TANT 4 - Complete fields
Hello there! These are notes for the fourth class of the course "Topics in algebra and number theory" held in Block 4 of the academic year 2017/18 at the University of Copenhagen.
In the previous lecture we have struggled with non-Archimedean places of a number field \( K \) and of a function field \( F(t) \). In particular we have proved that for every number field \( K \) the maps \[ \begin{aligned} \Sigma_K^{\infty} &\to \operatorname{Spec}(\mathcal{O}_K) \setminus \{ \mathbf{0} \} \\ [\phi] &\mapsto \{ x \in \mathcal{O}_K \mid \phi(x) < 1 \} \end{aligned} \qquad \text{and} \qquad \begin{aligned} \operatorname{Spec}(\mathcal{O}_K) \setminus \{ \mathbf{0} \} &\to \Sigma_K^{\infty} \\ \mathfrak{p} &\mapsto [\, \lvert \cdot \rvert_{\mathfrak{p}} \,] \end{aligned} \] are one the inverse of the other. Now we want to prove a similar statement for the Archimedean places \( \Sigma_{K,\infty} \). Recall that the Archimedean world is the world of \( \mathbb{R} \) and \( \mathbb{C} \), i.e. it is the world of the analysis that we are accustomed to.
For this reason to understand Archimedean places we need to start building an analytic theory of valued fields. In the second lecture we have defined absolute values over a field and we have seen that they carry with them a canonical topology. Now we want to define a category of valued fields.
Definition 1 The category of valued fields is the category whose objects are couples \( (K,\phi) \), where \( K \) is a field and \( \phi \colon K \to \mathbb{R}_{\geq 0} \) is an absolute value. A morphism \( (K,\phi) \to (L,\psi) \) in this category is given by any homomorphism of fields \( f \colon K \to L \) such that \( \psi(f(x)) = \phi(x) \) for all \( x \in K \).
Now it is natural to look for special objects in this category, e.g. for objects with universal properties. In the following section we will define such an object, the completion of a valued field \( (K,\phi) \), whose definition is related to the analytic properties of \( \phi \).
Definition 2 Let \( (K, \phi) \) be a valued field. We say that a sequence \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) converges to \( x_{\infty} \in K \) (with respect to \( \phi \)) if for every \( \varepsilon \in \mathbb{R}_{> 0} \) there exists \( n_{\varepsilon} \in \mathbb{N} \) such that for all \( n \geq n_{\varepsilon} \) we have that \( \phi(x_n - x_{\infty}) < \varepsilon \).
Exercise 3 Prove that if \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) converges to \( x_{\infty} \in K \) and to \( x_{\infty}' \in K \) then \( x_{\infty} = x_{\infty}' \).
In view of the previous exercise we will say that a sequence \( \{ x_n \} \) converges in \( K \) if it converges to some (and so, to one) \( x_{\infty} \in K \). We call \( x_{\infty} \) the limit of the sequence \( \{ x_n \} \).
Exercise 4 Prove that convergence respects sums and products. More precisely, prove that if \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) converges to \( x_{\infty} \in K \) and \( \{ y_n \}_{n \in \mathbb{N}} \subseteq K \) converges to \( y_{\infty} \in K \) then \( \{ x_n + y_n \}_{n \in \mathbb{N}} \) converges to \( x_{\infty} + y_{\infty} \) and \( \{ x_n \cdot y_n \}_{n \in \mathbb{N}} \) converges to \( x_{\infty} \cdot y_{\infty} \).
Exercise 5 Prove that every morphism of valued fields \( f \colon (K, \phi) \to (L,\psi) \) is sequentially continuous. More precisely prove that if \( \{ x_n \} \subseteq K \) converges to \( x_{\infty} \in K \) then \( \{ f(x_n) \} \subseteq L \) converges to \( f(x_{\infty}) \).
As we all learned some time ago, if a sequence \( \{ x_n \}_{n \in \mathbb{N}} \) converges then the \( x_n \)'s get "closer and closer" as \( n \to +\infty \), i.e. the sequence \( \{ x_n \}_{n \in \mathbb{N}} \) is a Cauchy sequence.
Definition 6 Let \( (K, \phi) \) be a valued field. We say that a sequence \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) is a Cauchy sequence if for every \( \varepsilon \in \mathbb{R}_{> 0} \) there exists \( n_{\varepsilon} \in \mathbb{N} \) such that for all \( n, m \geq n_{\varepsilon} \) we have that \( \phi(x_n - x_m) < \varepsilon \).
Exercise 7 Prove that Cauchy sequences respect sums and products. More precisely, prove that if \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) and \( \{ y_n \}_{n \in \mathbb{N}} \subseteq K \) are Cauchy sequences then \( \{ x_n + y_n \}_{n \in \mathbb{N}} \) and \( \{ x_n \cdot y_n \}_{n \in \mathbb{N}} \) are also Cauchy sequences.
Testing whether a sequence is Cauchy or not can be hard but it is doable in principle, whereas testing if a sequence has a limit or not can be really tough. For example we know that the sequences of rational numbers \[ x_n := \sum_{k=0}^{n} (-1)^{k}\frac{\left\lfloor\log_2(k + 2) \right\rfloor}{k + 2} \qquad \text{and} \qquad y_n := \sum_{k = 0}^{n} \frac{(-1)^{k}}{(2 k+1)^2} \] are Cauchy sequences with respect to the canonical Archimedean absolute value \( \lvert \cdot \rvert_{\infty} \colon \mathbb{Q} \to \mathbb{R}_{> 0} \) but no one knows if these sequences converge in \( \mathbb{Q} \) or not (see here and here).
The problem here is that \( (\mathbb{Q}, \lvert \cdot \rvert_{\infty}) \) is a valued field "with many holes" and so not all the Cauchy sequences converge. For example for every \( \alpha \in \mathbb{Q}_{> 0} \) we can define a sequence \( \{ x_n \}_{n \in \mathbb{N}} \) by setting \[ x_0 := 1 \qquad \text{and} \qquad x_{n+1} := \frac{1}{2} \left(x_n + \frac{\alpha}{x_n}\right). \] It is not difficult to prove that this sequence is always Cauchy for \( \lvert \cdot \rvert_{\infty} \). However, it converges in \( \mathbb{Q} \) if and only if \( \alpha \) is a square (so, for example, it does not converge if \( \alpha = 2 \)).
In order to avoid this sort of problems it would be nice if every Cauchy sequence was convergent. So nice that this property deserves a name.
Definition 8 A complete field is a valued field \( (K,\phi) \) such that any Cauchy sequence \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) converges.
Exercise 9 Prove that the field of real numbers \( (\mathbb{R},\lvert \cdot \rvert) \) is complete with respect to the canonical absolute value defined in Example 9 of the second lecture. Use in particular the definition of \( \mathbb{R} \) as the set of sequences \[ \mathbb{R} := \left\{ \varepsilon a_n a_{n - 1} \cdots a_0.a_{-1} a_{-2} \cdots a_{-m} \cdots \mid \varepsilon \in \{ \pm 1 \}, \ a_j \in \{ 0, \dots, 9 \} \right\} \setminus \mathcal{S}_{9} \] where \( \mathcal{S}_{9} \) is the set of sequences which are definitely equal to \( 9 \).
