3.5 Splitting of prime ideals
We now want to understand how prime ideals change as we go up in field extensions, by which we mean the following: In \(\mathbb {Z}\) we know that the prime ideals are all principal and generated by prime numbers, i.e, they are of the form \((p)\) for \(p\) a prime. Now, what happens when we extend our number field? Well if we now go from \((\mathbb {Q},\mathbb {Z})\) to \((\mathbb {Q}(\sqrt{-5}),\mathbb {Z}[\sqrt{-5}])\) then lets look at what happens to our prime ideals. In \(\mathbb {Z}[\sqrt{-5}]\), the ideal \((2)\) is no longer prime, since we will see that
where now \((2,1+\sqrt{-5})\) is a prime ideal in \(\mathbb {Z}[\sqrt{-5}]\). (You actually have enough to check this. You can compute its norm and see that it has norm \(2\))
Similarly, if we look at the ideal \((3)\) in \(\mathbb {Z}[\sqrt{-5}]\), we find that
where again each idea on the right is now prime.
Let \(K/F\) be a finite extension of number fields with rings of integers \(\mathcal{O}_K\) and \(\mathcal{O}_F\). If \(\mathfrak {P}\) is a non-zero prime ideal in \(\mathcal{O}_K\) and \(\mathfrak {p}\) is a non-zero prime ideal in \(\mathcal{O}_F\) then we say \(\mathfrak {P}\) lies over \(\mathfrak {p}\) (or equivalently \(\mathfrak {p}\) lies under \(\mathfrak {P}\)), if \(\mathfrak {P} \mid \mathfrak {p}\mathcal{O}_K\). Here \(\mathfrak {p}\mathcal{O}_K\) denotes the ideal generated by \(\mathfrak {p}\) in \(\mathcal{O}_K\).
Let \(K/F\) be a finite extension of number fields with rings of integers \(\mathcal{O}_K\) and \(\mathcal{O}_F\). If \(\mathfrak {P}\) is a non-zero prime ideal in \(K\) and \(\mathfrak {p}\) is a non-zero prime ideal in \(F\) then the following conditions are equivalent:
\(\mathfrak {P} \mid \mathfrak {p}\mathcal{O}_K\)
\(\mathfrak {P} \supset \mathfrak {p}\mathcal{O}_K\)
\(\mathfrak {P} \supset \mathfrak {p}\)
\(\mathfrak {P} \cap \mathcal{O}_F=\mathfrak {p}\)
\(\mathfrak {P} \cap F=\mathfrak {p}\).
Corollary 3.3.14 gives that \((1)\) and \((2)\) are equivalent. Since \(\mathfrak {P}\) is an ideal, if it contains \(\mathfrak {p}\) it contains \(\mathfrak {p}\mathcal{O}_K\), therefore \((2)\) and \((3)\) are equivalent. Since \(\mathfrak {P} \subset \mathcal{O}_K\) it follows that \((4)\) and \((5)\) are equivalent.
So it remains to prove \((3)\) and \((4)\) are equivalent. It obvious that \((4) \implies (3)\) so we just need the reverse. Note that \(\mathfrak {P} \cap \mathcal{O}_F\) contains \(\mathfrak {p}\) and is moreover an ideal in \(\mathcal{O}_F\). But now, since every prime ideal in a Dedekind domain is maximal, \(\mathfrak {p}\) is also maximal and therefore we must have \(\mathfrak {P} \cap \mathcal{O}_F=\mathfrak {p}\) or \( \mathfrak {P} \cap \mathcal{O}_F=\mathcal{O}_F\). But if \( \mathfrak {P} \cap \mathcal{O}_F=\mathcal{O}_F\) then \(1 \in \mathfrak {P}\) meaning \(\mathfrak {P}=\mathcal{O}_K\) which contradicts it being a prime ideal.
If we take \(F=\mathbb {Q}\) and \(K=\mathbb {Q}(\sqrt{3})\) then \((2,1+\sqrt{3})\) lies above \((2)\). To see this we just note that \((2,1+\sqrt{3}) \cap \mathbb {Z}=(2)\).
