How to show that this sequence is exact?

Question

Let $A$ be a local ring, $\mathfrak{m}$ its maximal ideal, $\mathfrak{a}$ a proper ideal. Set $\mathfrak{n}:=\mathfrak{m}/\mathfrak{a}$ and $k:=A/\mathfrak{m}$. Then this sequence of $k$-vector spaces is exact $$0\rightarrow (\mathfrak{m}^2+\mathfrak{a})/\mathfrak{m}^2 \rightarrow \mathfrak{m}/\mathfrak{m}^2\rightarrow \mathfrak{n}/\mathfrak{n}^2\rightarrow 0$$

Thanks for your help in advance.

Answer 1

Since $\mathfrak{n}=\mathfrak{m}/\mathfrak{a}$, we have $\mathfrak{n}^2=(\mathfrak{m}^2+\mathfrak{a})/\mathfrak{a}$. Thus $\mathfrak{n}/\mathfrak{n}^2\cong\mathfrak{m}/(\mathfrak{m}^2+\mathfrak{a})$. On the other hand, $\frac{\mathfrak{m}/\mathfrak{m}^2}{(\mathfrak{m}^2+\mathfrak{a})/\mathfrak{m}^2}\cong\mathfrak{m}/(\mathfrak{m}^2+\mathfrak{a})$. Thus, $\frac{\mathfrak{m}/\mathfrak{m}^2}{(\mathfrak{m}^2+\mathfrak{a})/\mathfrak{m}^2}\cong \mathfrak{n}/\mathfrak{n}^2$. Hence we have an exact sequence of $A$-modules $$0\rightarrow (\mathfrak{m}^2+\mathfrak{a})/\mathfrak{m}^2 \rightarrow \mathfrak{m}/\mathfrak{m}^2\rightarrow \mathfrak{n}/\mathfrak{n}^2\rightarrow 0.$$

Now since $\mathfrak{m}((\mathfrak{m}^2+\mathfrak{a})/\mathfrak{m}^2)=\mathfrak{m}(\mathfrak{m}/\mathfrak{m}^2)=\mathfrak{m}(\mathfrak{n}/\mathfrak{n}^2)=0$, we have $(\mathfrak{m}^2+\mathfrak{a})/\mathfrak{m}^2$,$\mathfrak{m}/\mathfrak{m}^2$ and $\mathfrak{n}/\mathfrak{n}^2$ are $k=A/\mathfrak{m}$-modules and the above sequence is an exact sequence of $A/\mathfrak{m}$-modules.

Answer 2

Consider the short exact sequence $$0\longrightarrow \mathfrak a\longrightarrow \mathfrak m\longrightarrow \mathfrak n \longrightarrow 0$$ and tensor it with $k$, to obtain the exact sequence $$\mathfrak a\otimes_Ak\longrightarrow \mathfrak m/\mathfrak m^2\longrightarrow \mathfrak n\otimes_Ak \longrightarrow 0.$$

Now, $\quad\mathfrak n\otimes_Ak=\mathfrak n/\mathfrak m\mathfrak n=(\mathfrak m/\mathfrak a)\big/(\mathfrak m\cdot\mathfrak m/\mathfrak a)=(\mathfrak m/\mathfrak a)\big/(\mathfrak m/\mathfrak a\cdot\mathfrak m/\mathfrak a)=\mathfrak n/\mathfrak n^2$.

On the other hand, the image of the left morphism is as well the image of $\mathfrak a$ in $=\mathfrak m/\mathfrak m^2$ by the canonical morphisms, as the following commutative diagram shows: \begin{matrix} \mathfrak a\mkern-1em &\xrightarrow{\hspace{2em}}&\mkern-1em\mathfrak m\\ \downarrow\mkern-1em & & \mkern-1em\downarrow \\ \mathfrak a/\mathfrak m\mathfrak a\mkern-1em &\longrightarrow &\mkern-1em\mathfrak m/\mathfrak m^2 \end{matrix}
This image is $\;(\mathfrak a+\mathfrak m^2)/\mathfrak m^2$, whence the short exact sequence in your post.