# Regularity and Short Exact Sequence

Suppose $0 \to M_1 \to M_2 \to M_3 \to 0$ is a short exact sequence of finitely generated graded $k[x_0,...,x_r]$-modules. Then show that $\mathrm{reg}(M_1) \leq\max(\mathrm{reg}(M_2),\mathrm{reg}(M_3)+1)$.

Since a short exact sequence gives a long exact sequence of local cohomology modules therefore we get a long exact sequence $$\cdots\to H_m^i(M_1)\to H_m^i(M_2) \to H_m^i(M_3)\to H_m^{i+1}(M_1)\to H_m^{i+1}{(M_2)}\to H_m^{i+1}(M_3)\to\cdots.$$

Am i on the right track? How to proceed from here?

Recall that $$\mathrm{reg}(M) := \sup\{\mathrm{end}(H^i_{R_+}(M)) + i : i ∈\mathbb N\}=\inf\{m \in\mathbb Z: H^{i}_{R_+}(M)_{n-i}=0\ \forall i \in \mathbb N\ \forall n > m\},$$ where $\mathrm{end}(N) := \sup\{n ∈ \mathbb Z: N_n\ne0\}$.
Set $d=\mathrm{reg}(M_1)$. If $d>\mathrm{reg}(M_2)$ and $d>\mathrm{reg}(M_3)+1$, then $H^i_{R_+}(M_2)_{d-i}=0\ \forall i\in\mathbb N$ and $H^{i-1}_{R_+}(M_3)_{d-i}=0\ \forall i\in\mathbb N^*$. It follows $H^i_{R_+}(M_1)_{d-i}=0\ \forall i\in\mathbb N$, a contradiction.