# All objects in $\bf{FinVect}$ have finite length

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.

• What are simple objects in your category? Can you show that any object $X$ in this category is of length $\dim(X)$? – Batominovski Nov 24 '18 at 12:35
• @Batominovski Thanks, that hint was enough - it feels quite obvious now – Soap Nov 25 '18 at 10:45

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