CHAPTER 3 Prime Splitting

In this chapter we will investigate how to explicitly factor ideals in rings of integers of number fields. A common theme will be considering ideals of one ring in another. Specifically, we will often have the following situation: K and L will be number fields with K ⊆ L (so O_{K} ⊆ O_{L}), 𝔞 will be an ideal of O_{K}, and we will consider the ideal 𝔞O_{L} of O_{L} generated by 𝔞. We will be especially interested in the case where 𝔞 = 𝔭 is a prime of O_{K}; determining how 𝔭O_{L} factors into primes of O_{K} (even though 𝔭 is prime in O_{K}, it doesn’t still have to be prime in O_{L}) will be the key to our factorization results.

One other construction which is sometimes useful is to take a prime 𝔓 of O_{L} and to consider the ideal 𝔓 ∩ O_{K} of O_{L}; this ideal is necessarily prime since there is an injection

O_{K}/𝔓 ∩ O_{K} ↪ O_{L}/𝔓

and subrings of integral domains are again integral domains.

Very often we will be considering the case of an extension K/ℚ, in which case the relevant ideals are of the form pO_{K} for rational primes p. Note that we could also write these ideals as (p), assuming that the ring the principal ideal is formed in is clear from context.

1. Example : Quadratic number fields

1.1. The prime ideals. Before we restrict to the case of quadratic number fields, we prove the following useful fact about prime ideals.

Lemma 1.1. Let K be a number field and let 𝔭 be a non-zero prime ideal of O_{K}. Then 𝔭 contains a rational prime.

Proof. By Lemma II.3.2 we know that 𝔭 contains a non-zero integer. Let n be the smallest positive integer in 𝔭; n is not 1 since 𝔭 ≠ O_{K}. Suppose that n is not prime in ℤ. Then n factors as ab for some a, b ∈ ℤ^{+}; since 𝔭 is a prime ideal it must contain at least one of a and b. But both of these factors are smaller than n, which contradicts the definition of n. Thus n must be prime, which proves the lemma.

Lemma 1.1 tells us that every prime of O_{K} contain a rational prime; it then follows from Lemma II.3.8 that all non-zero primes of O_{K} divide an ideal of the form pO_{K} for some prime p of ℤ. In particular, we can determine all primes of O_{K} simply by determining the factorization of these ideals pO_{K}. Our main results will be explicit determinations of these factorizations.

In the calculations below we will be working with polynomials both in ℤ[x] and in 𝔽_{p}[x]. We will write ḡ(x) for polynomials in 𝔽_{p}[x], and we will then let g(x) denote some choice of a polynomial in ℤ[x] reducing modulo p to ḡ(x); the precise choice of g(x) will never matter.

We now restrict to the case of quadratic number fields. Let K = ℚ(\sqrt{d}) be a quadratic number field, where d is a squarefree integer. Set α = \sqrt{d} if d ≡ 2, 3 (mod 4) and α = \frac{1+\sqrt{d}}{2} if d ≡ 1 (mod 4), so that O_{K} = ℤ[α]. Let f(x) be the minimal polynomial of α, so that f(x) = x^{2} − d if d ≡ 2, 3 (mod 4) and f(x) = x^{2} − x + \frac{1−d}{4} if d ≡ 1 (mod 4).

Let p be a rational prime. To determine the factorization of the ideal pO_{K} in O_{K} we will compute in an easier setting. Specifically, pO_{K} is the kernel of the map

O_{K} → O_{K}/pO_{K},

and we will find a second expression for this kernel. To do this, recall that O_{K} ≅ ℤ[x]/(f(x)), where under the isomorphism α corresponds to x. We now have

O_{K}/pO_{K} ≅ (ℤ[x]/f(x))/p ≅ ℤ[x]/(p, f(x)).

This last ring is isomorphic to

𝔽_{p}[x]/(f̄(x)),

and here we can finally compute easily.

There are three possibilities for the factorization of f̄(x) in 𝔽_{p}[x]. First of all, f̄(x) could be irreducible. Second, f̄(x) could factor as a product of distinct, monic linear (and therefore irreducible) polynomials. Third, f̄(x) could factor as the square of a single monic linear polynomial. We will consider all three cases separately.

Suppose first that f̄(x) is irreducible in 𝔽_{p}[x]. Then 𝔽_{p}[x]/(f̄(x)) is a field, so O_{K}/pO_{K} is as well. 𝔭O_{K} is therefore a prime ideal of O_{K}, by the definition of prime ideal, so it does not factor any further.

Before we do the next case, let us determine exactly what all of these maps are, since this will be important in determining the kernel. The sequence of maps is

\xymatrix{O_{K} \ar[r] \ar[d]^{≅} & O_{K}/pO_{K} \ar[d]^{≅} \\ ℤ[x]/(f(x)) \ar[r] & ℤ[x]/(p, f(x)) \ar[r]^{≅} & 𝔽_{p}[x]/(f̄(x))}

Both of the horizontal maps of the square are the natural quotient maps, and the two vertical isomorphisms send α to x. The last horizontal isomorphism simply sends x to x. (All maps always send 1 to 1, of course, which determines what happens to all of ℤ.) The ideal pO_{K} has now been expressed as the kernel of the map

O_{K} → 𝔽_{p}[x]/(f̄(x))

sending α to x.

Suppose now that f̄(x) factors as ḡ(x)h̄(x) in 𝔽_{p}[x], where ḡ(x) and h̄(x) are distinct, monic, linear polynomials. Then the Chinese remainder theorem (see Exercise 3.5) gives an isomorphism

𝔽_{p}[x]/(f̄(x)) ⟶^{≅} 𝔽_{p}[x]/(ḡ(x)) × 𝔽_{p}[x]/(h̄(x))

sending x to (x, x). Note that both of these factors are fields since ḡ(x) and h̄(x) are irreducible in 𝔽_{p}[x]; in fact, they are both isomorphic to 𝔽_{p}.

Consider now the composite map

O_{K} → 𝔽_{p}[x]/(ḡ(x)) × 𝔽_{p}[x]/(h̄(x))

sending α to (x, x), which still has kernel pO_{K}. The kernel into the first factor is the ideal (p, g(α)) of O_{K} (since α maps to x in this factor), and the kernel into the second factor is (p, h(α)). Thus the kernel of the map (which by construction is just pO_{K}) can also be written as

(p, g(α)) ∩ (p, h(α));

however, these ideals are easily seen to be relatively prime (see Exercise 3.3), so Exercise 3.4 shows that

(p, g(α)) ∩ (p, h(α)) = (p, g(α))(p, h(α)).

Furthermore, both of these ideals are prime, since as we saw above

O_{K}/(p, g(α)) ≅ 𝔽_{p}[x]/(ḡ(x)), O_{K}/(p, h(α)) ≅ 𝔽_{p}[x]/(h̄(x))

are fields. Thus we have determined the prime factorization

pO_{K} = (p, g(α))(p, h(α))

of pO_{K}. (Note that it does not matter which lifts of ḡ(x) and h̄(x) are chosen, since any two lifts differ by multiples of p and p lies in these ideals.)

The last case is the case where f̄(x) = ḡ(x)^{2} for some monic, linear polynomial ḡ(x) ∈ 𝔽_{p}[x]. In this case the above analysis does not quite work since the Chinese remainder theorem does not apply. However, it suggests strongly that

pO_{K} = (p, g(α))^{2},

and this we can check directly. We check the case where d ≡ 2, 3 (mod 4) and p ≠ 2; the other cases are similar. We have that f̄(x) = x^{2} − d is a square in 𝔽_{p}[x]. One checks easily that this implies that p divides d (since p ≠ 2), so we can take g(x) = x. Thus the claim is that

pO_{K} = (p, α)^{2}.

To check this, we simply compute

(p, α)(p, α) = (p^{2}, pα, α^{2}).

Since α^{2} = d is divisible by p, every generator of the ideal is divisible by p. Furthermore, since d is squarefree, p^{2} does not divide d; it follows that p is a linear combination of p^{2} and d, so that p lies in the ideal. Thus the ideal is simply pO_{K} as claimed.

We summarize our results in a proposition.