There is another classical Archimedean complete field: the field of complex numbers \( \mathbb{C} \) with the canonical absolute value. To prove that it is complete starting from the completeness of \( \mathbb{R} \) we will need the following lemma, which applies also to non-Archimedean absolute values.
Lemma 10 Let \( (K,\phi) \) be a complete valued field, and suppose that \( x^2 + 1 \in K[x] \) is irreducible (i.e. that \( -1 \) is not a square in \( K \)). Then \( L := K[x]/(x^2 + 1) = K(i) \) is a quadratic field extension of \( K \). \( L \) is also complete with respect to the valuation \( \psi \colon L \to \mathbb{R}_{\geq 0} \) defined by \( \psi(a + i b) := \sqrt{\phi(a^2 + b^2)} \), and the inclusion \( K \hookrightarrow L \) is a morphism of valued fields.
Proof The fact that \( L \) is a quadratic field extension of \( K \) is part of any course in Galois theory.
It is now trivial to observe that \( \psi \) satisfies the first two axioms of the definition of an absolute value, and that \( \phi = \psi \circ \iota \), where \( \iota \colon K \hookrightarrow L \). Showing that \( \psi \) satisfies the triangle inequality is rather difficult, and we will prove it for a general Galois extension \( K \subseteq L \) in one of the following lectures. If you can't wait you can look at Lemma 2.2 of these notes by Peter Stevenhagen.
Let now \( \{ a_n + i b_n \}_{n \in \mathbb{N}} \subseteq L \) be a sequence such that \( \{ a_n \} \) and \( \{b_n \} \) tend to zero with respect to \( \phi \). Then it is clear from the definition of \( \psi \) that \( \{ a_n + i b_n \} \) tends to zero with respect to \( \psi \).
Vice versa, let \( \{ a_n + i b_n \}_{n \in \mathbb{N}} \subseteq L \) be a sequence which tends to zero with respect to \( \psi \), and suppose by contradiction that \( \{ x_n \}_{n \in \mathbb{N}} \) does not tend to zero with respect to \( \phi \). Then there would exist a number \( C \in \mathbb{R}_{> 0} \) such that \( \phi(x_n) \geq C \) for all \( n \in \mathbb{N} \) and thus the sequence \( \{ 1 + (y_n/x_n)^2 \}_{n \in \mathbb{N}} \) would go to zero with respect to \( \phi \), because \( \phi(1 + (y_n/x_n)^2) = \phi(x_n)^{-2} \, \phi(x_n + y_n) \).
It is now not difficult to see that this implies that the sequence \( z_n := y_n/x_n \) is a Cauchy sequence. Since \( K \) is complete with respect to \( \phi \) this sequence converges to an element \( z_{\infty} \in K \). Using Exercise 4 we see then that \( z_{\infty}^2 = -1 \). But this contradicts the fact that \( x^2 + 1 \in K[x] \) is irreducible, and so it follows that \( \{ x_n \}_{n \in \mathbb{N}} \) actually tends to zero with respect to \( \phi \). We can make a similar argument to show that \( y_n \) also tends to zero with respect to \( \phi \).
Using again Exercise 4 we see now that a sequence \( \{ x_n + i y_n \}_{n \in \mathbb{N}} \subseteq L \) converges to \( x_{\infty} + i y_{\infty} \) with respect to \( \psi \) if and only if \( \{ x_n \} \) converges to \( x_{\infty} \) and \( \{ y_n \} \) converges to \( y_{\infty} \) with respect to \( \phi \). This implies in particular that \( L \) is complete with respect to \( \psi \), as we wanted to show. Q.E.D.
A question that now arises automatically is: how many complete fields do exist? The answer is... quite many! Indeed, the following theorem associates to every valued field \( (K,\phi) \) a complete field \( (K_{\Phi}, \Phi) \) which is the complete field "closest" to \( (K,\phi) \).
Definition 11 Let \( (K,\phi) \) be any valued field. We say that a complete valued field \( (L,\psi) \) is a completion of \( (K,\phi) \) if there is a morphism of valued fields \( \iota \colon (K,\phi) \to (L,\psi) \) such that for every complete valued field \( (M,\xi) \) and every morphism of valued fields \( f \colon (K,\phi) \to (M,\xi) \) there exists a unique morphism of valued fields \( f' \colon (L,\psi) \to (M,\xi) \) such that \( f = f' \circ \iota \).
Exercise 12 Prove that if \( (L_1,\psi_1) \) and \( (L_2,\psi_2) \) are two completions of the same valued field \( (K,\phi) \) then there exists a unique isomorphism of valued fields \( (L_1,\psi_1) \cong (L_2,\psi_2) \).
Theorem 13 Let \( (K,\phi) \) be a valued field. Then there exists a completion \( (L,\psi) \) of \( (K,\phi) \).
Proof Define \( R \) to be the set of all Cauchy sequences \( \{ x_n \} \subseteq K \). If we define \[ \{ x_n \} + \{ y_n \} := \{ x_n + y_n \} \qquad \text{and} \qquad \{ x_n \} \cdot \{ y_n \} := \{ x_n \cdot y_n \} \] then it is clear from Exercise 7 that \( R \) is a ring with respect to these "coordinate-wise" sum and product operations. Observe moreover that the subset \[ \mathfrak{m} := \left\{ \{x_n\} \in R \mid \lim_{n \to +\infty} \phi(x_n) = 0 \right\} \subseteq R \] is an ideal, because clearly if \( \{x_n\} \in \mathfrak{m} \) and \( \{y_n\} \in R \) then \( \phi(x_n y_n) = \phi(x_n) \phi(y_n) \to 0 \) as \( n \to + \infty \) because \( \{ y_n \} \) is a Cauchy sequence and thus \( \{ \phi(y_n) \}_{n \in \mathbb{N}} \subseteq \mathbb{R}_{\geq 0} \) is bounded.
Moreover \( \mathfrak{m} \) is a maximal ideal because if \( \{ x_n \} \notin \mathfrak{m} \) then there exists \( n_0 \in \mathbb{N} \) such that \( x_n \neq 0 \) for \( n \geq n_0 \). Thus if we define a sequence \( \{ y_n \} \subseteq K \) by putting \( y_n := 1 \) for all \( n < n_0 \) and \( y_n := 1/x_n \) for all \( n \geq n_0 \) we have that the sequence \( \{ x_n \cdot y_n - 1 \} \) is definitely zero, and thus it is zero in the quotient ring \( R / \mathfrak{m} \), i.e. \( [\{ x_n \cdot y_n \}] = 1 \) in \( R / \mathfrak{m} \).
What we have shown implies that \( \mathfrak{m} \) is a maximal ideal, i.e. that \( L := R / \mathfrak{m} \) is a field. Moreover the map \( K \hookrightarrow R \) which sends an element \( x \in K \) to the constant (Cauchy) sequence \( \{ x_n = x \} \in R \) induces an inclusion of fields \( \iota \colon K \hookrightarrow L \).
