(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, $$ \frac{M}{\Gamma_I(M)+\Gamma_J(M)} \cong \frac{\Gamma_I(M)+\Gamma_J(M)}{\Gamma_I(M)}.$$


2 Answers 2


(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.


In line -3, first isomorphism.

  • $\begingroup$ Can you be more specific? I'm not sure what you mean by line -3. $\endgroup$ Mar 24, 2014 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, 2014 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 $\Gamma_J(M)\cap \Gamma_I(M)=\Gamma_J(\Gamma_I(M))$. $\endgroup$ Mar 27, 2014 at 23:21
  • $\begingroup$ Take for instance examples with $\Gamma_I(M)=0 \neq \Gamma_J(M)$. $\endgroup$
    – Vinteuil
    Mar 28, 2014 at 11:32
  • $\begingroup$ Also, could you put your above comments into the actual answer? $\endgroup$ Mar 29, 2014 at 20:25

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.