Proposition 1.2. Let K = ℚ(\sqrt{d}) be a quadratic number field with d a squarefree integer and let

f(x) = \begin{cases}x^{2} − d & d ≡ 2, 3 (mod 4); \\ x^{2} − x + \frac{1−d}{4} & d ≡ 1 (mod 4). \end{cases}

Let p be a rational prime and let

f̄(x) = ḡ_{1}(x)^{e_{1}} ⋯ ḡ_{r}(x)^{e_{r}}

be the factorization of f̄(x) in 𝔽_{p}[x]. (Of course, we have r = 1 or 2 and e_{i} = 1 or 2.) Then the factorization of pO_{K} into primes of O_{K} is

pO_{K} = (p, g_{1}(α))^{e_{1}} ⋯ (p, g_{r}(α))^{e_{r}}.

We will say that pO_{K} is inert in K if it is a prime ideal in O_{K}; that it splits in K if it is a product of distinct prime ideals in O_{K}; and that it ramifies in K if it is the square of a prime ideal. Our above results show that pO_{K} is inert if and only if f(x) is irreducible modulo p; it splits if and only if f(x) factors into distinct linear factors; and it ramifies if and only if f(x) is the square of a linear polynomial. Determination of the ramification in a number field turns out to be of great importance, and our above analysis yields the following result.

Corollary 1.3. Let K = ℚ(\sqrt{d}) be a quadratic number field with d a squarefree integer. If d ≡ 1 (mod 4), then a prime of ℤ is ramified in O_{K} if and only if it divides d. If d ≡ 2, 3 (mod 4), then a prime of ℤ is ramified in O_{K} if and only if it is 2 or it divides d.

Proof. First take d ≡ 2, 3 (mod 4). Modulo 2 we have

x^{2} − d ≡ (x − d)^{2} (mod 2),

so p = 2 is always ramified. It was shown above that otherwise ramification occurs if and only if p divides d, which completes the analysis in this case.

When d ≡ 1 (mod 4), x^{2} −x+ \frac{1−d}{4} is never a square modulo 2, since all squares have no linear term. Thus p = 2 never ramifies. The fact that all p dividing d do ramifiy follows from the determination of the roots of x^{2} −x+ \frac{1−d}{4} by the quadratic formula; we leave the details to the reader.

Example 1.4. Let K = ℚ(\sqrt{−5}) and O_{K} = ℤ[\sqrt{−5}]. Let us factor the first few primes. First take p = 2. Then

x^{2} + 5 ≡ (x + 1)^{2} (mod 2),

so 2O_{K} ramifies:

2O_{K} = (2, \sqrt{−5} + 1)^{2}.

For p = 3, we have

x^{2} + 5 ≡ (x + 1)(x + 2) (mod 3),

so 3O_{K} splits as

3O_{K} = (3, \sqrt{−5} + 1)(3, \sqrt{−5} + 2).

(Note that these factorizations agree with those in Chapter 2, Section 1.4.) For p = 5,

x^{2} + 5 ≡ x^{2} (mod 5),

so

5O_{K} = (5, \sqrt{−5})^{2}.

Note that the second ideal is just the principal ideal (\sqrt{−5}), since \sqrt{−5} divides 5 in ℤ[\sqrt{−5}]. This illustrates the general fact that the above algorithm does not tell you whether or not the factors are principal ideals. Note also that 2O_{K} and 5O_{K} are the only primes which ramify in K, as we proved above that in this case that either p = 2 or p divides d = −5.

Continuing,

x^{2} + 5 ≡ (x + 3)(x + 4) (mod 7),

so

7O_{K} = (7, \sqrt{−5} + 3)(7, \sqrt{−5} + 4).

Next, x^{2} + 5 is irreducible in 𝔽_{11}[x], so 11O_{K} is still a prime ideal in ℤ[\sqrt{−5}]. For a final example, take p = 29. Then

x^{2} + 5 = (x + 13)(x + 16) (mod 29),

so

29O_{K} = (29, \sqrt{−5} + 13)(29, \sqrt{−5} + 16).

In this case, however, we also find the element factorization

29 = (3 − 2\sqrt{−5})(3 + 2\sqrt{−5}).

One can show with a little calculation that

(29, \sqrt{−5} + 13) = (3 − 2\sqrt{−5})
(29, \sqrt{−5} + 16) = (3 + 2\sqrt{−5}),

so we actually have the ideal factorization

(29) = (3 − 2\sqrt{−5})(3 + 2\sqrt{−5})

of (29) into principal ideals.

1.2. Quadratic forms. There are many deep and important connections between quadratic forms and the splitting of primes in quadratic fields. In this section we investigate some of the simplest; for more information on this subject, see [7].

We begin by refining our results of the previous section. Let K = ℚ(\sqrt{d}) and let p be a a rational prime. Recall that the behavior of the prime ideal pO_{K} was determined by the factorization of a certain polynomial in 𝔽_{p}[x]. The various behaviors of pO_{K} are captured well by the Legendre symbol.

Definition 1.5. Let p be an odd prime. We define the Legendre symbol

(\frac{·}{p}) : ℤ/pℤ → \{0, ±1\}

by

(\frac{a}{p}) = \begin{cases}1 & if a is a non-zero square modulo p; \\ 0 & if a ≡ 0 (mod p); \\ −1 & if a is not a square modulo p. \end{cases}

By abuse of notation we use the same symbol for a ∈ ℤ.

The fundamental properties of the Legendre symbol are in Exercise 3.9 and Exercise 3.10. It is possible to extend the definition of the Legendre symbol to include the case p = 2, but we will not do so here.

Proposition 1.6. Let K = ℚ(\sqrt{d}) be a quadratic number field and let p be an odd rational prime. Then the prime ideal pO_{K} splits in K if and only if (\frac{d}{p}) = 1; it

ramifies in K if and only if (\frac{d}{p}) = 0; and it is inert in K if and only if (\frac{d}{p}) = −1.

Proof. First suppose that d ≡ 2, 3 (mod 4). Then the behavior of pO_{K} is determined by the factorization of the polynomial x^{2} −d in 𝔽_{p}[x]. If d ≡ 0 (mod p), then this polynomial has the repeated factor x, so pO_{K} ramifies; if d is a non-zero square modulo p, then it splits into distinct linear factors, so pO_{K} splits; and if d is not a square modulo p, then it does not factor, so pO_{K} is inert. This is precisely the statement of the proposition in this case.

Now take d ≡ 1 (mod 4). This time the behavior is determined by the factorization of the polynomial x^{2} − x + \frac{1−d}{4} in 𝔽_{p}[x]. By the quadratic formula (which applies since p ≠ 2), this polynomial has roots

\frac{1 ± \sqrt{d}}{2},

from which the same analysis as above proves the proposition.

We now turn to quadratic forms. Recall that in the cases of ℚ(i),ℚ(\sqrt{−2}) and ℚ(ζ_{3}) = ℚ(\sqrt{−3}) the natural quadratic form to study were the “norm forms” x^{2} + y^{2}, x^{2} + 2y^{2} and x^{2} − xy + y^{2}. If d ≡ 2, 3 (mod 4), then the appropriate quadratic form is

N_{K/ℚ}(x + y\sqrt{d}) = x^{2} − dy^{2},

while if d ≡ 1 (mod 4), then it is

N_{K/ℚ} (x + y \frac{1 + \sqrt{d}}{2}) = x^{2} + xy + \frac{1 − d}{4}y^{2}.

(We get a different quadratic form for ℚ(\sqrt{−3}) here because we are using a different generator of the field; we nevertheless will obtain an equivalent result.) Write q_{K}(x, y) for the quadratic form attached to K. Then an integer n can be written as q_{K}(x, y) if and only if n is the norm of some element of O_{K}.

Note in particular that we can not at the moment study “natural” quadratic forms like x^{2} +3y^{2}; the correct quadratic form for ℚ(\sqrt{−3}) is x^{2} +xy+y^{2}. To study other quadratic forms one must work in certain subrings of O_{K}, where factorization is more complicated; we won’t go into it here.

The basic result is the following.

Proposition 1.7. Let K = ℚ(\sqrt{d}) be a quadratic number field and let q_{K}(x, y) be its norm quadratic form. Let p be a positive rational prime number. Then (at least) one of ±p is of the form q_{K}(x_{0}, y_{0}) for some x_{0}, y_{0} ∈ ℤ if and only if p splits (or ramifies) in K into principal ideals. If d < 0, then it is always p which is of this form, and never −p.

