What is a simple example for an exact sequence? Because in my opinion it seems that all the modules in the sequence should be equal to zero

Question

Let $M_i$ be R-modules and $f_i$ be homomorphisms of R-modules

If $\forall _n\ker f_n=\operatorname{im}f_{n-1}$, wouldn't that mean that for $$...\:\rightarrow M_{n-1}\:\rightarrow ^{f_{n-1}}\:M_n\:\rightarrow ^{f_n} \:M_{n+1}\:\rightarrow \:...$$ we'd get that $M_{n+1} = 0$? Because we'd basically have that $$f_n\left(\operatorname{im}f_{n-1}\right)=f_n\left(\ker f_n\right)=0$$ Or do I get something wrong here?

Answer 1

A simple and not entirely trivial example of an exact sequence is the exact sequence that you get from any homomorphism $f:M\to N$: $$ \ker(f)\xrightarrow{\quad\iota\quad} M\xrightarrow{\quad f\quad}N\xrightarrow{\quad\pi\quad} N/\operatorname{im}(f) $$ Here, $\iota$ is just the inclusion and $\pi$ the canonical projection. Note that $f\circ\iota=0$ because that's simply how the kernel of a morphism is defined. Furthermore, $\pi\circ f=0$ because for every $m\in M$, you have $f(m)\in\operatorname{im}(f)$, so $\pi(f(m))=0$.

I am sure you can come up with $M$, $N$, and $f$ so that not all of them are zero - I hope this helps to clear things up a bit.

Answer 2

$ 0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z} /n \mathbb{Z} \to 0$, where $\mathbb{Z} \to \mathbb{Z}, 1 \mapsto n$.

Answer 3

For an infinite exact sequence, consider the $\mathbb{Z}$-module $C_2\times C_2$, and the map $f\colon C_2\times C_2\to C_2\times C_2$ given by $f(a,b) = (b,0)$. Then $$\cdots \stackrel{f}{\longrightarrow} C_2\times C_2 \stackrel{f}{\longrightarrow} C_2\times C_2 \stackrel{f}{\longrightarrow} C_2\times C_2 \stackrel{f}{\longrightarrow}\cdots$$ is exact, since $\mathrm{Im}(f) = \mathrm{ker}(f) = C_2\times\{0\}$. Note that none of the modules are the zero module, and none of the maps are the zero map.

Answer 4

Here is the standard example of an exact sequence, corresponding to a homomorphism $f:M\longrightarrow N$ of $R$-modules: $$0\longrightarrow\ker f\xrightarrow{\;i\enspace} M\xrightarrow{\;f\enspace} N\xrightarrow{\;p\enspace}\operatorname{coker f}\longrightarrow 0$$ where $i$ is the canonical injection and $p$ the canonical surjection.

Answer 5

As a familiar example, why not

$$ 0 \to \mathbb{R} \to C^\infty(\mathbb{R}) \overset{\frac{d}{dx}}{\to} C^\infty(\mathbb{R}) \to 0 $$

This says that the kernel of the differentiation map is the space of constant functions, and every smooth function on $\mathbb{R}$ is the derivative of some function.

I learned this example a while ago this relevant MO post.

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I hope this helps ^_^