# How to show that the category of finite vector spaces is indecomposable?

In [1] it is said that the category $$\mathbf{FinVect}$$ of finite vector spaces is a tensor category. I am trying to convince myself that this is indeed the case.

One of the properties that $$\mathbf{FinVect}$$ must satisfy for this to hold is that it is indecomposable. I cannot think of a way to show that. Here are the definitions which are relevant for this problem:

Def: Let $$\mathbb C$$ be a locally finite $$\mathbb{K}$$-linear abelian rigid monoidal category. $$\mathbb{C}$$ is a multitensor category over $$\mathbb{K}$$ if $$\otimes:\mathbb{C}\times\mathbb{C}\rightarrow \mathbb{C}$$ is bilinear on morphisms. A multitensor category $$\mathbb{C}$$ is decomposable if it is equivalent to a direct sum of nonzero multitensor categories. If $$\mathbb{C}$$ is an indecomposable multitensor category and $$\text{End}(1)\cong \mathbb{K}$$ (as vector spaces), then $$\mathbb{C}$$ is a tensor category.

[1] Pavel Etingof et al. Tensor Categories. American Mathematical Society, 2015.

• Doesn't it follow from $\mathbb K$ being indecomposable ? Indeed take if $\mathbf{FinVect} \simeq C_1\times C_2$ then with $\mathbb K$ corresponding to, say $(a,b)$ we would have $\hom(\mathbb{K,K}) \cong \hom(a,a)\times \hom(b,b)$. So one of $a$ or $b$ is $0$ and then it should follow from there (and the generating character of $\mathbb K$) that $C_1$ or $C_2$ is $0$ – Max Sep 1 '19 at 13:15

If $$\mathbf{FinVect}\simeq C\times D$$ where $$C$$ and $$D$$ are both nonzero, let $$c$$ and $$d$$ be nonzero objects of $$C$$ and $$D$$, respectively. Then $$(c,0)$$ and $$(0,d)$$ are both nonzero objects of $$C\times D$$ (because their identity map is not equal to their zero map), but $$\operatorname{Hom}((c,0),(0,d))\cong \operatorname{Hom}(c,0)\times\operatorname{Hom}(0,d)\cong 0$$. This is a contradiction, since any two nonzero vector spaces have a nonzero linear map between them.