# Surjective homomorphism in Laurent polynomial ring.

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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It can be: $f(t)=t$ implies $B/I=0$ and so $\phi$ is the zero morphism. It is also possible that it fails to be surjective: $f(t)=t^2$ implies $B/I \cong \mathbb{C}[\epsilon]/(\epsilon^2)$ (dual numbers) whereas $\phi$ is still the zero morphism. –  Bill Cook Dec 15 '11 at 16:31
@Bill: I just put $a_i\ne 0$, that is a different question. Sorry about my failure! –  Binai Dec 15 '11 at 16:40
I don't think I buy your claim about the dual numbers; $t^2$ is a unit in $B.$ –  jspecter Dec 15 '11 at 16:42
@jspecter: You are right! –  Binai Dec 15 '11 at 16:51

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 $A$.
Good point! You gave me the full answer of a part of my problem. Actually, the problem which I am working has a stronger hypothesis that $a_i^k \ne a_j^k$ for $i\ne j$ and $k=1,2,3$. So, the counterexample that you gave is of the form $(t-1)(t+1)$ and therefore it is not satisfying my original hypothesis. I am very greatful about your answer and I am editing the original question with this variation. –  Binai Dec 16 '11 at 2:38