CHAPTER 1 Number Fields

1. Example : Quadratic number fields

Before we consider number fields in general, let us begin with the fairly concrete case of quadratic number fields. A quadratic number field is an extension K of ℚ of degree 2. The fundamental examples (in fact, as we shall see in a moment the only example) are fields of the form

ℚ(\sqrt{d}) = \{a + b\sqrt{d} | a, b ∈ ℚ\}

where d ∈ ℚ is not the square of another rational number.

There is an issue that arises as soon as we write down these fields, and it is important that we deal with it immediately: what exactly do we mean by \sqrt{d}? There are several possible answers to this question. The most obvious is that by \sqrt{d} we mean a specific choice of a complex square root of d. ℚ(\sqrt{d}) is then defined as a subfield of the complex numbers. The difficulty with this is that the notation “\sqrt{d}” is ambiguous; d has two complex square roots, and there is no algebraic way to tell them apart.

Algebraists have a standard way to avoid this sort of ambiguity; we can simply define

ℚ(\sqrt{d}) = ℚ[x]/(x^{2} − d).

There is no ambiguity with this notation; \sqrt{d} really means x, and x behaves as a formal algebraic object with the property that x^{2} = d.

This second definition is somehow the algebraically correct one, as there is no ambiguity and it allows ℚ(\sqrt{d}) to exist completely independently of the complex numbers. However, it is far easier to think about ℚ(\sqrt{d}) as a subfield of the complex numbers. The ability to think of ℚ(\sqrt{d}) as a subfield of the complex numbers also becomes important when one wishes to compare fields ℚ(\sqrt{d_{1}}) and ℚ(\sqrt{d_{2}}) for two different numbers d_{1} and d_{2}; the abstract algebraic fields ℚ[x]/(x^{2} − d_{1}) and ℚ[y]/(y^{2} −d_{2}) have no natural relation to each other, while these same fields viewed as subfields of ℂ can be compared more easily.

The best approach, then, seems to be to pretend to follow the formal algebraic option, but to actually view everything as subfields of the complex numbers. We can do this through the notion of a complex embedding; this is simply an injection

σ : ℚ[x]/(x^{2} − d) ↪ ℂ.

As we have already observed, there are exactly two such maps, one for each complex square root of d.

Before we continue we really ought to decide which complex number we mean by \sqrt{d}. There is unfortunately no consistent way to do this, in the sense that we can not arrange to have

\sqrt{d_{1}}\sqrt{d_{2}} = \sqrt{d_{1}d_{2}}

for all d_{1}, d_{2} ∈ ℚ. In order to be concrete, let us choose \sqrt{d} to be the positive square root of d for all d > 0 and \sqrt{d} to be the positive square root of −d times i for all d < 0. (There is no real reason to prefer these choices, but since it doesn’t really matter anyway we might as well fix ideas.)

With this choice, our two complex embeddings are simply

σ_{1} : ℚ[x]/(x^{2} − d) ↪ ℂ
σ_{2} : ℚ[x]/(x^{2} − d) ↪ ℂ

defined by

σ_{1}(a + bx) = a + b\sqrt{d};
σ_{2}(a + bx) = a − b\sqrt{d}.

Given any a + bx ∈ ℚ[x]/(x^{2} − d), we define its conjugates to be the images σ_{1}(a + bx) = a + b\sqrt{d} and σ_{2}(a + bx) = a − b\sqrt{d}.

Note that these maps have the same image. This gives us yet another way to view the ambiguity: we can take ℚ(\sqrt{d}) to be the subfield \{a + b\sqrt{d} | a, b ∈ ℚ\} of ℂ, and we remember that ℚ(\sqrt{d}) has an automorphism

a + b\sqrt{d} ↦ a − b\sqrt{d}.

This is the approach we will take; that is, we will regard ℚ(\sqrt{d}) as a subfield of ℂ via our choice of \sqrt{d}, but we always remember that \sqrt{d} is ambiguous, and thus that we have an automorphism of this field exchanging \sqrt{d} and −\sqrt{d}. From this point of view, the conjugates of an element a + b\sqrt{d} are a + b\sqrt{d} and a − b\sqrt{d}.

Let us now analyze these fields K = ℚ(\sqrt{d}). Note first that every α ∈ K has degree either 1 or 2 over ℚ, and it has degree 1 if and only if it is actually in ℚ. In particular, if α ∉ ℚ then we must have K = ℚ(α).

Let us now compute the norms and traces from K to ℚ. We take 1, \sqrt{d} as our basis for K over ℚ. Multiplication by α = a + b\sqrt{d} takes 1 to a + b\sqrt{d} and \sqrt{d} to bd + a\sqrt{d}, so the matrix for the linear transformation m_{α} is

[ \begin{matrix}a & bd \\ b & a \end{matrix} ] .

The characteristic polynomial of this matrix is

x^{2} − 2ax + (a^{2} − bd^{2}).

Thus

N_{K/ℚ}(α) = a^{2} − bd^{2}

and

Tr_{K/ℚ}(α) = 2a.

Note also that we have

N_{K/ℚ}(α) = (a + b\sqrt{d})(a − b\sqrt{d})

and

Tr_{K/ℚ}(α) = (a + b\sqrt{d}) + (a − b\sqrt{d}).

That is, the norm of α is the product of its conjugates and the trace of α is the sum of its conjugates. This follows immediately from the fact that the conjugates of α are the two roots of the characteristic polynomial of α.

It turns out that every quadratic field is of the form ℚ(\sqrt{d}) for some d ∈ ℚ. In fact, in the case of quadratic fields it is actually possible to give a complete classification, as described in the following theorem.

Theorem 1.1. Let K be a number field of degree 2. Then K is isomorphic to ℚ(\sqrt{d}) for a unique squarefree integer d ≠ 1.

Proof. First we will show that every extension of ℚ of degree 2 is isomorphic to one of the desired form. So let K/ℚ have degree 2 and choose a primitive element α for K, with minimal polynomial

f(x) = x^{2} + ax + b,

a, b ∈ ℚ. By the quadratic formula we have

α = \frac{−a ± \sqrt{a^{2} − 4b}}{2};

put differently,

(2α + a)^{2} = a^{2} − 4b.

