So far I'm thinking of using the evaluation at $ \sqrt{3} $ which is clearly a homomorphism to the ring $ \mathbb{Z}[\sqrt{3}] $ however I'm having trouble proving it to be a homomorphism to $ \mathbb{Z}_7[\sqrt{3}] $ using a method that isn't tedious brute force which I imagine was not what was intended. Did I simply pick the wrong homomorphism?

Also the question goes on to ask us to express the kernel in terms of generators for which I am completely lost so a nudge in the right direction for this would be appreciated.


However we approach this problem, we are trying to create a chain of homomorphism starting at $\mathbb{Z}[X]$ and ending at $\mathbb{Z}_7[\sqrt{3}]$.

If we start as you did with the evaluation map, we then need to find a homomorphism from $\mathbb{Z}[\sqrt{3}]$ to $\mathbb{Z}_7[\sqrt{3}]$. This is hard if we think about $\mathbb{Z}[\sqrt{3}]$ as a subring of $\mathbb{R}$, but easier if we think about it as $\mathbb{Z}[X]/(X^2-3)$, since then all we need is a homomorphism from $\mathbb{Z}[X]$ to $\mathbb{Z}_7[\sqrt{3}](=\mathbb{Z}_7[Y]/(Y^2-3))$ with kernel containing $(X^2-3)$ by the isomorphism theorem for rings. The map sending $X$ to $Y$ and $1$ to $1 \pmod{7}$ works. However, this way is needlessly complicated, because we have to deal with a lot of quotient rings. A much better way is to create a different chain.

Instead of looking for a composition $\mathbb{Z}[X]\rightarrow \mathbb{Z}[\sqrt{3}]\rightarrow \mathbb{Z}_7[\sqrt{3}]$, try looking for a composition $\mathbb{Z}[X]\rightarrow\mathbb{Z}_7[X]\rightarrow\mathbb{Z}_7[\sqrt{3}]$. This approach should be easier because we only have one polynomial quotient to deal with, the two maps we are composing are both from straight forward polynomial rings. This is much simpler to justify, I feel.

  • 1
    $\begingroup$ Okay, I now realise that I was the one thinking about $\mathbb{Z}[\sqrt{3}]$ wrong, I apologise. Starting is you do in the question is a perfectly natural and legitimate way to do this question, it just has the drawback of being more involved since we need to deal with additional quotient rings. Hopefully this answer is now clearer. $\endgroup$ – Tom Oldfield May 25 '14 at 14:46

Hint: try to find a ring homorphism with kernel the ideal $(7,x^2-3)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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