Allen Knutson's class

Answers to the final exam

2009-05-12T22:09:00.001-07:00

Here. I haven't graded things yet.

Final and review sessions

2009-05-01T05:30:00.000-07:00

Final: Tuesday May 12th, 11 AM-1:30 PM, in our usual room.
You may bring notes, HW, and stuff printed from the website, but not books.

Review sessions: Wednesday 6th, Monday 11th, at noon in the Malott lounge on the 5th floor. Be there or b².

Answers to HW #11

2009-04-29T07:45:00.000-07:00

1. Let R = F_p[x,y], and I generated by y(y-x^2).
Show there is no Frobenius splitting on R that compatibly splits I.
(Hint: if I were compatibly split, then so would various other ideals be, including one that isn't radical.)

We can colon out {y} to get {y-x^2}, or {y-x^2} to get {y}.
Adding those together, we get {y,y-x^2} = {y,x^2} (meaning the ideals; sorry about the lack of angle brackets).
These would all be compatibly split, hence radical, but {y,x^2} isn't.

2. Same problem, but I is generated by xy(x+y).

Colon out {x} to get {y(x+y)}, colon out {y(x+y)} to get {x},
add them together to get {x, y(x+y)} = {x, y^2}, same problem.

3. Let R be a graded ring, and phi a Frobenius splitting of it.
a) Show that phi is determined by its values on homogeneous elements.
b) For r homogeneous of degree k, define phi'(r) = the degree k/p part of phi(r), or 0 if k/p is not an integer.
For r not homogeneous, define phi'(r) = the sum of phi'(its homogeneous pieces).
Show that phi' is a Frobenius splitting.
c) Give a reasonable definition of a "graded Frobenius splitting".

a) This means, if phi_1 and phi_2 agree on all homogeneous elements, then they're supposed to be equal.
phi_1(anything) = phi_1(sum of homogeneous elements)
= sum of phi_1(those homogeneous elements), since phi_1 is additive
= sum of phi_2(those homogeneous elements), since phi_1 = phi_2 on them
= phi_2(sum of those homogeneous elements), since phi_2 is additive
= phi_2(original thing).

b) We have to check conditions (1),(2),(3) of a Frobenius splitting.
We get condition (1) for free, pretty much.
(3) is very easy: 1 is in R_0, so phi'(1) = the degree 0 part of phi(1) = the degree 0 part of 1 = 1.
(2) we only have to check homogeneous elements a,b, say of degrees j,k.
Then phi'(a^p b) = degree j+k/p part of phi(a^p b) = degree j+k/p part of a phi(b) = a*(degree k/p part of phi(b)) = a*phi'(b).

c) Call phi graded if its associated phi' is again phi.
Which is to say, phi(a homogeneous element of degree k) should be of degree k/p, and hence 0 if p doesn't divide k.

4. Let R be the subring of F_p[x] generated by x^2 and x^3, i.e. polynomials with no linear term. Show that R has no Frobenius splitting.

Note first that R is a graded ring, whose nth graded piece is multiples of x^n, unless n=1 in which case R_1=0.
Let phi be a splitting (for contradiction), and phi' its graded part, as constructed in #3.
Careful: you can't say phi'(x^p) = x phi'(1). That only holds for elements of R that are pth powers of elements of R.
x^3 = phi'(x^{3p}) = phi'(x^{2p} x^p) = x^2 phi'(x^p).
But phi'(x^p) should be degree 1, and the only thing there is 0.
So x^2 phi'(x^p) = x^2 * 0 = 0. Contradiction.
Hence there was no splitting.

5. Let R = C[a,b,c] / {ac} stupid blogger.com.
Let I be generated by {a,b}, as an ideal in R not just C[a,b,c].
a) Compute the Hilbert series H_R and H_{R/I}.
b) Show that H_{R/I} is not H_R times a polynomial.
c) Prove that the Hilbert Syzygy Theorem fails for this R and I; there is no finite graded resolution.

a) H_R = 1/(1-t)^3 * (1-t^2), H_{R/I} = H_{C[a,b,c]/{a,b}} = 1/(1-t).
b) The ratio is (1-t)^2 / (1-t^2) = (1-t)/(1+t).
The coefficients of its power series go 1,-2,+2,-2,+2,-2,... by the way.
c) If there were a finite graded resolution 0 -> ... R^powers -> ... -> R -> R/I -> 0, we could compute H_{R/I} = H_R * (1 - this + that ... ) where the finitely many terms in that alternating sum come from the terms in the resolution. But then H_{R/I} / H_R would be a polynomial.

(In fact there is a resolution that goes ... -> R^2 -> R^2 -> R^2 -> R -> R/I -> 0, with R^2s going back forever.)

6. Here is a simple program in Macaulay 2, a program to do (mostly) ring theory calculations.

Figure out what it's computing. Here's a comprehensive index of M2 commands. If you want to actually run M2 (so e.g. you can play with the code), here's how to get started.