Proof. Suppose first that p factors into principal ideals in O_{K}, say pO_{K} = (π)(π′). Then ππ′ is an associate of p, say ππ′ = up for some unit u. Thus

N(π) N(π′) = N(u) N(p) = ±p^{2}

by Lemma II.1.9. It follows that N(π) = ± N(π′) = ±p, which gives the desired expression for p.

Now suppose that ±p is of the form q_{K}(x_{0}, y_{0}) for some x_{0}, y_{0} ∈ ℤ. By the definition of q_{K}(x, y),

±p = N_{K/ℚ}(x_{0} + y_{0}α),

where α is \sqrt{d} or \frac{1+\sqrt{d}}{2}, as appropriate. This implies that

pO_{K} = (x_{0} + y_{0}α)(x_{0} + y_{0}ᾱ),

where ᾱ is the conjugate of α. Thus pO_{K} splits (or possibly ramifies) into principal ideals. The fact that p must be positive for d < 0 follows immediately from the fact that q_{K}(x, y) is positive definite in that case.

Corollary 1.8. With the above notation, ±p = q_{K}(x_{0}, y_{0}) for some x_{0}, y_{0} ∈ ℤ only if (\frac{d}{p}) = 0 or 1. The converse is also true if O_{K} is a PID.

In non-PID cases our answer is still far from complete. We will return to this question and give some surprising results in the next chapter.

2. Abstract factorization of primes

2.1. Factorization of rational primes. Before we extend the methods of Section 1 to more general number fields, it will be useful to give the abstract factorization results. We begin with the factorization of rational primes in number fields. Let K be a number field of degree n. If 𝔭 is a prime ideal of O_{K} and p is a rational prime, we say that 𝔭 lies above p if 𝔭 ∩ ℤ = pℤ. It is clear from Lemma 1.1 that every non-zero prime of O_{K} lies above a unique prime of ℤ, and it follows from Lemma II.3.8 that the primes of O_{K} lying above p are precisely those ideals occurring in the prime factorization of pO_{K}.

Let 𝔭 be a prime of O_{K} lying over p ∈ ℤ. Let e be the exact power of 𝔭 dividing pO_{K}. We call e the ramification index of 𝔭/p and write it as e(𝔭/p). The factorization of pO_{K} is thus

pO_{K} = ∏_{𝔭∩ℤ=pℤ} 𝔭^{e(𝔭/p)}.

We will also need a way to measure the relative “sizes” of ideals. The most natural way to do this is to consider the residue field O_{K}/𝔭, which we proved earlier is a finite field. Since it clearly has characteristic p, it must have order p^{f} for some f. We define the inertial degree f(𝔭/p) of 𝔭/p to be this integer f.

Example 2.1. Take K = ℚ(\sqrt{−5}). Our calculations in Example 1.4 yield the following values for e and f:

\begin{aligned}e((2, \sqrt{−5} + 1)/2) & = & 2 & f((2, \sqrt{−5} + 1)/2) & = & 1 \\ e((3, \sqrt{−5} + 1)/3) & = & 1 & f((3, \sqrt{−5} + 1)/3) & = & 1 \\ e((3, \sqrt{−5} + 2)/3) & = & 1 & f((3, \sqrt{−5} + 2)/3) & = & 1 \\ e((5, \sqrt{−5})/5) & = & 2 & f((5, \sqrt{−5})/5) & = & 1 \\ e((7, \sqrt{−5} + 3)/7) & = & 1 & f((7, \sqrt{−5} + 4)/7) & = & 1 \\ e((11)/11) & = & 1 & f((11)/11) & = & 2 \end{aligned}

The values of f can be computed using expressions for quotients of 𝔽_{p}[x]; for example,

O_{K}/(3, \sqrt{−5} + 1) ≅ 𝔽_{3}[x]/(x + 1) ≅ 𝔽_{3}.

It is useful to have a notion of the size of an ideal for non-prime ideals. Since in this case we can no longer isolate a specific rational prime of interest, we define the norm N′_{K/ℚ}(𝔞) of an ideal 𝔞 to be the size of the quotient ring O_{K}/𝔞; that this is finite follows easily from Lemma II.3.2 and Theorem II.2.22. (We will prove later that the ideal norm agrees with the absolute value of the usual element norm in the case that 𝔞 is a principal ideal; hopefully this notation should cause no confusion until then.) It follows immediately from the definition of inertial degree that if 𝔭 is a prime of O_{K} such that 𝔭 ∩ ℤ = pℤ, then

N′_{K/ℚ}(𝔭) = p^{f(𝔭/p)}.

A first indication that the ideal norm behaves like the usual norm is given by the following lemma.

Lemma 2.2. The ideal norm is multiplicative; that is,

N′_{K/ℚ}(𝔞𝔟) = N′_{K/ℚ}(𝔞) N′_{K/ℚ}(𝔟)

for any non-zero ideals 𝔞, 𝔟 of O_{K}.

Proof. Suppose first that 𝔞 and 𝔟 are relatively prime. Then by the Chinese remainder theorem (see Exercise 3.5) we have

O_{K}/𝔞𝔟 ≅ O_{K}/𝔞 × O_{K}/𝔟,

from which the lemma follows immediately. It will therefore be enough to prove that for any prime 𝔭 of O_{K} we have

N′_{K/ℚ}(𝔭^{m}) = N′_{K/ℚ}(𝔭)^{m};

the lemma will then follow from unique factorization of ideals and the relatively prime case.

Note that it is immediate from elementary group theory that

N′_{K/ℚ}(𝔭^{m}) = #(O_{K}/𝔭^{m}) = #(O_{K}/𝔭) · #(𝔭/𝔭^{2}) · #(𝔭^{2}/𝔭^{3}) ⋯ #(𝔭^{m−1}/𝔭^{m}).

(All quotients here are simply as abelian groups.) Thus it will suffice to show that

#(𝔭^{k}/𝔭^{k+1}) = #(O_{K}/𝔭)

for any k. To do this let γ be any element of 𝔭^{k} which does not lie in 𝔭^{k+1}; such a γ must exist since if the containment 𝔭^{k} ⊇ 𝔭^{k+1} were an equality it would violate unique factorization of ideals. We claim that the map

\xymatrix{O_{K}/𝔭 \ar[r] & 𝔭^{k}/𝔭^{k+1} \\ α \ar@{|->}[r] & γα}

is an isomorphism; we leave the details to the reader.

We can now give the fundamental relationship between the numbers e(𝔭/p), f(𝔭/p) and the degree n of K/ℚ.

Proposition 2.3. Let K be a number field of degree n and let p be a rational prime. Let

pO_{K} = 𝔭_{1}^{e_{1}} ⋯ 𝔭_{r}^{e_{r}}

be the factorization of pO_{K} into primes of O_{K}. (Thus e_{i} = e(𝔭_{i}/p).) Set f_{i} = f(𝔭_{i}/p). Then

∑_{i=1}^{r} e_{i}f_{i} = n.

Proof. Both pO_{K} and 𝔭_{1}^{e_{1}} ⋯ 𝔭_{r}^{e_{r}} have the same norm, so by Lemma 2.2
N′_{K/ℚ}(pO_{K}) = N′_{K/ℚ}(𝔭_{1})^{e_{1}} ⋯ N′_{K/ℚ}(𝔭_{r})^{e_{r}}
= p^{f_{1}e_{1}} ⋯ p^{f_{r}e_{r}}
= p^{f_{1}e_{1}+···+f_{r}e_{r}}.

On the other hand, we know that O_{K}/pO_{K} has p^{n} elements, since O_{K} is a free ℤmodule of rank n. Thus N′_{K/ℚ}(pO_{K}) = p^{n} and the proposition follows immediately.

Example 2.4. Let K be a quadratic number field and let p be a rational prime. We saw that there were three possibilities for the factorization of p: first of all, pO_{K} could still be prime, in which case f(pO_{K}/p) = 2 and e(pO_{K}/p) = 1. Next, pO_{K} could factor as 𝔭_{1}𝔭_{2} where f(𝔭_{1}/p) = f(𝔭_{2}/p) = 1 and e(𝔭_{1}/p) = e(𝔭_{2}/p) = 1. Lastly, pO_{K} could ramify as 𝔭^{2}, in which case f(𝔭/p) = 1 and e(𝔭/p) = 2. In all three cases we do indeed have the equality of Proposition 2.3.