Let \(K/F\) be a finite extension of number fields with rings of integers \(\mathcal{O}_K\) and \(\mathcal{O}_F\). Then every non-zero prime ideal \(\mathfrak {P}\subset \mathcal{O}_K\) lies over a unique non-zero prime ideal \(\mathfrak {p}\subset \mathcal{O}_F\).
Conversely, every non-zero prime ideal \(\mathfrak {p}\subset \mathcal{O}_F\) lies under at least one non-zero prime ideal \(\mathfrak {P}\subset \mathcal{O}_K\).
For the first part, we claim that \(\mathfrak {P}\cap \mathcal{O}_F\) is a prime ideal in \(\mathcal{O}_F\) (this is enough as from Proposition 3.5.2 (4) will give us uniqueness). For this, we first note that since \(1 \not\in \mathfrak {P}\), then \(\mathfrak {P}\cap \mathcal{O}_F\) wont be the whole ring. Moreover, by the proof of Proposition 3.2.4 we see that \(\mathfrak {P}\) contains \(N_{K/F}(a) \in \mathbb {Z}\) for all \(a \in \mathfrak {P}\). Since the norm of a non-zero element is non-zero, and \(\mathfrak {P}\) is non-zero, we get that \(\mathfrak {P}\cap \mathbb {Z}\) is non-zero and therefore so is \(\mathfrak {P}\cap \mathcal{O}_F\). So \(\mathfrak {P}\cap \mathcal{O}_F\) is a proper ideal, so we just need to prove it is prime. So if \(r,s \in \mathcal{O}_F\) with \(rs \in \mathfrak {P}\cap \mathcal{O}_F\) then \(rs \in \mathfrak {P}\), therefore either \(r \in \mathfrak {P}\) or \(s \in \mathfrak {P}\) since \(\mathfrak {P}\) is prime. From this the first part follows.
For the second part, it suffices to look check that \(\mathfrak {p}\mathcal{O}_K \neq \mathcal{O}_K\), since in this case, it will be divisible by some prime ideal, and by definition this prime ideal would lie over \(\mathfrak {p}\). So we are reduced to checking \(1 \not\in \mathfrak {p}\mathcal{O}_K.\) By Lemma 3.3.4 we can find \(x \in F \backslash \mathcal{O}_F\) such that \(x \mathfrak {p}\subset \mathcal{O}_F\). Then \(x\mathfrak {p}\mathcal{O}_K \subset \mathcal{O}_F\mathcal{O}_K=\mathcal{O}_K\). If \(1 \in \mathfrak {p}\mathcal{O}_K\) then \(x \in \mathcal{O}_K\), but then \(x\) is in \(\mathcal{O}_K \cap F=\mathcal{O}_F\) and this an algebraic integer, contradicting the fact that \(x \in F \backslash \mathcal{O}_F\).
Let \(K/F\) be a finite extension of number fields with rings of integers \(\mathcal{O}_K\) and \(\mathcal{O}_F\) and let \(\mathfrak {p}\) be a non-zero prime ideal in \(\mathcal{O}_F\). Then
We call the \(e_i\) the ramification indices and if needed we will denote them by \(e_{\mathfrak {P}_i|\mathfrak {p}}\).
Now, let \(\mathfrak {P}\) lie over \(\mathfrak {p}\). Recall that \(k_{\mathfrak {P}}:=\mathcal{O}_K/\mathfrak {P}\) and \(k_\mathfrak {p}:=\mathcal{O}_F/\mathfrak {p}\) are both finite fields, since \(\mathfrak {P},\mathfrak {p}\) are both maximal ideals and by Proposition 3.2.4 we know the quotient ring is always finite. These are called the residual fields attached to \(\mathfrak {P}\) and \(\mathfrak {p}\) respectively. Moreover, \(k_\mathfrak {p}\) is naturally a subfield of \(k_{\mathfrak {P}}\) (convince yourself of this). Therefore, \(k_{\mathfrak {P}}\) is a finite extension of \(k_{\mathfrak {p}}\) and \([k_{\mathfrak {P}}:k_{\mathfrak {p}}]\) is called the inertial degree or residue degree of \(\mathfrak {P}\) over \(\mathfrak {p}\), which will be denoted \(f_{\mathfrak {P}|\mathfrak {p}}\).