Thus K contains an element β = 2α + a of square a^{2} − 4b ∈ ℚ. Note also that a^{2} − 4b is not a square in ℚ, for otherwise f(x) would not be irreducible. It follows that β has degree 2 and thus is a primitive element for K. a^{2} − 4b may not be a squarefree integer, but one sees easily from unique factorization in ℤ that we can find some rational number c such that c^{2}(a^{2} − 4b) is a squarefree integer. cβ still generates K over ℚ, and it is now in the form we considered above. This shows that every extension of ℚ of degree 2 can be generated by the square root of a squarefree integer.

We now show that no two fields ℚ(\sqrt{d}) with d a squarefree integer (other than 1) are isomorphic. So let d_{1} and d_{2} be distinct squarefree integers and suppose that there is an isomorphism

φ : ℚ(\sqrt{d_{1}}) →^{≅} ℚ(\sqrt{d_{2}}).

We will show that d_{1} = d_{2}. Consider the element α = φ(\sqrt{d_{1}}) ∈ ℚ(\sqrt{d_{2}}). α has minimal polynomial x^{2} − d_{1}, so we read off that

N_{K_{2}/ℚ}(α) = −d_{1}

and

Tr_{K_{2}/ℚ}(α) = 0.

Writing α = a + b\sqrt{d_{2}}, our formulas for the norm and trace imply that a = 0 and b^{2}d_{2} = d_{1}. One now shows easily that the fact that d_{1} and d_{2} are squarefree integers implies that b = 1 and d_{1} = d_{2}, as claimed.

This sort of analysis does not work for any degree other than 2; even the cubic and quartic “formulas” are too complicated to use, and beyond that there aren’t any formulas at all.

2. Complex embeddings

A number field is a finite extension of the rational numbers ℚ. (This is not quite the same as the definitions given in [9] and [13], but it seems to be the most common definition.) We define the degree of a number field K to be the positive integer [K : ℚ]. The fundamental examples are fields of the form

ℚ[x]/(f(x))

where f(x) ∈ ℚ[x] is an irreducible polynomial. In fact, Proposition A.2.3 shows that every number field K is isomorphic to one of this form: simply choose a primitive element α ∈ K with minimal polynomial f(x) ∈ ℚ[x]. Then K = ℚ(α) and Lemma A.2.1 shows that K is isomorphic to ℚ[x]/(f(x)).

Let K and K′ be number fields and suppose that there is a homomorphism

φ : K → K′.

Then φ is automatically ℚ-linear: this is because it must send 1 to 1; it follows from the fact that it is an additive homomorphism that it must be the identity on all of ℤ, and it follows from the fact that it is a multiplicative homomorphism that it must be the identity on all of ℚ.

We now investigate complex embeddings of arbitrary number fields. That is, for a number field K we wish to determine all of the possible injections K ↪ ℂ. Recall that in the quadratic case we did this by exhibiting complex square roots. We will use the same method in the general case, although of course the polynomials of interest will now have larger degree.

Fix a number field K of degree n and choose a primitive element α ∈ K with minimal polynomial f(x) ∈ ℚ[x]. Since ℂ is algebraically closed, f(x) splits into n linear factors over ℂ; since f(x) is irreducible over ℚ, these linear factors must be distinct (see Problem 1.12), and thus f(x) has n distinct roots α_{1}, …, α_{n} ∈ ℂ.

For each root α_{i} we define a (necessarily ℚ-linear) map

σ_{i} : K ⟶^{≅} ℚ(α_{i}) ⊆ ℂ

sending α to α_{i}; that is,

σ_{i}(a_{0} + a_{1}α + a_{2}α^{2} + ⋯ + a_{n−1}α^{n−1}) = a_{0} + a_{1}α_{i} + a_{2}α_{i}^{2} + ⋯ + a_{n−1}α_{i}^{n−1}

where the a_{i} are all in ℚ. This map is well-defined since α_{i} satisfies f(x), it is injective since all non-zero maps of fields are injective, and it is surjective since α_{i} generates ℚ(α_{i}) over ℚ.

We have now embedded K as a subfield of ℂ in n distinct ways. (Note that we mean that the maps are distinct; the images of the embeddings could still be the same.) We claim that the σ_{i} are the only embeddings of K into ℂ. To see this, let σ : K ↪ ℂ be any such map. Then σ(α) must have the same minimal polynomial f(x) over ℚ as α; thus σ(α) must be one of the complex roots of f(x), which are precisely the α_{i}. Therefore σ(α) = α_{i} for some i, and since α generates K over ℚ, this implies that σ = σ_{i}. This proves the claim.

In particular, this implies that the embeddings σ_{i} are independent of the choice of primitive element α, since any other choice would yield n embeddings of K into ℂ which by the above argument must be the same as the σ_{i}. Combining all of this, we see that there are exactly n distinct embeddings of K into ℂ. We state this as a proposition.

Proposition 2.1. Let K be a number field of degree n. Then K has exactly n distinct complex embeddings.

Example 2.2. Consider the number field ℚ[x]/(x^{3} − 2). This has degree 3 over ℚ, so there should be three complex embeddings. These are determined by the three roots of x^{3} − 2 in ℂ. If we let α be the real cube root of 2 and let ζ be a third root of unity in ℂ, then these roots are α, ζα and ζ^{2}α. The three complex embeddings are then the three maps

ℚ[x]/(x^{3} − 2) → ℂ

sending x to α, ζα and ζ^{2}α respectively. Note that in contrast to the case of ℚ[x]/(x^{2} − 2) these maps have different images; for example, the first map has image inside of ℝ, while the other two do not.

Let α be an arbitrary element of K with minimal polynomial f(x) ∈ ℚ[x] of degree d. We define the conjugates of α to be the d complex roots of f(x); that is, they are simply the complex numbers which behave exactly the same as α does algebraically. Alternately, if τ_{1}, …, τ_{d} are the d complex embeddings of the subfield ℚ(α) of K (which is a number field since K is), the conjugates are precisely

τ_{1}(α), …, τ_{d}(α),

as is clear from the above discussion. In particular, if α is a primitive element for K, then its conjugates are the n complex numbers

σ_{1}(α), …, σ_{n}(α).

