Prime ideals in $\mathbb{Q}(\sqrt{-5})$ above rational prime ideals

Question

I'm really struggling to understand the concept of prime ideals lying above and below a given prime ideal. For example taking the extension $\mathbb{Q}(\sqrt{-5})\big/\mathbb{Q}$, how do we know $(2, 1+ \sqrt{-5})$ the only ideal in the bigger field lying above $(2)$? And why do $(3, 1+\sqrt{-5})$ and $(3, 1-\sqrt{-5})$ lie above $(3)$? (Sorry I appreciate that this is probably very elementary but I'm new to algebraic number theory and all the books I have just skip over this point assuming it is obvious to the reader which sadly it's not...)

Answer 1

Suppose that $K$ is a number field, by definition, this means that $K$ is a finite extension of $\mathbb{Q}$, and WLOG we can even assume that $K\subset \mathbb{C}$ with the condition that $[K:\mathbb{Q}]<\infty$.

Associated to any number field $K$, are the generalization of integers, called the ring of integers of $K$, these are the elements $x\in K$, so that $x$ satisfies a monic polynomial with integer coefficients. We denote the set of all such integral elements by $\mathcal{O}_K$. One can show that $\mathcal{O}_K$ is a sub-ring of $K$.

In algebraic number theory, one usually works with prime ideals, instead of prime elements (because the ring of integers usually does not have unique factorization). The prime ideals of $\mathbb{Z}$ are of the form $p\mathbb{Z}$ where $p$ is a prime number. We say that a prime ideal $\mathfrak{p} \subset \mathcal{O}_K$, lies above, or, divides $p$, if $\mathfrak{p}\cap \mathbb{Z} = p\mathbb{Z}$.

Answer 2

Probably you know that given a ring $R$ and an ideal $\mathfrak a$ there is a 1-1 correspondence between ideals containing $\mathfrak a$ (i.e. ideals above $\mathfrak a$) and ideals of $R/\mathfrak a$. Taking your ring of integers and a prime number $p$, the quotient is a quadratic (in your case) extension of a finite field. A polynomial ring over a field is a PID, so it is the quotient either by an irreducible polynomial, or by a polynomial that splits in two factors, usually different but occasionally equal, as in your case above 2. The prime ideals are generated by $p$ with the representative of a generator of a maximal ideal in these quotients.

Answer 3

To see how this works, it's most simple to appeal to the norm and the concept of "ramification." The most likely reason this was glossed over in your text is that you probably haven't proven these theorems yet, and I agree it's not immediately obvious why it's true otherwise.

So first let me start with the statement of Dedekind's Discriminant Theorem

Let $K/\Bbb Q$ be a finite field extension with ring of integers $\mathcal{O}_K$. Then the ideal $p\mathcal{O}_k$ is ramified iff $p|\Delta_K$, the discriminant of $K$.

Then since we know for quadratic fields, $K=\Bbb Q(\sqrt{m})$ with $m$ square-free that

$$\Delta_K=\begin{cases} m & m\equiv 1\mod 4 \\ 4m & m\equiv 2,3\mod 4\end{cases}.$$

Now, $-5\equiv 3\mod 4$ so that with $K=\Bbb Q(\sqrt{-5})$ we have $\Delta_K=-20$. Then we know that when we factor $(2)$ into prime ideals we have

$$(2)=\mathfrak{p}_1^{e_1}\ldots\mathfrak{p}_r^{e_r}$$

with some $e_i>1$. Since $N((2))=N(2)=4$ we see that $N(\mathfrak{p}_i)|4$, i.e. $N(\mathfrak{p}_i)\in\{2,4\}$. Assume WLOG $e_1>1$. Then $N(\mathfrak{p}_1^{e_1})>N(\mathfrak{p}_1)^2\ge 4$. So it must be that $e_1=2$ and $N(\mathfrak{p}_1)=2$, since any larger and the norm of the product is bigger than $4$, which is impossible. But then if we have some other prime $\mathfrak{p}_2$ in that factorization, i.e. $r>1$ then $N(\mathfrak{p}_2)\ge 2$ so that the norm of the product is at least $8$, impossible. So we conclude there is but one prime above $2$.

Now how do we conclude based on similar information that there are two primes (and no more) above $3$? Well, first we'd like to know that the ideals are actually above $3$.

In number theory if we have two integer rings $\mathcal{O}_K\subseteq\mathcal{O}_L$ the phrase "$\mathfrak{P}$ lies above $\mathfrak{p}$" means that $\mathfrak{p}\mathcal{O}_L=\mathfrak{P}^{e}\cdot\mathfrak{a}$ for some ideal $\mathfrak{a}\subseteq\mathcal{O}_L$. That is: $\mathfrak{P}$ appears in the prime factorization of the ideal $\mathfrak{p}\mathcal{O}_L\subseteq\mathcal{O}_L$.

In your case we can see that both of the given primes are indeed "above" $3$, so it suffices to see that there can be no more. Again, noting

$$(3,1+\sqrt{-5})(3,1-\sqrt{-5})|(3)$$

We see that $N((3,1\pm\sqrt{-5}))|N(3)=9$. But then if there are more primes in the factorization of $(3)$ we would get too big of a norm again, so there are at most $2$ primes above $3$, hence exactly $2$.

The same logic goes through other ideals, you can see the same construction works for all primes since their norms are squares, so there are always at most $2$ in a quadratic extension. Proving there are $2$ usually amounts to providing $2$ distinct ideals, proving there is only $1$ in the ramified case (i.e. primes dividing the discriminant) is exactly the same, and then for the others for which there is only one prime, but it is not ramified, things can get a bit trickier, and is usually more ad-hoc (unless you know a suitably advanced form of quadratic reciprocity).