Let \(F \subset K \subset L\) be number fields and \(p \subset \mathfrak {p}\subset \mathfrak {P}\) be prime ideals in \(\mathcal{O}_F \subset \mathcal{O}_K \subset \mathcal{O}_L\). Then
and
For the \(f\)’s this follows easily from the Tower Law 1.3.19. For the \(e\)’s the result is clear.
Let \(K\) be a number field with \([K:\mathbb {Q}]=n\) and \(p\) a prime number. If \(\mathfrak {p}_i\) are the prime ideals in \(\mathcal{O}_K\) dividing \(p\mathcal{O}_K\), then
with notation as in Definition 3.5.5.
Let \(e_i:=e_{\mathfrak {p}_i|p}\) and \(f_i:=f_{\mathfrak {p}_i|p}\). By definition we have
Now, lets take norms on both sides: By Proposition 3.4.5 we have \(N(p\mathcal{O}_K)=|N_{K/\mathbb {Q}}(p)|=p^n\).
So we have
So it remains to show that \(N(\mathfrak {p}_i)=p^{f_i}\). For this we note that \(\mathcal{O}_K/\mathfrak {p}_i\) is a finite extension of \(\mathbb {F}_p:=\mathbb {Z}/p\mathbb {Z}=:k_p\) of degree \([k_{\mathfrak {p}_i}/k_p]=f_i\). Therefore \(\mathcal{O}_K/\mathfrak {p}_i \cong \mathbb {F}_p^{f_i}\) (as a vector space) from which it follows that \(N(\mathfrak {p}_i)=p^{f_i}\).
Note that we have only defined this for primes \(\mathfrak {p}\) lying above a rational prime \(p \in \mathbb {Z}\). But these things make sense in more generality for primes \(\mathfrak {P}\) lying above a prime ideal \(\mathfrak {p}\) in \(\mathcal{O}_F\) for \(K/F\) a finite extension of number fields.
In this case one can again prove that
but the proof is a bit more involved. But if we assume the following proposition (which I wont prove)
Let \(K,F\) be a number fields with \([K:F]=n\). Let \(\mathfrak {a}\) be an ideal in \(\mathcal{O}_F\) and let \(\mathfrak {A}=\mathfrak {a}\mathcal{O}_K\), then
Then 5 is easy to prove from this.
Let \(K/F\) be an extension of number fields and \(\alpha \in \mathcal{O}_K\) a primitive element so that \(K \cong F(\alpha )\).
Let \(\mathfrak {p}\subset \mathcal{O}_F\) be a prime ideal and \(p\) the prime number such that \((p)=\mathfrak {p}\cap \mathbb {Z}\). Assume that \(p\) does not divide the index 1
Let \(m_{\alpha ,F}\) be the minimal polynomial of \(\alpha \) over \(F\) and let
be its reduction modulo \(\mathfrak {p}\). Now, write \(\overline{m}_{\alpha ,F}\) as a product of powers of irreducible polynomials (i.e factorize it)
Next, let \(m_i \in \mathcal{O}_F[x]\) be a polynomial whose reduction modulo \(\mathfrak {p}\) is \(\overline{m}_i\) and let
Then:
The ideals \(\mathfrak {P}_i\) are independent of choice of \(m_i\). (This is by construction)
The \(\mathfrak {P}_i\) are distinct prime ideals and they are precisely the prime ideals of \(\mathcal{O}_K\) lying over \(\mathfrak {p}\). Therefore
\[ \mathfrak {p}\mathcal{O}_K=\mathfrak {P}_1^{e_1}\dots \mathfrak {P}_r^{e_r} \]\(f_{\mathfrak {P}_i|\mathfrak {p}}=\deg (\overline{m}_i)\) and \(e_i=e_{\mathfrak {P}_i|\mathfrak {p}}\)
The result will follow from the following three claims:
Let \(f_i=\deg (\overline{m}_i)\). For each \(i\), either \(\mathfrak {P}_i=\mathcal{O}_K\) or \(\mathcal{O}_K/\mathfrak {P}_i\) is a field of size \(|\mathcal{O}_F/\mathfrak {p}|^{f_i}\).
