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This proposition is as follows: Let $$ 0\rightarrow L\rightarrow M\rightarrow N\rightarrow 0 $$ be a short exact sequence of R-modules. Let the following be projective resolutions of L and N respectively: $$ \cdots\rightarrow P_{1}\rightarrow P_{0}\rightarrow L\rightarrow 0 $$ $$ \cdots\rightarrow \bar{P}_{1}\rightarrow \bar{P}_{0}\rightarrow N\rightarrow 0 $$ The proposition claims that there is a projective resolution for M: $$ \cdots\rightarrow P_{1}\oplus\bar{P}_{1}\rightarrow P_{0}\oplus\bar{P}_{0}\rightarrow M\rightarrow 0 $$ that commutes with all the above exact sequences, as well as the obvious ones given by $$ 0\rightarrow P_n\rightarrow P_n\oplus\bar{P}_{n}\rightarrow\bar{P}_{n}\rightarrow 0 $$

The book constructs the maps in the projective resolution for M as follows: For the first map, first lift the map $\bar{P}_{0}\rightarrow N$ to $\Phi:\bar{P}_{0}\rightarrow M$ (this is possible because $M\rightarrow N\rightarrow 0$ is surjective and $\bar{P}_{0}$ is projective). Then, compose $P_{0}\rightarrow L$ and $L\rightarrow M$ to get a map $\psi:P_{0}\rightarrow M$, and define $\phi:P_{0}\oplus \bar{P}_{0}\rightarrow M$ as follows: $$ \phi(a,b)=\psi(a)+\Phi(b) $$ It is not hard to see that this makes the diagram commute. However, my problem is that I cannot see how the sequence given by the maps is exact at $P_{0}\oplus \bar{P}_{0}$.

I want to prove that if $\phi(a,b)\in ker(\phi)$, then $\phi(a,b)\in im(\alpha)$ where $\alpha:P_{1}\oplus \bar{P}_{1}\rightarrow P_{0}\oplus \bar{P}_{0}$. That means I need $\psi(a)$ to be zero so that $a$ is in the image of $P_{1}\rightarrow P_{0}$. However, there does not seem to be any reason for this to be true.

In particular, I have the following seeming counterexample. Suppose the initial exact sequence was $$ 0\rightarrow n\mathbb{Z}\rightarrow \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}\rightarrow 0 $$ and the projective resolutions are: $$ \cdots\rightarrow 0\rightarrow\mathbb{Z}\xrightarrow{n}n\mathbb{Z}\rightarrow 0 $$ $$ \cdots\rightarrow\mathbb{Z}\xrightarrow{\alpha}\mathbb{Z}\oplus\mathbb{Z}\xrightarrow{\phi}\mathbb{Z}\rightarrow 0 $$ $$ \cdots\rightarrow\mathbb{Z}\xrightarrow{n}\mathbb{Z}\xrightarrow{n\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\rightarrow 0 $$ Then we have $$ \phi(a,b)=an+b $$ $$ \alpha(c)=(0,nc) $$ by commutativity of the whole diagram (which unfortunately I do not know how to draw). However, $(-1,n)\in ker(\phi)$ but $(-1,n)\not\in im(\alpha)$. This seems to contradict exactness.

It would be great if someone could point out the flaws in my argument. Thanks very much in advance!

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What you're missing is that the map $P_1\oplus \bar{P}_1\to P_0\oplus \bar{P}_0$ should not be just defined to be the direct sum of the original maps $P_1\to P_0$ and $\bar{P}_1\to\bar{P}_0$. Instead, let $L_1$ denote the kernel of $P_0\to L$, let $N_1$ denote the kernel of $\bar{P}_0\to N$, and let $M_1$ denote the kernel of your map $\phi:P_0\oplus\bar{P}_0\to M$. Then there is a canonical short exact sequence $$0\to L_1\to M_1\to N_1\to 0$$ (exercise: construct the maps in this sequence and check that it is exact). Furthermore, $$\cdots\rightarrow P_{2}\rightarrow P_{1}\rightarrow L_1\rightarrow 0$$ and $$\cdots\rightarrow \bar{P}_{2}\rightarrow \bar{P}_{1}\rightarrow N_1\rightarrow 0$$ are projective resolutions of $L_1$ and $N_1$. So we can use the exact same procedure we used to construct $\phi$ to construct a surjection $\phi_1:P_1\oplus\bar{P}_1\to M_1$, which we then compose with the inclusion $M_1\to P_0\oplus\bar{P}_0$ to get the map $P_1\oplus \bar{P}_1\to P_0\oplus \bar{P}_0$, whose image is manifestly the kernel of $\phi$.

More generally, all of the maps $P_{n+1}\oplus \bar{P}_{n+1}\to P_n\oplus \bar{P}_n$ in the resolution of $M$ are constructed by iterating this process. These maps make the big diagram commute for the same reason that you saw that $\phi$ made the diagram commute.

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  • $\begingroup$ Thank you very much for clearing this up! I foolishly assumed that the map should be the canonical one, when in fact I should have simply repeated the initial process. Your answer is very clear and helpful. It is a pity I do not have enough reputation to change the vote count yet. $\endgroup$ – Houndoom Oct 13 '15 at 11:22

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