6
$\begingroup$

I had to do some homework problems involving the polynomial ring $R=\mathbb{Z}+x\,\mathbb{Q}[x]\subset\mathbb{Q}[x]$. This is an integral domain but not a UFD. Further, $x$ is not prime in $R$.

One of the problems was to describe to $R/(x)$.

Since $x$ is not a prime element, we know $(x)$ is not a prime ideal. So at the very least, $R/(x)$ is not an integral domain.... but what else can I say? This is perhaps something I should not admit, but problems of this form have always befuddled me. I know there's not any one "answer" they're looking for, but I never quite know what to say.

Anyway, this homework has been submitted already, so I am not including a homework tag. I'm just curious how you all would describe this particular quotient.

$\endgroup$

1 Answer 1

5
$\begingroup$

The key is to understand what that ideal $(x)$ looks like in the first place. So, clearly all polynomials in that ideal have degree at least 1, but what degree 1 polynomials can occur? For $f(x)*x$ to have degree 1, $f(x)$ must have a non-zero constant term. That term must be in $\mathbb{Z}$, and it is easy to see that you get any coefficients for higher degrees. So $(x)=\mathbb{Z}x + \mathbb{Q}x^2+\ldots+\mathbb{Q}x^n+\ldots$.

Thus, in the quotient, any coset is represented by some $n + ax$, $n\in \mathbb{Z},a\in \mathbb{Q}$, and another element $m+bx$ represents a different coset if and only if $n\neq m$ or $a-b\notin\mathbb{Z}$. In other words, the quotient is isomorphic to $\left(\mathbb{Z} + \mathbb{Q}/\mathbb{Z}x\right)/(x^2).$

$\endgroup$
2
  • $\begingroup$ This is quite a good description, haha. Thanks for the response $\endgroup$
    – Bey
    Jan 29, 2011 at 2:25
  • $\begingroup$ Sorry, what does it mean $\mathbb{Z}+\mathbb{Q}/\mathbb{Z}x$? $\mathbb{Q}/\mathbb{Z}$ is not a ring $\endgroup$ Feb 24, 2020 at 1:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .