(*I originally asked this question on Math.SE [here](https://math.stackexchange.com/q/716247/13487). As suggested on meta.MathOverflow ([posting an unanswered Math.SE question on MathOverflow](https://meta.mathoverflow.net/q/1569/36720)), 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,
$$ \frac{M}{\Gamma\_I(M)+\Gamma\_J(M)} \cong \frac{\Gamma\_I(M)+\Gamma\_J(M)}{\Gamma\_I(M)}.$$



local cohomology mayer-vietoris sequence
(*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*.


