the direct sum of injective modules need not be injective - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T20:31:20Z http://mathoverflow.net/feeds/question/103402 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103402/the-direct-sum-of-injective-modules-need-not-be-injective the direct sum of injective modules need not be injective Miss Independent 2012-07-28T18:27:36Z 2012-07-29T01:41:38Z <p>the Bass-Papp Theorem asserts that a commutative ring $R$ is Noetherian iff every direct sum of injective R-modules is injective. Thus every non-Noetherian ring carries a counterexample. </p> <p>If $$I_1 ⊊I_2⊊…⊊I_n⊊…$$ is an infinite properly ascending chain of ideals of $R$, then for all $n$ let $E_n=E(R/I_n)$ be the injective envelope of $R/I_n$, and let $E=\text{sum of } E_n$. Then $E$ is a direct sum of injective modules and that E is not itself injective. how to prove that $E$ it self is not injective????</p> http://mathoverflow.net/questions/103402/the-direct-sum-of-injective-modules-need-not-be-injective/103405#103405 Answer by Arturo Magidin for the direct sum of injective modules need not be injective Arturo Magidin 2012-07-28T20:46:29Z 2012-07-29T01:41:38Z <p>The standard proof that I am aware of is actually explicit in this regard.</p> <p>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.</p> <p>Let $I=\bigcup\limits_{i=1}^{\infty} I_n$. </p> <p>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). </p> <p>Since we have a map from $I$ to each $E(R/I_n)$, we obtain a map $f\colon I\to \mathop{\prod}\limits_{n=1}^{\infty}E(R/I_n)$ 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 $I\to E$. </p> <p>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 $e=\overline{f}(1)\in E$. </p> <p>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.</p> <p>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.</p>