-- What is the following code doing?
-- You can load it into Macaulay 2 by saying
-- load "hw.m2"
-- and run the subroutines yourself to figure out what they do.

syze = 2;
-- If you're willing to wait a while (overnight?), try changing this to syze=3

R = QQ[a_(1,1)..a_(syze,syze),b_(1,1)..b_(syze,syze)];

A = transpose genericMatrix(R,a_(1,1),syze,syze);
B = transpose genericMatrix(R,b_(1,1),syze,syze);


Generic matrices filled with variables


dp = M -> matrix apply(syze,i->apply(syze,j->(
       if (i==j) then M_(i,j) else 0)));
lp = M -> matrix apply(syze,i->apply(syze,j->(
       if (i>j) then M_(i,j) else 0)));


These take the diagonal part and strictly lower part of a matrix.


dec = I -> (
  print "old:";
  scan(flatten entries gens trim I, print); print "";
  cs = decompose I;
  scan(#cs, i->(print ("new in #" | toString(i+1) | ":");
       scan(select(flatten entries gens cs_i, g->(g%I != 0)), print);
       print ""; ))
  )


Take the ideal I. Decompose it as a product of prime ideals.
For each one of those, list the new generators contained in that larger ideal,
where "new" is checked by seeing if it reduces to 0 mod I.


I1 = ideal {lp(A*B), lp(B*A)}; -- what are these equations, in words?
dec(I1); -- what does this do?


These equations say that A*B and B*A are both required to be upper triangular
(that their strict lower triangles vanish).
[They always have the same eigenvalues, as is easy to see if A is invertible,
then use continuity. The eigenvalues are now on the diagonal, so there are
syze! ways to match up A*B's diagonal with B*A's diagonal. That's why this
thing decomposes into syze! pieces.]


C = A*B-B*A;
C = C - dp(C);
I2 = ideal C;  -- what are these equations, in words?
dec(I2);


C-dp(C) rips out the diagonal. Then we set the rest to 0. So the conditions are
that A and B almost commute -- A*B and B*A differ only on the diagonal.

[This turns out to break into 2 pieces. One is the piece where A and B
do in fact commute. There is exactly one other piece!
Moreover, the equations above are a Gr\"obner degeneration of these,
w.r.t. a certain weighting lambda on the variables.
If you find these equations interesting, you can read more
about this story here and here.]

Answers to HW #10

2009-04-22T13:38:00.000-07:00

1. A ring is called a division ring if R is a (not necessarily commutative) ring with unit (and 0 not equal to 1) such that for each r in R, there is an r' such that rr' = r'r = 1. The Quaternions is the set H = { a + bi + cj + dk : a,b,c,d are real numbers}. Define a multiplication by setting i^2 = j^2 = k^2 = -1, ij = k, jk = i, and ik = -j. Show that H is a division ring. [Hint: Try proving that the complex numbers is a field using the complex conjugate, and try changing the proof for the Quaternions H].

(a+bi+cj+dk)(a-bi-cj-dk) = a^2 + b^2 + c^2 + d^2.
If not all a,b,c,d are 0, we can divide through to find an inverse for a+bi+cj+dk.

2. Recall that given a nonzero ring R, one always has at least two ideals; the zero ideal, and the whole ring. If these are the only two ideals, R is called simple.
Show that the ring of nxn matrices over a field F is simple.

Let I be a nonzero ideal and M a nonzero element. Let m_ij be a nonzero matrix element. Gotta start somewhere.
Let e_ij denote the matrix with a 1 at (i,j) and 0 elsewhere.
Then for I to be a 2-sided ideal, it must contain e_{ai} M e_{jb}/m_ij, which one can calculate to be e_{ab} by matrix multiplication.
So I contains every e_{ab}, hence their linear combinations, which is all matrices.

For the rest of the problems, R is a commutative ring with unit.

3. Define the nilradical of R, denoted N(R) as the set of all nilpotent elements of R.
(a) Show that N(R) is an ideal.
(b) Given an ideal I of R, compute N(R/I). So, N(R/I) corresponds to an ideal of R containing I. What is this ideal?

Let pi denote the map R -> R/I.
So we want r such that pi(r) in N(R/I),
i.e. pi(r)^p = 0 in R/I for some p,
i.e. (r+I)^p = 0+I for some p,
i.e. r^p + I = 0+I for some p,
i.e. r^p in I for some p,
i.e. r in Rad(I).

4. Let F_p be the field of p elements, Z/pZ. Note that in this field, p = 0. Let R = F_p[x], where x is a variable. Show that the function phi : R --> R that sends an element f of R to f^p is a ring homomorphism. What is its kernel?
[It may help if you know that (f+g)^p = \sum_{i=0}^p (p choose i) f^i * g^(p-i), where (p choose i) is the binomial coefficient p!/(i!(p-i)!).]

(p choose i) = p(p-1)...(p-i+1)/i!. If i is more than 0, then the numerator contains p. If i is less than p, then the denominator doesn't. Hence in these cases, (p choose i) is a multiple of p. Which is 0 in R.

5. Let I be an ideal of R, and consider the R-module R/I. Show that the first syzygy of R/I can be chosen to be I. Note: This amounts to proving that the kernel of the R-module homomorphism R --> R/I is I.

The point is that we can generate R/I using 1+I, so the usual map R -> R/I is a good place to start the resolution.
Then k+I = 0+I iff k+i=0 for some i iff k=-i is in I.

6. Let R = F[x,y,z] where F is any field, and let I be the ideal generated by the monomials x^2,xy,xz,y^2,yz,z^2. Find the first syzygies (there are 8) among these generators (i.e. the relations), and find the second syzygies (there are 3) among these generators (which, by definition, are the relations among the relations).

