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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.

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    $\begingroup$ What are simple objects in your category? Can you show that any object $X$ in this category is of length $\dim(X)$? $\endgroup$ – Batominovski Nov 24 '18 at 12:35
  • $\begingroup$ @Batominovski Thanks, that hint was enough - it feels quite obvious now $\endgroup$ – Soap Nov 25 '18 at 10:45
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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 $X_k = \mathop{\mathrm{Span}}\{a_1,\ldots,a_k\}$.

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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 $\dim_\mathbb{K}(X)$.

Try induction on $\dim_\mathbb{K}(X)$.

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