\(\mathfrak {P}_i+\mathfrak {P}_j=\mathcal{O}_K\) whenever \(i \neq j\).
\(\mathfrak {p}\mathcal{O}_K\) divides \(\mathfrak {P}_1^{e_1}\dots \mathfrak {P}_r^{e_r}\).
From this we get the result as follows: By relabelling the \(\mathfrak {P}_i\) we can assume that \(\mathfrak {P}_1,\dots ,\mathfrak {P}_s \neq \mathcal{O}_K\) and \(\mathfrak {P}_{s+1},\dots ,\mathfrak {P}_r=\mathcal{O}_K\).
Now, from ((1)) it follows that \(\mathfrak {P}_1,\dots ,\mathfrak {P}_s\) are all prime ideals (since quotienting out by them gives a field). Moreover, by construction they contain \(\mathfrak {p}\), so lie above \(\mathfrak {p}\) and \(f_{\mathfrak {P}_i|\mathfrak {p}}=f_i\) (to see this just look at how the residue degrees are defined).
Next ((2)) tells us that \(\mathfrak {P}_1,\dots ,\mathfrak {P}_s\) are all distinct and ((3)) becomes
From this it follows that
with \(d_i \leq e_i\). Now, using 5 we have
but on the other hand
So comparing these gives \(r=s\), \(d_i=e_i\) for all \(i\).
So lets prove these three claims.
Note that each \(\overline{m}_i\) is an irreducible polynomial in \((\mathcal{O}_F/\mathfrak {p})[x]\). So if we let \(E:=\mathcal{O}_F/\mathfrak {p}\) (which is a field), then \(M_i:=E[x]/\overline{m}_i\) is again a field. We have a map
given by first reducing modulo \(\mathfrak {p}\) and then modulo \(\overline{m}_i\). This map is clearly surjective and the kernel is given by the ideal in \(\mathcal{O}_F[x]\) generated by \(\mathfrak {p}\) and \(m_i\). So we have an isomorphism
Alongside this, note that we have a ring homomorphism
given by evaluation at \(\alpha \) (i.e \(x \mapsto \alpha \)) and reducing modulo \(\mathfrak {P}_i\). By definition of \(\mathfrak {P}_i\) its clear that \((\mathfrak {p},m_i)\) is in the kernel. But from the above, we know \((\mathfrak {p},m_i)\) is a maximal ideal, so \(\ker (\phi )\) is either \((\mathfrak {p},m_i)\) or \(\mathcal{O}_F[x]\).
Next, we claim that \(\phi \) is surjective. To show this we need to show that \(\mathcal{O}_K=\mathcal{O}_F[\alpha ]+\mathfrak {P}_i\). It turns out that in fact something stronger is true, which is that \(\mathcal{O}_K=\mathcal{O}_F[\alpha ]+(p)\mathcal{O}_K\) (note that \(p\mathcal{O}_K \subset \mathfrak {P}_i\)). To prove this, we need to make use of the one thing we know, which is that \(p \nmid [\mathcal{O}_K:\mathcal{O}_F[\alpha ]]\). How do we use this fact? notice that the index 2 of \(\mathcal{O}_F[\alpha ]+p\mathcal{O}_K\) in \(\mathcal{O}_K\) must divide both of \([\mathcal{O}_K:\mathcal{O}_F[\alpha ]]\) and \([\mathcal{O}_K:p\mathcal{O}_K]\). But \([\mathcal{O}_K:p\mathcal{O}_K]\) is some power of \(p\) and by assumption \(p \nmid [\mathcal{O}_K:\mathcal{O}_F[\alpha ]]\), so these indexes are coprime and therefore
Therefore since \(\ker (\phi )\) is either is either or \(\mathcal{O}_F[x]\), it follows that either \(\mathcal{O}_K/\mathfrak {P}_i \overset {\sim }{\longrightarrow } M_i\) (which would be if the kernel is \((\mathfrak {p},m_i)\) ) or \(\mathfrak {P}_i=\mathcal{O}_K\) if the kernel is \(\mathcal{O}_F[x]\).
