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If R is an infinite direct product of fields, then R is an injective R-module...But I need an example of a quotient ring R, R/S, that is not injective ?? I feel that "if we take S as an infinite direct sum of fields then R/S may not be injective ??? But I couldn't show it...Please help me to find R/S which is not injective.

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I am not really sure I understand what you are asking for: Is there possibly a mix up between modules and rings? More precisely, do you actually want to consider R (the direct product of fields) as a ring? The most natural question along the lines of your question to me would be: for some ring R find an injective R-module M such that a quotient module M/N is not an injective R-Module. Do you want to know this, or something else? And, is the example of R you start with your attempt at finding an example, or do you need the assertion for this R? –  quid Apr 11 '11 at 10:17
    
I consider the ring R (product of fields) as a module over itself, and I search for a non-injective quotient module of this R. –  Selo Apr 11 '11 at 16:33
    
Thank you for the clarification. –  quid Apr 11 '11 at 17:18
    
I don't know if this helps, but the ideals of your product $R$ are in bijective correspondence with the set of ultrafilters on the indexing set $J$ used to create $R$ (thus $R = \prod_{j \in J} A_j$. ) This correspondence is given by $I \mapsto \{J \backslash supp(r) : r \in I\}$ where $supp( (r_j)_{j \in J} ) = \{j \in J : r_j \neq 0\}$ is the support of $r$. It preserves inclusions, so for instance maximal ideals of $R$ are in bijection with ultrafilters on $J$. So you could take some cyclic module $R/I$ and use this concrete (?) description of $R/I$ to try to see if it's injective or not. –  Konstantin Ardakov Apr 12 '11 at 12:47
    
Thanks, I will think about this. –  Selo Apr 14 '11 at 7:04

3 Answers 3

up vote 3 down vote accepted

(I just stumbled upon this question from a link that David White listed at a completely different question.)

I can give a simple argument that shows some noninjective quotient of $R$ does indeed exist. Perhaps folks already know this, but nobody seems to have said it.

A famous result of Osofsky states that a ring is semisimple if and only all of its cyclic modules are injective:

B. Osofsky, Rings all of whose finitely generated modules are injective, Pacific J. Math., 1964

(See the Theorem, p. 649.) Certainly this ring $R$ is not semisimple (as it's not even noetherian). So some quotient module $R/I$ must be noninjective.

I certainly think that the choice of $I$ above is as good a guess as any. If I have some time to think about whether it is indeed noninjective, I'll come back and edit this.

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By the way, since Lam's book is being reference heavily here, let me add that a presentation of Osofsky's result (based on her later work with P.F. Smith) is given in his "Lectures," Section 6D. –  Manny Reyes Sep 15 '11 at 14:59
    
...also, when I said "whether it is indeed noninjective," I was referring to $R/I$ rather than $I$ (which the vague sentence seems to suggest). –  Manny Reyes Sep 15 '11 at 23:59
    
Thank you very much Reyes..I got my answer, and I hope I will be able to see that my module $R/I$ is not injective? –  Selo Jun 16 '12 at 13:01

Your intuition that "if we take S as an infinite direct sum of fields then R/S may not be injective" is correct. This exact problem is addressed in T.Y. Lam's "Lectures on Modules and Rings" as Remark 3.11(C). I'll sketch the argument here:

Suppose $R = \prod_{j\in J}{A_j}$ for fields $A_j$ and an infinite index set $J$. Then as you pointed out above each $A_j$ is right self-injective, so $R$ is right self-injective. Let $I = \oplus_{j\in J}{A_j}$. Then $I$ fails to be injective. If it were injective then we'd have $R = I\oplus B$ for some right ideal $B\neq 0$. But for any $b=(b_j)\in B$ we have $b\cdot (0,\dots,0,1,0,\dots) = (0,\dots,0,b_j,0,\dots) \in (B \cap I) = 0$. This means all $b_j=0$, a contradiction because $B\neq 0$.

EDIT: As the comments make clear, this example fails because $I$ is not a quotient. I think I must have tried to answer this too early in the morning. I saw the idea to consider $\oplus_{j\in J} A_j$ and remembered something from Lam about it. I guess I should have checked this more closely.

Since no one else has given an answer, let me add some things which may help. First, $R$ is the classic example of a Von Neumann Regular ring which is not semisimple. This has several consequences:

1) Every $R$-module is divisible (Lam, Prop 3.18).

2) Any $R$-module of the form $R/\mathcal{m}$ for a maximal ideal $\mathcal{m}$ is injective (Lam, Theorem 3.72). In our case, this means copies of the field $k$ are injective $R$-modules. So any product of finitely many copies of $k$ is injective. This reduces the amount of modules you have to consider for the desired example.

3) Simple $R$-modules are injective (Lam, Corollary 3.73).

4) $R$ has weak dimension 0, so in particular $R$ is semihereditary and coherent, i.e. submodules of finitely generated projectives are projective, and finitely generated ideals are finitely presented. If $R$ was hereditary, i.e. had right global dimension less than 1, then we'd know there cannot be a quotient of an injective module which is not injective (Lam, Theorem 3.22). Being coherent often seems to help in some surprising way, and in particular may help if considering simple $R$-modules.

I'll keep thinking about this, and I'd be curious to see the solution. What evidence is there that injective quotients even exist at all in this context?

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\mathbb works as usual: $R=\prod_{j\in\mathbb N}\mathbb F_2$. –  Emil Jeřábek Apr 11 '11 at 13:34
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I am sorry. I don't see why $I=\oplus_{j\in \mathbb{N}} \mathbb{F}_2$ is a quotient of $R=\prod_{j\in \mathbb{N}} \mathbb{F}_2$. –  Jiangwei Xue Apr 11 '11 at 13:53
    
Thank you for your answer David, but I don't see too, why the submodule I is a quotient module of R ? –  Selo Apr 11 '11 at 19:18
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I can see why $I$ is not a quotient of $R$: if $\phi\colon R\to I$ is an epimorphism of $R$-modules, then $I= \phi(1)R$. But it is clear that $I$ is not a principal ideal. In fact, $I$ is not finitely generated. –  Frieder Ladisch Apr 11 '11 at 23:31
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I had another comment. My hope to use heredity does not work. According to "The concise handbook of algebra" page 221 (by Mikhalev and Pilz), an infinite product of fields is NOT hereditary. Thus, you know there must exist some factor module of an injective module which is not injective. We know that $R$ is self-injective because each field in the product is injective. So there is some hope that Selo's example works. I'd still like to see this problem resolved. –  David White May 16 '11 at 21:01

if S is not gorenstein then R/S is not injective

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Can you prove that such an ideal exists? –  Dag Oskar Madsen Jul 3 at 12:26

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