Incidentally H_{R/I} = 1 + 3t, which is a very easy calculation, and that's
= 1/(1-t)^3 (1 - 6 t^2 + 8 t^3 - 3 t^4).
So it would be nice (though it's not automatic) for the resolution to look as simple as that polynomial. And indeed it does.
Of the (6 choose 2) pairs, one only needs to look at pairs with gcd not 1.
That's where the 8 syzygies come from.

Macaulay 2, version 1.2
with packages: Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,
ReesAlgebra, SchurRings, TangentCone

i1 : R = QQ[x,y,z];

i2 : I = (ideal vars R)^2

2 2 2
o2 = ideal (x , x*y, x*z, y , y*z, z )

o2 : Ideal of R

i10 : syz gens I

o10 = {2} | -y 0 -z 0 0 0 0 0 |
{2} | x -z 0 -y 0 -z 0 0 |
{2} | 0 y x 0 0 0 0 -z |
{2} | 0 0 0 x -z 0 0 0 |
{2} | 0 0 0 0 y x -z 0 |
{2} | 0 0 0 0 0 0 y x |

6 8
o10 : Matrix R <--- R

i11 : ker oo

o11 = image {3} | z 0 0 |
{3} | x 0 -z |
{3} | -y 0 0 |
{3} | 0 z 0 |
{3} | 0 x 0 |
{3} | 0 -y z |
{3} | 0 0 x |
{3} | 0 0 -y |

HW #11, due Wed 4/29 (minor corrections to 3b and 5)

2009-04-22T13:21:00.000-07:00

1. Let R = F_p[x,y], and I generated by y(y-x^2).
Show there is no Frobenius splitting on R that compatibly splits I.
(Hint: if I were compatibly split, then so would various other ideals be, including one that isn't radical.)

2. Same problem, but I is generated by xy(x+y).

3. Let R be a graded ring, and phi a Frobenius splitting of it.
a) Show that phi is determined by its values on homogeneous elements.
b) For r homogeneous of degree k, define phi'(r) = the degree k/p part of phi(r), or 0 if k/p is not an integer.
For r not homogeneous, define phi'(r) = the sum of phi'(its homogeneous pieces).
Show that phi' is a Frobenius splitting.
c) Give a reasonable definition of a "graded Frobenius splitting".

4. Let R be the subring of F_p[x] generated by x^2 and x^3, i.e. polynomials with no linear term. Show that R has no Frobenius splitting.

5. Let R = C[a,b,c] / {ac} stupid blogger.com.
Let I be generated by {a,b}, as an ideal in R not just C[a,b,c].
a) Compute the Hilbert series H_R and H_{R/I}.
b) Show that H_{R/I} is not H_R times a polynomial.
c) Prove that the Hilbert Syzygy Theorem fails for this R and I; there is no finite graded resolution.

6. Here is a simple program in Macaulay 2, a program to do (mostly) ring theory calculations.

Figure out what it's computing. Here's a comprehensive index of M2 commands. If you want to actually run M2 (so e.g. you can play with the code), here's how to get started.

Wednesday 4/22

2009-04-22T13:18:00.000-07:00

A Frobenius splitting on a ring R containing F_p is a map phi : R->R such that (1) it's additive (2) phi(a^p b) = a phi(b) (3) phi(1) = 1.

Def: a compatibly split ideal I is one such that phi(I) stays in I.

Thm:
0) Frobenius split rings are reduced (they have no nilpotents).
1) If I is compatibly split, then R/I is split.
2) ...and therefore I is radical.
3) If I,J are split then I+J and I intersect J are split.
4) If I is split, then I:J is split.

Talk next Monday 4/27

2009-04-22T13:17:00.000-07:00

Next Monday David Vogan of MIT will be giving a talk aimed at undergrads that should be quite good.

Monday 4/20

2009-04-20T13:26:00.001-07:00

Exact sequences.
Graded modules and their Hilbert series.
Exact sequences of graded modules give an alternating sum formula for Hilbert series.

Let lambda be a weighting on the variables, a natural number for each x_i.
With this, we can generalize the notions of

degree,

homogeneous polynomial,

top-degree part of a polynomial, called init_lambda(p),

the initial ideal of an ideal,

homogenization of a lambda-inhomogeneous ideal.

Then the easy theorem:
homog_lambda(I) + ideal(y) = init_lambda(I) + ideal(y).

Stupid blogger.com is stealing my angle brackets again.

Note that on the LHS the familiar lambda is (1,1,1,1,...,1),
whereas on the RHS the familiar lambda is (N^n, N^{n-1}, ..., N) where N is very large. (That picks out the lex-first term of any fixed polynomial, once N is big enough.)

Answers to HW #9

2009-04-16T13:13:00.000-07:00

1. Let Delta_1, Delta_2 be simplicial complexes on {1..n}.
Show that Delta_1 intersect Delta_2, Delta_1 union Delta_2 are both simplicial complexes too.

A simplicial complex is a collection (A) of subsets of {1..n}, (B) closed under shrinkage. (A) is obvious for both union and intersection, so we turn to (B).
Let F be a subset of {1..n}, and G a subset of F.
If F is in the intersection, then F is in Delta_1 and Delta_2, so G is too, hence G is in the intersection.
If F is in the union, then F is in one of Delta_1 or Delta_2, so G is in that one too, hence G is in the union.