2.2. Localizations of integer rings. Before we extend the above results to the case of relative extensions we will need to introduce an additional piece of machinery. Very often in algebraic number theory it is convenient to work “one prime at a time”. More precisely, given a number field K with ring of integers O_{K} and given a prime 𝔭, we would like to find a larger subring of K in which the only non-zero prime ideal is 𝔭. We will construct such a ring in this section.

The definition of the ring is actually fairly simple. We define the local ring of O_{K} at 𝔭 to be the ring

O_{K,𝔭} = \{\frac{α}{β} ∈ K | α ∈ O_{K}, β ∈ O_{K} − 𝔭\} .

That is, O_{K,𝔭} consists of all elements of K which can be written as ratios of integers with the denominator not in 𝔭. (The terminology here is influenced by algebraic geometry. In fact, in a suitably general setting one can think of a ring of integers O_{K} as a curve, where the points are the non-zero prime ideals. In this setting the ring O_{K,𝔭} really is the ring of regular functions on the curve O_{K} at the point 𝔭.) One can check easily that O_{K,𝔭} is actually a ring.

Example 2.5. Let K = ℚ. Then the local ring ℤ_{(p)} is simply the subring of ℚ of rational numbers with denominator relatively prime to p. (We considered such rings in Chapter 2, Section 1.) Note that this ring ℤ_{(p)} is not the ring ℤ_{p} of p-adic integers; to get ℤ_{p} one must complete ℤ_{(p)}, which is a process which we will not go into here.

The usefulness of O_{K,𝔭} comes from the fact that it has a particularly simple ideal structure. Let 𝔞 be any proper ideal of O_{K,𝔭} and consider the ideal 𝔞 ∩ O_{K} of O_{K}. We claim that

𝔞 = (𝔞 ∩ O_{K})O_{K,𝔭};

that is, that 𝔞 is generated by the elements of 𝔞 in 𝔞 ∩ O_{K}. It is clear from the definition of an ideal that 𝔞 ⊇ (𝔞 ∩ O_{K})O_{K,𝔭}. To prove the other inclusion, let α be any element of 𝔞. Then we can write α = β/γ where β ∈ O_{K} and γ ∉ 𝔭. In particular, β ∈ 𝔞 (since β/γ ∈ 𝔞 and 𝔞 is an ideal), so β ∈ 𝔞∩O_{K}. Since 1/γ ∈ O_{K,𝔭}, this implies that α = β/γ ∈ (𝔞 ∩ O_{K})O_{K,𝔭}, as claimed.

We can use this fact to determine all of the ideals of O_{K,𝔭}. Let 𝔞 be any ideal of O_{K,𝔭} and consider the ideal factorization of 𝔞 ∩ O_{K} in O_{K}. Write it as

𝔞 ∩ O_{K} = 𝔭^{n}𝔟

for some n and some ideal 𝔟, relatively prime to 𝔭. We claim first that 𝔟O_{K,𝔭} = O_{K,𝔭}. To see this, note that 𝔟 is not contained in 𝔭 since 𝔟 is assumed to be relatively prime to 𝔭, and thus is not divisible by it. In particular, 𝔟 contains elements of O_{K} − 𝔭; these are units in O_{K,𝔭}, so 𝔟O_{K,𝔭} = O_{K,𝔭}.

We now find that

𝔞 = (𝔞 ∩ O_{K})O_{K,𝔭} = 𝔭^{n}𝔟O_{K,𝔭} = 𝔭^{n}O_{K,𝔭}

since 𝔟O_{K,𝔭} = O_{K,𝔭}. Thus every ideal of O_{K,𝔭} has the form 𝔭^{n}O_{K,𝔭} for some n; it follows immediately that O_{K,𝔭} is noetherian.

It is also now clear that 𝔭O_{K,𝔭} is the unique non-zero prime ideal in O_{K,𝔭}. Furthermore, the inclusion O_{K} ↪ O_{K,𝔭} induces an injection

O_{K}/𝔭 ↪ O_{K,𝔭}/𝔭O_{K,𝔭}

since 𝔭O_{K,𝔭} ∩O_{K} = 𝔭, as the reader can easily check. This map is also a surjection, since the residue class of α/β ∈ O_{K,𝔭} (with α ∈ O_{K} and β ∉ 𝔭) is the image of αβ^{−1} in O_{K}/𝔭, which makes sense since β is invertible in O_{K}/𝔭. Thus the map is an isomorphism. In particular, it is now abundantly clear that every non-zero prime ideal of O_{K,𝔭} is maximal.

To show that O_{K,𝔭} is a Dedekind domain, it remains to show that it is integrally closed in K. So let γ ∈ K be a root of a polynomial with coefficients in O_{K,𝔭}; write this polynomial as

x^{m} + \frac{α_{m−1}}{β_{m−1}} x^{m−1} + ⋯ + \frac{α_{0}}{β_{0}}

with α_{i} ∈ O_{K} and β_{i} ∈ O_{K} − 𝔭. Set β = β_{0}β_{1} ⋯ β_{m−1}. Multiplying by β^{m} we find that βγ is the root of a monic polynomial with coefficients in O_{K}. Thus βγ ∈ O_{K}; since β ∉ 𝔭, we have βγ/β = γ ∈ O_{K,𝔭} Thus O_{K,𝔭} is integrally closed in K.

Let us summarize our results in a proposition.

Proposition 2.6. Let K be a number field and let 𝔭 be a non-zero prime of O_{K}. Then O_{K,𝔭} is a Dedekind domain and every ideal of O_{K,𝔭} has the form 𝔭^{n}O_{K,𝔭} for some n ≥ 0. In particular, 𝔭O_{K,𝔭} is the only prime ideal of O_{K,𝔭}.

We have now shown that O_{K,𝔭} is a Dedekind domain with a unique non-zero prime ideal. Such a ring is called a discrete valuation ring or a DVR. These rings will be useful to us for the following reason.

Proposition 2.7. Let R be a discrete valuation ring. Then R is a principal ideal domain.

Proof. Let 𝔭 be the unique non-zero prime ideal of R. By unique factorization of ideals, every ideal of R has the form 𝔭^{n} for some n; thus it will suffice to show that 𝔭 itself is principal. Let π be any element in 𝔭 but not in 𝔭^{2}. By unique factorization of ideals, we have πR = 𝔭^{n} for some n ≥ 1. But we can not have n ≥ 2, since then π would lie in 𝔭^{2}. Thus πR = 𝔭, so 𝔭 is indeed principal.

A generator of the unique non-zero prime ideal of a DVR R is called a uniformizer. R has a particularly simple sort of unique factorization: every α ∈ R can be written as uπ^{n}, where u ∈ R^{∗} and n ≥ 0.

The usefulness of O_{K,𝔭} comes from the fact that it has all of the information about the prime 𝔭, but it has no other prime ideals to clutter things up. This makes O_{K,𝔭} much simpler than O_{K}, but also still useful for studying the prime 𝔭.

Example 2.8. Let p be a rational prime and consider the ring ℤ_{(p)}. The units of this ring are

ℤ_{(p)}^{∗} = \{\frac{m}{n} | (m, p) = (n, p) = 1\} .

The unique prime ideal of ℤ_{(p)} is (p), and every α ∈ ℤ_{(p)} can be written uniquely as

α = up^{n}

where u ∈ ℤ_{(p)}^{∗} and n ≥ 0.

Now let L/K be an extension of number fields of degree n. Let 𝔭 be a prime ideal of O_{K}; denote by O_{L,𝔭} the ring

\{\frac{α}{β} ∈ L | α ∈ O_{L}, β ∈ O_{K} − 𝔭\} .

O_{L,𝔭} is not quite a discrete valuation ring, since we allow only denominators in O_{K} − 𝔭, but it will suffice for our purposes. We will need to know two key facts about O_{L,𝔭}. First of all, note that 𝔭O_{L,𝔭} ∩ O_{L} = 𝔭O_{L}. Thus the natural inclusion O_{L} ↪ O_{L,𝔭} induces an injection

O_{L}/𝔭O_{L} ↪ O_{L,𝔭}/𝔭O_{L,𝔭}.

We claim that this map is an isomorphism. So let α/β represent a residue class in the range, where α ∈ O_{L} and β ∈ O_{K} − 𝔭. We know that O_{K}/𝔭 injects into O_{L}/𝔭O_{L}; thus β is invertible in O_{L}/𝔭O_{L}. It follows that the element αβ^{−1} is welldefined in O_{L}/𝔭O_{L}, and it maps to α/β. Thus the map is surjective, and therefore an isomorphism.