As with the quadratic case we would like to be able to think of number fields as specific subfields of the complex numbers. As we have just seen, we can do this in n different ways, where n is the degree of the number field K. In general, however, these embeddings have different images. Thus, although it is often useful to think of K in terms of these complex images, there is no single field that one can point to and say is the best choice for a complex version of K. We will always attempt to be careful about this point. For example, when we write ℚ(\sqrt[3]{2}), we do not mean to single out any of the three complex versions of it; if we wish to do so, we will make it explicit.

This sets up a slightly strange situation: whenever we say “let K be a number field”, we want to regard K independent of any complex embedding of K. On the other hand, our examples will usually involve specific subfields of ℂ in order to fix ideas. In particular, keep in mind that a subfield of ℂ can still have complex embeddings, just like any number field.

The one case where one can safely identify a number field with the images of its complex embeddings are when all of these complex embeddings are the same. In this case we will say that K is Galois (over ℚ). We will return to the theory of Galois extensions later.

3. Example : Cyclotomic fields

3.1. Cyclotomic polynomials. Before we define cyclotomic fields abstractly, let us work with subfields of the complex numbers. Recall that a complex number ζ is an m^{th} root of unity if ζ^{m} = 1; it is a primitive m^{th} root of unity if m is the smallest positive integer which works. The complex m^{th} roots of unity are precisely the numbers

e^{2πik/m}

for k = 0, 1, …, m − 1, and the primitive m^{th} roots of unity are those for which k and m are relatively prime. In particular, there are m complex m^{th} roots of unity and φ(m) complex primitive m^{th} roots of unity, where φ(m) is the Euler φ-function. (See Appendix B.)

Let ζ_{m} be a fixed complex primitive m^{th} root of unity. ζ_{m} is a root of x^{m} − 1, but for m > 1 this can not be its minimal polynomial, as it is not irreducible. We wish to determine the minimal polynomial f(x) ∈ ℚ[x] of ζ_{m}; we will do this by determining the complex roots of f(x).

Proposition 3.1. If p is a prime not dividing m, then ζ_{m}^{p} is a root of f(x).

Proof. f(x) divides x^{m} − 1 in ℚ[x]; thus we can write

x^{m} − 1 = f(x)g(x)

for some monic g(x) ∈ ℚ[x], and by Exercise 1.4 we actually have f(x), g(x) ∈ ℤ[x]. Since ζ_{m}^{p} is a root of x^{m} − 1, to show that it is a root of f(x) it will suffice to show that it is not a root of g(x).

So suppose that g(ζ_{m}^{p}) = 0. Let h(x) ∈ ℤ[x] be the monic polynomial g(x^{p}). Then h(ζ_{m}) = 0, so f(x) divides h(x) in ℚ[x]. Writing h(x) = f(x)q(x), Exercise 1.4 again shows that q(x) is actually in ℤ[x].

We now work modulo p. For any polynomial s(x) ∈ ℤ[x], we denote by s̄(x) its image in 𝔽_{p}[x] after reducing the coefficients modulo p. We have h̄(x) = f̄(x)q̄(x); also,

h̄(x) = ḡ(x^{p}) = ḡ(x)^{p}

by Exercise 1.15. Thus f̄(x) divides ḡ(x)^{p} in 𝔽_{p}[x]. Since 𝔽_{p}[x] is a unique factorization domain, this implies that f̄(x) and ḡ(x) have a monic common factor of positive degree, say r̄(x).

We have f̄(x)ḡ(x) = x^{m} − 1 ∈ 𝔽_{p}[x], so r̄(x)^{2} divides x^{m} − 1 in 𝔽_{p}[x]. By Exercise 1.11, this implies that r̄(x) divides mx^{m−1}. Since p does not divide m (this is the only place where we use that hypothesis), mx^{m−1} is a non-zero monomial, so r̄(x) must also be a non-zero monomial. But r̄(x) also divides x^{m} − 1; the only monic monomial with this property is 1, so r̄(x) = 1. This contradicts the fact that r̄(x) has positive degree, so the initial assumption that g(ζ_{m}^{p}) = 0 must be false. Thus f(ζ_{m}^{p}) = 0, which completes the proof.

Corollary 3.2. The conjugates of ζ_{m} are precisely the other primitive m^{th} roots of unity.

Proof. As before let f(x) be the minimal polynomial of ζ_{m}. Let ζ_{m}^{k} be any other primitive m^{th} root of unity. Then k is relatively prime to m, so it is divisible only by primes not dividing m. Write k = p_{1}p_{2} ⋯ p_{n}, with the p_{i} not necessarily distinct. Then Proposition 3.1 shows that ζ_{m}^{p_{1}} is a root of f(x). In particular, f(x) is also the minimal polynomial of ζ_{m}^{p_{1}}. Applying Proposition 3.1 with respect to the primitive m^{th} root of unity ζ_{m}^{p_{1}} shows that ζ_{m}^{p_{1}p_{2}} is also a root of f(x), and continuing in this way we see that ζ_{m}^{k} is a root of f(x). Thus all primitive m^{th} roots of unity are roots of f(x), and therefore conjugates of ζ_{m}.

To complete the proof we must show that ζ_{m} has no other conjugates. But if α is any other conjugate of ζ_{m}, then there is an isomorphism of ℚ(ζ_{m}) and ℚ(α) sending ζ_{m} to α; it follows that α must also be a primitive m^{th} root of unity, as claimed.

We now define the m^{th} cyclotomic polynomial Φ_{m}(x) ∈ ℤ[x] to be the minimal polynomial of the complex primitive m^{th} roots of unity. Since it is a minimal polynomial Φ_{m}(x) is irreducible, and our above arguments show that it has degree φ(m).

Since we have now shown that all primitive m^{th} roots of unity are essentially “the same” from the point of view of algebraic number theory, we might as well fix specific complex values for each ζ_{m}. Let us take

ζ_{m} = e^{2πi/m} ∈ ℂ

for all m. These roots of unity have the nice property that

ζ_{n}^{n/m} = ζ_{m}

whenever m divides n. (While it may appear to be true even if m doesn’t divide n, one then has all sorts of multiple-valued function stuff to worry about.) More generally, any choice of ζ_{m} with this compatibility would be fine, but we will stick with these for concreteness.

Corollary 3.2 gives the expression

Φ_{m}(x) = ∏_{\begin{matrix}1≤k<m \\ (k,m)=1 \end{matrix}}(x − ζ_{m}^{k}).

