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(I originally asked this question on Math.SE here. As suggested on meta.MathOverflow (posting an unanswered Math.SE question on MathOverflow), I've waited about a week before reposting it here. Note that although the original question has an answer posted, it didn't actually answer my question.)

As an assignment in my commutative algebra class, I need to prove the Mayer-Vietoris Sequence for local cohomology:

Let $R$ be a Noetherian ring, $I,J$ $R$-ideals, and $M$ an $R$-module. Show that there is a long exact sequence $$ \cdots \rightarrow H^i_{I+J}(M)\rightarrow H^i_I(M)\oplus H^i_J(M) \rightarrow H^i_{I\cap J}(M) \rightarrow H^{i+1}_{I+J}(M)\rightarrow\cdots. $$

I've reduced this to proving the following:

Let $R$ be a Noetherian ring, $I,J$ $R$-ideals, and $M$ an $R$-module such that $\Gamma_{I\cap J}(M)=M$. Then $M=\Gamma_I(M)+\Gamma_J(M)$.

However, in trying to prove this, I ended up "proving" that $M/(\Gamma_I(M)+\Gamma_J(M))\cong (\Gamma_I(M)+\Gamma_J(M))/\Gamma_I(M)$, which is certainly false in general---if it were true, then (assuming we've shown $M=\Gamma_I(M)+\Gamma_J(M)$) setting $J=0$, we'd get $\Gamma_I(M)=M$ for every ideal $I$ and every $R$-module $M$, which is clearly false.

Question: Where did I go wrong?

Here's my "proof" of the incorrect "fact":

Consider the short exact sequence $$ 0 \to \frac{\Gamma_I(M)}{\Gamma_I(M)\cap \Gamma_J(M)} \to \frac{M}{\Gamma_J(M)}\to \frac{M}{\Gamma_I(M)+\Gamma_J(M)}\to 0.\tag{1}$$ Notice that if $x\in M$, so that $(I\cap J)^n x=0$ for some $x\geq 0$, then $I^n(J^n x)\subseteq (IJ)^nx\subseteq (I\cap J)^n x=0$; hence, $J^nx \subseteq \Gamma_I(M)\subseteq \Gamma_I(M)+\Gamma_J(M)$. Thus, $$ \Gamma_J\left(\frac{M}{\Gamma_I(M)+\Gamma_J(M)}\right) = \frac{M}{\Gamma_I(M)+\Gamma_J(M)} \tag{2},$$ and so (by a previous homework set) $$ H^i_J\left(\frac{M}{\Gamma_I(M)+\Gamma_J(M)}\right) = 0 \quad \text{for all}\quad i>0 \tag{3}.$$ Also by a previous homework set, $$ \Gamma_J\left(\frac{M}{\Gamma_J(M)}\right)=0. \tag{4}$$ Hence, applying the long exact sequence of $H_J$ to (1), and using (2), (3), and (4), we get an exact sequence $$0\to \frac{M}{\Gamma_I(M)+\Gamma_J(M)} \to H^1_J\left(\frac{\Gamma_I(M)}{\Gamma_I(M)\cap \Gamma_J(M)}\right) \to H^1_J\left(\frac{M}{\Gamma_J(M)}\right)\to 0. $$ Hence, $$ \frac{M}{\Gamma_I(M)+\Gamma_J(M)} \cong \ker \left[H^1_J\left(\frac{\Gamma_I(M)}{\Gamma_I(M)\cap \Gamma_J(M)}\right) \to H^1_J\left(\frac{M}{\Gamma_J(M)}\right)\right].\tag{5}$$

Now, since $\Gamma_I(M)\cap \Gamma_J(M)=\Gamma_J(\Gamma_I(M))$, we get $$ \frac{\Gamma_I(M)}{\Gamma_I(M)\cap \Gamma_J(M)} = \frac{\Gamma_I(M)}{\Gamma_J(\Gamma_I(M))}.$$ From a previous homework assignment, if $N$ is any $R$-module, $R$ is Noetherian, and $J$ is an $R$-ideal, then for $i>0$, the map $H^i(N)\to H^i(N/\Gamma_I(N))$ induced by the projection $N\to N/\Gamma_I(N)$ is an isomorphism. Therefore, (5) becomes $$ \frac{M}{\Gamma_I(M)+\Gamma_J(M)} \cong \ker \left[H^1_J(\Gamma_I(M)) \to H^1_J(M)\right],$$ where $H^1_J(\Gamma_I(M)) \to H^1_J(M)$ is the natural map. But by the long exact sequence of $H_J$, $$ \ker \left[H^1_J(\Gamma_I(M)) \to H^1_J(M)\right] \cong \frac{\Gamma_J(M)}{\Gamma_J(\Gamma_I(M))} \cong \frac{\Gamma_I(M)+\Gamma_J(M)}{\Gamma_I(M)}.$$ Thus, $$ \frac{M}{\Gamma_I(M)+\Gamma_J(M)} \cong \frac{\Gamma_I(M)+\Gamma_J(M)}{\Gamma_I(M)}.$$

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(This is an elaboration on Vinteuil's answer. I did not accept Vinteuil's answer due to the lack of detail.)

So, it seems that I made a very foolish error: Consider the long exact sequence of $H_J$ that I refer to in the 4th-to-last line of my original post. Written out, it is $$\Gamma_J(M)\to \Gamma_J(M/\Gamma_I(M))\to H^1_J(\Gamma_I(M))\to H^1_J(M).$$ The correct thing to get from this is that the kernel of the last map is the image of the second, i.e. $$\ker\left[H^1_J(\Gamma_I(M))\to H^1_J(M)\right]=\operatorname{im}\left[\Gamma_J(M/\Gamma_I(M))\to H^1_J(\Gamma_I(M))\right].$$ However, what I did was set the kernel of the last map equal to the image of the first.

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In line -3, first isomorphism.

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  • $\begingroup$ Can you be more specific? I'm not sure what you mean by line -3. $\endgroup$ – Avi Steiner Mar 24 '14 at 14:16
  • $\begingroup$ $ \ker \left[H^1_J(\Gamma_I(M)) \to H^1_J(M)\right] \cong \frac{\Gamma_J(M)}{\Gamma_J(\Gamma_I(M))}$ $\endgroup$ – Vinteuil Mar 24 '14 at 14:32
  • $\begingroup$ Why is that incorrect? The long exact sequence $$\Gamma_J(M)\to \Gamma_J(M/\Gamma_I(M))\to H^1_J(\Gamma_I(M))\to H^1_J(M)$$ implies that $$\ker \left[H^1_J(\Gamma_I(M)) \to H^1_J(M)\right] \cong \frac{\Gamma_J(M)}{\Gamma_J(M)\cap \Gamma_I(M)}$$. But $\Gamma_J(M)\cap \Gamma_I(M)=\Gamma_J(\Gamma_I(M))$. $\endgroup$ – Avi Steiner Mar 27 '14 at 23:21
  • $\begingroup$ Take for instance examples with $\Gamma_I(M)=0 \neq \Gamma_J(M)$. $\endgroup$ – Vinteuil Mar 28 '14 at 11:32
  • $\begingroup$ Also, could you put your above comments into the actual answer? $\endgroup$ – Avi Steiner Mar 29 '14 at 20:25

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