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This question is similar to the question link.

Let $A= \mathbb C [t^2,t^{-2}]$ and $B= \mathbb C [t,t^{-1}]$. Given $r\in \mathbb Z_+$ and $f\in B$ with the form $f=(t-a_1)(t-a_2)\cdots(t-a_k)$, where $a_i\in \mathbb C\setminus \{0\}$ with $a_i^k\ne a_j^k$ $(i\ne j)$ for $k=1,2$, let $I$ be the ideal of $B$ generated by $f^r$. Define the natural map $\phi: A \to B/I$ by $t^k\mapsto \overline{t^k}$ for all $k\in \mathbb 2Z$.

QUESTION: Is the map $\phi$ surjective?

Notice that the link above mentioned is the particular case with $r=1$.

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1 Answer 1

Yes, $\phi$ is surjective.

$\newcommand{\CC}{\mathbb C}$ By the Chinese Remainder Theorem, $$B/I\simeq \prod_i \CC[t]/(t-a_i)^r, \quad A\twoheadrightarrow \prod_i \CC[t^2]/(t^2-a_i^2)^r$$ ($t^{-1}$ and $t^{-2}$ desappear because $a_i\ne 0$ and the second part uses the hypothesis $a_i\ne\pm a_j$ if $i\ne j$). So it is enough to show that $\CC[t^2]/(t^2-a_i^2)^r \to \CC[t]/(t-a_i)^r$ is surjective. Or, writting $a=a_i$, that $$ \psi: \CC[t^2]\to \CC[t]/(t-a)^r$$ is surjective. Let $C=\psi(\CC[t^2])$. This is a subring of the local Artinian ring $R:=\CC[t]/(t-a)^r$. Let $m=(t-a)R$ be the maximal ideal of $R$. As $t-a=(t^2-a^2)/2a-(t-a)^2/2a$, we get $$m \subseteq (t^2-a^2)\CC + m^2. \tag{#}$$ For any $n\ge 2$, taking the $n$th powers, we get $$m^n\subseteq (t^2-a^2)^n\CC +m^{n+1}.$$ So $$m\subseteq (t^2-a^2)\CC+ (t^2-a^2)^2\CC+\dots +(t^2-a^2)^n\CC + m^{n+1}\subseteq C+m^{n+1}.$$ Taking $n=r-1$, then $m\subseteq C$. As $\CC\subseteq C$ and $R=\CC+m$, we have $R=C$.

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@user2764: I don't see the problem when $r=1$ (then $m=0$ in $R$ and you only have to deal with constants). –  user18119 Apr 25 '12 at 5:29
Unless there is a very good reason, we prefer not to arbitrarily delete good content: the answers could well help other users in the future. –  robjohn Aug 9 '13 at 23:24

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