3.6 Embeddings and prime ideals
Lets now look at how embeddings interact with prime ideals. We begin with a definition you may have seen in your Galois theory course.
Let \(K/F\) be an extension of number fields. We say \(K\) is normal over \(F\) every embedding \(\sigma :K \to \mathbb {C}\) which fixes \(F\) has image again in \(K\). In other words the embedding \(\sigma \) is an automorphism \(\sigma :K \to K\) which fixes the elements of \(F\).
In particular, if \(K=F(\alpha )\) and \(K\) contains all of the conjugates of \(\alpha \), then \(K\) is normal.
Let \(K=\mathbb {Q}(\sqrt{2})\) and \(F=\mathbb {Q}\). Then since each embedding that fixes \(\mathbb {Q}\) sends \(\sqrt{2}\) to one of \(\pm \sqrt{2} \in K\), we see that \(K\) is normal.
If \(K=\mathbb {Q}(\sqrt[3]{2})\) then this is not normal, since one of the embeddings will send \(\sqrt[3]{2}\) to \(\zeta _3\sqrt[3]{2}\) (where \(\zeta _3\) is a non-trivial cube root of unity) which is a complex number and therefore not contained in \(K\).
Let \(K/F\) be a finite extension of number fields. Then there is a finite extension \(L/K\) such that \(L/F\) is normal as is \(L/K\).
If \(K=F(\alpha )\), then just set \(L=F(\alpha _1,\alpha _2,\dots ,\alpha _n)\) where \(\alpha _i\) are the conjugates of \(\alpha \).
If \(K/F\) is a normal extension, \(\mathfrak {a}\subset \mathcal{O}_K\) is an ideal and \(\sigma \) is an embedding of \(K\) fixing \(F\) (which in particular is an automorphism). Then we let \(\sigma (\mathfrak {a})\) be the ideal in \(K\) generated by the images of the elements of \(\mathfrak {a}\) under \(\sigma \).
Furthermore, we let \(\operatorname{Gal}(K/F)\) denote the set of embeddings of \(K\) which fix \(F\). Since \(K/F\) is normal, this can be made into a group by using composition as our group operation, i.e. \(\sigma _1,\sigma _2 \in \operatorname{Gal}(K/F)\) we let \((\sigma _1 \sigma _2)(x)= \sigma _1(\sigma _2(x))\). By Proposition 1.5.8 there are only \([K:F]\) such embeddings, so this group has size \([K:F]\) and is known as the Galois group.
Let \(K/F\) be a normal extension of number fields. Let \(\mathfrak {p}\) be a prime ideal in \(\mathcal{O}_F\) and \(\mathfrak {P},\mathfrak {P}'\) be two prime ideals of \(\mathcal{O}_K\) above \(\mathfrak {p}\). Then \(\sigma (\mathfrak {P})\) is again a prime ideal lying over \(\mathfrak {p}\), moreover there is some element \(\sigma \in \operatorname{Gal}(K/F)\), such that \(\sigma (\mathfrak {P})=\mathfrak {P}'\).
Let \(G:=\operatorname{Gal}(K/F)\). Note that since \(\mathfrak {P}\) is a prime ideal \(\mathcal{O}_K/\mathfrak {P}\) is a integral domain. Now, apply \(\sigma \) to this quotient. Since \(\sigma (\mathcal{O}_K)=\mathcal{O}_K\) (as \(K/F\) is normal) we get
and therefore \(\sigma (\mathfrak {P})\) is again a prime ideal. Moreover, since \(\mathfrak {P}\cap \mathcal{O}_F=\mathfrak {p}\) (this is what lying above means) and \(\sigma \) fixes \(F\) we see that \(\sigma (\mathfrak {P}) \cap \mathcal{O}_F=\mathfrak {p}\).
Now, suppose that \(\sigma (\mathfrak {P}) \neq \mathfrak {P}'\) for all \(\sigma \in G\). Then we can use the Chinese remainder theorem 3.3.21 to find some \(\alpha \in \mathcal{O}_K\) such that
and
for all \(\sigma \in G\).
