I(p, (y*((x^2+x)^2 + 3*y - y*x) - 3*F)^((x^2+x)^2 + 3*y - y*x)),
but as above we can change y*((x^2+x)^2 + 3*y - y*x) - 3*F = y*(x^2+x)^2 - y^2*x + 3*x^3 to
y*(x^2+x)^2 - x^3*x + 3*x^3 = x^2(y*(x+1)^2 - x^2 + 3*x) without changing the intersection
multiplicity, so I(p, F^((x^2+x)^2 + 3*y - y*x)) = 
I(p, (x^2(y*(x+1)^2 - x^2 + 3*x))^((x^2+x)^2 + 3*y - y*x)) =
I(p, (x^2)^((x^2+x)^2 + 3*y - y*x)) + I(p, (y*(x+1)^2 - x^2 + 3*x)^((x^2+x)^2 + 3*y - y*x)) =
2*1 + I(p, (y*(x+1)^2 - x^2 + 3*x)^((x^2+x)^2 + 3*y - y*x)).
But the tangent to (y*(x+1)^2 - x^2 + 3*x) at (0,0) is y+3*x, while the tangent
to ((x^2+x)^2 + 3*y - y*x) is 3*y. These are different so
I(p, (y*(x+1)^2 - x^2 + 3*x)^((x^2+x)^2 + 3*y - y*x)) = 1*1 = 1. Thus
I(p, F^G) = 2*2 + 2*1 + 1*1 = 7.
(c) This time F = -y^4 + x^4 + x^3 G = y^3*x^2 + y^5 + x^2.
Replace G by G + y*F = y^3*x^2 + x^2 + y*x^4 + y*x^3 = x^2*(y^3 + 1 + y*x^2 + y*x).
Since y^3 + 1 + y*x^2 + y*x does not vanish at p = (0,0), we can replace this new G by
x^2. Thus I(p, F^G) = I(p, F^(x^2)) = I(p, (y^4)^(y)) = 2*4 = 8.


Here are the same results, done by Macaulay2:


i8 : findIntMult(ideal(y^2 - x^5,y^2*x^2 - x^5 - y^5),ideal(x,y),R)

o8 = 10

i9 : findIntMult(ideal(y^2-x^3,(y^2+x^2)^2+3*x^2*y-y^3),ideal(x,y),R)

o9 = 7

i10 : findIntMult(ideal(y^4+x^4-x^3,y^3*x^2+y^5+x^2),ideal(x,y),R)

o10 = 8



Here are some useful Macaulay2 routines to compute intersection multiplicities.
Assuming R = k[x,y] is your ring, P = I(p) is the ideal of a point p in the plane,
and I = ideal(f,g) for some polynomials f and g in R which have no common factor 
vanishing at p, then mylocalize(I,P,R) gives the ideal Q_i
corresponding to p in the minimal primary decomposition 
I = Q₁^...^Q_r of I.  
Thus I(p, F^G) = dim_kR/Q_i, as discussed in class,
but the Macaulay2 command for "dim" in this case is degree
so I(p, F^G) = degree( R/mylocalize(I,P,R) ).



Now, mylocalize is not built in, it's actually a Macaulay2 script that I define
below. You can copy it from the web page for the homework and paste it into Macaulay2
(the whole thing all at once). For convenience we can also define a second script
that calls mylocalize to calculate I(p, F^G) in one step. 
Here are the scripts. Below I give a sample session using them.
findIntMult = (I,P,R) -> (
if(dim(R/mylocalize(I,P,R)) > 0 ) then
  print "The intersection multiplicity is infinite!!!!!!" else
  degree( R/mylocalize(I,P,R) ))

mylocalize = (I,P,R) -> (
L:=primaryDecomposition I;
j:=-1;
J:= I;
B:= ideal (matrix {{1}}**R);  -- Make B the ideal (1) but in current ring, not ZZ.
scan(#L, i->(
J=(L_i)*ideal (matrix {{1}}**R);  -- convert L_i into an ideal if it's of type MonomialIdeal
if( (P:J) == (ideal (matrix {{1}}**R)) ) then B=intersect(B,J);
));
B)


Sample Session:
Macaulay 2, version 0.9.2
--Copyright 1993-2001, D. R. Grayson and M. E. Stillman
--Singular-Factory 1.3c, copyright 1993-2001, G.-M. Greuel, et al.
--Singular-Libfac 0.3.2, copyright 1996-2001, M. Messollen

i1 : R = QQ[x,y]

o1 = R

o1 : PolynomialRing

i2 : findIntMult = (I,P,R) -> (
if(dim(R/mylocalize(I,P,R)) > 0 ) then
  print "The intersection multiplicity is infinite!!!!!!" else
  degree( R/mylocalize(I,P,R) ))

mylocalize = (I,P,R) -> (
L:=primaryDecomposition I;
j:=-1;
J:= I;
B:= ideal (matrix {{1}}**R);  -- Make B the ideal (1) but in current ring, not ZZ.
scan(#L, i->(
J=(L_i)*ideal (matrix {{1}}**R);  -- convert L_i into an ideal if it's of type MonomialIdeal
if( (P:J) == (ideal (matrix {{1}}**R)) ) then B=intersect(B,J);
));
B)

o2 = findIntMult

o2 : Function

i3 :                                         
o3 = mylocalize

o3 : Function

i4 : findIntMult(ideal(x^3,y^4),ideal(x,y),R)

o4 = 12

i5 : quit

Notes: 

(1) matrix {{1}} makes a 1x1 matrix whose entry is 1. But
ideal (matrix {{1}}) would be the unit ideal in ZZ, not R.
To get the unit ideal in R we need to tensor by R, hence I use
ideal (matrix {{1}}**R) in the script; "**" denotes tensoring.


(2) The colon in L:=primaryDecomposition I makes L into a local variable,
so you don't overwrite other L's you may have defined outside of the script.


(3) Now, primaryDecomposition I is a list of ideals. Thus L
is a list whose (i+1)st item is L_i (so L_0 is the first
item in the list); #L is the number of entries in the list, and 
scan(#L, i->( various commands go here ) ) makes i run from
0 to #L-1.


(4) P:J is the usual colon operation in commutative algebra; it is the
ideal of all f in R such that fJ is contained in P. Thus P:J = (1)
if and only if P contains J, so 
if( (P:J) == (ideal (matrix {{1}}**R)) ) then B=intersect(B,J)
intersects B with J if P contains J. Together with the scan statement
the effect is to make B the intersection of all ideals
in the minimal primary decomposition
of I which are contained in P.