Let now \( \Psi \colon R \to \mathbb{R}_{\geq 0} \) be defined as \( \Psi(\{x_n\}) := \lim_n \phi(x_n) \), which is well defined because \( R \) consists of Cauchy sequences. It is immediate to check that the composition of the inclusion \( K \hookrightarrow R \) and of \( \Psi \) is equal to \( \phi \) and it is also easy using the definition of \( \mathfrak{m} \) to show that \( \Psi \) descends to a map \( \psi \colon L \to \mathbb{R}_{\geq 0} \) which is an absolute value in view of Exercise 4.
We want now to prove that \( L \) is complete with respect to \( \psi \). To do so observe first of all that for every Cauchy sequence \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) and every \( \varepsilon > 0 \) there exists \( \alpha \in K \) such that \( \lim_{n \to +\infty} \phi(x_n - \alpha) < \varepsilon \). For example we can take \( \alpha = x_{n_{\varepsilon}} \), where \( n_{\varepsilon} \in \mathbb{N} \) is the integer associated to \( \varepsilon \) in the definition of a Cauchy sequence.
Let now \( \{ \{ x_n^j \}_n \}_j \subseteq L \) be a Cauchy sequence in \( L \), i.e. a sequence of sequences \( \{ x_n^j \}_n \subseteq K \) such that for all \( \varepsilon > 0 \) there exists \( j_0 \in \mathbb{N} \) such that for all \( j, l \geq n_0 \) we have that \( \lim_{n \to +\infty} \phi(x_n^j - x_n^l) < \varepsilon \). Since every sequence \( \mathbf{x}^j := \{ x_n^j \}_n \subseteq K \) is a Cauchy sequence we can pick elements \( a_j \in K \) such that \( \lim_{n \to +\infty} \phi( x_n^j - a_j ) < 1/j \). Observe now that the sequence \( \mathbf{x} := \{ a_j \}_{j \in \mathbb{N}} \subseteq K \) is a Cauchy sequence because for every \( \varepsilon > 0 \) we can choose \( j, l, n \in \mathbb{N} \) big enough so that \[ \phi(a_j - a_l) = \phi(a_j - x_n^j + x_n^l - a_l + x_n^j - x_n^l) \leq \phi(a_j - x_n^j) + \phi(x_n^l - a_l) + \phi(x_n^j - x_n^l) < \varepsilon \] in view of the definition of the \( a_j \)'s and of the fact that \( \lim_{n \to +\infty} \phi(x_n^j - x_n^l) < \varepsilon/3 \) for \( j \) and \( l \) big enough.
We finally have to show the universal property for \( (L,\psi) \). Let \( (M,\xi) \) be a complete valued field with a map of valued fields \( (K,\phi) \to (M,\xi) \). If this map is trivial there is nothing to prove, otherwise we have an injection \( f \colon K \hookrightarrow M \). Since \( M \) is complete with respect to \( \xi \) we see immediately that for every Cauchy sequence \( \{ x_n \} \subseteq K \) there exists the limit \( \lim_{n \to +\infty} f(x_n) \). This defines a map \begin{align} R &\to M \\ \{ x_n \} &\mapsto \lim_{n \to +\infty} \iota(x_n) \end{align} and it is not difficult to show that this factors through a map \( f' \colon L \to M \) because if two convergent sequences differ by a sequence which goes to zero they have the same limit. It is now immediate to see that \( f = f' \circ \iota \) and we only have to prove that \( f' \) is unique.
Let \( f'' \colon (L,\psi) \to (M,\xi) \) be another map of valued fields such that \( f = f'' \circ \iota \). Then using Exercise 5 we have for every \( \mathbf{x} = [\{ x_n \}] \in L \) that \[ f'(\mathbf{x}) - f''(\mathbf{x}) = \lim_{n \to +\infty} f(x_n) - f''(\mathbf{x}) = \lim_{n \to +\infty} f''(\iota(x_n)) - f''(\mathbf{x}) = f''\left( \lim_{n \to +\infty} \iota(x_n) - \mathbf{x} \right) = 0 \] and so \( f' = f'' \) as we wanted to prove. Q.E.D.
In view of the previous exercise and theorem we can speak of the completion of a valued field \( (K,\phi) \), which we will denote by \( (K_{\phi},\Phi) \).
Exercise 14 Prove that the completion of \( (\mathbb{Q}, \lvert \cdot \rvert_{\infty} ) \) is isomorphic to \( \mathbb{R} \) with the canonical absolute value, where the real numbers are defined as in Exercise 9.
Lemma 15 Every Archimedean absolute value \( \phi \colon \mathbb{Q} \to \mathbb{R}_{\geq 0} \) is equivalent to the canonical (Archimedean) absolute value \( \lvert \cdot \rvert_{\infty} \colon \mathbb{Q} \to \mathbb{R}_{\geq 0} \).
Proof Since \( \phi \) is Archimedean we know that there exists \( n \in \mathbb{N} \) with \( \phi(n) > 1 \), because otherwise \( \phi \) would be bounded on \( \mathbb{Z} \subseteq \mathbb{Q} \) and thus non-Archimedean (see Exercise 6 of the second lecture).
Let now \( a \in \mathbb{N}_{> 1} \) and for every \( k \in \mathbb{N} \) write \( n^k = \sum_{j = 0}^{m_k} c_{j,k} \, a^j \) where \( m_k \in \mathbb{N} \), \( c_j \in \{ 0, \dots, a - 1\} \) for all \( j \in \{ 0, \dots, m \} \) and \( c_m \neq 0 \). Applying the triangle inequality we get that \[ \phi(n^k) \leq \sum_{j = 0}^{m_k} \phi(c_{j,k}) \, \phi(a)^j \leq (m_k + 1) \, \max\{\phi(1),\dots,\phi(a - 1) \} \, \max(\phi(a)^{m_k}, 1) \] and using the fact that \( m_k \leq k \, \log(n)/\log(a) \) we get that \[ \phi(n) \leq \left( (k \, \log(n)/\log(a) + 1) \, \max\{\phi(1),\dots,\phi(a - 1) \} \, \max(\phi(a)^{k \, \log(n)/\log(a)}, 1) \right)^{1/k} \] which implies that \( \phi(n) \leq \max(\phi(a)^{\log(n)/\log(a)},1) \) for every integer \( a \in \mathbb{N}_{> 1} \).
Since \( \phi(n) > 1 \) this implies that \( \phi(a) > 1 \) for all \( a \in \mathbb{N}_{> 1} \). Thus we can use the same argument of the previous paragraph to show that for every couple of integers \( a,b \in \mathbb{N}_{> 1} \) we have that \( \phi(b) \leq \phi(a)^{\log(b)/\log(a)} \) and \( \phi(a) \leq \phi(b)^{\log(a)/\log(b)} \), which implies that \( \phi(a)^{1/\log(a)} = \phi(b)^{1/\log(b)} \). Thus \( \alpha := \phi(a)^{1/\log(a)} > 1 \) does not depend on \( a \) and we have that \( \phi(a) = \alpha^{\log(a)} = a^{\log(\alpha)} \) for all \( a \in \mathbb{N}_{> 1} \). Using the multiplicativity of \( \phi \) we see that this extends immediately to \( \mathbb{Q} \), and thus that \( \phi \) is equivalent to \( \lvert \cdot \rvert_{\infty} \). Q.E.D.
