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Let $R$ be a commutative Noetherian ring and $M$ an $R$-module. Let $0\rightarrow M\rightarrow E^{\bullet}$ be a minimal injective resolution of $M$ and $0\rightarrow M\rightarrow I^{\bullet}$ be an arbitrary injective resolution. Then how to construct an injective chain map from $E^{\bullet}$ to $I^{\bullet}$?

If we have $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} 0 &\ \ra{} & M & \ra{} & E^0& \ra{f_1} & E^1 & \ra{f_2} & E^2& \\ & & \da{=} & & \da{g_0} & & \da{g_1} & & \da{g_2} & & \\ 0 &\ \ra{} & M & \ra{} & I^0 & \ra{h_1} & I^1 & \ra{h_2} & I^2 & \ra{h_3} & I^3 \\ \end{array} $$ where $g_0$, $g_1$ and $g_2$ are all injections. If we can show $E^2/Im(f_2)$ can be embedded in $I^3$, then we can find $E^3\subseteq I^3$ to continue this diagram, but how to show $E^2/Im(f_2)$ can be embedded in $I^3$?

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2 Answers 2

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One can show that $(E^\bullet, f_\bullet)$ is a direct summand of $(I^\bullet, h_\bullet)$, i.e. $I^\bullet = E^\bullet \oplus J^\bullet$ for some (exact) complex $(J^\bullet, \phi_\bullet)$. $\DeclareMathOperator{\Im}{\operatorname{im}}$

It suffices to show that $\Im(f_2) = \ker(h_3g_2)$ (as then $E^2/\Im(f_2) = E^2/\ker(h_3g_2) \hookrightarrow I^3$, producing a split injection $g_3 : E^3 \hookrightarrow I^3$). We have by induction $g_2 = (\text{id},0) : E^2 \hookrightarrow I^2 = E^2 \oplus J^2$ and $h_2 : I^1 \to I^2$ given by $\begin{pmatrix} f_2 & 0 \\ 0 & \phi_2 \end{pmatrix}$. Then for $x \in E^2$, $x \in \ker(h_3g_2) \iff g_2(x) \in \ker h_3 \iff g_2(x) \in \Im(h_2) \iff \begin{pmatrix} x \\ 0 \end{pmatrix} = \begin{pmatrix} f_2 & 0 \\ 0 & \phi_2 \end{pmatrix} \begin{pmatrix} y_1 \\ y_1' \end{pmatrix}$ for some $y_1 \in E^1$, $y_1' \in J^1 \iff x \in \Im(f_2)$.

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  • $\begingroup$ I need to think about why $h_2$ is of the form in the answer and why $\phi_2$$(y_1^{'})\in J^2$. $\endgroup$
    – nick
    Aug 17, 2014 at 4:06
  • $\begingroup$ Thanks, the details are right! $\endgroup$
    – nick
    Aug 17, 2014 at 11:05
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It is well known that we can define some chain maps $u^{\bullet}:E^{\bullet}\to I^{\bullet}$ and $v^{\bullet}:I^{\bullet}\to E^{\bullet}$ which are identity on $M$. Thus we get a chain map $w^{\bullet}:E^{\bullet}\to E^{\bullet}$ which is identity on $M$, where $w^{\bullet}=v^{\bullet}u^{\bullet}$. Now using that $E^{\bullet}$ is minimal it's easy to prove (inductively) that $w^{\bullet}$ is a chain isomorphism and therefore $u^{\bullet}$ is injective.

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  • $\begingroup$ very very good answer, it instantly solves the question I posed. Thank you! $\endgroup$
    – nick
    Aug 17, 2014 at 4:03

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