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.

Question: Find a polynomial ring $R$ which is not an integral domain and an ideal $I$ such that $R/I$ is a field.

Answer: $R=\mathbb{Z}_6[x]$, $I=\langle 2,x\rangle$, $R/I$ is isomorphic to $\mathbb{Z}_2$.

What is exactly $I$ here? Is it a subset of $R$ such that it consists of elements $2\cdot a + x\cdot b$ such that $a \in \mathbb{Z}$ and $b\in\mathbb{Z}[x]$?

share|improve this question
4  
Well, $a$ should be in $\mathbb{Z}_6[x]$ as well. –  Alex Youcis Aug 14 '13 at 5:17
    
I've improved your question's formatting; you can see here how I edited your question. Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. Some MathJax advice: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. –  Zev Chonoles Aug 14 '13 at 5:22

1 Answer 1

up vote 2 down vote accepted

If $R$ is a ring, and $a_1,\ldots,a_n\in R$, then the ideal generated by $a_1,\ldots,a_n$, denoted $I=\langle a_1,\ldots,a_n\rangle$, is the smallest ideal in $R$ containing all of the $a_i$. It can be shown that this definition leads to the following characterization:

$$\langle a_1,\ldots,a_n\rangle=\{r_1a_1+\ldots+r_na_n\mid r_i\in R,\;\forall i=1,\ldots,n\}$$

See if you can show in the case of $I=\langle 2,x\rangle\subset\mathbb{Z}_6[x]$, that $I$ can be described as all polynomials whose constant term is even.

share|improve this answer
    
Well, calling the contant term "even" in this case can be a rather big stretch as we're not working within $\,\Bbb Z\,$. I'd rather say the free coefficient is either $\,0,2\;\text{ or }\;4\pmod 6\,$ –  DonAntonio Aug 14 '13 at 7:45

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.