Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is from a past qualifying exam.

Here is the question:

If $\alpha= (1+\sqrt{-19})/2$ then any ring homomorphism $f : \mathbb{Z}[\alpha] \rightarrow \mathbb{Z}_3$ is the zero map.

Here is what I thought of: $\mathbb{Z}_3$ is a field. So I was thinking of using something like $ \mathbb{Z}[\alpha]/(f(\alpha))$ is isomorphic to the field $\mathbb{Z}_3$. If I can somehow show that $f(\alpha)$ is the root of some irreducible polynomial over $\mathbb{Z_3}$ then I will have what I want? I also have that $\alpha(1-\alpha)=5$. So If I apply $f$ to this relation and then using the fact that $f$ is ring homomorphism I get $f(\alpha)\cdot (f(1)-f(\alpha))=f(5)$. I also know that $f(1)=1$. How do I proceed from here? Am I on the right track?

Can anybody please help me on this? Your time and answers are always appreciated.

Thanks.

share|improve this question
1  
This question can't be answered without knowing what $f$ is. –  Zev Chonoles Jun 1 '13 at 4:36
    
Maybe $f$ is any ring-homomorphism? –  awllower Jun 1 '13 at 4:37
    
Yes, $f$ is any ring homomorphism. This is how the question is stated. –  Jack Dawkins Jun 1 '13 at 4:38
3  
There is no unitary homomorphism as specified, i.e. carrying the multiplicative element of the one ring to that of the other. So the only homomorphism is the zero morphism, that takes everything in the domain to zero. You were trying to show that the map was onto, but it can’t be. Look at the whole situation again. –  Lubin Jun 1 '13 at 4:41
    
If you ever get around to studying algebraic number theory, you can reanswer this question as follows. The kernel of $f$ would have to be an ideal of index $3$. And there aren't any 'cause three is inert. –  Jyrki Lahtonen Jun 1 '13 at 5:41
add comment

2 Answers 2

up vote 4 down vote accepted

Using @Lubin's hint. Discuss what $f(\alpha)$ would be? If $f(1)=1$,then

  1. $f(\alpha)=0$? then $f(\alpha)(f(1)-f(\alpha))=f(5)$, so $0\cdot(1-0)=5=2$ !?
  2. $f(\alpha)=1$? then $1\cdot(1-1)=5$ !?
  3. $f(\alpha)=2$? then $2\cdot(1-2)=5$ !?
share|improve this answer
    
Thanks for your answer. But should it be $f(5)$ and not $5$ in your computations above??. Why would $f(5)=2$? –  Jack Dawkins Jun 1 '13 at 17:34
    
What is your $\mathbb{Z}_3$? Is $5=2$ in $\mathbb{Z}_3$?. $f(5)=f(4+1)=f(4)+f(1)=f(3+1)+f(1)=\cdots=f(1)+f(1)+f(1)+f(1)+f(1)=5f(1)=5$. –  wxu Jun 2 '13 at 1:41
    
Oh I got it :-). Thanks! –  Jack Dawkins Jun 2 '13 at 2:26
add comment

Hint: $\,\alpha\,$ is a root of $\,g(x) = x^2-x+5\,$ so its image is a root in $\,\Bbb Z_3,\,$ contra $\,g\,$ has no roots $\in \Bbb Z_3.$

share|improve this answer
    
Thank you this is helpful. –  Jack Dawkins Jun 2 '13 at 2:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.