The second fact we need is that O_{L,𝔭} is a free O_{K,𝔭}-module of rank n. The proof of this fact is exactly the same as the proof that O_{L} is a free O_{K}-module of rank n. One begins with a basis α_{1}, …, α_{n} for L/K with each α_{i} ∈ O_{L,𝔭}. Next, the discriminant Δ = Δ(α_{1}, …, α_{n}) lies in O_{K,𝔭} (using Lemma II.2.18). The same proof as in Proposition II.2.21 shows that

O_{L,𝔭} ⊆ \frac{1}{Δ}(O_{K,𝔭}α_{1} + … + O_{K,𝔭}α_{n}),

and we obviously have

O_{K,𝔭}α_{1} + ⋯ + O_{K,𝔭}α_{n} ⊆ O_{L,𝔭}.

Since O_{K,𝔭} is a PID, these two facts combine to show that O_{L,𝔭} is a free O_{K,𝔭}- module of rank n. (See Appendix C, Section 5.)

The fact we actually need from all of this is contained in the next proposition.

Proposition 2.9. Let L/K be an extension of number fields of degree n and let 𝔭 be a non-zero prime of O_{K}. Then

#(O_{L}/𝔭O_{L}) = (#(O_{K}/𝔭O_{K}))^{n}.

Proof. We saw above that O_{L,𝔭} is a free O_{K,𝔭}-module of rank n. Thus O_{L,𝔭}/𝔭O_{L,𝔭} is a free O_{K,𝔭}/𝔭O_{K,𝔭}-module of rank n as well. Therefore

#(O_{L,𝔭}/𝔭O_{L,𝔭}) = #(O_{K,𝔭}/𝔭O_{K,𝔭})^{n}.

But these residue rings are isomorphic to O_{L}/𝔭O_{L} and O_{K}/𝔭O_{K} respectively, so the proposition follows.

Corollary 2.10. Let L/K be an extension of number fields of degree n and let 𝔞 be a nonzero ideal of O_{K}. Then

N′_{L/ℚ}(𝔞O_{L}) = N′_{K/ℚ}(𝔞)^{n}.

Proof. Since each side of the desired equality is multiplicative in 𝔞, it will suffice to prove the result in the case that 𝔞 = 𝔭 is a prime of O_{K}. In this case the corollary is precisely Proposition 2.9.

Corollary 2.11. Let K be a number field of degree n and let α be in O_{K}. Then

N′_{K/ℚ}(αO_{K}) = | N_{K/ℚ}(α)|.

Proof. We assume a bit more Galois theory than usual for this proof. Assume first that K/ℚ is Galois. Let σ be an element of Gal(K/ℚ). It is clear that σ(O_{K})/σ(α) ≅ O_{K}/α; since σ(O_{K}) = O_{K}, this shows that

N′_{K/ℚ}(σ(α)O_{K}) = N′_{K/ℚ}(αO_{K}).

Taking the product over all σ ∈ Gal(K/ℚ), we have

N′_{K/ℚ}(N_{K/ℚ}(α)O_{K}) = N′_{K/ℚ}(αO_{K})^{n}.

Since N_{K/ℚ}(α) is a rational integer and O_{K} is a free ℤ-module of rank n,

O_{K}/ N_{K/ℚ}(α)O_{K}

will have order N_{K/ℚ}(α)^{n}; therefore

N′_{K/ℚ}(N_{K/ℚ}(α)O_{K}) = N_{K/ℚ}(αO_{K})^{n},

which completes the proof.

In the general case, let L be the Galois closure of K and set [L : K] = m. The above argument shows that

N′_{L/ℚ}(αO_{L}) = N_{L/ℚ}(α).

By Corollary 2.10 the first term is equal to N′_{K/ℚ}(αO_{K})^{m}, and it is easy to see that the second term is just N_{K/ℚ}(α)^{m}. This establishes the corollary.

From now on we will often write N_{K/ℚ} for both the ideal norm and the element norm; no confusion should result.

2.3. Relative factorizations. We now extend our earlier factorization results to arbitrary extensions of number fields. Let L/K be an extension of number fields of degree n. We first need to extend the notion of a prime of O_{L} lying over a prime of O_{K}.

Lemma 2.12. Let 𝔭 a non-zero prime of O_{K} and let 𝔓 be a non-zero prime of O_{L}. The following five conditions are equivalent.

1. 𝔓 divides 𝔭O_{L};

2. 𝔓 ⊇ 𝔭O_{L};
3. 𝔓 ⊇ 𝔭;
4. 𝔓 ∩ O_{K} = 𝔭;
5. 𝔓 ∩ K = 𝔭.

Furthermore, if any of the above are satisfied, then 𝔭 ∩ ℤ = 𝔓 ∩ ℤ.

Proof. See Exercise 3.7.

If 𝔭 and 𝔓 satisfy any of the equivalent conditions of Lemma 2.12, we say that 𝔓 lies over 𝔭 and that 𝔭 lies under 𝔓. Exercise 3.8 shows that every prime of O_{L} lies over a unique prime of O_{K}, and that every prime of O_{K} lies under at least one prime of O_{L}. Note also that by Lemma II.3.8 the primes lying over 𝔭 are precisely the primes occurring in the ideal factorization of 𝔭O_{L}.

Now, let 𝔭 and 𝔓 be as above and suppose that 𝔓 lies over 𝔭. We denote by e(𝔓/𝔭) the exact power of 𝔓 dividing 𝔭O_{L}; it is called the ramification index of 𝔓/𝔭. Thus we can write

𝔭O_{L} = ∏_{𝔓∩O_{K}=𝔭} 𝔓^{e(𝔓/𝔭)}.

Next, let p be the unique positive rational prime contained in 𝔭 and 𝔓. Then O_{K}/𝔭 and O_{L}/𝔓 are finite fields of characteristic 𝔭. Furthermore, the natural injection O_{K} ↪ O_{L} induces an injection

O_{K}/𝔭 ↪ O_{L}/𝔓,

since 𝔓 ∩ O_{K} = 𝔭 by Lemma 2.12. Thus O_{L}/𝔓 is an extension field of O_{K}/𝔭. We define the inertial degree f(𝔓/𝔭) to be the degree [O_{L}/𝔓 : O_{K}/𝔭] of this extension. Note that

N_{L/K}(𝔓) = N_{K/ℚ}(𝔭)^{f(𝔓/𝔭)}.

We can now state and prove our fundamental result.

Theorem 2.13. Let L/K be an extension of number fields of degree n and let 𝔭 be a prime of O_{K}. Let

𝔭O_{L} = 𝔓_{1}^{e_{1}} ⋯ 𝔓_{r}^{e_{r}}

be the factorization of 𝔭O_{L} into primes of O_{L}. Set f_{i} = f(𝔓_{i}/𝔭). Then

∑_{i=1}^{r} e_{i}f_{i} = n.

Proof. Taking ideal norms of both sides of the factorization of 𝔭O_{L}, we find that

N_{L/ℚ}(𝔭O_{L}) = N_{L/ℚ}(𝔓_{1})^{e_{1}} ⋯ N_{L/ℚ}(𝔓_{r})^{e_{r}}
= N_{K/ℚ}(𝔭)^{f_{1}e_{1}} ⋯ N_{K/ℚ}(𝔭)^{f_{r}e_{r}}

by the definition of the f_{i}. By Corollary 2.10 we know that N_{L/ℚ}(𝔭O_{L}) = N_{K/ℚ}(𝔭)^{n}, from which the theorem now follows immediately.

Let us finish this section with some additional facts and terminology. First of all, let M/L/K be number fields, let 𝔭_{K} be a prime of O_{K}, let 𝔭_{L} be a prime of O_{L} lying over 𝔭_{K}, and let 𝔭_{M} be a prime of O_{M} lying over 𝔭_{L}. Then clearly 𝔭_{M} lies over 𝔭_{K}, and it follows immediately from the definitions that we have

e(𝔭_{M}/𝔭_{K}) = e(𝔭_{M}/𝔭_{L})e(𝔭_{L}/𝔭_{K})

and

f(𝔭_{M}/𝔭_{K}) = f(𝔭_{M}/𝔭_{L})f(𝔭_{L}/𝔭_{K}).

