3 and onther time I forgot the terms

The standard proof that I am aware of is actually explicit in this regard.

As you note, assume that $I_1\subsetneq I_2\subsetneq\cdots$ is an infinite ascending chain of ideals of $R$, let $E(R/I_n)$ be the injective envelope of $R/I_n$ for each $n$, and let $$E=\bigoplus_{n=1}^{\infty}E(R/I_n)$$ be the their direct sum.

Let $I=\bigcup\limits_{i=1}^{\infty} I_n$.

For each $n$, let $f_n$ be the composition of the embedding $I\hookrightarrow R$ with the canonical map $R\to E(R/I_n)$ (map to the quotient, then embed into the envelope).

Since we have a map from $I$ to each $E(R/I_n)$, we obtain a map $f\colon I\to \mathop{\prod}\limits_{n=1}^{\infty}$ mathop{\prod}\limits_{n=1}^{\infty}E(R/I_n)$by the universal property of the product. In fact, the image of$f$lies in the direct sum, since for every$x\in I$there exists$n\in\mathbb{N}$such that$x\in I_m$for all$m\geq n$, hence the image of$x$is$0$in$R/I_n$. So we have a map$I\to E$. I claim that$f$does not extend to a module homomorphism$\overline{f}\colon R\to E$(which it would necessarily do if$E$were injective). Assume to the contrary that we have an extension$\overline{f}\colon R\to E$. Being a module homomorphism with domain the free module of rank$1$, it is completely determined by$\overline{f}(1)$, and so it has the form$\overline{f}(x) = xe$for all$x\in R$, where$e=\overline{f}(1)\in E$. Now, let$n_0$be a positive integer such that the$m$th component of$e$is$0$for all$m\geq n_0$. Let$x\in I_{n_0}\setminus I_{n_0-1}$. When we map$x$to$R/I_{n_0}$, we obtain a nonzero element; hence the$n_0$th component of$f(x)$must be nonzero (since$R/I_{n_0})$embeds into$E(R/I_{n_0})$). But$f(x) = \overline{f}(x) = xe$, and the$n_0$th component of$e$is$0$, hence so is that of$f(x)$, a contradiction. The contradiction arises from the assumption that we can extend the map$f\colon I\to E$to a module homomorphism$R\to E$. Hence no such extension exists, so$E$is not injective. 2 missing$I_n$in definition of$I$; rephrased slightly the proof The standard proof that I am aware of is actually explicit in this regard. As you note, assume that$I_1\subsetneq I_2\subsetneq\cdots$is an infinite ascending chain of ideals of$R$, let$E(R/I_n)$be the injective envelope of$R/I_n$for each$n$, and let $$E=\bigoplus_{n=1}^{\infty}E(R/I_n)$$ be the their direct sum. Let$I=\bigcup\limits_{i=1}^{\infty}$. I=\bigcup\limits_{i=1}^{\infty} I_n$.

For each $n$, let $f_n$ be the composition of the embedding $I\hookrightarrow R$ with the canonical map $R\to E(R/I_n)$ (map to the quotient, then embed into the envelope).

Since we have a map from $I$ to each $E(R/I_n)$, we obtain a map $f\colon I\to \mathop{\prod}\limits_{n=1}^{\infty}$ by the universal property of the product. In fact, the image of $f$ lies in the direct sum, since for every $x\in I$ there exists $n\in\mathbb{N}$ such that $x\in I_m$ for all $m\geq n$, hence the image of $x$ is $0$ in $R/I_n$. So we have a map $I\to E$.

Now suppose

I claim that $E$ is injective. Then the map $f\colon I\to E$ can be extended f$does not extend to a module homomorphism$\overline{f}\colon R\to E$(which it would necessarily do if$E$were injective). Assume to the contrary that we have an extension$\overline{f}\colon R\to E$. Being a module homomorphism with domain the free module of rank$1$, it is completely determined by$\overline{f}(1)$, and so it has the form$\overline{f}(x) = xe$for all$x\in R$, where$e=\overline{f}(1)$. e=\overline{f}(1)\in E$.

Now, let $n_0$ be a positive integer such that the $m$th component of $e$ is $0$ for all $m\geq n_0$. Let $x\in I_{n_0}\setminus I_{n_0-1}$. When we map $x$ to $R/I_{n_0}$, we obtain a nonzero element; hence the $n_0$th component of $f(x)$ must be nonzero (since $R/I_{n_0})$ embeds into $E(R/I_{n_0})$). But $f(x) = \overline{f}(x) = xe$, and the $n_0$th component of $e$ is $0$, hence so is that of $f(x)$, a contradiction.

The contradiction arises from the assumption that $E$ is injective; that is, we cannot can extend the map $f\colon I\to E$ to a module homomorphism $R\to E$. Hence no such extension exists, so $E$ is not injective.

1

The standard proof that I am aware of is actually explicit in this regard.

As you note, assume that $I_1\subsetneq I_2\subsetneq\cdots$ is an infinite ascending chain of ideals of $R$, let $E(R/I_n)$ be the injective envelope of $R/I_n$ for each $n$, and let $$E=\bigoplus_{n=1}^{\infty}E(R/I_n)$$ be the their direct sum.

Let $I=\bigcup\limits_{i=1}^{\infty}$.

For each $n$, let $f_n$ be the composition of the embedding $I\hookrightarrow R$ with the canonical map $R\to E(R/I_n)$ (map to the quotient, then embed into the envelope).

Since we have a map from $I$ to each $E(R/I_n)$, we obtain a map $f\colon I\to \mathop{\prod}\limits_{n=1}^{\infty}$ by the universal property of the product. In fact, the image of $f$ lies in the direct sum, since for every $x\in I$ there exists $n\in\mathbb{N}$ such that $x\in I_m$ for all $m\geq n$, hence the image of $x$ is $0$ in $R/I_n$. So we have a map $I\to E$.

Now suppose that $E$ is injective. Then the map $f\colon I\to E$ can be extended to a module homomorphism $\overline{f}\colon R\to E$. Being a module homomorphism with domain the free module of rank $1$, it is completely determined by $\overline{f}(1)$, and so it has the form $\overline{f}(x) = xe$, where $e=\overline{f}(1)$.

Now, let $n_0$ be a positive integer such that the $m$th component of $e$ is $0$ for all $m\geq n_0$. Let $x\in I_{n_0}\setminus I_{n_0-1}$. When we map $x$ to $R/I_{n_0}$, we obtain a nonzero element; hence the $n_0$th component of $f(x)$ must be nonzero (since $R/I_{n_0})$ embeds into $E(R/I_{n_0})$). But $f(x) = \overline{f}(x) = xe$, and the $n_0$th component of $e$ is $0$, hence so is that of $f(x)$, a contradiction.

The contradiction arises from the assumption that $E$ is injective; that is, we cannot extend the map $f\colon I\to E$ to a module homomorphism $R\to E$.