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Let $\mathbb{K}$ a finite extension of $\mathbb{Q}$ and $\mathcal{O}_\mathbb{K}$ its ring of integers. Assume $\mathcal{O}_\mathbb{K}=\mathbb{Z}[\alpha]$, that is generated as a ring by a single element $\alpha\in\mathbb{K}$. I am asked to show that, for every prime $p\in\mathbb{Z}$, there are at most $p$ places on $\mathbb{K}$ over $p$, that is at most $p$ ways to extend the non archimedean value $|\cdot|_p$ to $\mathbb{K}$.

I don't know how to start attacking the problem, any hint?

Edit: there was a typo in the original problem. The correct question was to prove that there are at most $p$ places on $\mathbb{K}$ over $p$ of residue degree 1, and the proof follows from Adam Hughes' answer.

Angelo Rendina
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  • It's possible I've misread your question, but it appears to be false. See my answer for a detailed discussion, and please comment if you think I've misunderstood something about what you've asked. – Adam Hughes Feb 10 '15 at 22:23
  • @AdamHughes I've double checked both the exercise text and your answer, and it absolutely look like you're right and the "proposition" is wrong. I'll ask my teacher. – Angelo Rendina Feb 10 '15 at 22:29
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    The correct statement is that there are at most $[K:\mathbf Q]$ ways to extend any absolute value from $\mathbf Q$ to $K$. – KCd Feb 10 '15 at 22:35
  • @KCd Which is kinda obvious since the number of such places is the number of irreducible factors in $\mathbb{Q}_p[X]$ of the minimal polynomial of $\mathbb{K}$ over $\mathbb{Q}$, whose degree is $[\mathbb{K}:\mathbb{Q}]$. – Angelo Rendina Feb 10 '15 at 22:38
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    Sure, if you know about $p$-adics and that extensions of $|\cdot|_p$ from $\mathbf Q$ to $K$ correspond to different field embeddings of $K$ into an algebraic closure of $\mathbf Q_p$ then the bound is obvious. It can be shown by other methods as well. – KCd Feb 10 '15 at 22:41
  • @AngeloRendina OK, that's definitely completely different, although you can get it from the early part of my answer: note the description of how primes above a fixed prime are exactly in 1-1 correspondence with irreducible factors of $p(x)\mod p$ as in my answer. – Adam Hughes Feb 14 '15 at 10:03

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Well we know that $\mathcal{O}_K$ is isomorphic to $\Bbb Z[x]/(p_\alpha(x))$ with $p_\alpha(x)$ the $\Bbb Z$-minimal polynomial for $\alpha$.

Then the number of primes over $p$ is determined by the splitting behavior of $p_{\alpha}(x)\mod p$. Since we see that if we have a prime $\mathfrak{p}|(p)$, then it comes from a factorization into (not necessarily unique) irreducibles

$$\overline{p}_\alpha(x)=\prod_{i=1}^r p_i(x)$$

via $\mathfrak{p}=(p_i(\alpha),p)$, since $\Bbb Z[\alpha]$ has trivial conductor in $\mathcal{O}_K$ (this is the critical fact).

But then your original result is false. Consider $K=\Bbb Q(\zeta_n)$ for a primitive $n^{th}$ root of unity, $\zeta_n$ then $\mathcal{O}_K=\Bbb Z[\zeta_n]$ by classical theory. We know that if we pick $n=p^{k}-1$ then consider the $n^{th}$ cyclotomic polynomial $\Phi_n(x)$, we know it will factor as the product of all irreducible polynomials of degree $k$ modulo $p$ since $\Bbb F_{p^k}$ is the unique finite field of order $p^k$ and consists of $0$ and the $(p^k-1)^{st}$ roots of $1$. So then if we can demonstrate more than $p$ such irreducible factors, we have disproven the statement. But we can count the number of irreducible polynomials of degree $q$, a prime not equal to $p$, namely ${p^q-p\over q}$ (I omit the proof, since it's not the subject of this post, but there are many available online, including here on MSE) so choose $p=2, q=5$ and we get $6$ distinct irreducibles modulo $p$, hence $6$ primes above $2$ in $\Bbb Z[\zeta_{p^q-1}]$, which is more than $2$.

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Adam Hughes
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    As stated in the post edit, there was a typo. However I'm accepting the answer since it both proves the correct question and disproves the wrong one. – Angelo Rendina Mar 02 '15 at 18:38