Theorem 16 Let \( (K,\phi) \) be a complete, Archimedean field. Then either \( (K,\phi) \cong (\mathbb{R}, \lvert \cdot \rvert ) \) or \( (K,\phi) \cong (\mathbb{C}, \lvert \cdot \rvert ) \), up to substituting \( \phi \) with an equivalent absolute value.
Proof Recall first of all the \( K \) has characteristic zero, because every absolute value on a field of positive characteristic is non-Archimedean (see Exercise 6 of the second lecture). Thus we have that \( \mathbb{Q} \hookrightarrow K \). If we can restrict \( \phi \) to \( \mathbb{Q} \) we get an Archimedean absolute value, which (thanks to the previous lemma) is equivalent to \( \lvert \cdot \rvert_{\infty} \). Moreover since \( K \) is complete by hypothesis we can use the previous exercise to obtain an inclusion \( \mathbb{R} \hookrightarrow K \).
Suppose now that \( -1 \in K \) is a square. Then we get an inclusion \( \iota \colon \mathbb{C} \hookrightarrow K \) and it is easy to see that this is automatically an inclusion of valued fields. Recall now that the polynomial \( t^n - 1 \in \mathbb{C}[t] \) can be written as \[ t^n - 1 = \prod_{j = 0 }^{n - 1} (t - \zeta_n^j) \qquad \text{where} \qquad \zeta_n := e^{\frac{2 \pi i}{n}} \in \mathbb{C} \] and thus using the inclusion \( \iota \) we can write \( x^n - y^n = (x - y) \prod_{j = 1}^{n - 1} (\zeta_n^j \, x - y) \) for every \( x,y \in K \). Thus using the multiplicativity of \( \phi \) we get that the identity \[ \left( \dagger \right) \qquad \phi(x - y) = \frac{\phi((x - z)^n - (y - z)^n)}{\prod_{j = 1}^{n - 1} \phi(\zeta_n^j \, (x - z) - (y - z))} \] holds for all \( x, y, z \in K \) for which \( \phi(\zeta_n^j \, (x - z) - (y - z)) \neq 0 \) for all \( j \in \{ 1,\dots, n - 1 \} \).
We want to show now that \( K = \iota(\mathbb{C}) \), and so suppose by contradiction that there exists \( \alpha \in K \setminus \iota(\mathbb{C}) \). Define now the continuous function \( f \colon \mathbb{C} \to \mathbb{R}_{> 0} \) as \( f(z) := \phi(z - \alpha) \) and let \( z_0 \in \mathbb{C} \) be a point of minimum for \( f \), and let \( r := f(z_0) \). Take now \( z \in \mathbb{C} \) such that \( \psi(z - z_0) < r \) and use the previous identity \( \left( \dagger \right) \) to show that \begin{align}
f(z) &= f(z_0)^n \, \phi\left(1 - \frac{(z - z_0)^n}{(\alpha - z_0)^n}\right) \, \prod_{j = 1}^{n - 1} f(\zeta_n^j \, (z - z_0) + z_0)^{-1} \leq f(z_0)^n \, \phi\left(1 - \frac{(z - z_0)^n}{(\alpha - z_0)^n}\right) \, r^{1 - n} \\ &\leq r \left(1 + \left(\frac{\phi(z - z_0)}{r}\right)^n \right)
\end{align} because \( \psi(\zeta_n) = 1 \) as it is easy to observe using the fact that \( \zeta_n^n = 1 \). Thus letting \( n \to +\infty \) and using the fact that \( f(z) \geq r \) we get that \( f(z) = r \) for all \( z \in \mathbb{C} \) with \( \phi(z - z_0) < r \). We can now go on by covering \( \mathbb{C} \) by balls of radius \( r \) to show that \( f \) is constant. But this is absurd because \( f(n) \geq \phi(n) (1 - \phi(n)/\phi(\alpha)) \) for all \( n \in \mathbb{N} \) and we know that \( \phi(n) \to +\infty \) as \( n \to +\infty \). Thus we have proved that \( K = \iota(\mathbb{C}) \), which shows that \( \iota \colon (\mathbb{C},\lvert \cdot \rvert) \to (K,\phi) \) is an isomorphism of valued fields.
Suppose now that \( -1 \in K \) is not a square. Then we can use what we have just proved and Lemma 10 to show that \( L := K[t]/(t^2 + 1) \) is isomorphic to \( \mathbb{C} \) as a valued field. But we know that \( \mathbb{R} \hookrightarrow K \) which implies that \( K \cong \mathbb{R} \) as valued fields. Q.E.D.
Definition 17 Let \( K \) be a number field. We say that two embeddings \( \sigma, \tau \colon K \to \mathbb{C} \) are equivalent, and we write \( \sigma \sim \tau \), if \( \sigma(x) = \overline{\tau(x)} \) for all \( x \in K \). Here \( \overline{z} = x - i y \) denotes the complex conjugate of any complex number \( z = x + i y \in \mathbb{C} \).
Corollary 18 Let \( K \) be a number field. Then the map \begin{align} \frac{\{\sigma \colon K \hookrightarrow \mathbb{C} \}}{\sim} &\to \Sigma_{K,\infty} \\ [\sigma] &\mapsto [ \lvert \cdot \rvert_{\sigma}] \end{align} is a bijection. Here \( \lvert \cdot \rvert_{\sigma} \) is the absolute value defined in Example 11 of the second lecture.
Proof It is easy to see that the map is well defined and injective, i.e. that \( \lvert \cdot \rvert_{\sigma} \) is equivalent to \( \lvert \cdot \rvert_{\tau} \) if and only if \( \sigma \sim \tau \). Moreover if \( \phi \) is an Archimedean absolute value on \( K \) then, up to substituting \( \phi \) with an equivalent absolute value, we have that \( K_{\phi} \cong \mathbb{R} \) or \( K_{\phi} \cong \mathbb{C} \) as valued fields. Thus the embedding \( K \hookrightarrow K_{\phi} \) is actually an embedding \( \sigma \colon K \to \mathbb{C} \) and \( \phi = \lvert \cdot \rvert_{\sigma} \) because as we have seen \( \phi \) is equal to the absolute value induced on \( K \) by the embedding \( K \hookrightarrow K_{\phi} \). Q.E.D.
Some references for this class are:
In the previous lecture we have struggled with non-Archimedean places of a number field \( K \) and of a function field \( F(t) \). In particular we have proved that for every number field \( K \) the maps \[ \begin{aligned} \Sigma_K^{\infty} &\to \operatorname{Spec}(\mathcal{O}_K) \setminus \{ \mathbf{0} \} \\ [\phi] &\mapsto \{ x \in \mathcal{O}_K \mid \phi(x) < 1 \} \end{aligned} \qquad \text{and} \qquad \begin{aligned} \operatorname{Spec}(\mathcal{O}_K) \setminus \{ \mathbf{0} \} &\to \Sigma_K^{\infty} \\ \mathfrak{p} &\mapsto [\, \lvert \cdot \rvert_{\mathfrak{p}} \,] \end{aligned} \] are one the inverse of the other. Now we want to prove a similar statement for the Archimedean places \( \Sigma_{K,\infty} \). Recall that the Archimedean world is the world of \( \mathbb{R} \) and \( \mathbb{C} \), i.e. it is the world of the analysis that we are accustomed to.
