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@ Fred: I think the following may be help you to give an example for (ITR) question:

1. Choose a non-Noetherian local ring $(A, \frak{m})$ such that $\frak{m}= \frak{m}^n$ for all $n$.

2. Let $E(k)$ be the injective envelope of $k = R/\frak{m}$.

3. Claim: if $E(k)$ is $\frak m$-torsion, then $E(k) = k$. Indeed, let $x \in E(k)$. There is $n$ such that $\frak{m}^n x = 0$. Hence $\frak{m} x = 0$. But $k \subseteq Rx$, so $x \in k$. So $k$ is injective.

Dear Fred, I give here that a such ring.

Let $k$ be a file and $Q^+$ be the semigroup of non-negative ration. Let $A = k[Q^+]$ be the semigroup ring, i.e

$A = { \sum_{\alpha}u_{\alpha}x^{\alpha}: u_{\alpha} \in k }$.

Let $\frak{n} = (x^{\alpha}: \alpha > 0)$ be the maximal ideal of $A$. Since for all $\alpha > 0$ we have $x^{\alpha} = x^{\alpha/2}x^{\alphax^{\alpha/2}x^{\alpha/2} \in \frak{n}^2$, so $\frak{n} = \frak{n}^2$.

Let $S = A_{\frak{n}}$. Let $I = (x^{\alpha}: \alpha > 1)$ be an ideal of $S$. Let $(R, \frak{m})$ be the quotient ring $S/I$. We have $\frak{m} = \frak{m}^2$ by above.

Claim: $k$ is not injective.

Proof of claim: We consider the ideal $(x) \subseteq R$. Notice that $(x) \cong k$ as $R$-modules. If $k$ is injective, then (x) is a direct summand of local ring $R$. It is a contradiction.

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@ Fred: I think the following may be help you to give an example for (ITR) question:

1. Choose a non-Noetherian local ring $(A, \frak{m})$ such that $\frak{m}= \frak{m}^n$ for all $n$.

2. Let $E(k)$ be the injective envelope of $k = R/\frak{m}$.

3. Claim: if $E(k)$ is $\frak m$-torsion, then $E(k) = k$. Indeed, let $x \in E(k)$. There is $n$ such that $\frak{m}^n x = 0$. Hence $\frak{m} x = 0$. But $k \subseteq Rx$, so $x \in k$. So $k$ is injective.

Dear Fred, I give here that a such ring.

Let $k$ be a file and $Q^+$ be the semigroup of non-negative ration. Let $A = k[Q^+]$ be the semigroup ring, i.e

$A = { \sum_{\alpha}u_{\alpha}x^{\alpha}: u_{\alpha} \in k }$.

Let $\frak{n} = (x^{\alpha}: \alpha > 0)$ be the maximal ideal of $A$. Since for all $\alpha >0$ we have $x^{\alpha} = x^{\alpha/2}x^{\alpha} \in \frak{n}^2$, so $\frak{n} = \frak{n}^2$.

Let $S = A_{\frak{n}}$. Let $I = (x^{\alpha}: \alpha > 1)$ be an ideal of $S$. Let $(R, \frak{m})$ be the quotient ring $S/I$. We have $\frak{m} = \frak{m}^2$ by above.

Claim: $k$ is not injective.

Proof of claim: We consider the ideal $(x) \subseteq R$. Notice that $(x) \cong k$ as $R$-modules. If $k$ is injective, then (x) is a direct summand of local ring $R$. It is a contradiction.

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@ Fred: I think the following may be help you to give an example for (ITR) question:

1. Choose a non-Noetherian local ring $(A, \frak m)$ frak{m})$such that$\frak m= \frak{m}= \frak m^n$frak{m}^n$ for all $n$.

2. Let $E(k)$ be the injective envelope of $k = R/\frak m$R/\frak{m}$. 3. Claim: if$E(k)$is$\frak m$-torsion, then$E(k) = k$. Indeed, let$x \in E(k)$. There is$n$such that$\frak m^n \frak{m}^n x = 0$. Hence$\frak m \frak{m} x = 0$. But$k \subseteq Rx$, so$x \in k$. So$k\$ is injective.

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