Wednesday, April 15, 2009

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!