Answers to HW #11

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

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)

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

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

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

Monday 4/20

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

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

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].

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.

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?

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)!).]

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.

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).

Have fun!

Answers to HW #8

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

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

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

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

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

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.