By construction the \(\overline{m}_i\) are distinct irreducible polynomials in \(E[x]\). Therefore for \(i \neq j\) we can find \(\overline{g},\overline{h}\) such that \(\overline{m}_i\overline{g}+\overline{m}_j\overline{h}=1\) (in \(E[x]\)). This mean that we can find \(g,h \in \mathcal{O}_F[x]\) such that \(m_ig+m_jh \equiv 1 \pmod\mathfrak {p}\). If we now simply evaluate at \(x=\alpha \) we see that
and therefore
Since \(\mathfrak {P}_i=(\mathfrak {p},m_i(\alpha ))\) it follows that \(\mathfrak {P}_1^{e_1}\dots \mathfrak {P}_r^{e_r}\) is contained in
But this means (by Corollary 3.3.14) that \(\mathfrak {P}_1^{e_1}\dots \mathfrak {P}_r^{e_r}\) is divisible by \((\mathfrak {p},m_1(\alpha )^{e_1}\dots m_r(\alpha )^{e_r})\).
But note that
Therefore
thus \((\mathfrak {p},m_1(\alpha )^{e_1}\dots m_r(\alpha )^{e_r})=\mathfrak {p}\mathcal{O}_K\) which gives the result.
This theorem is really useful for us. It will make it really easy to understand how prime ideals change as we extend our field. For clarity, lets just see what this theorem says when our bottom field (\(F\)) is simply \(\mathbb {Q}\).
Let \(K\) be a number field of degree \(n\) over \(\mathbb {Q}\) and let \(\alpha \in \mathcal{O}_K\) be a primitive element so that \(K\cong \mathbb {Q}(\alpha )\).
Let \(p\) be a prime number not dividing \([\mathcal{O}_K:\mathbb {Z}[\alpha ]]\) and let
be the reduction modulo \(p\) of the minimal polynomial \(m_\alpha \) of \(\alpha \). Then in \(\mathcal{O}_K\) the ideal \((p)\) factorizes as
where \(\mathfrak {p}_i=(p,m_i(\alpha ))\) and \(f_{\mathfrak {p}_i|p}=\deg (\overline{m}_i)\) and \(e_i=e_{\mathfrak {p}_i|p}\).
Let \(K\) be a number field and \(p\) a prime number. We say
\(p\) is ramified in \(K\), if there exists a \(\mathfrak {p}|p\) such that \(e_{\mathfrak {p}|p}{\gt}1\). Otherwise we say \(p\) is unramified.
\(p\) is totally ramified if there is a \(\mathfrak {p}|p\) such that \(e_{\mathfrak {p}|p}=[K:\mathbb {Q}]\).
\(p\) is inert if \(p\) is unramified and there exists a unique prime \(\mathfrak {p}\) lying above \(p\) (which will have \(f_{\mathfrak {p}|p}=[K:\mathbb {Q}]\)).
\(p\) is split if it is unramified and for some \(\mathfrak {p}|p\) we have \(f_{\mathfrak {p}|p}=1\). If it is unramified and for all \(\mathfrak {p}|p\) we we have \(f_{\mathfrak {p}|p}=1\), we say \(p\) is totally split.
Going back to \(\mathbb {Z}[\sqrt{-5}]\), we saw that the ideal \((2)\) is no longer prime, since we will see that
where now \((2,1+\sqrt{-5})\) is a prime ideal in \(\mathbb {Z}[\sqrt{-5}]\). Therefore \((2)\) is totally ramified in \(\mathbb {Z}[\sqrt{-5}]\).
Similarly, if we look at the ideal \((3)\) in \(\mathbb {Z}[\sqrt{-5}]\), we find that
where again each idea on the right is now prime. So \((3)\) totally splits in \(\mathbb {Z}[\sqrt{-5}]\).