For this reason to understand Archimedean places we need to start building an analytic theory of valued fields. In the second lecture we have defined absolute values over a field and we have seen that they carry with them a canonical topology. Now we want to define a category of valued fields.
Definition 1 The category of valued fields is the category whose objects are couples \( (K,\phi) \), where \( K \) is a field and \( \phi \colon K \to \mathbb{R}_{\geq 0} \) is an absolute value. A morphism \( (K,\phi) \to (L,\psi) \) in this category is given by any homomorphism of fields \( f \colon K \to L \) such that \( \psi(f(x)) = \phi(x) \) for all \( x \in K \).
Now it is natural to look for special objects in this category, e.g. for objects with universal properties. In the following section we will define such an object, the completion of a valued field \( (K,\phi) \), whose definition is related to the analytic properties of \( \phi \).
Analytic properties of absolute values
Since we are interested in the analytic theory of valued fields the first place to start from will be the concept of limit. The definition looks familiar, at least to anyone who was born after Weierstrass...Definition 2 Let \( (K, \phi) \) be a valued field. We say that a sequence \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) converges to \( x_{\infty} \in K \) (with respect to \( \phi \)) if for every \( \varepsilon \in \mathbb{R}_{> 0} \) there exists \( n_{\varepsilon} \in \mathbb{N} \) such that for all \( n \geq n_{\varepsilon} \) we have that \( \phi(x_n - x_{\infty}) < \varepsilon \).
Exercise 3 Prove that if \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) converges to \( x_{\infty} \in K \) and to \( x_{\infty}' \in K \) then \( x_{\infty} = x_{\infty}' \).
In view of the previous exercise we will say that a sequence \( \{ x_n \} \) converges in \( K \) if it converges to some (and so, to one) \( x_{\infty} \in K \). We call \( x_{\infty} \) the limit of the sequence \( \{ x_n \} \).
Exercise 4 Prove that convergence respects sums and products. More precisely, prove that if \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) converges to \( x_{\infty} \in K \) and \( \{ y_n \}_{n \in \mathbb{N}} \subseteq K \) converges to \( y_{\infty} \in K \) then \( \{ x_n + y_n \}_{n \in \mathbb{N}} \) converges to \( x_{\infty} + y_{\infty} \) and \( \{ x_n \cdot y_n \}_{n \in \mathbb{N}} \) converges to \( x_{\infty} \cdot y_{\infty} \).
Exercise 5 Prove that every morphism of valued fields \( f \colon (K, \phi) \to (L,\psi) \) is sequentially continuous. More precisely prove that if \( \{ x_n \} \subseteq K \) converges to \( x_{\infty} \in K \) then \( \{ f(x_n) \} \subseteq L \) converges to \( f(x_{\infty}) \).
As we all learned some time ago, if a sequence \( \{ x_n \}_{n \in \mathbb{N}} \) converges then the \( x_n \)'s get "closer and closer" as \( n \to +\infty \), i.e. the sequence \( \{ x_n \}_{n \in \mathbb{N}} \) is a Cauchy sequence.
Definition 6 Let \( (K, \phi) \) be a valued field. We say that a sequence \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) is a Cauchy sequence if for every \( \varepsilon \in \mathbb{R}_{> 0} \) there exists \( n_{\varepsilon} \in \mathbb{N} \) such that for all \( n, m \geq n_{\varepsilon} \) we have that \( \phi(x_n - x_m) < \varepsilon \).
Exercise 7 Prove that Cauchy sequences respect sums and products. More precisely, prove that if \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) and \( \{ y_n \}_{n \in \mathbb{N}} \subseteq K \) are Cauchy sequences then \( \{ x_n + y_n \}_{n \in \mathbb{N}} \) and \( \{ x_n \cdot y_n \}_{n \in \mathbb{N}} \) are also Cauchy sequences.
Testing whether a sequence is Cauchy or not can be hard but it is doable in principle, whereas testing if a sequence has a limit or not can be really tough. For example we know that the sequences of rational numbers \[ x_n := \sum_{k=0}^{n} (-1)^{k}\frac{\left\lfloor\log_2(k + 2) \right\rfloor}{k + 2} \qquad \text{and} \qquad y_n := \sum_{k = 0}^{n} \frac{(-1)^{k}}{(2 k+1)^2} \] are Cauchy sequences with respect to the canonical Archimedean absolute value \( \lvert \cdot \rvert_{\infty} \colon \mathbb{Q} \to \mathbb{R}_{> 0} \) but no one knows if these sequences converge in \( \mathbb{Q} \) or not (see here and here).
The problem here is that \( (\mathbb{Q}, \lvert \cdot \rvert_{\infty}) \) is a valued field "with many holes" and so not all the Cauchy sequences converge. For example for every \( \alpha \in \mathbb{Q}_{> 0} \) we can define a sequence \( \{ x_n \}_{n \in \mathbb{N}} \) by setting \[ x_0 := 1 \qquad \text{and} \qquad x_{n+1} := \frac{1}{2} \left(x_n + \frac{\alpha}{x_n}\right). \] It is not difficult to prove that this sequence is always Cauchy for \( \lvert \cdot \rvert_{\infty} \). However, it converges in \( \mathbb{Q} \) if and only if \( \alpha \) is a square (so, for example, it does not converge if \( \alpha = 2 \)).
In order to avoid this sort of problems it would be nice if every Cauchy sequence was convergent. So nice that this property deserves a name.
Definition 8 A complete field is a valued field \( (K,\phi) \) such that any Cauchy sequence \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) converges.
Exercise 9 Prove that the field of real numbers \( (\mathbb{R},\lvert \cdot \rvert) \) is complete with respect to the canonical absolute value defined in Example 9 of the second lecture. Use in particular the definition of \( \mathbb{R} \) as the set of sequences \[ \mathbb{R} := \left\{ \varepsilon a_n a_{n - 1} \cdots a_0.a_{-1} a_{-2} \cdots a_{-m} \cdots \mid \varepsilon \in \{ \pm 1 \}, \ a_j \in \{ 0, \dots, 9 \} \right\} \setminus \mathcal{S}_{9} \] where \( \mathcal{S}_{9} \) is the set of sequences which are definitely equal to \( 9 \).
There is another classical Archimedean complete field: the field of complex numbers \( \mathbb{C} \) with the canonical absolute value. To prove that it is complete starting from the completeness of \( \mathbb{R} \) we will need the following lemma, which applies also to non-Archimedean absolute values.
Lemma 10 Let \( (K,\phi) \) be a complete valued field, and suppose that \( x^2 + 1 \in K[x] \) is irreducible (i.e. that \( -1 \) is not a square in \( K \)). Then \( L := K[x]/(x^2 + 1) = K(i) \) is a quadratic field extension of \( K \). \( L \) is also complete with respect to the valuation \( \psi \colon L \to \mathbb{R}_{\geq 0} \) defined by \( \psi(a + i b) := \sqrt{\phi(a^2 + b^2)} \), and the inclusion \( K \hookrightarrow L \) is a morphism of valued fields.
