Talk:Injective module

message: This is an excellent article. Thanks. Ewjw 16:45, 24 April 2006 (UTC)
thread_title: Thanks

message: Why is the injective dimension useful to know?  - 129.100.182.244 18:23, 18 May 2007 (UTC)
replies: text: Basically injective modules are very well behaved, and their injective dimension is 0.  Injective dimension measures how far from injective a module is.  If A ≤ B with B and B/A injective, then A is nearly injective itself (it has injective dimension 1), and so many arguments for A merely have to combine arguments for B and B/A, both of which are supposed to be easy because they are injective modules.  For instance, over the ring of integers, every quotient module of an injective module is injective, and every module is a submodule of an injective, so every module is either injective of has injective dimension 1.  This means the infinitely long exact sequence of the Ext functor simplifies to 0 → Hom(M/N,A) → Hom(M,A) → Hom(N,A) → Ext(M/N,A) → Ext(M,A) → Ext(N,A) → 0.  The global dimension of a ring is the max of all the injective dimensions of its modules, and gives a global idea of how easy it is to understand the ring.  Rings which have finite injective dimension over themselves are particularly well behaved, and are called (Iwanaga–)Gorenstein rings when they are noetherian. JackSchmidt (talk) 19:17, 27 June 2008 (UTC)

text: Also, low injective dimension is used to prove every module over a Dedekind domain with bounded torsion splits.  Most other uses of dimensions I've seen are either global or projective. JackSchmidt (talk) 06:19, 28 June 2008 (UTC)
thread_title: Why is injective dimension useful?

message: In the paragraph Commutative examples, I believe there is something fishy. It says:
The [injective hull of k as a k[x]-module] is easily described as k(x)/k[x], but the form of the elements is quite transparent: The module has a basis consisting of "inverse monomials", that is x^−n for n = 1, 2, …. Multiplication by scalars is as expected, and multiplication by x behaves normally except that x·x^−1 = 0. The endomorphism ring is simply the ring of formal power series.
I have two concerns about this sentence. 
First, I don't think k(x)/k[x] has a basis (I assume the sentence refers to a k-basis) as described. This would be a basis of k[x,x^-1]/k[x], and the canonical embedding of k[x,x^-1] in k(x) is not an isomorphism. In other words, this family of elements does not span k(x) as a k-vector space.
Second, an injective hull of k should come with an injective map from k to the injective hull. What is this map? I don't see a natural nonzero map of k[x]-module from k to k(x)/k[x] nor to k[x,x^-1]/k[x]. I agree that k(x)/k[x] is injective (because k[x] is integral and a PID), but I don't see why it is the hull of k.  —Preceding unsigned comment added by 86.212.240.153 (talk) 13:34, 11 October 2010 (UTC)
thread_title:  injective hull of k as a k[x]-module 

message: "Specifically, if Q is a submodule of some other module"
Do you mean "Specifically, if Q is an (injective) submodule of some other module"  — Preceding unsigned comment added by 89.11.142.202 (talk) 23:18, 27 June 2011 (UTC)
thread_title:  Tiny thing with potential for confusion 

message: Substituted at 02:14, 5 May 2016 (UTC)
thread_title: Assessment comment