# Characteristic of residue field

Let $\mathcal{O}=\mathbb{Z}[\omega]$ be the ring of algebraic integers in $\mathbb{Q}(\omega)$. It can be shown that $\mathcal{O}$ has a maximal ideal $\mathfrak{m}$ generated by $1-\omega$ (see my previous questions).

Let $\mathcal{O}_{\mathfrak{m}}$ denote $\mathcal{O}$ localised at $\mathfrak{m}$.

Why does the residue field $\mathcal{O}_m/(1-\omega)\mathcal{O}_{\mathfrak{m}}$ have characteristic 3?

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Is it simply because $3=(1-\omega)(2+\omega)\in (1-\omega)\mathcal{O}_{\mathfrak{m}}$? – Clinton Boys Nov 22 '11 at 6:07
That's the reason, more or less; see my answer below. – Zev Chonoles Nov 22 '11 at 6:30
I added a link to a previous question of yours that seemed to be the most relevant one. If I got it wrong, please correct. I just think that having the link here makes it easier for new readers of your question to understand the context. Also, $\omega$ is, indeed, often used to denote the third primitive root of unity with a positive imaginary component, but this is not universal, so you may consider making this clear. Having the link may be sufficient? – Jyrki Lahtonen Nov 22 '11 at 6:49
You seem to have grasped the main point. A factorization like the one you gave shows that $3=0$ in the quotient ring. The order of $1$ in the additive group of the quotient ring must thus be a divisor of $3$ - and there aren't too many alternatives. – Jyrki Lahtonen Nov 22 '11 at 6:56

Prove that the composition $$\mathbb{Z}\hookrightarrow\mathcal{O}\hookrightarrow\mathcal{O}_{\frak{m}}\twoheadrightarrow\mathcal{O}_{\frak{m}}/(1-\omega)\mathcal{O}_{\frak{m}}$$ has a kernel of $3\mathbb{Z}$, and so there is an injective ring homomorphism $\mathbb{Z}/3\mathbb{Z}\to \mathcal{O}_{\frak{m}}/(1-\omega)\mathcal{O}_{\frak{m}}$. Thus, the field $\mathcal{O}_{\frak{m}}/(1-\omega)\mathcal{O}_{\frak{m}}$ is an extension of $\mathbb{F}_3$, and therefore must be of characteristic 3.
@ZevChonoles: Wouldn't \twoheadrightarrow be better? – Arturo Magidin Jul 8 '12 at 18:36