However, this formula is not very useful for actually computing the Φ_{m}(x) by hand. For this we have the following result, which gives an expression for Φ_{m}(x) entirely in terms of integer arithmetic.

Proposition 3.3. We have

x^{m} − 1 = ∏_{d|m} Φ_{d}(x)

and

Φ_{m}(x) = ∏_{d|m}(x^{d} − 1)^{μ(m/d)}

where μ is the Mobius function.

Proof. The first equality is clear since each side has exactly the same complex roots; namely, each m^{th} root of unity is a root of exactly one of the Φ_{d}(x) with d dividing m. The second equality comes from Mobius inversion of the first. See Example B.2.5.

Using this formula we see immediately that for any prime p,

Φ_{p}(x) = \frac{x^{p} − 1}{x − 1} = x^{p−1} + x^{p−2} + +x^{p−3} ⋯ + x + 1.

The first few cyclotomic polynomials are

Φ_{1}(x) = x − 1
Φ_{2}(x) = x + 1
Φ_{3}(x) = x^{2} + x + 1
Φ_{4}(x) = x^{2} + 1
Φ_{5}(x) = x^{4} + x^{3} + x^{2} + x + 1
Φ_{6}(x) = x^{2} − x + 1
Φ_{7}(x) = x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1
Φ_{8}(x) = x^{4} + 1

Φ_{9}(x) = x^{6} + x^{3} + 1
Φ_{10}(x) = x^{4} − x^{3} + x^{2} − x + 1
Φ_{11}(x) = x^{10} + x^{9} + x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1
Φ_{12}(x) = x^{4} − x^{2} + 1

There are many patterns which can be found among the cyclotomic polynomials; we leave these to the reader. (We will at least point out that it is not true that every Φ_{m} has only coefficients ±1 and 0, although Φ_{105} is the first which violates this.)

3.2. Abstract cyclotomic fields. We define the m^{th} cyclotomic field to be the field

ℚ[x]/(Φ_{m}(x))

where Φ_{m}(x) is the m^{th} cyclotomic polynomial. ℚ[x]/(Φ_{m}(x)) has degree φ(m) over ℚ since Φ_{m}(x) has degree φ(m). The roots of Φ_{m}(x) are just the primitive m^{th} roots of unity, so the complex embeddings of ℚ[x]/(Φ_{m}(x)) are simply the φ(m) maps

σ_{k} : ℚ[x]/(Φ_{m}(x)) ↪ ℂ,

1 ≤ k < m, (k, m) = 1, where

σ_{k}(x) = ζ_{m}^{k},

ζ_{m} being our fixed choice of primitive m^{th} root of unity. Note that ζ_{m}^{k} ∈ ℚ(ζ_{m}) for every k; it follows that ℚ(ζ_{m}) = ℚ(ζ_{m}^{k}) for all k relatively prime to m. In particular, the images of the σ_{i} coincide, so ℚ[x]/(Φ_{m}(x)) is Galois over ℚ. This means that we can write ℚ(ζ_{m}) for ℚ[x]/(Φ_{m}(x)) without much fear of ambiguity; we will do so from now on, the identification being ζ_{m} ↦ x. One advantage of this is that one can easily talk about cyclotomic fields being extensions of one another, or intersections or compositums; all of these things take place considering them as subfields of ℂ.

We now investigate some basic properties of cyclotomic fields. The first issue is whether or not they are all distinct; to determine this, we need to know which roots of unity lie in ℚ(ζ_{m}). Note, for example, that if m is odd, then −ζ_{m} is a 2m^{th} root of unity. We will show that this is the only way in which one can obtain any non-m^{th} roots of unity.

Lemma 3.4. If m divides n, then ℚ(ζ_{m}) is contained in ℚ(ζ_{n}).

Proof. Since ζ_{n}^{n/m} = ζ_{m}, we have ζ_{m} ∈ ℚ(ζ_{n}), so the result is clear.

Lemma 3.5. If m and n are relatively prime, then

ℚ(ζ_{m}, ζ_{n}) = ℚ(ζ_{mn})

and

ℚ(ζ_{m}) ∩ ℚ(ζ_{n}) = ℚ.

(Recall that ℚ(ζ_{m}, ζ_{n}) is the compositum of ℚ(ζ_{m}) and ℚ(ζ_{n}).)

Proof. One checks easily that ζ_{m}ζ_{n} is a primitive mn^{th} root of unity, so that ℚ(ζ_{mn}) ⊆ ℚ(ζ_{m}, ζ_{n}). Furthermore, by Lemma A.3.3,

[ℚ(ζ_{m}, ζ_{n}) : ℚ] ≤ [ℚ(ζ_{m}) : ℚ][ℚ(ζ_{n}) : ℚ] = φ(m)φ(n) = φ(mn);

3. EXAMPLE : CYCLOTOMIC FIELDS 13

since [ℚ(ζ_{mn}) : ℚ] = φ(mn), this implies that

ℚ(ζ_{m}, ζ_{n}) = ℚ(ζ_{mn}).

We now have a field diagram

[figure: ℚ(ζ_{m}, ζ_{n}); ℚ(ζ_{m}); φ(m); ℚ(ζ_{n}); φ(n); ℚ(ζ_{m}) ∩ ℚ(ζ_{n}); ℚ]

We know that ℚ(ζ_{m}, ζ_{n}) has degree φ(mn) over ℚ, so we must have

[ℚ(ζ_{m}, ζ_{n}) : ℚ(ζ_{m})] = φ(n)

and

[ℚ(ζ_{m}, ζ_{n}) : ℚ(ζ_{n})] = φ(m).

Now Lemma A.3.3 shows that

[ℚ(ζ_{m}) : ℚ(ζ_{m}) ∩ ℚ(ζ_{n})] ≥ φ(m)

and thus that ℚ(ζ_{m}) ∩ ℚ(ζ_{n}) = ℚ.

Proposition 3.6. For any m and n,

ℚ(ζ_{m}, ζ_{n}) = ℚ(ζ_{[m,n]})

and

ℚ(ζ_{m}) ∩ ℚ(ζ_{n}) = ℚ(ζ_{(m,n)});

here [m, n] and (m, n) denote the least common multiple and the greatest common divisor of m and n, respectively.