Next return to the case of an extension L/K of degree n and let 𝔭 be a prime of O_{K}. Let

𝔭O_{L} = 𝔓_{1}^{e_{1}} ⋯ 𝔓_{r}^{e_{r}}

be the factorization of 𝔭O_{L} into primes of O_{L}. Set f_{i} = f(𝔓_{i}/𝔭). If any of the e_{i} are not equal to 1, then we say that 𝔭 ramifies in L/K. (It is an important fact that only finitely many primes ramify in an extension, and which primes these are and how badly they ramify is an essential invariant of the extension.) If r = 1 and e_{1} = n (so that f_{1} = 1), then 𝔭 is said to be totally ramified in L/K:

𝔭O_{L} = 𝔓^{n}.

If r = 1 and e_{1} = 1 (so that f_{1} = n), we say that 𝔭 is inert or remains primes in L/K; this is the case where 𝔭O_{L} is still prime. Lastly, if e_{i} = f_{i} = 1 for all i, we say that 𝔭 splits completely in L/K:

𝔭O_{L} = 𝔓_{1} ⋯ 𝔓_{n}.

2.4. Factorization in Galois extensions. Let L/K be a Galois extension. The presence of automorphisms of K causes factorizations to behave much more regularly than in arbitrary extensions, for the simple reason that if two primes are mapped to each other by an element of Gal(L/K), then the primes must have isomorphic residue fields. The key fact is the following.

Lemma 2.14. Let L/K be a Galois extension and let 𝔭 be a prime of O_{K}. Let 𝔓_{1}, …, 𝔓_{r} be the primes of L lying over 𝔭. Then Gal(L/K) acts transitively on this set of primes; that is, for any i and j, there exists σ ∈ Gal(L/K) such that σ(𝔓_{i}) = 𝔓_{j}.

Proof. Fix distinct primes 𝔓 and 𝔓′ lying over 𝔭. Suppose that

σ(𝔓) ≠ 𝔓′

for all σ ∈ Gal(L/K). Using this hypothesis, by the Chinese remainder theorem (see Exercise 3.6), we can find α ∈ O_{L} such that

α ≡ 0 (mod 𝔓′)

and

α ≡ 1 (mod σ(𝔓))

for all σ ∈ Gal(L/K). Consider

N_{L/K}(α) = ∏_{σ∈Gal(L/K)} σ(α) ∈ O_{K}.

Since α ∈ 𝔓′, this norm must lie in 𝔓′ ∩ O_{K} = 𝔭.

On the other hand, since α ≡ 1 (mod σ(𝔓)) for all σ, we also have α ∉ σ(𝔓); thus

σ^{−1}(α) ∉ 𝔓

for any σ ∈ Gal(L/K). Since as σ runs through Gal(L/K), σ^{−1} also runs through Gal(L/K), we find that

N_{L/K}(α) = ∏_{σ∈Gal(L/K)} σ^{−1}(α).

Since none of the factors lie in 𝔓 and 𝔓 is prime, this implies that N_{L/K}(α) ∉ 𝔓. Thus N_{L/K}(α) ∉ 𝔓 ∩ O_{K} = 𝔭. But thus is a contradiction, which proves the lemma.

Corollary 2.15. Let L/K be a Galois extension of degree n and let 𝔭 be a prime of O_{K}. Let

𝔭O_{L} = 𝔓_{1}^{e_{1}} ⋯ 𝔓_{r}^{e_{r}}

be the factorization of 𝔭 in O_{L}, and set f_{i} = f(𝔓_{i}/𝔭). Then

f_{1} = f_{2} = ⋯ = f_{r}

and

e_{1} = e_{2} = ⋯ = e_{r}.

In particular, re_{i}f_{i} = n for all i.

Proof. If r = 1 then the corollary is trivial, so we assume that r ≥ 2. We will prove that e_{1} = e_{2} and f_{1} = f_{2}; the general case is the same. By Lemma 2.14 we can find σ ∈ Gal(L/K) such that σ(𝔓_{1}) = 𝔓_{2}. Applying σ to our factorization, and using the fact that σ(𝔭) = 𝔭 since σ fixes K, we find that

𝔭O_{L} = σ(𝔓_{1})^{e_{1}}σ(𝔓_{2})^{e_{2}} ⋯ σ(𝔓_{r})^{e_{r}}
= 𝔓_{2}^{e_{1}}σ(𝔓_{2})^{e_{2}} ⋯ σ(𝔓_{r})^{e_{r}}.

Furthermore, if σ(𝔓_{i}) = 𝔓_{2}, then 𝔓_{i} = σ^{−1}(𝔓_{2}) = 𝔓_{1}; thus σ(𝔓_{i}) ≠ 𝔓_{2} for i ≠ 1. Therefore 𝔓_{2}^{e_{1}} is the only factor of 𝔓_{2} occurring in this factorization of 𝔭O_{K}; by unique factorization of ideals we now see that e_{1} must equal e_{2}.

The fact that f_{1} = f_{2} is immediate from the fact that σ induces an isomorphism

O_{L}/𝔓_{1} ≅ O_{L}/𝔓_{2}.

3. Explicit factorization of ideals

3.1. Factorization of primes. Let K be a number field of degree n. For this section we make the additional hypothesis that O_{K} = ℤ[α] for some α ∈ O_{K}, with minimal polynomial f(x) ∈ ℤ[x]. (We have in mind the case K = ℚ(ζ_{m}), in which case this hypothesis is satisfied.) The general case is somewhat more complicated and we will not treat it here. Let p be a prime of ℤ. We wish to explicitly determine the factorization of the ideal pO_{K} of O_{K}.

We will mimic the method we used in the quadratic case. Let

f̄(x) = ḡ_{1}(x)^{e_{1}} ⋯ ḡ_{r}(x)^{e_{r}}

be the factorization of f̄(x) into irreducibles in 𝔽_{p}[x]. (As usual we will write g_{i}(x) for any lift of ḡ_{i}(x) to ℤ[x].) Let f_{i} be the degree of ḡ_{i}(x); we have ∑ e_{i}f_{i} = n.

We claim that each ideal

𝔭_{i} = (p, g_{i}(α))

of O_{K} is prime, and that

pO_{K} = 𝔭_{1}^{e_{1}} ⋯ 𝔭_{r}^{e_{r}}

We further claim that

f(𝔭_{i}/p) = f_{i}.

Some of these assertions are immediate. First of all,

O_{K}/𝔭_{i} = ℤ[α]/(p, g_{i}(α))
≅ ℤ[x]/(f(x), p, g_{i}(x))
≅ 𝔽_{p}[x]/(f̄(x), ḡ_{i}(x))
≅ 𝔽_{p}[x]/(ḡ_{i}(x))

since ḡ_{i}(x) divides f̄(x) in 𝔽_{p}[x]. Since ḡ_{i}(x) is an irreducible polynomial of degree f_{i}, 𝔽_{p}[x]/(ḡ_{i}(x)) is a field of order p^{f_{i}}, which shows both that 𝔭_{i} is prime and that f(𝔭_{i}/p) = f_{i}. Note also that it follows from Exercise 3.3 that the 𝔭_{i} are relatively prime.

Let us now relate these to the factorization of pO_{K}. We will determine the kernel of the quotient map

O_{K} → O_{K}/pO_{K},

which of course is just pO_{K}, in a different way. Note that

O_{K}/pO_{K} = ℤ[α]/pℤ[α] ≅ ℤ[x]/(p, f(x)) ≅ 𝔽_{p}[x]/(f̄(x)).

The Chinese remainder theorem shows that

𝔽_{p}[x]/(f̄(x)) ≅ 𝔽_{p}[x]/(ḡ_{1}(x)^{e_{1}}) × ⋯ × 𝔽_{p}[x]/(ḡ_{r}(x)^{e_{r}});

thus we can consider the map O_{K} → O_{K}/pO_{K} as the map

O_{K} → 𝔽_{p}[x]/(ḡ_{1}(x)^{e_{1}}) × ⋯ × 𝔽_{p}[x]/(ḡ_{r}(x)^{e_{r}}

sending α to (x, …, x). The kernel into each factor is just (p, g_{i}(α)), so the kernel of the map (which is just 𝔭O_{K}) is

(p, g_{1}(α)^{e_{1}}) ∩ ⋯ ∩ (p, g_{r}(α)^{e_{r}}).