Lastly, if we look at \((11)\) we will see that this remains prime in \(\mathbb {Z}[\sqrt{-5}]\) and is therefore inert.
Lets now do some examples:
Let \(K=\mathbb {Q}(\sqrt{6})\) (so here \(\alpha =\sqrt{6}\)). From Theorem 2.1.11 we have \(\mathcal{O}_K=\mathbb {Z}[\sqrt{6}]\) and therefore \([\mathcal{O}_K:\mathbb {Z}[\alpha ]]=1\) so Theorem 3.5.10 works for any prime number. Now, note that \(m_\alpha =x^2-6\).
Lets use this to find how the ideal \(2\mathcal{O}_K\) factorizes in \(\mathbb {Z}[\sqrt{6}]\). From now on I will just denote \(p\mathcal{O}_K\) by \((p)\). First step is to reduce \(m_\alpha \) modulo \(2\). Which gives \(\overline{m}_\alpha (x)=x^2=\overline{m}_1(x)^2\) (if you cant see why we only have one \(\overline{m}_i\) on the right hand side, just look back at the theorem and note that the \(\overline{m}_i\) are distinct by construction).
Next, we need to find some \(m_1(x)\) whose reduction modulo \((2)\) agrees with \(\overline{m}_1(x)\). For this lets just be lazy and take \(m_1(x)=x\) (this might be a good point to comeback to and try and convince yourself the the choice of \(m_1(x)\) wont make a difference to then end result, i.e. see what happens if we take \(m_1(x)=x+2\)). The theorem then says that
This means \((2)\) ramifies in \(\mathcal{O}_K\).
Ok lets do some more examples. To speed things up here is a table of values of \(m_\alpha (x)\) for some small \(x\) which is useful for figuring out how \(m_\alpha \) factorizes modulo \(p\).
|
\(x\) |
\(m_\alpha (x)=x^2-6\) |
|
\(0\) |
\(-6\) |
|
\(\pm 1\) |
\(-5\) |
|
\(\pm 2\) |
\(-2\) |
|
\(\pm 3\) |
\(3\) |
|
\(\pm 4\) |
\(10\) |
|
\(\pm 5\) |
\(19\) |
From this table we see that:
\begin{align*} & m_\alpha (x)=x^2 \pmod3\\ & m_\alpha (x)=(x-1)(x+1) \pmod5\\ & m_\alpha (x)=(x^2-6) \pmod7\\ & m_\alpha (x)=(x^2-6) \pmod{11}\\ \end{align*}From this it follows that:
\begin{align*} & (3)=\mathfrak {p}_3^2:=(3,\sqrt{6})^2 \\ & (5)=\mathfrak {p}_5\mathfrak {p}_5’:=(5,\sqrt{6}-1)(5,\sqrt{6}+1)\\ & (7)=\mathfrak {p}_7:=(7)\\ & (11)=\mathfrak {p}_{11}:= (11)\\ \end{align*}So we see that \((3)\) ramified in \(\mathcal{O}_K\), \((5)\) splits and \((7),(11)\) are inert in \(\mathcal{O}_K\).
BUT WAIT, THERE’S MORE (insert relevant meme): The theorem tells us the norm of the ideals \(\mathfrak {p}\). Specifically, the theorem says that \(f_{\mathfrak {p}|p}=\deg (\overline{m}_i)\) and by Theorem 3.5.7 we saw \(N(\mathfrak {p})=p^{f_{\mathfrak {p}|p}}\). So, in this case we get
\begin{align*} & N(\mathfrak {p}_2)=2 \\ & N(\mathfrak {p}_3)=3 \\ & N(\mathfrak {p}_5)=5^1 \text{ and } N(\mathfrak {p}_5’)=5^1 \\ & N(\mathfrak {p}_7)=7^2 \\ & N(\mathfrak {p}_{11})=11^2 \\ \end{align*}We can also use this factorize other ideals in \(\mathcal{O}_K\).
