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…)
[Math] Prime ideals in $\mathbb{Q}(\sqrt{-5})$ above rational prime ideals
algebraic-number-theorymaximal-and-prime-idealsnumber theory
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There are several ways to deduce that these are ideals are prime. The easiest might be to just compute the quotient:
$$\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5}) \cong \mathbb Z[X]/(X^2+5,2,1+X) = \mathbb Z[X]/(2,1+X) \cong \mathbb Z/2\mathbb Z$$
But you can also use some theory (And this somehow fits to your norm approach). Whenever we have an quadratic integer ring and an integer prime number $p \in \mathbb Z$, then the ideal $(p) \subset \mathcal O_K$ behaves in three ways:
- $(p)$ is prime.
- $(p) = \mathfrak p^2$ for some prime ideal.
- $(p) = \mathfrak p_1 \mathfrak p_2$ for two different prime ideals.
Together with the fact, that there exists a unique prime ideal factorization, we get the following corollary: Whenever we have $(p)=IJ$ for some ideals $I,J$, then $I$ and $J$ are necessarily prime (If one of them would factor into primes, $(p)$ would factor into at least $3$ primes).
This is exactly the situation that the Dedekind-Kummer theorem treats. This is given in the following two theorems from Keith Conrad's Factoring after Dedekind.
Theorem 1 (Dedekind). Let $K$ be a number field and $\alpha \in O_K$ such that $K= \mathbb{Q}(\alpha)$. Let $f(T)$ be the minimal polynomial of $\alpha$ in $\mathbb{Z}[T]$. For any prime $p$ not dividing $[O_K:\mathbb{Z}[\alpha]]$, write $$ f(T) \equiv \pi_1(T)^{e_1}\cdots \pi_g(T)^{e_g} \pmod{p} $$ where the $\pi_i(T)$ are distinct monic irreducibles in $\mathbb{F}_p[T]$. Then $ (p) = p O_K$ factors into prime ideals as $$ \newcommand{\p}{\mathfrak{p}} (p) = \p_1^{e_1} \cdots \p_g^{e_g} $$ where there is a bijection between the $\p_i$ and $\pi_i(T)$ such that $N(\p_i) = p^{\deg(\pi_i)}$. In particular, this applies for all $p$ if $O_K=\mathbb{Z}[\alpha]$.
He further describes how to obtain generators for the $\p_i$ in Theorem 8.
Theorem 8. In the notation of Theorem 1, when $\p_i$ is the prime ideal corresponding to $\pi_i(T)$ we have the formula $\p_i = (p,\Pi_i(\alpha))$ where $\Pi_i(T)$ is any polynomial in $\mathbb{Z}[T]$ that reduces mod $p$ to $\pi_i(T)$ mod $p$.
Returning to your problem, the minimal polynomial of $\sqrt{-5}$ is $T^2 + 5$. How does this factor mod $3$? What does this tell you about the primes ideals in the factorization of $(3)$ and their generators?
Best Answer
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
Then since we know for quadratic fields, $K=\Bbb Q(\sqrt{m})$ with $m$ square-free that
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
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 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
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).