I am trying to solve some exercises from last year's exam, and there is this exercise I am stuck. Can anyone help me please?

Let $\alpha$ be a root of $f(X) = X^4+4X+2$, and $K = Q(\alpha)$. show that $[O_K:\mathbb Z[\alpha]]\le 2^4$. Show that the inclusion $\mathbb Z[\alpha]\rightarrow O_K$ induces isomorphisms from $\mathbb Z[\alpha]/\alpha \mathbb Z[\alpha] \rightarrow O_K/\alpha O_K$ and $\mathbb Z[\alpha]/2 \mathbb Z[\alpha] \rightarrow O_K/2O_K$.

The first part is clear, following from $disc(f) = 2^8. 19$ and $disc(f) = disc(O_K).[O_K:\mathbb Z[\alpha]]^2$. It's also clear that the induced morphisms are injective. I am having problem with showing why they are surjective. Also, how do we find how the prime 2 factors in $O_K$? 2 might divide index, and we will not be able to use Dedekind's theorem.


$f$ is Eisenstein for $p=2$, so $(2)$ is totally ramified in $O_K$; in fact, as $N_{K/\mathbb Q}(\alpha)=f(0)=2$, we get $(2)=(\alpha)^4$. Since both $\mathbb Z[\alpha]/\alpha\mathbb Z[\alpha]$ and $O_K/\alpha O_K$ have 2 elements, the morphism $\mathbb Z[\alpha]/\alpha\mathbb Z[\alpha]\to O_K/\alpha O_K$ is iso (it has to send 0 to 0 and 1 to 1).

We have $\mathbb Z[\alpha]/2 \mathbb Z[\alpha]\cong \mathbb F_2[X]/(X^4)$ ($X^4$ is the reduction of $f$ mod $2$) and the map $\mathbb Z[\alpha]/2 \mathbb Z[\alpha] \rightarrow O_K/2O_K$, i.e. $\mathbb F_2[X]/(X^4)\to O_K/2O_K$, is given by $X\mapsto\alpha$. The map is injective, as the image of $X^3$ is $\alpha^3\notin 2O_K=(2)=(\alpha)^4$; since both rings have the same number of elements (or dimension over $\mathbb F_2$), injectivity implies isomorphism. [I feel I made it more complicated than necessary]

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  • $\begingroup$ Maybe so, but still more efficiently than I would have done it. $\endgroup$ – Lubin Dec 23 '14 at 4:20

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