Say \(\mathfrak {a}\) is an ideal, then we can first calculate \(N(\mathfrak {a})\) (which if \(\mathfrak {a}\) is principal we know how to do, but in general this might be hard) and factorize it.
For each prime number \(p\) appearing, we then factor \((p)\) in \(\mathcal{O}_K\) as we did above.
Note that Corollary 3.4.10 tells us that there are only finitely many prime ideals of norm \(p\), we can find them all by looking at the factorization of \((p)\).
Next we use Proposition 3.3.14, tells us that \(\mathfrak {p}| \mathfrak {a}\) if and only if the generators of \(\mathfrak {p}\) are in \(\mathfrak {a}\).
Lets do this for \(\mathfrak {a}=(12+7\sqrt{6})\). To find the norm of this ideal we use Proposition 3.4.5 this gives
From the above calculation we see that \(\mathfrak {p}_2\) is the only prime ideal of norm \(2\) and \(\mathfrak {p}_3\) is the only prime ideal of norm \(3\) and that we have two ideals of norm \(5\). This immediately gives us that there are only 3 ideals of norm \(150\) and they are
So we need to check with one of these is \(\mathfrak {a}\). First we see that since \(\mathfrak {p}_5\mathfrak {p}_5'=(5)\) and \(12+7\sqrt{6}\) is not a multiple of \(5\), it follows it can’t be the one with a \(\mathfrak {p}_5\mathfrak {p}_5'\) term. Now, note that \(12+7\sqrt{6}=5+7(1+\sqrt{6})\) therefore its contained in \(\mathfrak {p}_5'\) and therefore
Lets do a cubic. Let \(K=\mathbb {Q}(\alpha )\) where \(\alpha \) is a root of \(m_\alpha (x)=x^3+10x+1\). This is irreducible since it is irreducible in \(\mathbb {F}_{13}\) (just check it doesnt have a root). By Corollary 2.2.26 we have that
which is a prime number, in particular it is square-free, so this is an integral basis by Corollary 2.2.9. So \(\mathcal{O}_K=\mathbb {Z}[\alpha ]\). So we can freely apply Theorem 3.5.10.
Lets see how some prime split in \(K\):
We have
therefore
Thus \((2)\) is unramified, and the residue degrees are \(f_{\mathfrak {p}_2|2}=1\) and \(f_{\mathfrak {p}_2'|2}=2\).
Lets look at \((29)\). Here we see that
meaning
Thus \((29)\) is also unramified and in particular it is totally split. So the residue degrees are all \(1\).
Lastly, lets look at \((4027)\). Here we see
therefore
Thus \((4027)\) is ramified. \(e_{\mathfrak {p}_{4027}|4027}=2\), \(e_{\mathfrak {p}_{4027}'|4027}=1\) and the residue degrees are all \(1\).
Let \(K\) be a number field and \(\alpha \in \mathcal{O}_K\) a primitive element with \(m_\alpha \) Eisenstein at \(p\). Then we have
with \(\mathfrak {p}=(p,\alpha )\).
If \(\alpha \) is Eisenstein at \(p\) then \(\frac{\mathcal{O}_K}{\mathbb {Z}[\alpha ]}\) has no element of order \(p\) by Lemma 2.2.21 and therefore the index \([\mathcal{O}_K:\mathbb {Z}[\alpha ]]\) is coprime to \(p\). So we can apply Theorem 3.5.10 which, since \(m_\alpha \) is Eisenstein at \(p\), gives \(\overline{m}_\alpha (x)=x^{[K:\mathbb {Q}]}\), from which the result follows.
Lets see what this tells us for quadratic extensions:
Note that if \(n\) is a square-free integer, then in \(\mathbb {Q}(\sqrt{n})\) there are three possibilities for how primes decompose (this follows from Theorem 3.5.7). These being
Let \(n\) be a square-free integer, \(K:=\mathbb {Q}(\sqrt{n})\) and \(p\) a prime number.
If \(p\mid n\) then \((p)=(p,\sqrt{n})^2\).
