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

It's true.  If $A$ has no nilpotent elements, then it is semisimple.  By the Artin-Wedderburn theorem, it must be a direct sum of matrix algebras over division rings.  The only commutative such containing $\mathbb{C}$ is $\mathbb{C}$ itself.
This is probably overkill in that there is probably a more direct purely commutative proof.