2. Show H_{Delta_1 union Delta_2} = H_{Delta_1} + H_{Delta_2} - H_{the intersection}.

We computed H_Delta = \sum_{F in Delta} (t/(1-t))^|F|.
On the left hand side we sum over each F in the union, once.
On the right we sum over each F in the union either 1+0-0, 0+1-0, or 1+1-1 times, depending on whether it is in Delta_1, Delta_2, or both.

3. Take the two simplicial complexes from the last homework. Compute their Hilbert series (meaning, that of the associated Stanley-Reisner ideals) using the formula we have for general monomial ideals. Confirm that it matches the answer we get from the formula specifically for SR ideals.

The first one is I generated by {e, af, bd}. The usual monomial formula is 1/(1-t)^6 (1 -t-t^2-t^2 + t^3+t^3+t^4 - t^5). The fact that the LCMs are all products says that this is a complete intersection, i.e. the numerator factors, so we could also say 1/(1-t)^6 (1-t)(1-t^2)(1-t^2). Then that simplifies to 1/(1-t)^3 (1+t)^2.
Meanwhile, the formula specifically for S-R ideals gives us
1 + 5(t/(1-t)) + 8(t/(1-t))^2 + 4(t/(1-t))^3, for the 1 empty face, 5 vertices, 8 edges, and 4 triangles.
Multiply both sides by (1-t)^3, and the first calculation gives (1+t)^2, whereas the second gives (1-t)^3 + 5t(1-t)^2 + 8t^2(1-t) + 4t^3, and these are indeed the same.

The second ideal is generated by abc, so the general monomial formula is 1/(1-t)^4 (1-t^3) = 1/(1-t)^3 (1+t+t^2). The S-R formula is 1 + 4(t/(1-t)) + 6(t/(1-t))^2 + 3(t/(1-t))^3. Multiplying again by (1-t)^3, we get 1+t+t^2 vs. (1-t)^3 + 4t(1-t)^2 + 6t^2(1-t) + 3t^3, which again match.

4. Let I be a homogeneous ideal. But let's homogenize it again, anyway! Relate H_I and H_{homog(I)}.

Pick a homogeneous Gr\"obner basis. Then homogenizing it does exactly nothing to the basis; it only puts it into a ring with one more variable.
We can compute the Hilbert series from the leading terms of the Gr\"obner basis, as 1/(1-t)^{# variables} * something depending on those terms.
So the only difference between the two calculations is the number of variables.
Hence H_{homog(I)} = 1/(1-t) H_I.

HW #10, Due Wed. 4/22

2009-04-15T16:48:00.000-07:00

Answers to HW #8

2009-04-11T15:59:00.000-07:00

1. Let I be a radical ideal in R, and homog(I) its homogenization in R[y].
Show that homog(I) is radical.

Let f^2 be in homog(I). Let f_0 be the lowest-degree part of f, so f_0^2 is the lowest-degree part of f^2. Since f^2 is in homog(I) a homogeneous ideal, each degree component of it, e.g. f_0^2, is in homog(I).
Then there are two cases: f_0 is in homog(I) or not. If it is, replace f with f-f_0 and start over. That must terminate since the number of terms goes down. Eventually we get to the other case, that f_0 is not in homog(I). Replace f with f_0 and start over.

Okay, now f^2 is in homog(I), but f isn't, and f is homogeneous. Hence homog(dehomog(f)) = f / y^k, where k is the largest power of y that divides f. Let g = f / y^k; it is homogeneous and not a multiple of y. Hence homog(dehomog(g)) = g.

Okay, now (g y^k)^2 is in homog(I). Dehomogenizing, we learn dehomog(g)^2 is in I. Hence dehomog(g) is in I, since I is radical. Hence homog(dehomog(g)) = g is in homog(I). Therefore f = g y^k is in homog(I).

2. Let I be a radical homogeneous ideal in R[y].
Show that dehomog(I) is radical.

Let f^2 be in dehomog(I). Then f^2 = dehomog(g) for some homogeneous g in I. So homog(f^2) = g / y^k for the largest y^k dividing g.
Hence y^k homog(f^2) is in I. If k is odd, multiply by y again, staying in I.
Hence y^{k or k+1} homog(f)^2 is in I.
Since I is radical, y^{ceiling(k/2)} homog(f) is in I.
Dehomogenizing, we learn f is in I.

3. Let I be an ideal in R such that homog(I) is radical. Show that I is radical.

I = dehomog(homog(I)). Now apply #2.

4. Give an example of I in R[y], homogeneous but not radical, such that dehomog(I) is radical.

The simplest is I generated by y^2.

Geometrically, the idea is this. PV(I) lives in projective space. Being nonradical means it has some fuzz on it somewhere (that the generators of the radical would shave off). Dehomogenizing means intersecting PV(I) with affine space, throwing away the stuff at infinity. So we need to arrange for all the fuzz to be at infinity. The ideal y^2 cuts out exactly the hyperplane at infinity, but with fuzz.

5. Let I be generated by xy, x+y-1. Find a term order such that init(homog(I)) is radical.
Conclude that I is radical.

homog(I) is generated by xy, x+y-z (here the new variable is z).
With the right lex order, the initial terms are xy and -z.
Since those have gcd=1, this is a Gr\"obner basis whose initial terms are squarefree.
Hence I is radical.

