Macaulay 2 on-line!

It's still semi-secret, but you can run Macaulay 2 on-line if you didn't manage to install it.

Answers to HW #4

"1. Let r be a positive irrational number. Write x^a y^b < x^c y^d if a+rb < c+rd.
Show that this defines a monomial order."

We need to check that it's a total order (this uses the irrationality), that any decreasing sequence of monomials terminates (this uses the positivity), that it's transitive (really easy), and that a < b implies ac < bc (really easy).

"2. Consider the vector space of 2x4 matrices M, with entries (m_ij), and let X = {M : M's rows are linearly dependent}.
For i < j, both from 1 to 4, let p_ij be the 2x2 determinant using columns i,j of M.
What's the relation of p_ij to X?"

Each p_ij vanishes on X. More specifically, X is the vanishing set of the ideal generated by the six p_ij.

"3. (continuing 2) Lex-order the variables m_11, m_12, m_13, m_14, m_21, m_22, m_23, m_24.
What are the leading terms of the six guys p_ij?"

p_ij = m_1i m_2j - m_1j m_2i. The lex-ordering sees the m_1i term first, if i < j. So the leading term is m_1i m_2j.

"4. (continuing) Show that the p_ij are a Gr\"obner basis, by computing S-pairs."

If the LCM of the two leading terms is just the product, the S-pair quickly reduces to 0.
So the cases to consider are S(p_ij, p_ik), S(p_ik, p_jk), where i < j < k.
(Secretly, I'm thinking about 2xn matrices, not just 2x4, here.)

I like to do this by starting with LCM - LCM, reduce each one using a different generator, get to something less obviously in I, then reduce that until stuck.

Case 1. Start with m_1i m_2j m_2k - m_1i m_2j m_2k,
-> m_2i m_1j m_2k - m_2i m_2j m_1k
-> m_2i m_2j m_1k - m_2i m_2j m_1k
= 0

Case 2. Start with m_1i m_1j m_2k - m_1i m_1j m_2k
-> m_2i m_1j m_1k - m_1i m_2j m_1k
-> m_2i m_1j m_1k - m_2i m_1j m_1k
= 0

HW #6

Book problems (p48) 3.1, 3.3, 3.6.

#4. Let I = < f >, J = < g > be principal ideals in C[x_1...x_n].
a) Figure out what I intersect J is, given that it's supposed to give the union of the vanishing sets.
b) Follow the computation of (tI + (1-t)J) intersect C[x_1..x_n], explicitly, to see that you get the same answer as you did in part (a).

#5. Get Macaulay 2 running on some computer, to the extent that you can do the following computation. I've put [editorial comments] in the below to explain what's going on.

----------------------------------------------------------------------------
Macaulay2, version 1.4
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra, TangentCone


i1 : R = QQ[m_(1,1)..m_(2,3)];

[R is the ring of polynomials with rational coefficients in six variables m_(i,j)]

i2 : M = transpose genericMatrix(R,m_(1,1),3,2)

[M is the matrix with the matrix entries m_(i,j) ]

i3 : I = ideal {det M_{0,1}, det M_{1,2}}

[M_{list} is the submatrix using only those columns, numbered 0,...,width-1 as usual in computer science. Here we're asking that columns 0 and 1, and columns 1 and 2, be linearly dependent.]

i4 : cs = decompose I

[We did this example in class a couple of weeks ago; this ideal I is not a prime ideal, but the intersection of two prime ideals. In one of them column 1 vanishes, in the other columns 0 and 2 are linearly dependent. "decompose" finds the two.]

i5 : intersect cs

[This just recovers I -- it is indeed the intersection of the two.]

Feb 28

The geometry of (tI + (1-t)J) intersect C[x_1...x_n]: it's the projection of (V(I) x 1) union (V(J) x 0), hence V(I) union V(J).

Example: we used this rule to compute the ideal vanishing on the set {(0,0), (1,1)}.

Colon ideals.

V(I : J) = the closure of V(I) \ V(J).
The definition of I : f^{infinity} (which we'll compute next time as an elimination).
Theorem: f is in the radical of I <=> I : f^{infinity} contains 1.