# 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?

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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

## 2 Answers

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.

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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.

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Dear @Martin, This is really nice! –  Keenan Kidwell Jan 16 '13 at 18:46
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
In a local jacobson ring, there is only one prime ideal, and artinian = noetherian + zero-dimensional. –  Martin Brandenburg Jan 21 '13 at 22:58
Basically I use the same argument which you suggest. –  Martin Brandenburg Jan 22 '13 at 8:47