M953 Homework 3 Solutions

Due Friday, February 15, 2002

Macaulay2 is a symbolic algebra program to do commutative algebra. To use Macaulay2, logon to a Linux machine. Using a terminal window, type in M2. At the beginning of each line you input the program puts i5: (or i#:, # being the number of the current input). Each output line starts with o# . Anything from "--" to the end of a line is a comment and will be ignored by Macaulay2. Documentation is on-line at http://www.math.uiuc.edu/Macaulay2/Manual.

To get started, you must define the polynomial ring you will be using. Here are some examples:
i1 : R = ZZ/101[x,y,z] -- polynomials in x, y, and z over the integers mod the prime 101
i1 : R = QQ[x,y] -- polynomials in x and y over the rationals

Assuming you've already defined your ring with variables x and y, here's how to define an ideal named I:
i2 : I = ideal(x,y,x*y-x^2, y^2-x^3)
Here's how to find the square of I:
i3 : I^2 -- or use (ideal(x,y,x*y-x^2, y^2-x^3))^2
Note that the output for I^2 is not a minimal set of generators. Here's how to get a minimal set:
i4 : mingens(I^2) -- the output is an array, with each generator an entry in the array
Given an ideal, here's how to find the ideals of the irreducible components of V(I), with output given as a list of ideals:
i5 : decompose(ideal(x*y-x^2, y^2-x^3)) -- or decompose(I) or decompose I if I is the ideal
Here's how to find the radical of I:
i6 : radical I -- or mingens radical I if you want simpler output
Here's a useful operation: given ideals I and J, I:J is the ideal of all f in your ring R such that fg is in I for all g in J. So, to find the least power of x-y that is in ideal(x*y-x^2, y^2-x^3), keep trying ideal(x*y-x^2, y^2-x^3) : ideal((x-y)^n) until you find an n such that the output is the ideal (1) (you may want to use mingens to simplify the output):
i7 : mingens (ideal(y-x^2, y^2-x^3) : ideal((x-y)^1))
Here's how to intersect a list of ideals:
i8 : intersect(I,J)
When you're done, just quit:
i7 : quit

So here's the homework:

[1] In Problem [1](b)(i) from Homework 2, use Macaulay2 to find the minimum n and m that work. (Pick a polynomial ring over a field of your choice.)

Solution: We see from below, over the rationals, that m = 3 and n = 2, since m = 3 is the least m such that ideal(y+x*y-x^2,y^2+x*y^2-x^3) : ideal(x^m) = (1) and n = 2 is the least n such that ideal(y+x*y-x^2,y^2+x*y^2-x^3) : ideal(y^2) = (1):
[bharbour@math ~]$ M2
Macaulay 2, version 0.9
--Copyright 1993-2001, D. R. Grayson and M. E. Stillman
--Singular-Factory 1.3b, 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 : I = ideal(y+x*y-x^2,y^2+x*y^2-x^3)

               2               3      2    2
o2 = ideal (- x  + x*y + y, - x  + x*y  + y )

o2 : Ideal of R

i3 : mingens (I : ideal(x))

o3 = | y x2 |

             1       2
o3 : Matrix R  <--- R

i4 : mingens (I : ideal(x^2))

o4 = | y x |

             1       2
o4 : Matrix R  <--- R

i5 :   mingens (I : ideal(x^3))

o5 = | 1 |

             1       1
o5 : Matrix R  <--- R

i6 : mingens (I : ideal(y))

o6 = | y x |

             1       2
o6 : Matrix R  <--- R

i7 : mingens (I : ideal(y^2))

o7 = | 1 |

             1       1
o7 : Matrix R  <--- R

[2] Given an algebraic set X, a theorem in commutative algebra says that Radical(I(X)) is the intersection of the ideals of the irreducible components of X. Demonstrate this in Macaulay2 with X = V(x*y-x^2, y^2-x^3) in A2 by comparing Radical(I(X)) with the intersection of the ideals of the irreducible components of X.

