Ring of formal power series finitely generated as algebra?

I'm asked if the ring of formal power series is finitely generated as a $K$-algebra. Intuition says no, but I don't know where to start. Any hint or suggestion?

• You mean formal power series? – Siméon Jan 6 '13 at 13:47
• Try to write $1+x+x^2+x^3+\cdots$ as a finite linear combination? – Hui Yu Jan 6 '13 at 13:55
• @HuiYu yes, you can write it as $1\times (1+x+x^2+...)$. – Louis La Brocante Jan 6 '13 at 13:56
• formal series, right sorry – user55354 Jan 6 '13 at 13:56
• If $K$ is a field, then show that $K[[x]]$ has uncountable dimension as a $K$-vector space, while any finitely-generated $K$-algebra has at most countable dimension. – Zhen Lin Jan 6 '13 at 14:04

Let $A$ be a non-trivial commutative ring. Then $A[[x]]$ is not finitely generated as a $A$-algebra.
Indeed, observe that $A$ must have a maximal ideal $\mathfrak{m}$, so we have a field $k = A / \mathfrak{m}$, and if $k[[x]]$ is not finitely-generated as a $k$-algebra, then $A[[x]]$ cannot be finitely-generated as an $A$-algebra. So it suffices to prove that $k[[x]]$ is not finitely generated. Now, it is a straightforward matter to show that the polynomial ring $k[x_1, \ldots, x_n]$ has a countably infinite basis as a $k$-vector space, so any finitely-generated $k$-algebra must have an at most countable basis as a $k$-vector space.
However, $k[[x]]$ has an uncountable basis as a $k$-vector space. Observe that $k[[x]]$ is obviously isomorphic to $k^\mathbb{N}$, the space of all $\mathbb{N}$-indexed sequences of elements of $k$, as $k$-vector spaces. But it is well-known that $k^\mathbb{N}$ is of uncountable dimension: see here, for example.
Finitely generated $k$-algebras are Jacobson, hence finitely generated local $k$-algebras are artinian, hence finitely generated local $k$-domains are fields. Well, $k[[x]]$ is not a field.
• I don't understand your claim that finitely-generated local $k$-algebras are artinian, but it's certainly true that a local Jacobson domain must be a field. (Because then the unique maximal ideal = Jacobson radical = nilradical = 0.) – Zhen Lin Jan 17 '13 at 9:59