Proof. Write m = p_{1}^{e_{1}} ⋯ p_{k}^{e_{k}} and n = p_{1}^{f_{1}} ⋯ p_{k}^{f_{k}} where the p_{i} are distinct primes. (We allow e_{i} or f_{i} to be zero.) By Lemma 3.5 we have

ℚ(ζ_{m}) = ℚ(ζ_{p_{1}^{e_{1}}} )ℚ(ζ_{p_{2}^{e_{2}}} ) ⋯ ℚ(ζ_{p_{k}^{e_{k}}})

and

ℚ(ζ_{n}) = ℚ(ζ_{p_{1}^{f_{1}}} )ℚ(ζ_{p_{2}^{f_{2}}} ) ⋯ ℚ(ζ_{p_{k}^{f_{k}}}).

Thus

ℚ(ζ_{m}, ζ_{n}) = ℚ(ζ_{p_{1}^{e_{1}}} ) ⋯ ℚ(ζ_{p_{k}^{e_{k}}})ℚ(ζ_{p_{1}^{f_{1}}} ) ⋯ ℚ(ζ_{p_{k}^{f_{k}}})
= ℚ(ζ_{p_{1}^{e_{1}}} )ℚ(ζ_{p_{1}^{f_{1}}} ) ⋯ ℚ(ζ_{p_{k}^{e_{k}}})ℚ(ζ_{p_{k}^{f_{k}}})
= ℚ(ζ_{p_{1}^{max{e_{1},f_{1}}}}) ⋯ ℚ(ζ_{p_{k}^{max{e_{k},f_{k}}}})
= ℚ(ζ_{p_{1}^{max{e_{1},f_{1}}}···p_{k}^{max{e_{k},f_{k}}}})
= ℚ(ζ_{[m,n]});

the third equality uses Lemma 3.4, the fourth uses Lemma 3.5 and the last uses a standard expression for least common multiples. An entirely similar computation shows that ℚ(ζ_{m}) ∩ ℚ(ζ_{n}) = ℚ(ζ_{(m,n)}).

Corollary 3.7. If m is even, then the only roots of unity in ℚ(ζ_{m}) are the m^{th} roots of unity. If m is odd, then the only roots of unity in ℚ(ζ_{m}) are the 2m^{th} roots of unity.

Proof. Suppose that ζ_{n} ∈ ℚ(ζ_{m}). Then ζ_{m}ζ_{n} is a [m, n]^{th} root of unity, so ℚ(ζ_{[m,n]}) ⊆ ℚ(ζ_{m}). Thus

φ([m, n]) ≤ φ(m).

One easily shows that this can happen only if m is odd and n divides 2m, or if m is even and n divides m. This proves the corollary.

Corollary 3.8. If m < n and ℚ(ζ_{m}) = ℚ(ζ_{n}), then m is odd and n = 2m.

4. Galois theory of number fields

Let K be a Galois extension of ℚ of degree n. Recall that this means that if σ_{1}, …, σ_{n} denote the complex embeddings of K, then the σ_{i} all have the same image in ℂ. Let us denote this image by K_{0} for the remainder of this section. We wish to reinterpret the complex embeddings as automorphisms of K. To do this, fix one embedding, say σ_{1} : K → K_{0}. Consider the n maps

σ_{1}^{−1} ∘ σ_{i} : K → K.

These maps are all automorphisms of K (that is, isomorphisms from K to K) since the σ_{i} are all isomorphisms from K to K_{0}.

We claim that in fact these are all of the automorphisms of K. So suppose that σ : K → K is any automorphism of K. Then σ_{1} ∘ σ : K → K_{0} ↪ ℂ is a complex embedding of K, and thus equals one of the σ_{i}. Thus σ = σ_{1}^{−1} ∘ σ_{i}, as claimed.

In general, if M is any sort of object, then the set of automorphisms of M form a group with composition as the group law; this is because the composition of two automorphisms and the inverse of an automorphism are again automorphisms. We define the Galois group Gal(K/ℚ) of K over ℚ to be the group of automorphisms of K; our above arguments show that as a set Gal(K/ℚ) is just the maps σ_{1}^{−1} ∘ σ_{i} : K → K. Note in particular that

(σ_{1}^{−1} ∘ σ_{i}) ∘ (σ_{1}^{−1} ∘ σ_{j})

and

(σ_{1}^{−1} ∘ σ_{i})^{−1} = σ_{i}^{−1} ∘ σ_{1}

are again of the form σ_{1}^{−1} ∘ σ_{k} for some k, although it is not at all clear which k it is.

Note that Gal(K/ℚ) has order n; even if K is not Galois one could still consider the automorphisms of K, but the above construction no longer works and it is somewhat harder to determine how many automorphisms there are.

When one actually computes Galois groups, it is usually much simpler to consider the fields as subfields of ℂ. So let K be a Galois number field which is also a subfield of ℂ. The automorphisms of K are now simply its complex embeddings σ_{i} : K → K ⊆ ℂ. (With our earlier notation, we really are just considering the case where σ_{1} is the identity map.) Note in particular that σ_{i} ∘ σ_{j} and σ_{i}^{−1} are also complex embeddings of K, although it is not immediately clear which.

To determine which, let α be a primitive element for K over ℚ and let α_{1} = α, α_{2}, …, α_{n} be its conjugates, so that the complex embeddings of K are given by σ_{i}(α) = α_{i}. We can now determine σ_{i} ∘ σ_{j} simply by determining for which k we have

σ_{i} ∘ σ_{j}(α) = α_{k};

we then have σ_{i} ∘ σ_{j} = σ_{k}.

Example 4.1. Let d be a squarefree integer (other than 1) and consider the field ℚ(\sqrt{d}). This has the two embeddings σ_{1} and σ_{2} characterized by

σ_{1}(\sqrt{d}) = \sqrt{d}

and

σ_{2}(\sqrt{d}) = −\sqrt{d}.

We find that

σ_{2}σ_{2}(\sqrt{d}) = σ_{2}(−\sqrt{d}) = −σ_{2}(\sqrt{d}) = \sqrt{d};

that is, σ_{2}^{2} = σ_{1}. This confirms that

Gal(ℚ(\sqrt{d})/ℚ) ≅ ℤ/2ℤ

as it must be; σ_{1} is the identity element and σ_{2} is the non-trivial element.

