3
$\begingroup$

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?

  • 1
    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.2011-11-22

1 Answers 1

5

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.

  • 0
    @ZevChonoles: [Detexify](http://detexify.kirelabs.org/classify.html) is a good way to find out if there is a simple command for a particular symbol like that.2012-07-08