Let $K$ be a number field such that $\mathcal{O}_K= \mathbb{Z}[\alpha]$ for some $\alpha$ algebraic integer. Prove that there are infinitely many primes $\mathcal{P} \subset \mathcal{O}_K$, such that $f(\mathcal{P}/p)=1$, where $(p)=\mathcal{P} \cap \mathbb{Z}$. HINT: Use that if $f (x) \in \mathbb{Z} [x]$ is not constant, there are infinitely primes $p$ for which $f (x)$ has a root modulo $p$.

Any help is appreciated!

  • 2
    $\begingroup$ Could you please explain what you mean by the notation $f(\mathcal{P}/p)$? $\endgroup$ – Rankeya Nov 27 '12 at 5:53
  • $\begingroup$ ramification index? $\endgroup$ – Gerry Myerson Nov 27 '12 at 6:23
  • $\begingroup$ That's what I'd think myself, @GerryMyerson, yet after that he uses the same letter $\,f\,$ to talk about an integer polynomial...?? $\endgroup$ – DonAntonio Nov 27 '12 at 10:13
  • $\begingroup$ @Don, it wouldn't be the first time someone used one symbol in two very different ways in the course of two sentences. But some clarification from OP is definitely in order. $\endgroup$ – Gerry Myerson Nov 27 '12 at 11:40
  • $\begingroup$ @Rankeya, $f(\mathcal{P}/p)$ is the inertial degree of $\mathcal{P}$ over $(p)$. $\endgroup$ – P. M. O. Nov 27 '12 at 16:01

Hint: The inertial degree is the degree of the field extension $\mathbb{Z}/(p) = \mathbb{F}_p \subset \mathbb{Z}[\alpha]/\mathcal{P}$.

Given what you already know, the only sensible choice of the polynomial $f$ is to take $f$ to be the minimal polynomial of $\alpha$ over $\mathbb{Q}$, which since $\alpha$ is integral, will actually have coefficients in $\mathbb{Z}$.

Now there are infinitely map primes $p$ for which $f(x) \equiv 0 (mod p)$ has a solution $r_p$ in $\mathbb{Z}$. Consider the ideal $(\alpha - r_p, p) \subset \mathbb{Z}[\alpha]$.

We would like $(\alpha - r_p, p)$ to be a prime ideal. To show this, observe that we have $\mathbb{Z}[\alpha]/(\alpha - r_p, p) \cong \mathbb{Z}[x]/(f(x), x - r_p , p) = \mathbb{Z}[x]/(x-r_p,p)$.

Why does the last equality hold (hint divide $f(x)$ by $x - r_p$ and deduce that the constant remainder is a multiple of $p$ from the fact that $f(r_p) \equiv 0 (mod \hspace{1mm} p)$)?

What is the ring $\mathbb{Z}[x]/(x - r_p , p) $ isomorphic to?

Now can you conclude that $(\alpha - r_p, p) \cap \mathbb{Z} = (p)$.

I think my solution is okay (can someone confirm this?) given that in your question $\mathcal{O}_K = \mathbb{Z}[\alpha]$. Things would have been slightly more complicated if $\mathcal{O}_K$ was just the ring of integers of $K = \mathbb{Q}(\alpha)$ for an algebraic interger $\alpha$.

  • $\begingroup$ What is the ring $\mathbb{Z}[x]/(x−r_p,p)$ isomorphic to? $\mathbb{Z}_p$ $\endgroup$ – P. M. O. Nov 28 '12 at 0:27
  • $\begingroup$ That is correct. $\endgroup$ – Rankeya Nov 28 '12 at 2:09
  • $\begingroup$ @Rankeya See my answer below for the general case. $\endgroup$ – user38268 Jan 16 '13 at 17:25

Let $K = \Bbb{Q}(\alpha)$; write $f(x)$ for the minimal polynomial of $\alpha$ over $\Bbb{Q}$. Our task now is to create infinitely many prime ideals $P$ in $\mathcal{O}_K$ such that $f(P|p) = 1$ where $p$ is the prime in $\Bbb{Z}$ lying under $P$. Now we know that there are infinitely many primes $p$ for which $f$ has a root mod $p$ . Thus for each such prime $p$ is an integer $n_p$ such that

$$g(x)(x - n_p) = f(x) \pmod{p}.$$

Then we claim that for all but finitely many primes $p$ the ideal $P = (p, \alpha- n_p)$ is a prime ideal. When $p \nmid n$ where $n = \left[\mathcal{O}_K : \Bbb{Z}[\alpha]\right]$, it is shown in pg.3 of the document "Factoring After Dedekind" that the map $$\overline{\iota} : \Bbb{Z}[\alpha]/(p) \to \mathcal{O}_K/(p)$$ induced by inclusion $\iota: \Bbb{Z}[\alpha] \to \mathcal{O}_K$ is an isomorphism. Now consider the diagram

enter image description here

where the bottom row is an isomorphism by the first isomorphism theorem. The bottom left and right rings are each isomorphic respectively to $\Bbb{Z}[\alpha]/(p,\alpha - n_p)$ and $\mathcal{O}_K/(p,\alpha - n_p)$ by the third isomorphism theorem. Thus the primality of the ideal $(p,\alpha - n_p)$ in $\mathcal{O}_K$ will follow immediately from the fact that \begin{eqnarray*} \Bbb{Z}[\alpha]/(p,\alpha - n_p) &\cong& \bigg(\Bbb{Z}[x]/(f(x)) \bigg) \bigg/ \bigg((p,x - n_p)/(f(x))\bigg)\\ &\cong& \Bbb{Z}[x] / (p, x -n_p) \\ &\cong& \Bbb{Z}/p\Bbb{Z}[x]/ (x- \overline{n_p}) \\ &\cong& \Bbb{Z}/p\Bbb{Z}.\end{eqnarray*}

Note that my answer in some ways is a generalisation of Rankeya's because we don't assume that $\mathcal{O}_K = \Bbb{Z}[\alpha]$.


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