Question: I want to show that every object $X$ of $\bf{FinVect}$ has finite length. *i.e.* there is a sequence of monos
$$
 0=X\_0 \hookrightarrow X\_1 \hookrightarrow ... \hookrightarrow X\_{n-1} \hookrightarrow X\_n = X
$$
such that $\forall i,{X\_i}/{X\_{i-1}}$ is simple.

I think this is supposed to be a trivial case, but I do not see how. If we use just the zero map $0\xrightarrow{0} X$ it does not work, since $X/\{0\}\cong X$ need not be simple.

Answer: 
The simple objects in $\mathbf{FinVect}$ are exactly the 1-dimensional vector spaces. Take any ordered basis $(a\_1,\ldots,a\_n)$ of $X$ (which exists, since $X$ is a finite dimensional vector space), and take $X\_k = \mathop{\mathrm{Span}}\{a\_1,\ldots,a\_k\}$.
**Hints.** Let $\mathbb{K}$ be the base field. The simple objects of $\mathbf{C}:=\mathbf{FinVect}(\mathbb{K})$ are $1$-dimensional $\mathbb{K}$-vector spaces, i.e., they are all isomorphic to $\mathbb{K}$. This shows that, if $X\in \mathbf{C}$ has a filtration 
$$0=X\_0\hookrightarrow X\_1 \hookrightarrow X\_2\hookrightarrow \ldots \hookrightarrow X\_{l-1}\hookrightarrow X\_l=X\tag{\*}$$
in which $X\_i/X\_{i-1}$ are simple (whence $1$-dimensional), then $$\dim\_\mathbb{K}(X)=\dim\_\mathbb{K}(X\_0)+\sum\_{i=1}^l\,\dim\_\mathbb{K}\left(X\_i/X\_{i-1}\right)=0+\sum\_{i=1}^l\,1=l\,$$
For convenience, we call (\*) a *composition series* of $X$. Ergo, you have the following statement: "if $X\in\mathbb{C}$ has a composition series of length $l$, then $\dim\_\mathbb{K}(X)=l$." How would you show that the converse of this statement is also true? That is, show that any composition series of $X\in\mathbf{C}$ exists and is of length $\dim\_\mathbb{K}(X)$.

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>  Try induction on $\dim\_\mathbb{K}(X)$.
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