I want to prove $\dim R=1$. So I think I have to prove that R has only 1 prime ideal. That's why I ask this question
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I'm a bit confused: if you're going to mod out the ring $\mathbb{Z}_2[x]$ by an ideal $I$ , then we should have $I \subseteq \mathbb{Z}_2[x]$. The problem with setting $I=(2x)$ is that $2x=0$ in $\mathbb{Z}_2[x]$, so you are essentially modding out by 0. – Ethan Dec 09 '16 at 05:56

1@Ethan It is the localization, i.e. all fractions with odd denominator. – MooS Dec 09 '16 at 06:17
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Prime ideals of a quotient $R/I$ are prime ideals of $R$, that contain $I$.
$P \supset (2x)$ is equivalent to $2 \in P$ or $x \in P$.
We have $R/(2) = \mathbb F_2[x]$ and $R/(x)= \mathbb Z_{(2)}$, hence the prime ideals are given by
$$(2),(x) ~ \text{ and } ~ (2,f)$$ where $f \in \mathbb Z[X]$ is irreducible modulo $2$. In particular there are infinitely many of them. Though the dimension is $1$, because all chains are of the form
$$(2) \subset (2,f) ~ \text{ or } ~ (x) \subset (2,x).$$
MooS
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1It should be well known that nonzero prime ideals of $K[x]$ corredpond to irreducible polynomials. This is a PID. – MooS Dec 09 '16 at 07:52

Oh sorry, i know that. Just a little confused with "f irreducible module 2". But I think I got it too. Thank u – T C Dec 09 '16 at 08:27