6. Draw the simplicial complex associated to the ideal generated by {e, af, bd} in C[a,b,c,d,e,f].

It's a solid square cut into four triangles with a,b,f,d at the corners and c in the middle.

7. Draw the solid abc triangle, and put a vertex d in the middle, with new edges connected to a,b,c. (The red, but not blue, lines in this picture.) What's the corresponding Stanley-Reisner ideal, and its Hilbert series?

This wasn't at all well specified, for which I apologize. If you didn't manage to read my mind, but made clear what complex you were working with, that's good enough.

What I'd intended to indicate was a union of three solid triangles, abd & bcd & cad. So the only faces missing are abcd and abc itself. In particular the S-R ideal should be generated by abc.

Then the Hilbert series is 1/(1-t)^4 * (1-t^3).

HW #9, due Wednesday 4/15

2009-04-08T11:45:00.000-07:00

1. Let Delta_1, Delta_2 be simplicial complexes on {1..n}.
Show that Delta_1 intersect Delta_2, Delta_1 union Delta_2 are both simplicial complexes too.

2. Show H_{Delta_1 union Delta_2} = H_{Delta_1} + H_{Delta_2} - H_{the intersection}.

3. Take the two simplicial complexes from the last homework. Compute their Hilbert series (meaning, that of the associated Stanley-Reisner ideals) using the formula we have for general monomial ideals. Confirm that it matches the answer we get from the formula specifically for SR ideals.

4. Let I be a homogeneous ideal. But let's homogenize it again, anyway! Relate H_I and H_{homog(I)}.

~~...more to come~~

Hints

2009-04-07T21:10:00.000-07:00

If p is a homogeneous polynomial in C[x_1,...,x_n,y], then homog(dehomog(p)) = p / y^k, where k is taken largest possible. (Namely, it's taken to be the smallest power occurring among the terms of p.)

If J is a homogeneous ideal in C[x_1,..,x_n,y], then the map dehomog: J -> dehomog(J) is onto.

If I is an ideal in C[x_1,..,x_n], then the map homog: I -> homog(I) is not onto; its image is the homogeneous polynomials in homog(I) that are not multiples of y.

Feel free to use these without proving them (though they're easy).

Monday 4/6

2009-04-06T13:45:00.001-07:00

The Hilbert series of SR(Delta) is sum_{faces F} (t/(1-t))^|F|.
Hence the Hilbert polynomial, in degree d, is sum_F (d-1 choose |F|-1).
Hence the Hilbert dimension is the # of elements of the largest face.

Something I really should have proved:
Thm.
If g_1,..,g_m is a Gr\"obner basis of I using a graded term order,
then homog(g_1),...,homog(g_m) is a Gr\"obner basis of homog(I)
using a graded term order in which y is cheap.

Pf. homog(g_i) has the same leading term as g_i (since it's of highest degree in g_i, so doesn't get any powers of y on it).
If we follow the reduction of S(homog(g_i), homog(g_j)),
at each step its dehomogenization matches that of the reduction S(g_i,g_j),
so we don't get stuck and we do reduce it to 0.

Corollary. If yf is in homog(I), then f is in homog(I).
Proof. If f isn't in homog(I), then after reducing for a while using one of the Gr\"obner bases supplied above it gets stuck.
Hence yf is also stuck, because the leading terms in {homog(g_i)} don't involve y. So yf isn't in homog(I).

Thm. If I is prime, then homog(I) is also prime.
Pf. Start with a,b s.t. ab in homog(I). Reduce to the case that a = homog(a_2), b = homog(b_2), using the Corollary. Then use a = homog(dehomog(a)) etc.

HW #8, due Wednesday 4/8

2009-04-01T10:41:00.000-07:00

You may use the following result (very similar to one proved in class):
Let J be a homogeneous ideal. If for every homogeneous f one
has f^2 in J => f in J, then J is radical.

1. Let I be a radical ideal in R, and homog(I) its homogenization in R[y].
Show that homog(I) is radical.

2. Let I be a radical homogeneous ideal in R[y].
Show that dehomog(I) is radical.

3. Let I be an ideal in R such that homog(I) is radical. Show that I is radical.

4. Give an example of I in R[y], homogeneous but not radical, such that dehomog(I) is radical.

5. Let I be generated by xy, x+y-1. Find a term order such that init(homog(I)) is radical.
Conclude that I is radical.

6. Draw the simplicial complex associated to the ideal generated by {e, af, bd} in C[a,b,c,d,e,f].

(Grr. Blogger is eating my less-than and more-than symbols with which to generate ideals.)

7. Draw the solid abc triangle, and put a vertex d in the middle, with new edges connected to a,b,c. (The red, but not blue, lines in this picture.) What's the corresponding Stanley-Reisner ideal, and its Hilbert series?

Monday 3/30

2009-04-01T10:25:00.000-07:00

The homogenization and dehomogenization of polynomials and ideals.
Geometrically, homogenization corresponds to taking the closure in projective space, whereas dehomogenization corresponds to intersecting a projective set with affine space.
We did x^2 = y^2 + 1 as an example, then dehomogenized using x,
which turned a hyperbola into a circle.

Midterm answers

2009-03-31T20:15:00.000-07:00

Here they are.

The midterm

2009-03-25T13:25:00.001-07:00

I handed out the midterm. If you didn't get one email me.
It's due in Anna's basement office Friday 2:30-3:30. If you can't come then email her.
Once you start it, you have three hours (consecutive!), during which you can't talk to other people about it nor touch a computer, nor use textbooks.
You may use stuff from the course webpage you've printed out, and anything else you've written.
Other details are on the front page. They shouldn't be too surprising.

Monday 3/23

2009-03-25T13:22:00.000-07:00

Went back to proving that Hilb dim = homogeneous Krull dim.

Projective geometry.
Points, lines, and conics in the projective plane.
The embedding of affine space in projective space.
The three types of conics all look the same in the projective plane;
it's just a question of how they intersect the line at infinity.

The homogenization of a polynomial using a new variable.

Answers to HW #7

2009-03-13T10:18:00.000-07:00

1. Let I be a homogeneous ideal in a polynomial ring R, and r a homogeneous element of degree k.
Let J = I + < r >.
Show that
a) For each n, h_J(n) is at least h_I(n) - h_I(n-k).
b) If they are equal for all n, then r is not a zero divisor.

a) Consider the map (R/I)_{n-k} -> (R/I)_n given by multiplication by r.
We figured out in class that the quotient by the image is (R/J)_n.
So dim (R/I)_n = dim (R/J)_n + dim r*(R/I)_{n-k}.
The dimension of r*(R/I)_{n-k} is at most the dimension of (R/I)_{n-k}, so we get the inequality claimed.
b) If they're equal, then dim r*(R/I)_{n-k} = dim (R/I)_{n-k}, so the multiply-by-r map has no kernel. Which means r is not a zero divisor.

2. A list {r_1, r_2, ..., r_m} is called a regular sequence if each r_j is not a zero divisor in R/< r_1, ..., r_{j-1} >.
If {b,c} is a regular sequence, show that {c,b} is a regular sequence.
Oops: I had meant b,c to be homogeneous. (It's true even if they're not, but don't bother with that.)

We showed in class that if r is not a zero divisor in R/I, and is homogeneous of degree k, then H_{I+< r >}(t) = H_I(t) * (1-t^k).
So if {b,c} is a regular sequence, with degrees k,k', then H_{< b,c >}(t) = H_0(t) (1 - t^k) (1 - t^k').
Certainly c is not a zero divisor in R. So H_{< c >}(t) = H_0(t) (1 - t^k').
By the previous question, H_{< c,b >}(t) is coefficientwise at least H_0(t) (1 - t^k') (1 - t^k), with equality iff b is not a zero divisor in R/< c >.
But since < b,c > = < c,b >, we know this is an equality -- so b is not a zero divisor.

3. Let p(n) be a polynomial of degree d, and k a number. Show that q(n) = p(n) - p(n-k) is a polynomial of degree d-1.

First we check it for p(n) = n^d. Then q(n) = n^d - (n^d - d n^{d-1} k + ...), canceling the first term in the binomial series but not the next term, k d n^{d-1}.
Now say p(n) = a n^d + terms of degree at most d-1.
Then q(n) = (a k d n^{d-1} + terms of degree at most d-2) + (terms of degree at most d-2), which is indeed of degree d-1.

4. For p(n) a polynomial, let Delta p be the polynomial with values (Delta p)(n) = p(n) - p(n-1).
Notice that if p only takes integer values (when fed integers), then Delta p does so too.
a) Show that Delta {n choose k} = {n n-1 choose k-1}.

(n choose k) - (n-1 choose k) counts all k-element subsets of {1..n} minus all k-element subsets of {1..n-1}. What's left over is k-element subsets of {1..n} that include n. By ripping out the element n, those subsets correspond 1:1 with (k-1)-element subsets of {1..n-1}. The number of those is {n-1 choose k-1}.

b) Show that every polynomial p(n) is a linear combination of the polynomials {n choose k}, where k goes from 0 to degree p.

If p is constant it's clear, which will be the base of our induction.

Let d = degree(p).
Then p(n) = c n^d + a polynomial of degree < d.
Note that d! {n choose d} = n^d + a polynomial of degree < d.
So p(n) - c d! {n choose d} = a polynomial of degree < d.
By induction on d, the RHS is a linear combination as desired. Now add c d! {n choose d} to both sides.

c) Give an example of a polynomial with noninteger coefficients that nonetheless always produces integers.

{n choose 2} = n(n-1)/2.

d) Show that every integer-valued polynomial p(n) is a linear combination with integer coefficients of the polynomials {n choose k}, where k goes from 0 to degree p.
(In particular, this applies to Hilbert polynomials.)

Write p(n) = sum_{k=0}^{degree p} c_k {n choose k} using part (b).
Then (Delta^i p)(n) = sum_{k=0}^{degree p} c_k {n-i choose k-i}.
Obviously, all the terms with k < i vanish for any n.
If n = i, then the terms with k > i vanish too.
So (Delta^i p)(i) = c_i.
Notice that when q is integer-valued, so is Delta q, so Delta^i p only takes on integer values. Hence each c_i is an integer.

5. Let I in C[x_1...x_5] be generated by {x_i x_j - x_k x_l}, for all i,j,k,l such that i+j=k+l.
a) Show this is a Gr\"obner basis with respect to lex order.

First notice that we can leave out the generators with i > j since they're the same as the ones with i and j switched. Then we can leave out those with i=k or j=k, since the generator is then 0.

