# Why $\mathrm{Ext}^n_{R[t]}(X, Y)\simeq\mathrm{Ext}^n_R(X, Y)\oplus\mathrm{Ext}^{n-1}_R(X, Y)$?

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...

• 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 $M \to 0$. – user14972 May 1 '18 at 3:31
• @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 $_{R[t]}Y\simeq _RY\oplus _RY$ and \begin{align}\operatorname{Ext}&_{R[t]}^n(X,Y)\\\simeq&\operatorname{Hom}_{D(R-\mathsf{Mod})}(X\oplus X,(Y\oplus Y)[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... – josephz May 1 '18 at 5:31
• @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 $R[t]\otimes_R X\to X\to R[t]\otimes_RX\overset{\times t}{\to}R[t]\otimes_RX$? It seems works... – josephz May 1 '18 at 8:38
• 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 $$0\to P\to C\to P\to 0$$ 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 $Hom(C,Y)=Hom(P,Y)\oplus Hom(P,Y)$. – Roland May 1 '18 at 8:50
• 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). – Roland May 1 '18 at 9:13

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 $$R[t]\otimes_RX\overset{\times t}{\to}R[t]\otimes_RX\to X\to R[t]\otimes_RX$$ 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 $Y$.