Solution: Here we see that Radical(I(X)) and the intersection of the ideals of the irreducible components of X (working over the rationals) both give the same ideal, namely ideal(x-y, y2-y):
i9 : J = ideal(x*y-x^2, y^2-x^3)

               2           3    2
o9 = ideal (- x  + x*y, - x  + y )

o9 : Ideal of R

i10 : mingens radical J

o10 = | x-y y2-y |

              1       2
o10 : Matrix R  <--- R

i11 : mingens intersect decompose J

o11 = | x-y y2-y |

              1       2
o11 : Matrix R  <--- R
[3] Do problems #25(b) on p. 17 and #31 on p. 20. Use Macaulay2 (over a field of your choice) and compare its output with your answer.

Solution: First, consider #25(b) on p. 17. Since (y^2+x)(y^2-x) = y^4-x^2 and (y^2+x)(y - x)(y + x) = y^4-x^2*y^2+x*y^2-x^3, we know V(y^4-x^2, y^4-x^2*y^2+x*y^2-x^3) is the union of V(y^2+x) and V(y^2-x, (y - x)(y + x)). Now, V(y^2+x) is irreducible: it is infinite (in fact, a sideways parabola) and y^2+x is irreducible (since it has degree 1 regarded as a polynomial in x with coefficients that may involve y). Thus, by Corollary 2 on p. 19 of Fulton's book, V(y^2+x) is irreducible. Next, V(y^2-x, (y - x)(y + x)) is the intersection of the degenerate hyperbola (y - x)(y + x) = 0 with y^2-x = 0, which just gives three points, (0,0), (1,1) and (1, -1). Since (0,0) is already in V(y^2+x), the irreducible components of V(y^4-x^2, y^4-x^2*y^2+x*y^2-x^3) are: V(y^2+x), {(1,1)} and {(1, -1)}.

Here is what Macaulay2 gives:
i12 : decompose ideal(y^4-x^2, y^4-x^2*y^2+x*y^2-x^3)

o12 = {ideal (x - 1, y - 1), ideal (x - 1, y + 1), ideal(y  + x)}

o12 : List
And here is a graph:

Now consider #31(a) on p. 20. First we have V(y^2-x*y-x^2*y+x^3). But y^2-x*y-x^2y+x^3 = (y - x)(y - x^2), so by Cor. 3 on p. 20 (over the complexes) or by Cor. 2 on p. 19 (over the reals) the irreducible components of V(y^2-x*y-x^2y+x^3) are the line V(y - x) and the parabola V(y - x^2).

Here is what Macaulay2 gives:
i13 : decompose ideal(y^2-x*y-x^2*y+x^3)

o13 = {ideal(x - y), ideal(x  - y)}

o13 : List
And here is a graph:

Next there is #31(b): consider first V(x^3+x-x^2*y-y). Since x^3+x-x^2*y-y = (x^2 + 1)(x - y), the irreducible components of V(x^3+x-x^2*y-y) are the three lines V(x - y) and V(x - i) and V(x + i) over the complexes, and just the one line V(x - y) over the reals.

Here is what Macaulay2 gives:
i14 : decompose ideal(x^3+x-x^2*y-y)

o14 = {ideal(x  + 1), ideal(x - y)}

o14 : List
Finally, consider V(y^2-x^3+x). Clearly, y^2-x^3+x = 0 has infinitely many solutions over the reals (or the complexes): for any value of y, x^3 - x - y^2 has some root in x since it has odd degree in x. Thus by Corollary 2 on p. 19 V(y^2-x^3+x) is irreducible as long as y^2-x^3+x is. But y^2 + (x - x^3) is irreducible by "Eisenstein's criterion" (since k[x] is a domain, and x - x^3 is in the prime ideal (x), but not in (x^2)).

Here is what Macaulay2 gives:
i15 : decompose ideal(y^2-x^3+x)

                3    2
o15 = {ideal(- x  + y  + x)}

o15 : List
And here is a graph: