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Let $K/\Bbb Q$ be a finite extension and $N>1$ an integer. What are some conditions on $N$ such that we can find an ideal $I \subset \mathcal O_K$ such that $\mathcal O_K / I \cong \Bbb Z / N \Bbb Z$ ? We can assume that $K / \Bbb Q$ is Galois if it makes the answer easier.

It is sufficient to assume that $N = p^r$ for some prime number $p$. Let $P$ a prime of $K$ above $p$. If its inertia index $f(P/p)$ is $1$, then $\mathcal O_K / P^r$ has cardinality $p^r$, but I don't see why the natural map $$\Bbb Z / p^r \Bbb Z \to \mathcal O_K / P^r$$ should be injective (hence an isomorphism). Thank you!

Alphonse
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  • The kernel of the map $\mathbb Z\to\mathcal O_K/P^r$ is $P^r\cap\mathbb Z$. – Mathmo123 Sep 29 '17 at 03:07
  • @Mathmo123 : yes, thank you! But $P^r \cap \Bbb Z = (P \cap \Bbb Z)^r$ might not be true in general : if $P \cap \Bbb Z = (p) \subset \Bbb Z$ and $p \mathcal O_K = P^2$ with $[K : \Bbb Q] = 2$, then $P^2 \cap \Bbb Z = (p) \neq (p)^2$. Maybe, if $p$ splits in $K$, then it works. What do you think? – Alphonse Sep 29 '17 at 06:01
  • @Mathmo123 : would you have any idea if $p$ splits in $K$ ? Thank you so much! – Alphonse Oct 15 '17 at 09:59
  • This might help : https://math.stackexchange.com/questions/2577145/intersection-of-powers-of-prime-ideals-with-subring – Alphonse Dec 23 '17 at 11:14

1 Answers1

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I can provide a general proof.

Let $L/K$ be a finite extension of number fields, and $J \subset O_K$ be an ideal. Assume that for any prime $\mathfrak p$ dividing $J$ in $O_K$, there is a prime $P \subset O_L$ above $\mathfrak p$ such that $e(P/\mathfrak p) = f(P/\mathfrak p)=1$ (e.g. $\mathfrak p$ is totally split in $L$).

Then there is an ideal $I \subset O_L$ such that $O_L/I \cong O_K/J$ as rings (via the natural map $O_K \to O_L/I$).

Proof : fix a prime $P = P(\mathfrak p) \subset O_L$ above each prime divisor $\mathfrak p \mid J$, such that $e(P/\mathfrak p) = f(P/\mathfrak p)=1$, as in the hypothesis. Let $I$ be the product ideal $$I := \prod_{\mathfrak p \mid J} P(\mathfrak p)^{v_{\mathfrak p}(J)} \subset O_L.$$ We have $$|O_L / I| = \prod_{\mathfrak p \mid J} |O_L / P(\mathfrak p)|^{v_{\mathfrak p}(J)} = \prod_{\mathfrak p \mid J} |O_K / \mathfrak p|^{v_{\mathfrak p}(J) f(P(\mathfrak p)/\mathfrak p)} = \prod_{\mathfrak p \mid J} |O_K / \mathfrak p|^{v_{\mathfrak p}(J)} = |O_K/J| $$ Therefore, if I show that the natural morphism $\phi : O_K/J \to O_L/I$ is injective, it will be an isomorphism.

The kernel of $O_K \to O_L/I$ is $$O_K \cap I = \{x \in O_K : \forall \mathfrak p \mid J, v_{P(\mathfrak p)}(x) \geq v_{\mathfrak p}(J)\}$$ and $v_{P(\mathfrak p)}\vert_{K} = e(P(\mathfrak p) / \mathfrak p) v_{\mathfrak p} = v_{\mathfrak p}$. Therefore $O_K \cap I = J$, which means that $\phi$ is injective.

Alphonse
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  • Why does $v_{P(\mathfrak p)}\vert_{K} = e(P(\mathfrak p) / \mathfrak p) v_{\mathfrak p}$ hold? Let $x \in O_K$. We want $v_P(x) = e(P/p)v_{\mathfrak p}(x)$. This is easy if $\mathfrak p = (p) \subset O_K$ is principal, because we may write $x = p^r k$ for some $k$ coprime with $p$ (with $r = v_{\mathfrak p}(x) \in \Bbb N$), and then $$v_P(x) = r v_P(p) = r e(P/p) v_{\mathfrak p}(p) = e(P/p) v_{\mathfrak p}(x).$$ – Alphonse Dec 26 '17 at 17:23