quotient of the ring of integers by a prime ideal

Question

Let $\mathcal{P}$ be a prime ideal of the ring of integers $\mathcal{O}_{K}$ of a field $K$. Since $\mathcal{O}_{K}$ is a Dedekind domain therefore we can say that $\mathcal{O}_{K}/ \mathcal{P}$ is a field. My question is, what is the structure of this field?

More precisely, I want to know that whether all these fields are finite or not. If some fields are not finite then are those fields isomorphic to some number fields?

If $K = \mathbb{Q}(\alpha)$ and $f(x)$ be the minimal polynomial of $\alpha$ in $K$ then we know that, $\mathcal{O}_{K}$ is isomorphic to $\mathbb{Z}[x]/(f(x))$ and any prime ideal $\mathcal{P}$ is of the form $(g(x),p)$ for each prime integer $p$ and for each irreducible factor $g(x)$ of $f(x)$ mod $p$. But I couldn’t deduce the structure of the quotient ring (which is actually a field).

Answer 1

Here is a proof that $\mathcal{O}_K/\mathcal{P}$ is finite for any nonzero prime ideal $\mathcal{P}$. Let $n$ be the degree of $K$.

First of all, for every nonzero $m \in \mathbb{Z}$, there are isomorphisms of groups $$\mathcal{O}_K/m\mathcal{O}_K \cong \mathbb{Z}^n/m\mathbb{Z}^n \cong (\mathbb{Z}/m \mathbb{Z})^n.$$ The second one is easy to see. For the first one, choose an integral basis $\beta_1,...,\beta_n$ of $K$ and write each element in $\mathcal{O}_K$ as $k_1\beta_1+\dots+k_n\beta_n$ with $k_i \in \mathbb{Z}$. Then, the association $$\mathcal{O}_K \to \mathbb{Z}^n/m\mathbb{Z}^n,\phantom{aa}k_1\beta_1+\dots+k_n\beta_n \mapsto (k_1,...,k_n) + m\mathbb{Z}^n$$ gives an onto homomorphism. It is not hard to see that $m\mathcal{O}_K$ is its kernel.

The above shows that $\mathcal{O}_K/m\mathcal{O}_K$ is finite for any nonzero $m \in \mathbb{Z}$.

Now, let's consider $\mathcal{O}_K/\mathcal{P}$. It is not hard to see that $\mathcal{P}$ contains a nonzero $m \in \mathbb{Z}$. (Take for example any nonzero $\alpha \in \mathcal{P}$ and consider its norm $N_K(\alpha)$.) Hence, $m\mathcal{O}_K \subseteq \mathcal{P}$. It follows that $$\mathcal{O}_K/m\mathcal{O}_K \to \mathcal{O}_K/\mathcal{P},\phantom{aa}\alpha+m\mathcal{O}_K \mapsto \alpha +\mathcal{P}$$ is a well defined map. It is obviously onto.

Note that exactly the same argument shows that $\mathcal{O}_K/\mathcal{I}$ is finite for any nonzero ideal $\mathcal{I}$.

For a book which contains a version of the above proof and which is also a great introduction to algebraic number theory, I recommend Number Fields by Daniel Marcus.

Answer 2

First of all, you mean $\mathcal P$ a nonzero prime ideal. Anyways, $\mathcal P \cap \mathbb Z$ is therefore a nonzero prime ideal in $\mathbb Z$. Let's say it equals $(p)$. We say that the prime $\mathcal P$ "lies over" $(p)$. We therefore get an induced map of fields $\mathbb Z/(p) \longrightarrow \mathcal O_K/\mathcal P$. Thus, $\mathcal O_K$ is a field extension of $\mathbb Z/(p) = \mathbb F_p$. It's a finite extension because $\mathcal O_K$ is finite over $\mathbb Z$.

So what particular extension is this? It is known that finite fields are classified completely by their degree over the prime subfield, so we need only determine what the degree $[\mathcal O_K/\mathcal P : \mathbb F_p]$ is. It's a general fact of Dedekind domains that ideals factor uniquely into products of prime ideals. Indeed, let's write $p \mathcal O_K = \prod_{i=1}^g \mathcal P_i^{e_i}$ where the $\mathcal P_i$ are distinct primes. One of them, say $\mathcal P_1$, will be our original prime $\mathcal P$.

Now, by the same reasoning as for $\mathcal P$, for each $i$, $\mathcal O_K/\mathcal P_i$ will be a finite extension of $\mathbb F_p$. Let's denote by $f_i$ the degree $[\mathcal O_K/\mathcal P_i : \mathbb F_p]$. The relationship here is that $\sum_{i=1}^g e_i f_i = [K : \mathbb Q]$. This at least gives some constraint on the degree $f_1$ we're after.

If $K/\mathbb Q$ is Galois we can say even more. In that case, the $e_i$ are all the equal, so we let $e = e_i$. The $f_i$ are also all equal, so we let $f = f_i$. Then the above formula simplifies to $[K : \mathbb Q] = e f g$. To explicitly find the values of all of these is not always easy. The Kummer-Dedekind theorem lets you compute these by performing a factorization in $\mathbb F_p[x]$. This is a powerful way to understand $\mathcal O_K/\mathcal P = \mathbb F_{p^f}$.

I haven't proved anything, of course. You could look at the algebraic number theory books by Neukirch or Lang, for instance, to get the details.