If \(n\) is odd then
\[ (2)= \begin{cases} (2,1+\sqrt{n})^2, & \text{if } n \equiv 3 \pmod4 \hspace{0.5cm} (3.3) \\ \left(2,\frac{1+\sqrt{n}}{2}\right)\left(2,\frac{1-\sqrt{n}}{2}\right) & \text{if } n \equiv 1 \pmod8 \hspace{0.5cm} (3.4)\\ (2) & \text{if } n \equiv 5 \pmod8 \hspace{0.5cm} (3.5) \end{cases} \]If \(p\) is odd and \(p \nmid n\), then
\[ (p)=\begin{cases} (p,a+\sqrt{n}) (p,a-\sqrt{n}) & \text{if } n \equiv a^2 \pmod p \hspace{0.5cm} (3.6)\\ (p) & \text{if } \left( \frac{n}{p}\right)=-1 \end{cases} \]
Moreover, in (3.4) and (3.6) the ideals appearing are distinct.
We will defer the proof that the ideals in (3.4) and (3.6) are distinct to Corollary 3.6.12.
For (1), note that \((p,\sqrt{n})^2=(p^2,p\sqrt{n},n)\), which if \(p \mid n\) is contained in \((p)\). Conversely, \((p^2,p\sqrt{n},n)\) contains the GCD of \(p^2\) and \(n\), which is \(p\), so we get \((p) \subset (p,\sqrt{n})^2\), therefore we get the equality we are after.
We will split the rest of proof into two cases, which together give the result. The first is if \(n \equiv 2,3 \pmod4\) and the second if \(n \equiv 1 \pmod4\).
From Theorem 2.1.11 we know that if \(n \equiv 2,3 \pmod4\) then \(\mathcal{O}_K=\mathbb {Z}[\sqrt{n}]\). In this case Theorem 3.5.10 (using \(\alpha =\sqrt{n}\) and \(m_\alpha =x^2-n\)) gives (3) and (3.3) (which is the only one that applies in this case).
Now, we look at the case \(n \equiv 1 \pmod4\). In this case, Theorem 2.1.11 tells us that \(\mathcal{O}_K=\mathbb {Z}[\frac{1+\sqrt{n}}{2}]\). Note that this means the index \([\mathcal{O}_K:\mathbb {Z}[\sqrt{n}]]=2\) but \(p\) is odd, so it doesn’t divide this index. So Theorem 3.5.10 gives (3).
So, we are left with (3.4) and (3.5) of (2). Here we cannot apply Theorem 3.5.10. So lets assume \(n \equiv 1 \pmod8\), then
Now, since each of these generators is divisible by \(2\), we get that
Conversely, \(\left(4,1-\sqrt{n},1+\sqrt{n},\frac{1-n}{4}\right)\) contains the \(\gcd (4,1-\sqrt{n})=2\) which gives the reverse inclusion.
Finally, lets assume \(n \equiv 5 \pmod8\). In this case lets consider \(\mathcal{O}_K/(2)\). By Theorem 3.5.7 its enough to show that \(\mathcal{O}_K/(2)\) is a field not isomorphic to \(\mathbb {Z}/2\mathbb {Z}\) (since this means the residue degree is \(2\) and therefore, since we are in a quadratic field, it must be inert). To show they are not isomorphic, consider
By construction, this has a root in \(\mathcal{O}_K\) and therefore has a root in \(\mathcal{O}_K/(2)\) (just reduce your root modulo \(2\)). On the other hand, since \(n \equiv 5 \pmod8\) we have \(m_\alpha (x)=x^2+x+1 \pmod2\) which by Exercise 1.2.19 we know is irreducible in \(\mathbb {Z}/2\mathbb {Z}\) and therefore has no root in \(\mathbb {Z}/2\mathbb {Z}\). So \(\mathcal{O}_K/(2)\) is an extension of \(\mathbb {F}_2\) by a root of \(x^2+x+1\), so its a field, which also goes by the name \(\mathbb {F}_4\).
Using this theorem together with the quadratic reciprocity law, which says
(with \(p,q\) primes) we can easily find how primes decompose in quadratic extensions.