Proof The fact that \( L \) is a quadratic field extension of \( K \) is part of any course in Galois theory.
It is now trivial to observe that \( \psi \) satisfies the first two axioms of the definition of an absolute value, and that \( \phi = \psi \circ \iota \), where \( \iota \colon K \hookrightarrow L \). Showing that \( \psi \) satisfies the triangle inequality is rather difficult, and we will prove it for a general Galois extension \( K \subseteq L \) in one of the following lectures. If you can't wait you can look at Lemma 2.2 of these notes by Peter Stevenhagen.
Let now \( \{ a_n + i b_n \}_{n \in \mathbb{N}} \subseteq L \) be a sequence such that \( \{ a_n \} \) and \( \{b_n \} \) tend to zero with respect to \( \phi \). Then it is clear from the definition of \( \psi \) that \( \{ a_n + i b_n \} \) tends to zero with respect to \( \psi \).
Vice versa, let \( \{ a_n + i b_n \}_{n \in \mathbb{N}} \subseteq L \) be a sequence which tends to zero with respect to \( \psi \), and suppose by contradiction that \( \{ x_n \}_{n \in \mathbb{N}} \) does not tend to zero with respect to \( \phi \). Then there would exist a number \( C \in \mathbb{R}_{> 0} \) such that \( \phi(x_n) \geq C \) for all \( n \in \mathbb{N} \) and thus the sequence \( \{ 1 + (y_n/x_n)^2 \}_{n \in \mathbb{N}} \) would go to zero with respect to \( \phi \), because \( \phi(1 + (y_n/x_n)^2) = \phi(x_n)^{-2} \, \phi(x_n + y_n) \).
It is now not difficult to see that this implies that the sequence \( z_n := y_n/x_n \) is a Cauchy sequence. Since \( K \) is complete with respect to \( \phi \) this sequence converges to an element \( z_{\infty} \in K \). Using Exercise 4 we see then that \( z_{\infty}^2 = -1 \). But this contradicts the fact that \( x^2 + 1 \in K[x] \) is irreducible, and so it follows that \( \{ x_n \}_{n \in \mathbb{N}} \) actually tends to zero with respect to \( \phi \). We can make a similar argument to show that \( y_n \) also tends to zero with respect to \( \phi \).
Using again Exercise 4 we see now that a sequence \( \{ x_n + i y_n \}_{n \in \mathbb{N}} \subseteq L \) converges to \( x_{\infty} + i y_{\infty} \) with respect to \( \psi \) if and only if \( \{ x_n \} \) converges to \( x_{\infty} \) and \( \{ y_n \} \) converges to \( y_{\infty} \) with respect to \( \phi \). This implies in particular that \( L \) is complete with respect to \( \psi \), as we wanted to show. Q.E.D.
Completions
We have just seen how important complete fields are. In particular we can restrict the category of all valued fields to the category of complete valued fields, and we have a forgetful functor (i.e. an inclusion) \[ \{ \text{complete valued fields} \} \hookrightarrow \{ \text{valued fields} \}. \]A question that now arises automatically is: how many complete fields do exist? The answer is... quite many! Indeed, the following theorem associates to every valued field \( (K,\phi) \) a complete field \( (K_{\Phi}, \Phi) \) which is the complete field "closest" to \( (K,\phi) \).
Definition 11 Let \( (K,\phi) \) be any valued field. We say that a complete valued field \( (L,\psi) \) is a completion of \( (K,\phi) \) if there is a morphism of valued fields \( \iota \colon (K,\phi) \to (L,\psi) \) such that for every complete valued field \( (M,\xi) \) and every morphism of valued fields \( f \colon (K,\phi) \to (M,\xi) \) there exists a unique morphism of valued fields \( f' \colon (L,\psi) \to (M,\xi) \) such that \( f = f' \circ \iota \).Exercise 12 Prove that if \( (L_1,\psi_1) \) and \( (L_2,\psi_2) \) are two completions of the same valued field \( (K,\phi) \) then there exists a unique isomorphism of valued fields \( (L_1,\psi_1) \cong (L_2,\psi_2) \).
Theorem 13 Let \( (K,\phi) \) be a valued field. Then there exists a completion \( (L,\psi) \) of \( (K,\phi) \).
Proof Define \( R \) to be the set of all Cauchy sequences \( \{ x_n \} \subseteq K \). If we define \[ \{ x_n \} + \{ y_n \} := \{ x_n + y_n \} \qquad \text{and} \qquad \{ x_n \} \cdot \{ y_n \} := \{ x_n \cdot y_n \} \] then it is clear from Exercise 7 that \( R \) is a ring with respect to these "coordinate-wise" sum and product operations. Observe moreover that the subset \[ \mathfrak{m} := \left\{ \{x_n\} \in R \mid \lim_{n \to +\infty} \phi(x_n) = 0 \right\} \subseteq R \] is an ideal, because clearly if \( \{x_n\} \in \mathfrak{m} \) and \( \{y_n\} \in R \) then \( \phi(x_n y_n) = \phi(x_n) \phi(y_n) \to 0 \) as \( n \to + \infty \) because \( \{ y_n \} \) is a Cauchy sequence and thus \( \{ \phi(y_n) \}_{n \in \mathbb{N}} \subseteq \mathbb{R}_{\geq 0} \) is bounded.
Moreover \( \mathfrak{m} \) is a maximal ideal because if \( \{ x_n \} \notin \mathfrak{m} \) then there exists \( n_0 \in \mathbb{N} \) such that \( x_n \neq 0 \) for \( n \geq n_0 \). Thus if we define a sequence \( \{ y_n \} \subseteq K \) by putting \( y_n := 1 \) for all \( n < n_0 \) and \( y_n := 1/x_n \) for all \( n \geq n_0 \) we have that the sequence \( \{ x_n \cdot y_n - 1 \} \) is definitely zero, and thus it is zero in the quotient ring \( R / \mathfrak{m} \), i.e. \( [\{ x_n \cdot y_n \}] = 1 \) in \( R / \mathfrak{m} \).
What we have shown implies that \( \mathfrak{m} \) is a maximal ideal, i.e. that \( L := R / \mathfrak{m} \) is a field. Moreover the map \( K \hookrightarrow R \) which sends an element \( x \in K \) to the constant (Cauchy) sequence \( \{ x_n = x \} \in R \) induces an inclusion of fields \( \iota \colon K \hookrightarrow L \).
Let now \( \Psi \colon R \to \mathbb{R}_{\geq 0} \) be defined as \( \Psi(\{x_n\}) := \lim_n \phi(x_n) \), which is well defined because \( R \) consists of Cauchy sequences. It is immediate to check that the composition of the inclusion \( K \hookrightarrow R \) and of \( \Psi \) is equal to \( \phi \) and it is also easy using the definition of \( \mathfrak{m} \) to show that \( \Psi \) descends to a map \( \psi \colon L \to \mathbb{R}_{\geq 0} \) which is an absolute value in view of Exercise 4.
