If R is an infinite direct product of fields, then R is an injective Rmodule...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.

(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. 


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 selfinjective, so $R$ is right selfinjective. 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? 


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

