Show that $A \cong \mathbb{C}^n$ with A a commutative algebra 
Let A be a commutative algebra of finite dimension, and if $A$ has no nilpotent elements other than $0$, is true that $A \cong  \mathbb{C}^n$ ? 

The question emerge to my mind, I thought that the finite dimension tell us that the scheme is Artinian (geometrically dimension 0).
I think the pattern is just a meeting of $n$ points but I have not managed to prove it.
Someone can enlighten me please ?
Thanks

An artinian ring $A$ has only finitely many prime ideals, which are all maximal. Thus, by the Chinese remainder theorem, 
$$A / \mathfrak{n} \cong A / \mathfrak{m}_1 \times \cdots \times A / \mathfrak{m}_r$$
where $\mathfrak{m}_1, \ldots, \mathfrak{m}_r$ are the distinct prime/maximal ideals of $A$ and $\mathfrak{n}$ is the nilradical/Jacobson radical. 
In particular, if the only nilpotent element of $A$ is $0$, then $A$ is a product of finitely many fields. Moreover, if $A$ is a finite $\mathbb{C}$-algebra, then each $A / \mathfrak{m}_i$ must also be a finite $\mathbb{C}$-algebra, hence, must be (isomorphic to) $\mathbb{C}$ itself.