Now,
On the other hand \(\alpha \not\in \sigma (\mathfrak {P})\) by construction, and therefore \(\sigma ^{-1}(\alpha ) \not\in \mathfrak {P}\). But we have
This gives us a contradiction, since we have just seen that the left hand side is in \(\mathfrak {p}\). But the right hand is not contained in \(\mathfrak {P}\) since \(\sigma ^{-1}(\alpha ) \not\in \mathfrak {P}\), but \(\mathfrak {p}\subset \mathfrak {P}\).
Let \(K/F\) be a normal extension of number fields and let \(\mathfrak {P}\) and \(\mathfrak {P}'\) be two primes lying above the prime \(\mathfrak {p}\). Then
Lets start by factoring \(\mathfrak {p}\) in \(\mathcal{O}_K\). We have
Now, by Theorem 3.6.6, if we apply \(\sigma \) to this equation we get
but then by the fact that for any \(i,j\) we can find \(\sigma \) such that \(\mathfrak {P}_i=\sigma (\mathfrak {P}_j)\) and uniqueness of factorization, we get that
which gives the first part of the result.
For the second part, we note that by the proof of Theorem 3.6.6 we have seen that
from which the result follows.
Let \(n\) be an integer and \(\zeta _n\) an \(n\)-th root of unity. Let \(K=\mathbb {Q}(\zeta _n)\) and \(p\) a prime number. Then
where \(\mathfrak {p}_i\) are the primes of \(\mathcal{O}_K\) over \(p\), which moreover all have the same inertial degree.
This follows from Corollary 3.6.7 by noting that \(K\) is a normal extension.
Next we have a theorem that tells us about which primes will ramify in an extension.
Let \(p\) be a prime number and \(K\) a number field. If \(p\) is ramified in \(\mathcal{O}_K\) then \(p \mid \Delta (\mathcal{O}_K)\).
Let \(\mathfrak {p}\) be a prime ideal in \(\mathcal{O}_K\) over \(p\) such that \(e_{\mathfrak {p}|p}{\gt}1\). Then we can write \((p)=\mathfrak {p}\mathfrak {a}\) where \(\mathfrak {a}\) is an ideal divisible by all prime ideals in \(\mathcal{O}_K\) which are above \(p\).
Now, let \(\alpha _1,\dots ,\alpha _n\) be a integral basis for \(\mathcal{O}_K\). Since \(\mathfrak {a}\) properly contains \((p)\) we can find some \(\alpha \in \mathfrak {a}\backslash (p)\). Then writing
with \(m_i \in \mathbb {Z}\) we see that since \(\alpha \not\in (p)\), one of the \(m_i\) must not be divisible by \(p\). So after relabelling, we can assume \(p \nmid m_1\). Now, similar to Proposition 2.2.4 we have that
Since \(p \nmid m_1\) its enough to show that \(p \mid \Delta (\alpha ,\alpha _2,\dots ,\alpha _n)\).
Now, let \(\sigma _1,\dots ,\sigma _n\) denote the embeddings of \(K\) into \(\mathbb {C}\). Let \(L\) be a finite extension of \(K\) such that \(L/\mathbb {Q}\) is normal (which we can do by Proposition 3.6.4). Let \(G:=\operatorname{Gal}(L/\mathbb {Q})\). Then the elements of \(G\) are just the embeddings of \(L\) extending the embeddings of \(K\) (if you dont remember what this means go to Definition 1.5.5)
Since \(\alpha \) is contained in \(\mathfrak {a}\), its contained in every prime ideal in \(\mathcal{O}_K\) over \(p\). It follows that \(\alpha \) is also in every prime ideal \(\mathfrak {P}\) in \(\mathcal{O}_L\) containing \(p\), since each such prime contains \(p\) and \(\mathfrak {P}\cap \mathcal{O}_K=\mathfrak {p}\) is again a prime ideal which contains \(p\), hence \(\mathfrak {p}\) lies over \(p\) and therefore contains \(\alpha \). Fix \(\mathfrak {P}\) a prime ideal of \(\mathcal{O}_L\) over \(p\). Then we claim that for any \(\sigma \in G\), \(\sigma (\alpha ) \in \mathfrak {P}\). To see this note that \(\sigma ^{-1}(\mathfrak {P})\) is again a prime ideal in \(\mathcal{O}_L\) over \(p\) and thus contains \(\alpha \). In particular, \(\sigma _i(\alpha ) \in \mathfrak {P}\) for all \(i\). Therefore, by Proposition 2.2.16 \(\Delta (\alpha ,\alpha _2,\dots ,\alpha _n) \in \mathfrak {P}\). But since the discriminant is an integer, we have \(\Delta (\alpha ,\alpha _2,\dots ,\alpha _n) \in \mathfrak {P}\cap \mathbb {Z}=p\mathbb {Z}\) and therefore \(p\) divides the discriminant.
