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Let $R$ be a non zero ring with identity. Clearly $R[x]$ embeds in $R[[x]]$. Is it true that for any $n$, one can embed $R[x_1,...,x_n]$ in $R[[x]]$ ?

flag	asked Dec 29 at 19:39 tvector 385●7

Peter Michor · Answer 1 · 2012-12-29 19:54:25Z UTC

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Yes. Take algebraically independent power series $a_1(x),\dots, a_n(x)$ and map $x_i$ to $a_i$. For example, one can take $a_1(x) = x$, $a_2(x) = \exp(x)$, $a_3(x) = \exp(\exp(x))$, ...

answered Dec 29 at 19:54

Peter Michor
5,577●11●18

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If there is some objection that coefficients $1/n!$ might not exist in $R$, one can always modify by considering things like $\sum_n x^{10^n}$ as giving nonalgebraic functions, a la Liouville. – Todd Trimble Dec 29 at 20:20

Alternatively: since the natural map $M \otimes_{\mathbf{Z}} \mathbf{Z}[\![x]\!] \rightarrow M[\![x]\!]$ is injective for any $\mathbf{Z}$-module $M$ (use direct limit considerations), it suffices to treat $R = \mathbf{Z}$ (then tensor against any $\mathbf{Z}$-algebra $R$ and compose with the injection $R \otimes_{\mathbf{Z}} \mathbf{Z}[\![x]\!] \rightarrow R[\![x]\!]$). Thus, we just have to check ${\rm{Frac}}(\mathbf{Z}[\![x]\!])$ has infinite transcendence degree over $\mathbf{Q}$ (then chase "denominators", etc.), and this is immediate by uncountability. – ayanta Dec 29 at 22:42

Polynomial Embedding

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