Example 4.2. Consider the field ℚ(\sqrt{2}, \sqrt{3}). This field has degree 4 over ℚ, with complex embeddings characterized by

σ_{1}(\sqrt{2}) = \sqrt{2}, σ_{1}(\sqrt{3}) = \sqrt{3}
σ_{2}(\sqrt{2}) = −\sqrt{2}, σ_{2}(\sqrt{3}) = \sqrt{3}
σ_{3}(\sqrt{2}) = \sqrt{2}, σ_{3}(\sqrt{3}) = −\sqrt{3}
σ_{4}(\sqrt{2}) = −\sqrt{2}, σ_{4}(\sqrt{3}) = −\sqrt{3}

One computes easily that each of σ_{2}, σ_{3} and σ_{4} have square σ_{1} and that the product of any two of them is the third, so that Gal(ℚ(\sqrt{2}, \sqrt{3})/ℚ) is isomorphic to ℤ/2ℤ × ℤ/2ℤ.

Example 4.3. Consider the cyclotomic field ℚ(ζ_{m}). This has φ(m) complex embeddings σ_{k} (for (k, m) = 1), where σ_{k}(ζ_{m}) = ζ_{m}^{k}. We compute

σ_{k}σ_{j}(ζ_{m}) = σ_{k}(ζ_{m}^{j}) = σ_{k}(ζ_{m})^{j} = ζ_{m}^{jk};

if jk ≡ l (mod m), then this shows that σ_{k}σ_{j} = σ_{l}. In particular, we obtain a map

Gal(ℚ(ζ_{m})/ℚ) → (ℤ/mℤ)^{∗}

sending σ_{k} to the class of k in (ℤ/mℤ)^{∗}; the above calculation shows that this is a group homomorphism. It is also clearly bijective by our characterization of the σ_{k}. Thus we have obtained an isomorphism

Gal(ℚ(ζ_{m})/ℚ) ⟶^{≅} (ℤ/mℤ)^{∗}.

Note that if ζ = ζ_{m}^{i} is any m^{th} root of unity in ℚ(ζ_{m}), then
σ_{k}(ζ) = σ_{k}(ζ_{m}^{i}) = σ_{k}(ζ_{m})^{i} = ζ_{m}^{ki} = ζ^{k}.

This means that the above isomorphism is completely canonical, in the sense that the automorphism corresponding to k ∈ (ℤ/mℤ)^{∗} has the effect of exponentiation by k on any m^{th} root of unity in ℚ(ζ_{m}). Note also that Gal(ℚ(ζ_{m})/ℚ) is abelian; in some sense this is the fact which will make all of the applications in this class work.

Example 4.4. For a non-abelian example, let \sqrt[4]{2} be the positive real fourth root of 2 and consider the field ℚ(\sqrt{−1}, \sqrt[4]{2}). The conjugates of \sqrt[4]{2} are \sqrt[4]{2}, \sqrt{−1}\sqrt[4]{2}, −\sqrt[4]{2}, −\sqrt{−1}\sqrt[4]{2}. This field has degree 8 over ℚ, with embeddings σ_{0}, …, σ_{7} characterized by

σ_{i}(\sqrt[4]{2}) = \sqrt{−1}^{i} \sqrt[4]{2}

and

σ_{i}(\sqrt{−1}) = \begin{cases}\sqrt{−1} & i = 0, 1, 2, 3; \\ −\sqrt{−1} & i = 4, 5, 6, 7. \end{cases}

(To see that this field has degree 8, it is enough to show that [ℚ(\sqrt[4]{2}) : ℚ] = 4 and that \sqrt{−1} ∉ ℚ(\sqrt[4]{2}). The first of these follows from Exercise 1.8 and the second can be done in the same way as Exercise 1.16.)

We compute easily

σ_{i}σ_{j}(\sqrt{−1}) = \begin{cases}\sqrt{−1} & both i, j ∈ \{0, 1, 2, 3\} or both i, j ∈ \{4, 5, 6, 7\}; \\ −\sqrt{−1} & otherwise. \end{cases}

On the other hand,

σ_{i}σ_{j}(\sqrt[4]{2}) = σ_{i}(\sqrt{−1}^{j} \sqrt[4]{2})

= σ_{i}(\sqrt{−1})^{j}σ_{i}(\sqrt[4]{2})

= \begin{cases}\sqrt{−1}^{j}\sqrt{−1}^{i} \sqrt[4]{2} & i ∈ \{0, 1, 2, 3\}; \\ (−\sqrt{−1})^{j}\sqrt{−1}^{i} \sqrt[4]{2} & i ∈ \{4, 5, 6, 7\}; \end{cases}

= \begin{cases}\sqrt{−1}^{i+j} \sqrt[4]{2} & i ∈ \{0, 1, 2, 3\}; \\ \sqrt{−1}^{i−j} \sqrt[4]{2} & i ∈ \{4, 5, 6, 7\}. \end{cases}

For example,

σ_{3}σ_{5}(\sqrt{−1}) = −\sqrt{1}

and

σ_{3}σ_{5}(\sqrt[4]{2}) = \sqrt{−1}^{8} \sqrt[4]{2} = \sqrt[4]{2},

so σ_{3}σ_{5} = σ_{4}. On the other hand,

σ_{5}σ_{3}(\sqrt{−1}) = −\sqrt{−1}

and

σ_{5}σ_{3}(\sqrt[4]{2}) = \sqrt{−1}^{2} \sqrt[4]{2} = −\sqrt[4]{2}

so σ_{5}σ_{3} = σ_{6}. Thus Gal(ℚ(\sqrt[4]{2}, i)/ℚ) is non-abelian; with a little squinting one discovers that it is isomorphic to the dihedral group of order 8.

5. Relative extensions

5.1. Relative embeddings. Let L and K be two number fields such that L ⊇ K; set n = [L : ℚ], m = [K : ℚ], d = [L : K] = n/m. We wish to relate the complex embeddings of L to those of K. Let us fix an embedding

σ : K ↪ ℂ

and determine how many complex embeddings of L restrict to σ on K. (Such an embedding of L is said to extend σ.)

