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This question is similar to the question link, with a stronger hypothesis.

Let $A= \mathbb C [t^2,t^{-2}]$ and $B= \mathbb C [t,t^{-1}]$. Consider $f\in B$ 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 $k=1,2$ and let $I$ be the ideal generated by $f$ in $B$. Define the map $\phi: A \to B/I$ by $t^k\mapsto \overline{t^k}$.

QUESTION: Is the map $\phi$ surjective?

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up vote 3 down vote accepted

$\newcommand\CC{\mathbb{C}}$Since $a_1$, $\dots$, $a_n$ are different, the map $\psi:\CC[t^{\pm1}]/(f)\to\CC^n$ (the codomain being the direct product of $n$ copies of $\CC$) given by $$\psi(\overline g)=(g(a_1),\dots,g(a_n))$$ is an isomorphism of rings. The image of the composition $\CC[t^{\pm2}]\to\CC[t^{\pm1}]\to\CC[t^{\pm1}]/(f)\xrightarrow{\;\psi\;}\;\CC^n$ contains the $n$ vectors $$(1,\dots,1), \qquad (a_1^2,\dots,a_n^2), \qquad (a_1^4,\dots,a_n^4), \qquad\dots,\qquad (a_1^{2(n-1)},\dots,a_n^{2(n-1)}),$$ because they are in fact the elements $\psi(\overline1)$, $\psi(\overline{t^2})$, $\psi(\overline{t^4})$, $\dots$, $\psi(\overline{t^{2(n-1)}})$. Since the numbers $a_1^2$, $\dots$, $a_n^2$ are different, these $n$ vectors are linearly independent: this follows from the usual Vandermonde argument. Therefore, the image of that composition is surjective.

This implies immediately that your map is surjective.

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Hi, could you give a look at my other question in link. Thanks, – Binai Apr 19 '12 at 14:23

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