Furthermore, Exercise 3.3 shows that all of these ideals are pairwise relatively prime, so that by Exercise 3.4 the kernel is just the product

(p, g_{1}(α)^{e_{1}}) ⋯ (p, g_{r}(α)^{e_{r}}).

This shows that

pO_{K} = (p, g_{1}(α)^{e_{1}}) ⋯ (p, g_{r}(α)^{e_{r}}).

However, these factors are not yet primes for any i such that e_{i} > 1.

It remains to “pull out” the e_{i}. First, 𝔭_{i}^{e_{i}} divides (p, g_{i}(α)^{e_{i}}). To see this, note that every generator of

𝔭_{i}^{e_{i}} = (p, g_{i}(α))^{e_{i}}

is divisible by p except for g_{i}(α)^{e_{i}}. This shows that every generator of 𝔭_{i}^{e_{i}} lies in (p, g_{i}(α)^{e_{i}}), so 𝔭_{i}^{e_{i}} itself is contained in (p, g_{i}(α)^{e_{i}}). Lemma II.3.8 now gives the asserted division.

We now know that 𝔭_{1}^{e_{1}} ⋯ 𝔭_{r}^{e_{r}} divides pO_{K}. The norm of the first term is

p^{f_{1}e_{1}} ⋯ p^{f_{r}e_{r}} = p^{n};

this is also the norm of pO_{K}. This implies that the two ideals must be equal, since if one ideal contains another and has the same norm they must be equal. This completes the proof of the explicit factorization of pO_{K}.

Example 3.1. Let α be a root of the polynomial f(x) = x^{3} + 2x + 1 and let K = ℚ(α). Exercise 3.18 shows that O_{K} = ℤ[α], so we may apply the methods of this section. Let us factor some small rational primes. Note that factoring f(x) modulo primes is easy, since if f(x) has any factors then it will have roots; this is no longer true for degree 4 and higher.

When p = 2, we find that

x^{3} + 2x + 1 ≡ (x + 1)(x^{2} + x + 1) (mod 2),

so

2O_{K} = (2, α + 1)(2, α^{2} + α + 1).

Here the first factor has inertial degree 1 and the second factor has inertial degree 2. For p = 3, f(x) is irreducible in 𝔽_{3}[x], so 3O_{K} factors as (3, f(α)) = (3); that is, 3O_{K} remains prime. The reader can easily check that 5O_{K} and 7O_{K} also remain prime. For p = 11, we find that

x^{3} + 2x + 1 ≡ (x + 2)(x^{2} − 2x + 6) (mod 11),

so

11O_{K} = (11, α + 2)(11, α^{2} − 2α + 6).

13O_{K} also remains prime, while

x^{3} + 2x + 1 ≡ (x − 3)(x − 5)(x − 9) (mod 17),

so 17O_{K} splits completely as

17O_{K} = (17, α − 3)(17, α − 5)(17, α − 9).

Since K has discriminant −59, Exercise 3.16 shows that 59 will be the only prime which ramifies. One finds that

x^{3} + 2x + 1 ≡ (x − 14)^{2}(x − 31) (mod 59),

so

59O_{K} = (59, α − 14)^{2}(59, α − 31).

3.2. Factoring cyclotomic polynomials. The methods of the previous section give us a computational procedure for determining prime splitting in many number fields, but it becomes difficult to carry out in practice as soon as the degree of the number field becomes large. Luckily, in the important case of cyclotomic fields we can give a good description of the factorization of cyclotomic polynomials, even if it is difficult to write down the actually factors. The key result is the following lemma, which says that the m^{th} cyclotomic polynomial is the “universal” polynomial for testing if an element of a field is a primitive m^{th} root of unity.

Lemma 3.2. Let m be a positive integer and let K be a field of characteristic not dividing m. Let α be an element of K. Then Φ_{m}(α) = 0 if and only if α is a primitive m^{th} root of unity.

Proof. Recall that

x^{m} − 1 = ∏_{d|m} Φ_{d}(x).

Since this factorization is in ℤ[x], it also makes sense in K[x]. Note also that x^{m} −1 has no multiple roots in K, as follows immediately from the derivative test. (This is the only place where we will use the assumption on the characteristic of K.)

Suppose first that α is a primitive m^{th} root of unity. Then α is a root of x^{m} −1, so it must be a root of some Φ_{d}(x) with d dividing m. Suppose that α is a root of Φ_{d}(x) with d < m. Then, since Φ_{d}(x) divides x^{d} − 1, α^{d} = 1. This contradicts the fact that α is a primitive m^{th} root of unity, so α must be a root of Φ_{m}(x).

Conversely, suppose that Φ_{m}(α) = 0. Since Φ_{m}(x) divides x^{m} − 1, this implies that α^{m} = 1; that is, α is an m^{th} root of unity. Suppose that α is actually a primitive d^{th} root of unity for some divisor d of m with d < m. Then the argument in the first half of the proof shows that Φ_{d}(α) = 0. But then α would be a double root of x^{m} − 1, which is a contradiction since x^{m} − 1 does not have multiple roots. Thus α is a primitive m^{th} root of unity.

Let K = ℚ(ζ_{m}) be a cyclotomic field and let p be a rational prime. Let 𝔭 be any prime of O_{K} = ℤ[ζ_{m}] lying over p. We wish to determine e = e(𝔭/p) and f = f(𝔭/p). Note that by Corollary 2.15 these numbers are independent of the choice of prime 𝔭. Put differently, in 𝔽_{p}[x] Φ_{m}(x) factors as

Φ_{m}(x) = (g_{1}(x) ⋯ g_{r}(x))^{e}

where deg g_{i} = f for all i and efr = φ(m).

We begin with the case that p does not divide m. Since x^{m} −1 has no repeated factors in 𝔽_{p}[x], Φ_{m}(x) doesn’t either; in particular, we must have e = 1. Thus we are left to determine f and r. Before we do the general case, we examine the case f = 1 to illustrate the idea. If f = 1, then Φ_{m}(x) splits into linear factors in 𝔽_{p}[x]; thus Φ_{m}(x) has roots in 𝔽_{p}. By Lemma 3.2, this implies that 𝔽_{p} has primitive m^{th} roots of unity. But 𝔽_{p}^{∗} is a cyclic group of order p − 1, so it has elements of exact order m if and only if m divides p − 1; that is, if and only if

p ≡ 1 (mod m).

The above argument is reversible, so we have shown that a rational prime p splits completely in ℚ(ζ_{m}) if and only if p does not divide m and p ≡ 1 (mod m).

In the general case we must go to an extension of 𝔽_{p} to find a primitive m^{th} root of unity. Let g(x) be one of the irreducible factors of Φ_{m}(x) in 𝔽_{p}[x]; g(x) has degree f. Let α be a root of g(x) and set F = 𝔽_{p}(α) ≅ 𝔽_{p}[x]/(g(x)); this is an extension of 𝔽_{p} of degree f. Note that α is a primitive m^{th} root of unity since it satisfies g(x) and thus Φ_{m}(x). Furthermore, F is clearly the smallest extension of 𝔽_{p} containing a primitive m^{th} root of unity (since it is just 𝔽_{p} adjoined a m^{th} root of unity), so we have shown that f is the degree of the smallest extension of 𝔽_{p} containing a primitive m^{th} root of unity.

Let us now determine this extension in another way. Let F_{i} be the unique extension of 𝔽_{p} of degree i. Then F_{i}^{∗} is cyclic of order p^{i} − 1, so it contains a primitive m^{th} root of unity if and only if m divides p^{i} − 1. Thus the smallest extension of 𝔽_{p} containing a primitive m^{th} root of unity will be F_{i}, where i is the smallest positive integer such that

p^{i} ≡ 1 (mod m);

that is, i is the order of p in (ℤ/mℤ)^{∗}. Combining this with our earlier arguments, we obtain the following result.

Proposition 3.3. Let p be a rational prime not dividing m and let 𝔭 be a prime of ℤ[ζ_{m}] lying over p. Then e(𝔭/p) = 1, f(𝔭/p) is the order of p in (ℤ/mℤ)^{∗}, and there are exactly φ(m)/f(𝔭/p) primes of ℤ[ζ_{m}] lying over p.

Example 3.4. Let K = ℚ(ζ_{5}). The behavior of a rational prime p in O_{K} is determined entirely by the residue class of p in (ℤ/5ℤ)^{∗}. If p ≡ 1 (mod 5) (for example, p = 11), then p splits completely in O_{K}. If p ≡ 4 (mod 5), then p splits into 2 prime factors, each with inertial degree 2. If p ≡ 2, 3 (mod 5), then p remains prime in O_{K}.

