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

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 $b\in\mathbb{Z}[x]$?

-
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

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.
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