Let \(K\) be a number field with \(K=\mathbb {Q}(\alpha )\) for \(\alpha \) a primitive element and let \(f \in \mathbb {Z}[x]\) be any monic polynomial such that \(f(\alpha )=0\). If \(p\) is a prime number such that \(p \nmid N_{K/\mathbb {Q}}(f'(\alpha ))\) then \(p\) is unramified.
There are only finitely many primes of \(\mathbb {Z}\) which ramify in a number field \(K\).
Going back to the cyclotomic example, one can determine the ramification and the inertial degree completely in this case.
Let \(n\) be a positive integer and \(\zeta _n\) an \(n\)-th root of unity. Let \(K=\mathbb {Q}(\zeta _n)\) and \(p\) a prime number. Write \(n=p^km\) with \(p \nmid m\) and set \(e=\varphi (p^k)\) (where \(\varphi \) is Euler’s Totient function). Lastly, let \(f\) be the (multiplicative) order of \(p\) modulo \(m\).
Then
(so \(e=e_{\mathfrak {p}_i|p}\)) and moreover \(f=f_{\mathfrak {p}_i|p}\).
We start by letting \(\alpha =\zeta ^m\) and \(\beta =\zeta ^{p^k}\). So \(\alpha \) is a \(p^k\)-th root of unity and \(\beta \) is a \(m\)-th root of unity. We will prove the theorem by seeing how \((p)\) factorizes in \(\mathbb {Q}(\alpha )\) and in \(\mathbb {Q}(\beta )\) and then combining the result.
With the above notation. In \(\mathbb {Q}(\alpha )\) we have
with \((1-\alpha )\) a prime ideal.
First note that \([\mathbb {Q}(\alpha ):\mathbb {Q}]=\varphi (p^k)\) since this is the degree of the \(p^k\) cyclotomic polynomial (See Lemma 2.3.1).
Next, from Exercise 2.3.6 we have \(p=u(1-\alpha )^{\varphi (p^k)}\), this tells us that (as ideals in \(\mathbb {Z}[\alpha ]\))
Theorem 3.5.7 then gives the result.
In \(\mathbb {Q}(\beta )\), \(p\) is unramified and for each \(\mathfrak {p}\) over \(p\), we have \(f_{\mathfrak {p}|p}=f\).
Using Corollary 3.6.10 with \(f(x)=x^m-1\), we need to check if \(p\) divides
But \(p \nmid m\) and \(\beta \) is a unit, so its norm is \(\{ \pm 1\} \) therefore \(p\) is unramified.
So in \(\mathbb {Z}[\beta ]\) we have
Moreover, since \(\mathbb {Q}(\beta )/\mathbb {Q}\) is normal we know that \(f_{\mathfrak {p}_i|p}=f_{\mathfrak {p}_j|p}\). So let \(\mathfrak {p}:=\mathfrak {p}_1\). Its enough to check that \(f_{\mathfrak {p}|p}=f\) (where \(f\) is as above- the multiplicative order of \(p\) modulo \(m\)).
Let \(\mathbb {F}_{\mathfrak {p}}:=\mathbb {Z}[\beta ]/\mathfrak {p}\). First, recall that \(f_{\mathfrak {p}|p}=[\mathbb {F}_\mathfrak {p}: \mathbb {F}_p]\). So \(f_{\mathfrak {p}|p}=h\) for some \(h\). We want to show that \(h=f\). We will do this in two steps, by first showing \(h \geq f\) and then \(h \leq f\).
