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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 $a_i\in \mathbb C\setminus \{0\}$ 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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    @jspecter: You are right!2011-12-15

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Consider the example where $f=1-t^2$.

To show that the map is not surjective, we can make the following observations. Let $\sigma:B\to B$ be the unique algebra automorphism such that $\sigma(t)=-t$. Then $A\subseteq B$ is precisely the subalgebra of $\sigma$-fixed elements. Since my $f=1-t^2$ is fixed by $\sigma$, the ideal $(f)$ is $\sigma$-invariant, and therefore $\sigma$ induces an action on $B/(f)$ and the canonical map $B\to B/(f)$ is $\sigma$-equivariant.

The composition $A\to B\to B/(f)$ is then $\sigma$-equivariant and therefore its image is contained in the subset of $B/(f)$ of $\sigma$-fixed elements.

Now the class $\bar t$ of $t$ in $B/(f)$ is not zero ($t$ is a unit in $B$ and $(f)$ is a proper ideal) and it is not fixed under $\sigma$. It follows that $\bar t$ is not in the image of $A$.

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    Thank you for helping me to do in the better way! I was not realizing the importance of this hypothesis which was inserted in the new question [link](http://math.stackexchange.com/q/91910/2764). It was very helpful to understand what happens if I take out this hypothesis on my original problem. See the other question if you can!2011-12-16