Choose a primitive element α for L/K and let f(x) ∈ K[x] be its minimal polynomial. Let g(x) = σ(f(x)) ∈ ℂ[x]. Since ℂ is algebraically closed and g(x) is irreducible in K[x], g(x) has d distinct roots α_{1}, …, α_{d} in ℂ. For each such root we can define a map

τ_{i} : L ↪ ℂ

to be σ on K and to send α to α_{i}. This procedure yields exactly d distinct embeddings of L into ℂ, all of which restrict to σ on K. Explicitly, we have

τ_{i}(a_{0}+a_{1}α+a_{2}α^{2}+⋯+a_{n−1}α^{n−1}) = σ(a_{0})+σ(a_{1})α_{i}+σ(a_{2})α_{i}^{2}+⋯+σ(a_{n−1})α^{n−1}

where the a_{i} are all in K.

We can actually conclude that these are all of the embeddings of L into ℂ extending σ by a counting argument. Specifically, for any of the m complex embeddings σ′ of K the above procedure yields d complex embeddings of L restricting to σ′ on K. In this way we can therefore obtain md = n distinct complex embeddings of L. But these are then all of the n complex embeddings of L; this implies that each embedding of K has exactly d extensions to L, as if it had any more then we would obtain too many complex embeddings of L.

Summarizing our work to this point, we have shown that each complex embedding of K extends to d complex embeddings of L. In the case that K = ℚ, σ must be the unique embedding of ℚ into ℂ, and this all reduces to our original discussion of complex embeddings.

We extend some of our earlier terminology to this situation. Given α ∈ L with minimal polynomial f(x) over K of degree e, we say that the σ-K-conjugates of α are the e (distinct) complex roots of σ(f(x)). Continuing to let τ_{1}, …, τ_{d} be the extensions of σ to L, we find that each σ-K-conjugate of α occurs precisely d/e times among the numbers τ_{1}(α), …, τ_{d}(α). To see this, fix a σ-K-conjugate α_{1} of α and consider the embedding ρ : K(α) ↪ ℂ given by σ on K and sending α to α_{1}. By the above discussion applied to L/K(α), there are exactly [L : K(α)] = d/e embeddings of L extending ρ, which means that there are exactly d/e embeddings of L extending σ and sending α to α_{1}, as claimed. In particular, in the case K = ℚ we find that each conjugate of α appears exactly d/e times among the images of α under the complex embeddings of L.

Example 5.1. Let K = ℚ(\sqrt{2}), L = ℚ(\sqrt{2}, \sqrt{3}). Let σ_{1} : K ↪ ℂ be the complex embedding

σ_{1}(a + b\sqrt{2}) = a + b\sqrt{2}.

The two extensions τ_{1}, τ_{2} : L ↪ ℂ of σ to L are given by

τ_{1}(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}
τ_{2}(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a + b\sqrt{2} − c\sqrt{3} − d\sqrt{6}.

Similarly, the two embeddings extending the other embedding σ_{2} of K are

τ_{3}(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a − b\sqrt{2} + c\sqrt{3} − d\sqrt{6}
τ_{2}(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a − b\sqrt{2} − c\sqrt{3} + d\sqrt{6}.

Let α = \sqrt{2} + \sqrt{3} ∈ K. The σ_{1}-K-conjugates of α are τ_{1}(α) = \sqrt{2} + \sqrt{3} and τ_{2}(α) = \sqrt{2} − \sqrt{3}. The σ_{2}-K-conjugates of α are τ_{3}(α) = −\sqrt{2} + \sqrt{3} and τ_{4}(α) = −\sqrt{2} − \sqrt{3}. Together these give the four conjugates of α.

5.2. Relations to characteristic polynomials. Our next goal is to relate complex embeddings to norms and traces. We first work with characteristic polynomials.

Lemma 5.2. Let L/K be an extension of number fields of degree d and let α be a primitive element for L/K. Let σ be a fixed complex embedding of K and let τ_{1}, …, τ_{d} be the extensions of σ to L. Let g(x) ∈ K[x] be the characteristic polynomial (and thus the minimal polynomial) of α for L/K. Then

σ(g(x)) = ∏_{i=1}^{d}(x − τ_{i}(α)) ∈ ℂ[x].

Proof. We know that the τ_{i} are constructed by taking the complex roots α_{1}, …, α_{d} of σ(g(x)) and mapping α to each α_{i}; that is, we have τ_{i}(α) = α_{i}. Thus the τ_{i}(α) are precisely the complex roots of σ(g(x)), which is the statement of the lemma.

Proposition 5.3. Let L/K be an extension of number fields of degree d and let α be an arbitrary element of L. Let σ be a fixed complex embedding of K and let τ_{1}, …, τ_{d} be the extensions of σ to L. Let g(x) be the characteristic polynomial of α for L/K. Then

σ(g(x)) = ∏_{i=1}^{d}(x − τ_{i}(α)).

Proof. Let α have minimal polynomial f(x) of degree e over K and consider the tower of fields L/K(α)/K. Let ρ_{1}, …, ρ_{e} be the extensions of σ to K(α). By Lemma 5.2 we know that

σ(f(x)) = ∏_{i=1}^{e}(x − ρ_{i}(α)).

By Corollary A.4.4 we know that g(x) = f(x)^{d/e}. We also know, from the discussion of the previous section, that each σ-K-conjugate ρ_{i}(α) of α occurs exactly d/e times among τ_{1}(α), …, τ_{d}(α). Combining all of these facts yields the proposition.

The next result gives the fundamental connection between embeddings and norms and traces.

Corollary 5.4. Let L/K be an extension of number fields of degree d. Let σ : K ↪ ℂ be a complex embedding of K and let τ_{1}, …, τ_{d} be the d complex embeddings of L extending σ. Then for any α ∈ K,

σ(N_{L/K} α) = τ_{1}(α) ⋯ τ_{d}(α)

and

σ(Tr_{L/K} α) = τ_{1}(α) + ⋯ + τ_{d}(α).

Proof. This is immediate from Proposition 5.3 and the definitions of the norm and trace in terms of characteristic polynomials.

For convenience, let us restate our main results in the case of an extension K/ℚ.

Corollary 5.5. Let K be a number field of degree n with complex embeddings σ_{1}, …, σ_{n}. Let α be an element of K with characteristic polynomial g(x). Then

g(x) = ∏_{i=1}^{n}(x − σ_{i}(α)).