Consider the subset of the form x_i x_j - x_{floor{(i+j)/2}} x_{ceiling{(i+j)/2}}, where floor(x) is the greatest integer below x, ceiling(x) the least above.
Call this second term RHS(i,j) for now. Then any other relation x_i x_j - x_k x_l can be written as

x_i x_j - RHS(i,j) - (x_k x_l - RHS(k,l))

so this subset already generates the ideal. In particular, if it's a Gr\"obner basis, then adding the other generators won't break that.

At this point all of our generators have i at most j. But the i=j generators and the i=j-1 generators are x_i x_i - x_i x_i and x_i x_j - x_i x_j, i.e. zero. So we were already leaving those out.

When we compute the S-polynomial of two of these generators (ignoring the pairs with relatively prime initial terms, since we can), we get a cubic. So let's see what the reduction algorithm does to an arbitrary cubic monomial x_a x_b x_c, with a at most b at most c. (Note that a,b,c may include repeats.)

If c-a > 1, then the reduction algorithm trades x_a x_c either for x_d^2 or x_d x_{d+1}, depending on c-a being even or odd. In particular it gives us another cubic monomial x_a' x_b' x_c', where a+b+c = a'+b'+c'. It only gets stuck when a=b=c, or a=b=c-1, or a+1=b=c. We can even predict which case we'll get to, depending on the value of a+b+c mod 3 (0 if a=b=c, 1 if a=b=c-1, 2 if a+1=b=c).

Now let's see what happens to an S-polynomial x_a x_b x_c - x_d x_e x_f under reduction. Observe that a+b+c=d+e+f. We've just figured out that the reduction algorithm will replace either term with the same cubic monomial, and then they'll cancel. So the S-polynomials reduce to zero, making this a Gr\"obner basis.

b) Find the reduced Gr\"obner basis.

It's the subset we said above -- x_i x_j - RHS(i,j), where i < j-1.
The leading coefficients are all 1.
Since each generator is homogeneous quadratic, the only way this set would fail to be a reduced Gr\"obner basis is if some term of some generator was a constant multiple of the leading term of another generator. But the left terms have i < j-1, and the RHS terms have i'=j' or i'=j'-1.

c) Decompose V(the initial ideal).

Two variables can be nonzero at the same time only if they're adjacent, like x_3 and x_4, otherwise their product is the leading term of one of the generators.
So the components are the 12-plane, the 23-plane, the 34-plane, and the 45-plane. (And so on, if we had more variables.)

d) Compute the Hilbert polynomial.

The easiest thing is to compute directly the number of standard monomials of degree n. We've already figured out that a monomial is standard if it uses one variable, or two adjacent variables.
The number of monomials x_i^k x_{i+1}^{n-k}, for k not 0 or n, is n-1.
So the total is 5 (for the monomials x_i^n, i from 1 to 5) plus 4(n-1) (just counted, where i goes from 1 to 4), or 4n+1.

Wednesday 3/11

2009-03-13T10:05:00.000-07:00

Def: the Hilbert dimension of a homogeneous ideal is 1 + degree of the Hilbert polynomial, or 0 if the Hilbert polynomial is 0.

Theorem. HilbDim(I) = HilbDim(Rad(I)).
Proof. If I is not radical, we can add some homogeneous x whose square is in I.
Then do nullity plus rank on the map (R/I)_n -> (R/I)_{n+k} that multiplies by x.

Theorem. Let I be homogeneous.
1) If I is a prime ideal, and J (homogeneous) properly contains I, then HilbDim(J) < HilbDim(I).
2a) If I is not prime, then there exists J (homogeneous) properly containing it, with HilbDim(J) = HilbDim(I).
2b) That J can be taken to be prime.

Proof.
1) Even adding one element lowers the HilbDim by 1, by a calculation last week.
2a) If I is not radical, use the previous theorem.
Otherwise let a,b not in I, ab in I, and check that the map R/I -> R/I+< a > \oplus R/I+< b > is 1:1. Do nullity plus rank on that.
2b) Repeat 2a and use ACC.

We didn't get to

Theorem. HilbDim(I) = homogeneous KrullDim(I) (which is at most KrullDim(I)).
Proof. In a chain of homogeneous prime ideals, the HilbDim must drop at each step by (1) of the last theorem. That gives "HilbDim is at most homogeneous KrullDim".

Conversely, we can alternate the following two steps: increase I to a prime ideal without changing the HilbDim (using (2b)), and then increase it using one new homogeneous generator, lowering the HilbDim by 1, by the calculation last week. We get stuck only when we get to whole ring. This constructs a long sequence of prime ideals, giving "HilbDim is at least homogeneous KrullDim".

Monday 3/9

2009-03-09T14:26:00.000-07:00

Domains, prime ideals, Krull dimension.
Next time: Krull dimension = degree(Hilbert polynomial)+1 for homogeneous ideals.

HW #7, due Wednesday 3/11 (corrected)

2009-03-05T12:34:00.000-08:00

1. Let I be a homogeneous ideal in a polynomial ring R, and r a homogeneous element of degree k.
Let J = I + < r >.
Show that
a) For each n, h_J(n) is at least h_I(n) - h_I(n-k).
b) If they are equal for all n, then r is not a zero divisor.