We want now to prove that \( L \) is complete with respect to \( \psi \). To do so observe first of all that for every Cauchy sequence \( \{ x_n \}_{n \in \mathbb{N}} \subseteq K \) and every \( \varepsilon > 0 \) there exists \( \alpha \in K \) such that \( \lim_{n \to +\infty} \phi(x_n - \alpha) < \varepsilon \). For example we can take \( \alpha = x_{n_{\varepsilon}} \), where \( n_{\varepsilon} \in \mathbb{N} \) is the integer associated to \( \varepsilon \) in the definition of a Cauchy sequence.
Let now \( \{ \{ x_n^j \}_n \}_j \subseteq L \) be a Cauchy sequence in \( L \), i.e. a sequence of sequences \( \{ x_n^j \}_n \subseteq K \) such that for all \( \varepsilon > 0 \) there exists \( j_0 \in \mathbb{N} \) such that for all \( j, l \geq n_0 \) we have that \( \lim_{n \to +\infty} \phi(x_n^j - x_n^l) < \varepsilon \). Since every sequence \( \mathbf{x}^j := \{ x_n^j \}_n \subseteq K \) is a Cauchy sequence we can pick elements \( a_j \in K \) such that \( \lim_{n \to +\infty} \phi( x_n^j - a_j ) < 1/j \). Observe now that the sequence \( \mathbf{x} := \{ a_j \}_{j \in \mathbb{N}} \subseteq K \) is a Cauchy sequence because for every \( \varepsilon > 0 \) we can choose \( j, l, n \in \mathbb{N} \) big enough so that \[ \phi(a_j - a_l) = \phi(a_j - x_n^j + x_n^l - a_l + x_n^j - x_n^l) \leq \phi(a_j - x_n^j) + \phi(x_n^l - a_l) + \phi(x_n^j - x_n^l) < \varepsilon \] in view of the definition of the \( a_j \)'s and of the fact that \( \lim_{n \to +\infty} \phi(x_n^j - x_n^l) < \varepsilon/3 \) for \( j \) and \( l \) big enough.
We finally have to show the universal property for \( (L,\psi) \). Let \( (M,\xi) \) be a complete valued field with a map of valued fields \( (K,\phi) \to (M,\xi) \). If this map is trivial there is nothing to prove, otherwise we have an injection \( f \colon K \hookrightarrow M \). Since \( M \) is complete with respect to \( \xi \) we see immediately that for every Cauchy sequence \( \{ x_n \} \subseteq K \) there exists the limit \( \lim_{n \to +\infty} f(x_n) \). This defines a map \begin{align} R &\to M \\ \{ x_n \} &\mapsto \lim_{n \to +\infty} \iota(x_n) \end{align} and it is not difficult to show that this factors through a map \( f' \colon L \to M \) because if two convergent sequences differ by a sequence which goes to zero they have the same limit. It is now immediate to see that \( f = f' \circ \iota \) and we only have to prove that \( f' \) is unique.
Let \( f'' \colon (L,\psi) \to (M,\xi) \) be another map of valued fields such that \( f = f'' \circ \iota \). Then using Exercise 5 we have for every \( \mathbf{x} = [\{ x_n \}] \in L \) that \[ f'(\mathbf{x}) - f''(\mathbf{x}) = \lim_{n \to +\infty} f(x_n) - f''(\mathbf{x}) = \lim_{n \to +\infty} f''(\iota(x_n)) - f''(\mathbf{x}) = f''\left( \lim_{n \to +\infty} \iota(x_n) - \mathbf{x} \right) = 0 \] and so \( f' = f'' \) as we wanted to prove. Q.E.D.
In view of the previous exercise and theorem we can speak of the completion of a valued field \( (K,\phi) \), which we will denote by \( (K_{\phi},\Phi) \).
Exercise 14 Prove that the completion of \( (\mathbb{Q}, \lvert \cdot \rvert_{\infty} ) \) is isomorphic to \( \mathbb{R} \) with the canonical absolute value, where the real numbers are defined as in Exercise 9.
Complete Archimedean fields
We are now ready to prove the main result of this lecture, which will allow us to characterize the set of Archimedean places of a number field. To do so we need firstly a lemma about Archimedean absolute values on \( \mathbb{Q} \).Lemma 15 Every Archimedean absolute value \( \phi \colon \mathbb{Q} \to \mathbb{R}_{\geq 0} \) is equivalent to the canonical (Archimedean) absolute value \( \lvert \cdot \rvert_{\infty} \colon \mathbb{Q} \to \mathbb{R}_{\geq 0} \).
Proof Since \( \phi \) is Archimedean we know that there exists \( n \in \mathbb{N} \) with \( \phi(n) > 1 \), because otherwise \( \phi \) would be bounded on \( \mathbb{Z} \subseteq \mathbb{Q} \) and thus non-Archimedean (see Exercise 6 of the second lecture).
Let now \( a \in \mathbb{N}_{> 1} \) and for every \( k \in \mathbb{N} \) write \( n^k = \sum_{j = 0}^{m_k} c_{j,k} \, a^j \) where \( m_k \in \mathbb{N} \), \( c_j \in \{ 0, \dots, a - 1\} \) for all \( j \in \{ 0, \dots, m \} \) and \( c_m \neq 0 \). Applying the triangle inequality we get that \[ \phi(n^k) \leq \sum_{j = 0}^{m_k} \phi(c_{j,k}) \, \phi(a)^j \leq (m_k + 1) \, \max\{\phi(1),\dots,\phi(a - 1) \} \, \max(\phi(a)^{m_k}, 1) \] and using the fact that \( m_k \leq k \, \log(n)/\log(a) \) we get that \[ \phi(n) \leq \left( (k \, \log(n)/\log(a) + 1) \, \max\{\phi(1),\dots,\phi(a - 1) \} \, \max(\phi(a)^{k \, \log(n)/\log(a)}, 1) \right)^{1/k} \] which implies that \( \phi(n) \leq \max(\phi(a)^{\log(n)/\log(a)},1) \) for every integer \( a \in \mathbb{N}_{> 1} \).
Since \( \phi(n) > 1 \) this implies that \( \phi(a) > 1 \) for all \( a \in \mathbb{N}_{> 1} \). Thus we can use the same argument of the previous paragraph to show that for every couple of integers \( a,b \in \mathbb{N}_{> 1} \) we have that \( \phi(b) \leq \phi(a)^{\log(b)/\log(a)} \) and \( \phi(a) \leq \phi(b)^{\log(a)/\log(b)} \), which implies that \( \phi(a)^{1/\log(a)} = \phi(b)^{1/\log(b)} \). Thus \( \alpha := \phi(a)^{1/\log(a)} > 1 \) does not depend on \( a \) and we have that \( \phi(a) = \alpha^{\log(a)} = a^{\log(\alpha)} \) for all \( a \in \mathbb{N}_{> 1} \). Using the multiplicativity of \( \phi \) we see that this extends immediately to \( \mathbb{Q} \), and thus that \( \phi \) is equivalent to \( \lvert \cdot \rvert_{\infty} \). Q.E.D.
