How to find the prime ideals of $\mathbb{Q}(\sqrt{2})$ in order to find ideals of norm $20$ (not prime).

Question

Very new to algebraic number theory, so I was just wondering if someone could clarify how to do this?

Essentially I would like to proceed by finding elements of norm 20. So $20=2^2\times5$ so I guess I would want to figure out what ideals have norm $2$ and $5$?

Answer 1

If you are interested in finding ideals with a prescribed norm, you should use that the norm is multiplicative, hence if you want to find $I\subset \mathbb Z[\sqrt 2]$ with $N(I)=20$, it suffices to find primes $\mathfrak q_1,\mathfrak q_2$ with norm $2$ (or with norm $4$ if there is no prime of norm $2$) and $5$ and then set $I=\mathfrak q_1^2\mathfrak q_2$ (resp. $\mathfrak q_1\mathfrak q_2$ in the second case).

Now quite generally, let $p$ be a prime number and suppose you are interested in identifying the primes $\mathfrak q$ living over $p$ (i.e. contracting to $(p)$ along the inclusion $\mathbb Z\to\mathbb Z[\sqrt 2]$). For this a very convenient tool is the following result due to Kummer and Dedekind:

Let $K$ be a number field and let $p$ be a prime number. Suppose that $\mathcal O_K=\mathbb Z[a]$ for some $a\in\mathcal O_K$. Let $f\in\mathbb Z[X]$ be the minimal polynomial of $a$. Let $\bar f=\prod_i \bar g_i^{e_i}$ be the prime decomposition in $\mathbb F_p[X]$ and lift the $\bar g_i$ to monic polynomials $g_i$ in $\mathbb Z[T]$. Then $p\mathcal O_K=\prod_i (p,g_i(a))^{e_i}$ is the factorization of $p\mathcal O_K$ into prime ideals in $\mathcal O_K$.

You can find this result for example as Theorem 3.41 in Milne's notes.

Applying this to the situation you are interested in, you will find $a=\sqrt 2$ and hence $f=X^2-2$. Modulo $2$ this gives you $\bar f=\bar X^2$, hence the ideal $(2,\sqrt 2)=(\sqrt 2)$ is prime in $\mathbb Z[\sqrt 2]$ and is the unique prime ideal lying over $2$. Modulo $5$, on the other hand, you find that $\bar f=\bar X^2-2$ is irreducible since it has no roots in $\mathbb F_5$, hence the unique prime ideal lying over $5$ is $(5)$.

Now the only thing left to do is to compute the norm of these ideals. For this you need to know the degrees of the residue field extensions. But since $2$ ramifies in $\mathbb Z[\sqrt 2]$ you already know that $N(\sqrt 2)=2$. Now $5$ does not ramify but also has a unique prime ideal in its fiber, hence the degree of residue field extensions must be $2$ and we find $N(5)=5^2=25$. Hence you have shown that in fact no ideal of norm $20$ can exist in $\mathbb Z[\sqrt 2]$.

Answer 2

Given a quadratic field $K$ with ring of integers $A$, the (unique) decomposition of a rational prime $p$ into a product of prime ideals in $A$ is classically known. For reasons of degrees of ramification and inertia, there are 3 and only 3 possible cases : 1) $p$ is inert, i.e. $p$ remains prime, i.e. $pA$ is a prime ideal $P$ ; 2) $p$ is ramified, i.e. $pA$ is of the form $P^2$ ; 3) $p$ splits, i.e. $pA=P.P'$, the product of 2 distinct conjugate prime ideals. The characterization of these 3 cases can be conveniently expressed in terms of quadratic residue symbols, see e.g. D. Marcus, "Number Fields", chap.3, thm. 25.

In your case, where $K=\mathbf Q(\sqrt 2)$ and $A=\mathbf Z[\sqrt 2]$ is a PID, the mentioned theorem tells you that an odd prime $p$ is a norm iff $p$ splits, iff $(\frac 2 p)=+1$. The "complementary formula" for Legendre symbols reads $(\frac 2 p)= (-1)^{(p^2 -1)/8}$, so you can check e.g. that $3,5,11...$ are not norms, whereas $7,23,31...$ are. Explanation : Let $H=$ {$\bar1,\bar3,\bar5,\bar7$} be the group of invertible elements of $\mathbf Z/8\mathbf Z$. The map $H \to (\pm 1)$ defined by $\bar p \to (\frac 2 p)$ is a surjective homomorphism with kernel $H'=$ {$\bar1,\bar7$}.