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.]