Furthermore,

N_{K/ℚ} α = σ_{1}(α) ⋯ σ_{n}(α)

and

Tr_{K/ℚ} α = σ_{1}(α) + ⋯ + σ_{n}(α).

5.3. Relative Galois extensions. Let L/K be an extension of number fields of degree d. Fix a complex embedding σ of K with image K_{0}, and let τ_{1}, …, τ_{d} be the extensions of σ to L. If the τ_{i} all have the same image L_{0} in ℂ, we will say that L is Galois over K. (We will check in a moment that this definition is independent of the choice of σ.)

Let us define the Galois group Gal(L/K) to be the group of K-linear automorphisms of L; that is, it is the group of automorphisms of L which fix every element of K. As with Galois groups over ℚ, we can describe Gal(L/K) in terms of embeddings. Specifically, fix the embedding τ_{1} and consider the d maps

τ_{1}^{−1} ∘ τ_{i} : L → L.

These are automorphisms of L, since the τ_{i} are all isomorphisms; furthermore, they are the identity on K, since both τ_{i} and τ_{1} act on K as σ. Thus we have exhibited d K-linear automorphisms of L.

We claim that all K-linear automorphisms of L are of this form. So suppose that τ : L → L is another such automorphism. Then τ_{1} ∘ τ : L → L_{0} is a complex embedding of L. Furthermore, it is simply σ on K, since τ is the identity on K. Thus τ_{1} ∘ τ must be one of the τ_{i}, so that τ = τ_{1}^{−1} ∘ τ_{i}, as claimed.

Notice now that the definition of Gal(L/K) made no mention of σ. In particular, let σ′ : K → ℂ be another complex embedding of K with extensions τ′_{1}, …, τ′_{d}. We claim that the τ′_{i} all have the same image. To see this, note that for every ρ ∈ Gal(L/K), τ′_{1} ∘ ρ is a complex embedding of L which extends σ. The d different elements of Gal(L/K) yield d different such embeddings, all with the same image τ′_{1}(L); these must be nothing more than τ′_{1}, …, τ′_{d}, since those are all of the embeddings of L which extend σ. In particular, this shows that the property of L being Galois over K is independent of the choice of embedding of K.

As before, one can actually compute Gal(L/K) by considering L and K as specific subfields of ℂ and then considering the action on σ-K-conjugates of generators.

Example 5.6. Take L = ℚ(\sqrt{2}, \sqrt{3}) and K = ℚ(\sqrt{2}). We computed Gal(L/ℚ) in Example 4.2. Of the four automorphisms given there, σ_{1} and σ_{3} are the identity on K, so we can identify

Gal(L/K) = {σ_{1}, σ_{3}} ≅ ℤ/2ℤ.

Notice in particular that if L/ℚ is Galois, then Gal(L/K) is a subgroup of Gal(L/ℚ). The main theorem of Galois theory is a generalization of this fact.

Theorem 5.7. Let L/K be a Galois extension of number fields. There is a bijective correspondence between subgroups of Gal(L/K) and subfields of L, given by

H ⊆ Gal(L/K) → L^{H} = {x ∈ L; h(x) = x for all h ∈ H}
{σ ∈ Gal(L/K); σ|_{L′} = id} ← L′.

This correspondence is inclusion reversing, and L is Galois over each subfield L^{H} with Galois group Gal(L/L^{H}) = H. Lastly, L^{H} is Galois over K if and only if H is a normal subgroup of Gal(L/K), in which case Gal(L^{H}/K) ≅ Gal(L/K)/H.

Corollary 5.8. Let L/K be a Galois extension and let α be an element of L. Then α ∈ K if and only if σ(α) = α for all σ ∈ Gal(L/K).

Proof. The fact that σ(α) = α for all α ∈ K and all σ ∈ Gal(L/K) is part of the definition of the Galois group. Conversely, by Theorem 5.7, the subfield K of L must correspond to the largest subgroup of Gal(L/K); that is, it corresponds to the entire group G = Gal(L/K), and thus by the definition of the Galois correspondence we find that K = L^{G}, as claimed.

We conclude with the promised strengthening of Lemma A.3.3.

Lemma 5.9. Let M/K be an extension of number fields and let L_{1} and L_{2} be subfields of M containing K. Suppose that L_{2} is Galois over L_{1} ∩ L_{2}. Then L_{1}L_{2} is Galois over L_{1} and

Gal(L_{1}L_{2}/L_{1}) ≅ Gal(L_{2}/L_{1} ∩ L_{2}).

In particular,

[L_{1}L_{2} : L_{1}] = [L_{2} : L_{1} ∩ L_{2}]

and

[L_{1}L_{2} : L_{2}] = [L_{1} : L_{1} ∩ L_{2}].

Proof. Set d = [L_{2} : L_{1} ∩ L_{2}]. Let σ be an element of Gal(L_{2}/L_{1} ∩ L_{2}). We define an automorphism σ̃ of L_{1}L_{2} to act as σ on L_{2} and to be the identity on L_{1}; one checks easily that this is well-defined, since σ is the identity on L_{1}∩L_{2}. Applying this construction to every element of Gal(L_{2}/L_{1} ∩ L_{2}), we obtain d automorphisms of L_{1}L_{2} fixing L_{1}.

Let now τ : L_{1} ↪ ℂ be a complex embedding and let ρ : L_{1}L_{2} ↪ ℂ extend τ. Then the d maps ρ ∘ σ̃ are all distinct complex embeddings of L_{1}L_{2} extending τ, and they all have the same image. By Lemma A.3.3,

[L_{1}L_{2} : L_{1}] ≤ d,

so the existence of these embeddings implies both that

[L_{1}L_{2} : L_{1}] = d

and that L_{1}L_{2} is Galois over L_{1}, since we have exhibited d automorphisms of L_{1}L_{2} over L_{1}.

Our map σ ↦ σ̃ can now be interpreted as a map

Gal(L_{2}/L_{1} ∩ L_{2}) → Gal(L_{1}L_{2}/L_{1}).

One checks immediately that the map restricting an automorphism of L_{1}L_{2} to L_{2} gives an inverse map, so they must both be isomorphisms. This proves everything but the last equality of the lemma; this follows from Lemma A.1.1 and the second to last equality.