2. A list {r_1, r_2, ..., r_m} is called a regular sequence if each r_j is not a zero divisor in R/< r_1, ..., r_{j-1} >.
If {b,c} is a regular sequence, show that {c,b} is a regular sequence.
Oops: I had meant b,c to be homogeneous. (It's true even if they're not, but don't bother with that.)

3. Let p(n) be a polynomial of degree d, and k a number. Show that q(n) = p(n) - p(n-k) is a polynomial of degree d-1.

4. For p(n) a polynomial, let Delta p be the polynomial with values (Delta p)(n) = p(n) - p(n-1).
Notice that if p only takes integer values (when fed integers), then Delta p does so too.
a) Show that Delta {n choose k} = {n n-1 choose k-1}.
b) Show that every polynomial p(n) is a linear combination of the polynomials {n choose k}, where k goes from 0 to degree p.
c) Give an example of a polynomial with noninteger coefficients that nonetheless always produces integers.
d) Show that every integer-valued polynomial p(n) is a linear combination with integer coefficients of the polynomials {n choose k}, where k goes from 0 to degree p.
(In particular, this applies to Hilbert polynomials.)

5. Let I in C[x_1...x_5] be generated by {x_i x_j - x_k x_l}, for all i,j,k,l such that i+j=k+l.
a) Show this is a Gr\"obner basis with respect to lex order.
b) Find the reduced Gr\"obner basis.
c) Decompose V(the initial ideal).
d) Compute the Hilbert polynomial.

Answers to HW #6 (tiny correction to #1)

2009-03-04T18:29:00.001-08:00

1. Let I be a radical ideal, and {g_1..g_m} a reduced Gr\"obner basis for it.
Show that each g_i is squarefree, i.e. is not divisible by f^2 for any polynomial f of degree > 0.

If g_i = f^2 h, then h g_i = f^2 h^2 is in I too, so fh is in I (by the assumption I = Rad(I)). Hence some init(g_j) divides init(fh). By the assumption deg f > 0, j is not i. Since init(g_j) divides init(fh), it divides init(g f^2 h) = init(g_i). But then our list is not a reduced Gr\"obner basis, contradiction.

2. Let I be an ideal such that for all f, if f^2 is in I, then f is in I. Show that I is radical.

If I is not radical, then there exists g not in I, and a number N, such that g^N is in I. Let k be the least number such that g^k is in I. (So k > 1, and is at most N.) Let f = g^{k-1}. Then f^2 = g^{2(k-1)} = g^k g^{k-2}, which only makes sense because k is at least 2. In particular, f is not in I, but f^2 is in I.

3. Let I be a monomial ideal generated by squarefree monomials. Show that I is radical.
Hint: show that if p is in I, then init(p) and p-init(p) are in I. Use this to show that it is enough to test the condition in question #2 when f is a monomial.

4. Let I be an ideal with a Gr\"obner basis {f_1,...,f_n}, such that each init(f_i) is a squarefree monomial. Show that I is radical.

#3 is actually the special case of #4 where each f_i = init(f_i). So we'll just do #4.

By #2 it's enough to check that f^2 in I => f in I.
So say f is a polynomial such that f^2 is in I. If f=0 we're done; otherwise we can talk about init(f).
Since f^2 is in I and we have a Gr\"obner basis, some init(f_i) | init(f^2) = init(f)^2.
Since init(f_i) is squarefree, it already divides init(f).
So we can replace f by f - m f_i where m = init(f)/init(f_i), canceling the leading term of f. This new guy has (f - mf_i)^2 = f^2 + f_i(-2m + f_i), so again in I, so we can run the reduction algorithm again. Since it only gets stuck when f=0, we see that f reduces to 0, which means f is in I.

5. Let I and J be two radical ideals.
a) Show that I intersect J is radical.
b) Give an example where I+J (which concatenates their generators) is not radical.

a) If f^N is in I intersect J, then f^N is in I and f^N is in J, hence f is in I and J, hence f is in I intersect J.
b) My favorite example is I = < y >, J = < y-x^2 >. (We can see that J is radical by taking a term order for which y is the leading term and applying #4.)

6. Let I = < ac,bc,bd,ae,de > inside C[a,b,c,d,e]. Decompose V(I) as a union of subspaces. Make it minimal, so no subspace in your list contains another.

If a=0, we have bc,bd,de=0 left.
If b=0, we have de=0 left, so either d=0 or e=0. So far {a=b=d=0} union {a=b=e=0}.
If not b=0, we have c=d=0. This adds {a=c=d=0}.
If not a=0, we have c=e=0, leaving bd=0, so either b=0 or d=0: {c=e=b=0} union {c=e=d=0}.

In the end, the solutions are the union of five 2-planes in 5-space.

Wednesday 3/4

2009-03-04T13:17:00.000-08:00

Thm. The Hilbert function of R/I is eventually a polynomial, the "Hilbert polynomial".

Def. Zero divisor in a ring.

Thm: If R/I contains a nonzero divisor r of degree k, then one can write down the Hilbert series and polynomial of R/(I + ) in terms of that of R/I.

Ex. If r,s are of degree A and B in C[x_1..x_n], then r is automatically a nonzero divisor, but s might be a zero divisor in R/< r >. If it's not, then the Hilbert series is 1/(1-t)^n * (1-t^A) * (1-t^B).
(Remember a HW problem that asked for a Hilbert series that looked like that?)