(\(h \geq f\)): Let \(\overline{\beta }\) denote the image of \(\beta \) in \(\mathbb {F}_{\mathfrak {p}}\). We will first show that \(\overline{\beta }\) has order \(m\) in \(\mathbb {F}_{\mathfrak {p}}\). Assume for contradiction this is not the case, then \(\overline{\beta }^{m/l} \equiv 1 \pmod\mathfrak {p}\) (i.e. \(\mathfrak {p}\mid (\beta ^{m/l}-1)\)) for some prime factor \(l\) of \(m\). Now, \(\beta ^{m/l}\) is an \(l\)-th root of unity. Then, by Exercise 2.3.6 we see that \(\beta ^{m/l}-1\) divides \(l\) and \(l\) is a factor of \(m\). Thus \(\mathfrak {p}\mid m\). This gives a contradiction as \(\mathfrak {p}\mid p\) and \(\gcd (m,p)=1\).
So \(\overline{\beta }\) has order \(m\) in \(\mathbb {F}_{\mathfrak {p}}\). By Lagrange’s theorem this means \(m\mid N(\mathfrak {p})-1\) and note that since \(\mathfrak {p}\) is a prime ideal over \(p\), we have \(N(\mathfrak {p})=p^h\). Thus \(p^h \equiv 1 \pmod m\). But the order of \(p\) modulo \(m\) is \(f\), so we have \(h \geq f\).
(\(h \leq f\)): Since \(p^f \equiv 1 \pmod m\) it follows that \(\beta ^{p^f} = \beta \). Now, note that
This means that for all \(\gamma \in \mathbb {Z}[\beta ]\) we have
(also using FLT). Now since \((p) \subset \mathfrak {p}\) we see that
Therefore every non-zero element of \(\mathbb {F}_{\mathfrak {p}}\) is a root of \(x^{p^f}-x\). Now, since we are in a field, any polynomial of degree \(d\) has at most \(d\) roots. This polynomial has \(p^h\) roots (as this is the size of \(\mathbb {F}_{\mathfrak {p}}\)) therefore \(p^h \leq p^f \implies h \leq f\). This gives the result.
Lets now put these two lemmas together and finish proving the theorem.
Lemma 3.6.14 and Theorem 3.5.7 tells us that
Similarly, Lemma 3.6.15 gives
Now, if \(\mathfrak {P}_i\) denote the prime ideals of \(\mathbb {Z}[\zeta _n]\) over \(p\), then we know that
Lastly, using Proposition 3.5.6 together with
gives
from which the result follows.
This theorem is sometimes called the cyclotomic reciprocity law, since it allows us to quickly find out how primes factor in cyclotomic fields, just like quadratic reciprocity is used for quadratic fields. For general number fields, we know of no such simple description.
Lets use this theorem to see how primes factorize in \(\mathbb {Q}(\zeta _5)\).
|
\(p \mod 5\) |
Order of \(p \mod m\) |
Factorization of \((p)\) |
Norms |
|
\(0\) |
- |
\((p)=\mathfrak {p}^4\) |
\(N(\mathfrak {p})=p\) |
|
\(1\) |
\(1\) |
\((p)=\mathfrak {p}_1\mathfrak {p}_2\mathfrak {p}_3\mathfrak {p}_4\) |
\(N(\mathfrak {p}_i)=p\) |
|
\(2\) |
\(4\) |
\((p)\) |
\(N((p))=p^4\) |
|
\(3\) |
\(4\) |
\((p)\) |
\(N((p))=p^4\) |
|
\(4\) |
\(2\) |
\((p)=\mathfrak {p}_{1}\mathfrak {p}_2\) |
\(N(\mathfrak {p}_i)=p^2\) |
We can use this to say things about how primes factor in extensions of \(\mathbb {Q}(\zeta _5)\). For example, in \(\mathbb {Q}(\zeta _{55})\) we see that
with \(f_{\mathfrak {p}_i|p}=1\).
Describe the factorization of the following ideals into prime ideals in \(\mathbb {Q}(\zeta _{55})\):
Let \(\zeta _7\) be a \(7\)-th root of unity and \(K=\mathbb {Q}(\zeta _7)\). Complete the following table describing the decomposition of ideals \((p)\) (with \(p\) a prime number) in \(\mathcal{O}_K\).
|
\(p \mod 7\) |
Order of \(p \mod m\) |
Factorization of \((p)\) |
Norms |
|
\(0\) |
|||
|
\(1\) |
|||
|
\(2\) |
|||
|
\(3\) |
|||
|
\(4\) |
|||
|
\(5\) |
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|
\(6\) |