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?
Asked
Active
Viewed 2,401 times
7
-
You mean formal power series? – Siméon Jan 06 '13 at 13:47
-
Try to write $1+x+x^2+x^3+\cdots$ as a finite linear combination? – Hui Yu Jan 06 '13 at 13:55
-
5@HuiYu yes, you can write it as $1\times (1+x+x^2+...)$. – Louis La Brocante Jan 06 '13 at 13:56
-
formal series, right sorry – user55354 Jan 06 '13 at 13:56
-
2If $K$ 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 06 '13 at 14:04
-
By looking at the degree, you see the only possible generator is $x$. But it does not generate the algebra. – Hui Yu Jan 06 '13 at 14:15
-
@Hui Yu: Are you sure that this sketch of proof works? For example, what is the degree of a formal power series? – Martin Brandenburg Jan 16 '13 at 18:11
2 Answers
13
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.
-
This is a nice answer. However, a subalgebra of A[[x]] can be finitely generated. Am I right ? – MAS Sep 27 '20 at 07:30
7
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.
Martin Brandenburg
- 181,922
-
-
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 09:59
-
In a local jacobson ring, there is only one prime ideal, and artinian = noetherian + zero-dimensional. – Martin Brandenburg Jan 21 '13 at 22:58
-