Theorem 16 Let \( (K,\phi) \) be a complete, Archimedean field. Then either \( (K,\phi) \cong (\mathbb{R}, \lvert \cdot \rvert ) \) or \( (K,\phi) \cong (\mathbb{C}, \lvert \cdot \rvert ) \), up to substituting \( \phi \) with an equivalent absolute value.
Proof Recall first of all the \( K \) has characteristic zero, because every absolute value on a field of positive characteristic is non-Archimedean (see Exercise 6 of the second lecture). Thus we have that \( \mathbb{Q} \hookrightarrow K \). If we can restrict \( \phi \) to \( \mathbb{Q} \) we get an Archimedean absolute value, which (thanks to the previous lemma) is equivalent to \( \lvert \cdot \rvert_{\infty} \). Moreover since \( K \) is complete by hypothesis we can use the previous exercise to obtain an inclusion \( \mathbb{R} \hookrightarrow K \).
Suppose now that \( -1 \in K \) is a square. Then we get an inclusion \( \iota \colon \mathbb{C} \hookrightarrow K \) and it is easy to see that this is automatically an inclusion of valued fields. Recall now that the polynomial \( t^n - 1 \in \mathbb{C}[t] \) can be written as \[ t^n - 1 = \prod_{j = 0 }^{n - 1} (t - \zeta_n^j) \qquad \text{where} \qquad \zeta_n := e^{\frac{2 \pi i}{n}} \in \mathbb{C} \] and thus using the inclusion \( \iota \) we can write \( x^n - y^n = (x - y) \prod_{j = 1}^{n - 1} (\zeta_n^j \, x - y) \) for every \( x,y \in K \). Thus using the multiplicativity of \( \phi \) we get that the identity \[ \left( \dagger \right) \qquad \phi(x - y) = \frac{\phi((x - z)^n - (y - z)^n)}{\prod_{j = 1}^{n - 1} \phi(\zeta_n^j \, (x - z) - (y - z))} \] holds for all \( x, y, z \in K \) for which \( \phi(\zeta_n^j \, (x - z) - (y - z)) \neq 0 \) for all \( j \in \{ 1,\dots, n - 1 \} \).
We want to show now that \( K = \iota(\mathbb{C}) \), and so suppose by contradiction that there exists \( \alpha \in K \setminus \iota(\mathbb{C}) \). Define now the continuous function \( f \colon \mathbb{C} \to \mathbb{R}_{> 0} \) as \( f(z) := \phi(z - \alpha) \) and let \( z_0 \in \mathbb{C} \) be a point of minimum for \( f \), and let \( r := f(z_0) \). Take now \( z \in \mathbb{C} \) such that \( \psi(z - z_0) < r \) and use the previous identity \( \left( \dagger \right) \) to show that \begin{align}
f(z) &= f(z_0)^n \, \phi\left(1 - \frac{(z - z_0)^n}{(\alpha - z_0)^n}\right) \, \prod_{j = 1}^{n - 1} f(\zeta_n^j \, (z - z_0) + z_0)^{-1} \leq f(z_0)^n \, \phi\left(1 - \frac{(z - z_0)^n}{(\alpha - z_0)^n}\right) \, r^{1 - n} \\ &\leq r \left(1 + \left(\frac{\phi(z - z_0)}{r}\right)^n \right)
\end{align} because \( \psi(\zeta_n) = 1 \) as it is easy to observe using the fact that \( \zeta_n^n = 1 \). Thus letting \( n \to +\infty \) and using the fact that \( f(z) \geq r \) we get that \( f(z) = r \) for all \( z \in \mathbb{C} \) with \( \phi(z - z_0) < r \). We can now go on by covering \( \mathbb{C} \) by balls of radius \( r \) to show that \( f \) is constant. But this is absurd because \( f(n) \geq \phi(n) (1 - \phi(n)/\phi(\alpha)) \) for all \( n \in \mathbb{N} \) and we know that \( \phi(n) \to +\infty \) as \( n \to +\infty \). Thus we have proved that \( K = \iota(\mathbb{C}) \), which shows that \( \iota \colon (\mathbb{C},\lvert \cdot \rvert) \to (K,\phi) \) is an isomorphism of valued fields.
Suppose now that \( -1 \in K \) is not a square. Then we can use what we have just proved and Lemma 10 to show that \( L := K[t]/(t^2 + 1) \) is isomorphic to \( \mathbb{C} \) as a valued field. But we know that \( \mathbb{R} \hookrightarrow K \) which implies that \( K \cong \mathbb{R} \) as valued fields. Q.E.D.
Definition 17 Let \( K \) be a number field. We say that two embeddings \( \sigma, \tau \colon K \to \mathbb{C} \) are equivalent, and we write \( \sigma \sim \tau \), if \( \sigma(x) = \overline{\tau(x)} \) for all \( x \in K \). Here \( \overline{z} = x - i y \) denotes the complex conjugate of any complex number \( z = x + i y \in \mathbb{C} \).
Corollary 18 Let \( K \) be a number field. Then the map \begin{align} \frac{\{\sigma \colon K \hookrightarrow \mathbb{C} \}}{\sim} &\to \Sigma_{K,\infty} \\ [\sigma] &\mapsto [ \lvert \cdot \rvert_{\sigma}] \end{align} is a bijection. Here \( \lvert \cdot \rvert_{\sigma} \) is the absolute value defined in Example 11 of the second lecture.
Proof It is easy to see that the map is well defined and injective, i.e. that \( \lvert \cdot \rvert_{\sigma} \) is equivalent to \( \lvert \cdot \rvert_{\tau} \) if and only if \( \sigma \sim \tau \). Moreover if \( \phi \) is an Archimedean absolute value on \( K \) then, up to substituting \( \phi \) with an equivalent absolute value, we have that \( K_{\phi} \cong \mathbb{R} \) or \( K_{\phi} \cong \mathbb{C} \) as valued fields. Thus the embedding \( K \hookrightarrow K_{\phi} \) is actually an embedding \( \sigma \colon K \to \mathbb{C} \) and \( \phi = \lvert \cdot \rvert_{\sigma} \) because as we have seen \( \phi \) is equal to the absolute value induced on \( K \) by the embedding \( K \hookrightarrow K_{\phi} \). Q.E.D.
Conclusions and references
In this lecture we have:- defined the notion of convergence, Cauchy sequence and completeness for a valued field \( (K, \phi) \);
- associated to any valued field \( (K,\phi) \) a universal, complete valued field \( (K_{\phi}, \Phi) \);
- proved that all complete, Archimedean valued fields are isomorphic either to \( \mathbb{R} \) or to \( mathbb{C} \) as topological fields;
- concluded the proof of Ostrowski's theorem which characterizes the set \( \Sigma_K = \Sigma_K^{\infty} \cup \Sigma_{K,\infty} \) for any number field \( K \).
Some references for this class are:
- Section 2 of these notes by Stevenhagen;
- Chapter 3 of the book "Local Fields" by Cassels;
- Section 4.4.1 of the book "Number theory" by Borevich and Shafarevich.
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