(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 |
