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In my algebraic number theory lecture, we came up with a certain example which I unfortunately only understand to some extent. I would be very grateful if some of you could either explain to me the solution given in class or show me any other possible way to solve this task.

We were given the number field $K= \mathbb{Q}(a)$ with its ring of integers $O_K$ where the minimal polynomial of the primitive element $a$ is given by $f(x)=x^5-x+1$. The claim is that the class number of $K$ is $1$.

In the following, I was able to compute $d_k= 2869$ and by using the Minkowski bound that in every class $C \in Cl_K$, there exists an ideal $A$ such that the Norm of $A$ is either $1,2$ or $3$. Now I assume that the class number is greater than $1$, which gives $N(A)$ either $2$ or $3$.

And up to this point, everything is fine. However, the proof continues:

Let $p$ be a prime ideal of $O_K$ with $p \cap \mathbb{Z}=$p$\mathbb{Z}$ with p $\in \{2,3\}$. If [p] is not trivial, then the norm of $p$ has to be p as it has to divide $4$ or $9$ (Why?). This gives us the isomorphism $O_K/p \simeq \mathbb{Z}/$ p$ \mathbb{Z}$ (Why?). We know that $f(a)=0$ which gives that $a(a^4-1)=-1 \neq 0$. Looking at this equation modulo p, then gives us that the case that p is either $2$ or $3$ is not possible.

Thank you very much for your help!

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    I'm not sure why they reference norms $4$ and $9$, only ideals of norms bounded by $3$ are candidates for representatives, if none exist, then there's no problem. It seems they're just going after arbitrary ideals above $2$ or $3$ at first, but that's clearly pointless since all you need to do is show none of the two prime norms exist, which the (impossible) congruence $\alpha^5-\alpha\equiv -1\pmod{2,3}$ shows. – 2017-01-24
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    Thank you for your answer - only considering the prime norms also makes more sense to me. – 2017-01-24
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    Keep in mind it's **not** prime norms that are relevant here necessarily, if you had the Minkowski bound to be $4.12345$ or something like that, you would have to consider primes above $2$ of norm $4$ (if such existed). The **norm** is bounded for a representative, so while the **ideals** are obviously primes that we are checking, the **norm** of the ideal may not itself be prime in general. – 2017-01-24

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The norm of the ideal $P$ is, by definition, the cardinality of $\mathcal{O}_K/P$, hence $$ p=N(P)=|\mathcal{O}_K/P|. $$ So $\mathcal{O}_K/P$ has degree $1$ over $\mathbb{F}_p$, hence $\mathcal{O}_K/P\cong \mathbb{F}_p$.

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    Thank you for your answer - that helps a lot! – 2017-01-24