Exa3

1. Specialise one section of the proof of the main theorem
on dimension to establish
  dim_k m/m^2 >= dim A  for a local ring A,m,k

Remark. m/m^2 is a vector space over k = A/m. In the geometric
case it is the dual of the tangent space to a variety, and
dim m/m^2 = dim A is the condition for nonsingularity.


2. Let A be a ring and A -> S^-1A the localisation of A w.r.t.
multiplicative set S. Write out the proof that prime ideals of A
disjoint from S map bijectively to primes ideals of S^-1P.
[easy, but necessary background]

3. Use the main theorem on dimension for local rings to prove that
dim A/(x) = dim A - 1 for A a Noetherian ring and x in A a
nonzerodivisor.
[the only issue is to pass from local to the whole of A]

4. Define the _height_ of a prime ideal P of A to be the Krull
dimension dim A_P of the local ring A_P. Prove that this is the
maximum length of all chains
  P0 < P1 < .. Pn = P of prime ideals contained in P.
[Most textbooks define height as the max length of chains, and
deduce that it equals dim A_P. This is an easy exc with the
above hint.]

5. Check that ht P = 0 means that P is a minimal prime. The minimal
prime ideals correspond to the irreducible components of Spec A.
Recall that
  nilrad(A) = intersection of prime ideals
            = intersection of minimal prime ideals.
If a prime ideal P contains a nonzerodivisor x, prove that ht P >= 1.
[Easy: [A&M] Cor. 11.17.]

6. For A a Noetherian ring and x in A, let P be a prime that is
minimal among prime ideals containing x. Use the main theorem on
dimension to prove that ht P <= 1. (See [A&M] Cor. 11.17, and
[Ma], Theorem 13.5. The result is called Krull's Hauptidealsatz.)

7. In the same way, if I = (a1,.. ar) and P is a minimal prime
divisor of I then ht P <= r.

Here "prime divisors of I" means P is an associated primes of A/I,
that is P in Ass A/I. This means that there is some y in A/I such
that P = Ann y, so that A*y iso A/P < A/I. [This needs primary
decomposition and Ass M. See for example [UCA], Chap 7.]

8. Verify Nagata's example of a Noetherian integral domain with
dim = infty [A&M, p. 126, Ex 4]. Start from A = k[x1,.. xn,..].
Cover NN by disjoint intervals Ii = [ai, bi] of length bi-ai that
is unbounded as i increases. Each ideal
  Pi = (x_j | j in Ii)
is prime. Define S in A as the complemement of the union of the
Pi. Check S is multiplicative.

Now the localisation of S^-1A has each local rings A_{Pi} of A at
the individual primes Pi as a further localisation. These have
dimension bi-ai unbounded, so dim S^-1A = infty. Nothing strange
or difficult here.

Now why on earth should S^-1A be Noetherian? An element of S^-1A
is a/s with s notin Pi for any n. This is invertible unless a in
Pi for some Pi. Use this to show that any ideal of S^-1A other
than (1) is S^-1I where I is an ideal contained in Pi (inside A).
([A&M] say use their Ex 7.9, p.85, but I can't follow their hint,
and I don't see why they need to make it so complicated.)