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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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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
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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
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1 Answer

up vote 2 down vote accepted

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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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
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Please do not change the question into something different. It was impossible to satisfy your original hypothesis because you did not tell anyone about it... The correct thing to do is to ask another question, possibly linking to this one. –  Mariano Suárez-Alvarez Dec 16 '11 at 2:43
    
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. 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! –  Binai Dec 16 '11 at 3:10
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