# Is $k[x][[h]]$ finitely generated as $k[[h]]$-algebra?

Is $k[x][[h]]$ finitely generated as an algebra over $k[[h]]$, where $k$ is a field, and $xh=hx$.

## migrated from mathoverflow.netApr 3 '15 at 16:30

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• It seems that in general people don't think this is the right level for this site... However, these sort of issues can be confusing at first (e.g. that k[x][[h]] \neq k[[h]][x]). In brief I believe the answer to your question is "no." If you have trouble working out a proof I'd try to prove that k[[x]] is not finitely generated over k (hint consider cardinality as vector space over k) and then adapt an argument from that case... – user36931 Apr 3 '15 at 14:42
• I’m not sure, @user36931, that your proposed strategy is the right one, but that may be the kind of approach for OP to use. – Lubin Apr 3 '15 at 16:49

Let $F_1, \ldots, F_m \in K[x][[h]]$. Let $d_j$ be the maximum $x$-degree of the coefficient of $h^j$ in $F_1, \ldots, F_m$ and let $\delta_n = \max \{d_0,\ldots, d_n\}$. In any term $f F_1^{a_1}\ldots, F_m^{a_m}$ where $f \in k[[h]]$, the $x$-degree of the coefficient of $h^n$ is at most $(a_1+\cdots+a_m)\delta_n$. Let $\epsilon_n = (n+\delta_n)^2$. Given any $a \in \mathbb{N}$, we have $\epsilon_n > a \delta_n$ for all $n$ sufficiently large. Hence
$$x^{\epsilon_0} + x^{\epsilon_1}h + \cdots + x^{\epsilon_n}h^n + \cdots$$
is not in the $k[[h]]$ algebra generated by $F_1, \ldots, F_m$.