For some explicit examples, we consider the primes 3, 7, 11, 19. For p = 3, 7 the above argument shows that Φ_{5}(x) = x^{4} + x^{3} + x^{2} + x + 1 is irreducible modulo p, so 3O_{K} and 7O_{K} are both prime ideals of O_{K}. For p = 19, we find that

x^{4} + x^{3} + x^{2} + x + 1 ≡ (x^{2} + 5x + 1)(x^{2} + 15x + 1) (mod 19),

so

(19) = (19, ζ_{5}^{2} + 5ζ_{5} + 1)(19, ζ_{5}^{2} + 15ζ_{5} + 1).

Lastly, modulo 11 we have

x^{4} + x^{3} + x^{2} + x + 1 = (x + 2)(x + 6)(x + 7)(x + 8) (mod 11),

so

(11) = (11, ζ_{5} + 2)(11, ζ_{5} + 6)(11, ζ_{5} + 7)(11, ζ_{5} + 8).

The ramified case works out somewhat differently. We will only consider the case of the splitting of pO_{K} in ℚ(ζ_{p}), which is by far the most important.

We must determine the factorization of

Φ_{p}(x) = \frac{x^{p} − 1}{x − 1} = x^{p−1} + x^{p−2} + ⋯ + 1

in 𝔽_{p}[x]. By Exercise I.1.15, we see that

x^{p} − 1 ≡ (x − 1)^{p} (mod p),

so

Φ_{p}(x) = (x − 1)^{p−1}

in 𝔽_{p}[x]. Thus

pO_{K} = (p, ζ_{p} − 1)^{p−1}.

Furthermore,

O_{K}/(p, ζ_{p} − 1) = ℤ[ζ_{p}]/(p, ζ_{p} − 1) ≅ ℤ/pℤ,

so (p, ζ_{p} − 1) is prime with inertial degree 1. Thus pO_{K} is totally ramified. Note that we actually already had a better form of this result; see Exercise 2.13.

3.3. Applications to quadratic fields. There are some very interesting applications of the arithmetic of cyclotomic fields to quadratic fields. Consider the field ℚ(ζ_{p}) for some odd prime p. Recall that this is a Galois extension of ℚ with Galois group isomorphic to (ℤ/pℤ)^{∗}, where the automorphism corresponding to a ∈ (ℤ/pℤ)^{∗} is σ_{a} characterized by

σ_{a}(ζ_{p}) = ζ_{p}^{a}.

Since (ℤ/pℤ)^{∗} is cyclic of order p − 1, it contains a unique subgroup of index 2, consisting of all of the squares in (ℤ/pℤ)^{∗}. Denote by S the corresponding subgroup of Gal(ℚ(ζ_{p})/ℚ). Let K be the fixed field of S; that is, K is the subfield of ℚ(ζ_{p}) of elements fixed by all of S. Galois theory tells us that [K : ℚ] = 2; thus K is a quadratic field. It remains to determine which quadratic field it is.

We can do this by considering ramfication. Recall that p is totally ramified in ℚ(ζ_{p}); that is, there is a unique prime 𝔓 of ℚ(ζ_{p}) lying over p, and (p) = 𝔓^{p−1}.

Let 𝔭 be any prime of K lying over p. Then 𝔓 lies over 𝔭 (since 𝔓 is the only prime of K lying over p) and

e(𝔓/p) = e(𝔓/𝔭)e(𝔭/p);

since e(𝔓/p) = p − 1 and ramification indices are bounded by the degrees of the extensions, this implies that e(𝔓/𝔭) = \frac{p−1}{2} and e(𝔭/p) = 2. In particular, 𝔭 is the only prime of K lying over p, and it is totally ramified.

Let 𝔔 be any other prime of ℚ(ζ_{p}), let 𝔮 be the prime of K which it lies over, and let q be the prime of ℤ which it lies over. A similar argument, using the fact that e(𝔔/q) = 1, shows that e(𝔮/q) = 1, so that q is not ramified in K. We conclude that p is the only prime of ℤ which ramifies in K.

Now, we have already determined the ramification in every quadratic field, and the only quadratic field in which only p ramifies is ℚ(\sqrt{εp}), where ε = ±1 is such that

εp ≡ 1 (mod 4).

(See Corollary 1.3.) We can take ε = (−1)^{(p−1)/2}. We have therefore established the following distinctly non-obvious fact.

Proposition 3.5. The field ℚ(ζ_{p}) contains the quadratic field ℚ(\sqrt{εp}), where ε = (−1)^{(p−1)/2}. In particular, \sqrt{εp} can be written as a rational linear combination of p^{th} roots of unity.

We are now in a position to prove the celebrated quadratic reciprocity law.

Theorem 3.6 (Quadratic Reciprocity). Let p and q be distinct, positive odd primes. Then

(\frac{p}{q}) (\frac{q}{p}) = (−1)^{\frac{p−1}{2} \frac{q−1}{2}} .

Proof. We showed above that \sqrt{εp} ∈ ℚ(ζ_{p}). Denote this element by τ. Consider the automorphism σ_{q} ∈ Gal(ℚ(ζ_{p})/ℚ); it is defined by σ_{q}(ζ_{p}) = ζ_{p}^{q}. Since the conjugates of τ are simply ±τ, we must have

σ_{q}(τ) = ±τ.

Furthermore, letting S be the subgroup of Gal(ℚ(ζ_{p})/ℚ) defined above, σ_{q}(τ) = τ if and only if σ_{q} ∈ S. (This is because ℚ(τ) is the fixed field of S by definition.) Under the identification of Gal(ℚ(ζ_{p})/ℚ) and (ℤ/pℤ)^{∗}, S corresponds to the subgroup of squares; combining all of this, we see that σ_{q}(τ) = τ if and only if q is a square in (ℤ/pℤ)^{∗}; that is,

σ_{q}(τ) = (\frac{q}{p}) τ.

Now let 𝔮 be a prime of O_{K} lying over q. Write τ = a_{0} + a_{1}ζ_{p} + ⋯ + a_{p−2}ζ_{p}^{p−2} with a_{i} ∈ ℤ. (Note that τ is visibly an algebraic integer.) Using that σ_{q}(ζ_{p}) = ζ_{p}^{q} and a^{q} = a for all a ∈ 𝔽_{q}, we find that

σ_{q}(τ) = a_{0} + a_{1}ζ_{p}^{q} + a_{2}ζ_{p}^{2q} + ⋯ + a_{p−2}ζ_{p}^{(p−2)q}
≡ a_{0}^{q} + a_{1}^{q}ζ_{p}^{q} + a_{2}^{q}ζ_{p}^{2q} + ⋯ + a_{p−2}^{q}ζ_{p}^{(p−2)q} (mod 𝔮)
≡ (a_{0} + a_{1}ζ_{p} + a_{2}ζ_{p}^{2} + ⋯ + a_{p−2}ζ_{p}^{p−2})^{q} (mod 𝔮)
≡ τ^{q} (mod 𝔮).

3. EXPLICIT FACTORIZATION OF IDEALS 71

Combining this with our other expression for σ_{q}(τ) yields

(\frac{q}{p}) τ ≡ τ^{q} (mod 𝔮).

Since 𝔮 is prime and we clearly have τ ∉ 𝔮, we can cancel τ modulo 𝔮; we conlude that

(\frac{q}{p}) ≡ τ^{q−1} ≡ (εp)^{(q−1)/2} (mod 𝔮).

By Exercise 3.9, this shows that

(\frac{q}{p}) ≡ (\frac{εp}{q}) (mod 𝔮).

By definition, this means that

(\frac{q}{p}) − (\frac{εp}{q}) ∈ 𝔮;

since (\frac{q}{p}) and (\frac{εp}{q}) are integers, this difference is actually contained in 𝔮∩ℤ = qℤ. In fact, (\frac{q}{p}) and (\frac{εp}{q}) are just ±1, so the difference is certainly smaller than ±q. It follows that we actually have an equality

(\frac{q}{p}) = (\frac{εp}{q}) .

The fact that

(\frac{ε}{q}) = (\frac{(−1)^{(p−1)/2}}{q} ) = (−1)^{\frac{p−1}{2} \frac{q−1}{2}}

completes the proof.