Math 615, Winter 2011
Problem Set #5: Solutions
1. (a) Since K is a eld, the map F : K K splits over K: note that F (K) may be a
proper subeld of K. Let be a splitting of this map of K-vector spaces. Extend to
on the polynomial ring R as follows: if c K
The structure theory of complete local rings
Introduction
In the study of commutative Noetherian rings, localization at a prime followed by completion at the resulting maximal ideal is a way of life. Many problems, even some that
seem global, can be attac
Formal power series rings, inverse limits, and I-adic completions of rings
Formal semigroup rings and formal power series rings
We next want to explore the notion of a (formal) power series ring in nitely many
variables over a ring R, and show that it is
Regular rings and nite projective resolutions
Recall that a local ring (R, m, K) is regular if its embedding dimension, dimK (m/m2 ),
which may also be described as the least number of generators of the maximal ideal m, is
equal to its Krull dimension. Th
The local nature of an element of a ring or module
Let R be any commuative ring, and let M be any R-module. Very possibly, M = R.
Consider a Zariski open cover of X = Spec (R) by sets of the form D(fi ) = cfw_P Spec (R) :
fi P , Here, i runs through an in
Math 615, Winter 2011
Problem Set #1 : Solutions
1. The number of copies of ER (R/P ) in the direct sum decomposition does not change
when we localize at P , since ERP (NP ) ER (N )P . It follows that we may assume that
=
(R, P ) is local. Then for any R-
Math 615, Winter 2011
Problem Set #4: Solutions
n
1. Since dim (M/xM ) < dim (M ) = n, we know from a class theorem that HI (M/xM ) =
0. Consequently, the three terms of degree n in the long exact sequence for local cox
homology coming from the short exac
Math 615, Winter 2011
Problem Set #2: Solutions
1. Let J = (f1 , . . . , fn )R, and map g : M M n by m (f1 m, . . . , fn m). Then
g
0 AnnM J M M n is exact. Since is an exact, we get an exact sequence
n g
M
M (AnnM J) 0. Since g (u1 un ) = f1 u1 + fn un
Math 615, Winter 2011
Problem Set #3: Solutions
1. The exact sequence 0 R/(I J) R/I R/J R/(I + J) 0 yields that
i1
i1
i1
i
i
i
Hm (R/I) Hm (R/J) Hm (R/(I + J) Hm (R/(I J) Hm (R/I) Hm (R/J)
is exact. If i d h, the second and last terms are 0, and, hence, s
Exact Sequences with a at cokernel and a sketch of properties of Tor
We give two arguments to prove the following fact:
Lemma. Let 0 N M F 0 be an exact sequence of R-modules such that F is
R-at. Then for every R-module A, 0 A R N A R M A R F 0 is exact.