2
$\begingroup$

I am stuck in a step of this problem

Suppose that $R$ is a ring and $X$ and $Y$ are $R$-modules. If $X$ and $Y$ are regarded as $R[t]$(the polynomial ring over $R$)-modules through the ring homomorphism $R[t]\to R$, $t\mapsto 0$, then $\operatorname{Ext}^n_{R[t]}(X, Y)\simeq\operatorname{Ext}^n_R(X, Y)\oplus\operatorname{Ext}^{n-1}_R(X, Y).$


Here is what I have for the moment:

Let us consider the exact sequence $$0\longrightarrow R[t]\overset{\times t}{\longrightarrow}R[t]\longrightarrow R\longrightarrow 0.$$ Tensoring with $X$ produces an exact (since $R_R$ is flat) sequence of $R[t]$-modules $$0\longrightarrow R[t]\otimes_R X\overset{\times t}{\longrightarrow}R[t]\otimes_R X\longrightarrow X\longrightarrow 0.$$ Then applying the derived functor $R\operatorname{Hom}_{R[t]}(-,Y)$ to it we obtain a long exact sequence \begin{align} 0&\to\operatorname{Hom}_{R[t]}(X,Y)\to\operatorname{Hom}_{R[t]}(R[t]\otimes_RX,Y)\overset{\times t}{\to}\operatorname{Hom}_{R[t]}(R[t]\otimes_RX,Y)\\ &\to\operatorname{Ext}_{R[t]}^1(X,Y)\to\operatorname{Ext}_{R[t]}^1(R[t]\otimes_RX,Y)\overset{\times t}{\to}\operatorname{Ext}_{R[t]}^1(R[t]\otimes_RX,Y)\\ &\to\operatorname{Ext}_{R[t]}^2(X,Y)\to\operatorname{Ext}_{R[t]}^2(R[t]\otimes_RX,Y)\to\cdots \end{align} Since $t$ annihilates $Y$ and $$\operatorname{Ext}_{R[t]}^n(R[t]\otimes_RX,Y)=\operatorname{Hom}_{D(R[t]-\mathsf{Mod})}((R[t]\otimes_RX)[-n],Y),$$ it can be seen that the morphisms marked with $\times t$ are actually all $0$. Moreover by choosing a projective resolution of $X$ it can be shown that $\operatorname{Ext}_{R[t]}^n(R[t]\otimes_RX,Y)\simeq\operatorname{Ext}_{R}^n(X,Y)$. Thus far the long exact sequence can be tore into exact sequences \begin{align} 0&\to\operatorname{Hom}_{R[t]}(X,Y)\to\operatorname{Hom}_{R}(X,Y)\to 0\\ 0&\to\operatorname{Hom}_{R}(X,Y)\to\operatorname{Ext}_{R[t]}^1(X,Y)\to\operatorname{Ext}_{R}^1(X,Y)\to 0\\ 0&\to\operatorname{Ext}_{R}^1(X,Y)\to\operatorname{Ext}_{R[t]}^2(X,Y)\to\operatorname{Ext}_{R}^2(X,Y)\to 0\\ &\cdots \end{align}


And I am stuck here, as not being able to show that these short exact sequences split. So I would like to ask for some hints of what to do next, and thanks in advance...

$\endgroup$
  • $\begingroup$ I'm just brainstorming, but I wonder if identifying $R[t]$-mod with the full subcategory of $\mathrm{Ch}(R)$ of two-term complexes of the form $M \to M$ (the differential is the action of multiplcation-by-$t$) suggests anything. When the differential is zero, such a complex is the biproduct of $0 \to M$ and $M \to 0$. $\endgroup$ – user14972 May 1 '18 at 3:31
  • $\begingroup$ @Hurkyl So intuitively $ _{R[t]}Y=(\cdots\to 0\to Y\overset{\times t}{\to} Y\to 0\to\cdots)$ while the first $Y$ takes the 0-th entry and the second $Y$ is seated at the 1-st entry. Since the differential is actually zero, then $ _{R[t]}Y\simeq _RY\oplus _RY[1]$ and \begin{align}\operatorname{Ext}&_{R[t]}^n(X,Y)\\\simeq&\operatorname{Hom}_{D(R-\mathsf{Mod})}(X\oplus X[1],(Y\oplus Y[1])[n])\\\simeq&\operatorname{Ext}_R^n(X,Y)\oplus \operatorname{Ext}_R^{n-1}(X,Y)\oplus\cdots\end{align} Did I understand your comment correctly? But yet I cannot see how the short exact sequences split either... $\endgroup$ – josephz May 1 '18 at 5:31
  • 1
    $\begingroup$ @Roland Oh, or I just rotate the distinguished triangle $X\to R[t]\otimes_R X\overset{\times t}{\to}R[t]\otimes X\to X[-1]$ in $D(R[t]-\mathsf{Mod})^{\mathrm{op}}$ to $R[t]\otimes_R X[1]\to X\to R[t]\otimes_RX\overset{\times t}{\to}R[t]\otimes_RX$? It seems works... $\endgroup$ – josephz May 1 '18 at 8:38
  • 1
    $\begingroup$ Actually, you can probably avoid the use of derived category. Let $P$ be a projective resolution of $R[t]\otimes_R X$ and $C$ be the cone of $P\overset{\times t}\to P$. The short exact sequence of projectives $$0\to P\to C\to P[1]\to 0$$ may be used to compute the long exact sequence $Ext$. Now apply $Hom(.,Y)$, you have $Hom(C,Y)$ the cone of $Hom(P,Y)\overset{\times t}\to Hom(P,Y)$ so you should be able to prove that $Hom(C,Y)=Hom(P,Y)\oplus Hom(P[1],Y)$. $\endgroup$ – Roland May 1 '18 at 8:50
  • 1
    $\begingroup$ Sure they are, and in my opinion the derived category way is more conceptual : a long exact sequence which gives split short exact sequences may come from a zero map in the derived category (this is what jumped into my mind when I read your post). $\endgroup$ – Roland May 1 '18 at 9:13
1
$\begingroup$

This is a summary of the comments above, which addresses the problem why those short exact sequences split. Let us start with the exact sequence $$0\to R[t]\otimes_RX\overset{\times t}{\to}R[t]\otimes_RX\to X\to 0.$$ It induces a distinguished triangle $$R[t]\otimes_RX\overset{\times t}{\to}R[t]\otimes_RX\to X\to R[t]\otimes_RX[1]$$ in $D(R[t]\operatorname{-}\mathsf{Mod})$. Applying $R\mathrm{Hom}_{R[t]}(-,Y)$ to it, we arrive at a distinguished triangle $(\ast)$: $$R\mathrm{Hom}_{R[t]}(R[t]\otimes_RX, Y)[-1]\to R\mathrm{Hom}_{R[t]}(X, Y)\to R\mathrm{Hom}_{R[t]}(R[t]\otimes_RX, Y)\overset{\times t}{\to}R\mathrm{Hom}_{R[t]}(R[t]\otimes_RX, Y)$$ and it remains to verify that $\times t$ is the zero morphism. To this end, pick a projective resolution $P\overset{qis}{\twoheadrightarrow} R[t]\otimes_RX$ (that is, $P\in K^-(R[t]\operatorname{-}\mathsf{Mod})$ with $P^{-i}$ projective). It follows that $$R\mathrm{Hom}_{R[t]}(R[t]\otimes_RX, Y)\simeq\operatorname{Hom}^\bullet_{R[t]}(P,Y)$$ and therefore it can be seen that the morphism $\times t$ in $(\ast)$ is $0$, since $t$ annihilates $Y$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.