TANT 3 - Ostrowski's theorems

Hello there! These are notes for the third 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 defined the set of places $\Sigma_K$ of a field $K$ which can be divided into Archimedean and non-Archimedean places (and yes, the trivial place). Moreover we have seen as an example that if we have a Dedekind domain $R$ and a prime ideal $\mathfrak{p} \subseteq R$ we can define a non-Archimedean absolute value $\lvert \cdot \rvert_{\mathfrak{p}}$ on the field $K := \operatorname{Frac}(R)$ by setting $\lvert x \rvert_{\mathfrak{p}} := c^{-\operatorname{ord}_{\mathfrak{p}}(x)}$ for some constant $c \in \mathbb{R}_{> 1}$, where $\operatorname{ord}_{\mathfrak{p}}(x)$ is the maximum power of $\mathfrak{p}$ which divides $x R$.

In this lecture we will see that this construction is actually almost a bijection when $R = F[t]$ for some field $F$ and $R = \mathcal{O}_K$ for some number field $K$. The reason why this is not a bijection tout court is because on number fields we have the Archimedean absolute values given by embeddings $K \hookrightarrow \mathbb{C}$ and for function fields $K = F(t)$ we have the absolute value $\lvert \cdot \rvert_{\infty}$ and we have absolute values which are not trivial on $F$.


To define this bijection we will need the following observation, which is part of the many strange things that happen in the non-Archimedean world.

 Remark 1  From now on all non-Archimedean absolute values are assumed to be non-trivial.

 Lemma 2  If $\phi \colon K \to \mathbb{R}_{\geq 0}$ is a non-Archimedean absolute value then the closed unit ball $A_{\phi} := \{ x \in K \mid \phi(x) \leq 1 \}$ is a subring of $K$ and the open unit ball $\mathfrak{m}_{\phi} := \{ x \in K \mid \phi(x) < 1 \}$ is its unique maximal ideal. Moreover $A_{\phi}^{\times} = A_{\phi} \setminus \mathfrak{m}_{\phi}$ and $A_{\phi}$ is a discrete valuation ring if and only if $\phi(K^{\times}) \subseteq \mathbb{R}_{> 0}$ is a discrete subgroup.

 Proof   Clearly $0, 1 \in A_{\phi}$ and if $x, y \in A_{\phi}$ then $x \, y \in A_{\phi}$. The key fact here is that the non-Archimedean property $\phi(x+y) \leq \max(\phi(x),\phi(y))$ tells us that if two elements $x, y \in A_{\phi}$ are in the unit ball then also their sum $x + y \in A_{\phi}$ is in the unit ball! It is also immediately clear that $\mathfrak{m}_{\phi}$ is an ideal, thanks again to the non-Archimedean property, and that $A_{\phi}$ is a valuation ring, i.e. for every $x \in K$ either $x \in A_{\phi}$ or $x^{-1} \in A_{\phi}$ (and in particular $K = \operatorname{Frac}(A_{\phi})$). Think about how different this is from the unit balls of $\mathbb{R}$ and $\mathbb{C}$ for instance.

Observe now that if $x \in A_{\phi}^{\times}$ this means that there exists $y \in A_{\phi}^{\times}$ such that $x y = 1$, which gives us that $\phi(x) \phi(y) = 1$ which would be impossible if $x \in \mathfrak{m}_{\phi}$, because in that case $\phi(x) < 1$ and $\phi(y) \geq 1$. Thus we get that $A_{\phi}^{\times} \subseteq A_{\phi} \setminus \mathfrak{m}_{\phi}$. Vice versa if $x \in A_{\phi} \setminus \mathfrak{m}_{\phi}$ then in particular $x \neq 0$ and thus there exists $y \in K$ such that $x y = 1$. But this implies that $1 = \phi(x) \phi(y) = \phi(y)$ because $\phi(x) = 1$, and thus we have that $y \in A_{\phi}$ and $x \in A_{\phi}^{\times}$ (and also $y \in A_{\phi}^{\times}$).

We have hence proved that $A_{\phi}^{\times} = A_{\phi} \setminus \mathfrak{m}_{\phi}$, and this implies immediately that the quotient $\kappa_{\phi} := A_{\phi} / \mathfrak{m}_{\phi}$ is a field. Moreover if $\mathfrak{n} \subseteq A_{\phi}$ is a maximal ideal then $A^{\times} \subseteq A \setminus \mathfrak{n}$ and thus $\mathfrak{n} \subseteq \mathfrak{m}_{\phi}$ which implies that $\mathfrak{n} = \mathfrak{m}_{\phi}$, i.e. that $\mathfrak{m}_{\phi}$ is the unique maximal ideal of $A_{\phi}$.

Suppose now that $A_{\phi}$ is a discrete valuation ring. This implies that $\mathfrak{m}_{\phi} = \pi \, A_{\phi}$ for some $\pi \in A_{\phi}$. If we use multiple times the fact that $A_{\phi} = A_{\phi}^{\times} \cup \mathfrak{m}_{\phi}$ we get that $$A_{\phi} = A_{\phi}^{\times} \cup \pi \, A_{\phi} = A_{\phi}^{\times} \cup \pi \, (A_{\phi}^{\times} \cup \pi \, A_{\phi}) = \dots = \bigcup_{j = 0}^{+\infty} \pi^j \, A_{\phi}^{\times}.$$ Thus since $A_{\phi}$ is a valuation ring we get that for every $x \in K^{\times}$ there exists $u \in A_{\phi}^{\times}$ and $k \in \mathbb{Z}$ such that $x = u \, \pi^k$. This implies in particular that $\phi(x) = \phi(u) \, \phi(\pi)^k = \phi(\pi)^k$ and thus that $\phi(K^{\times}) = \phi(\pi)^{\mathbb{Z}}$ is a discrete subgroup of $\mathbb{R}_{> 0}$.

Vice versa suppose that $\phi(K^{\times})$ is a discrete subgroup of $\mathbb{R}_{> 0}$. Then we know (see Exercise 3) that $\phi(K^{\times}) = \alpha^{\mathbb{Z}}$ for some $\alpha \in \mathbb{R}_{> 0}$. In particular we can take $\alpha < 1$ and we can choose $\pi \in K^{\times}$ such that $\phi(\pi) = \alpha$. Clearly $\pi \in \mathfrak{m}_{\phi}$ and if $x \in K^{\times}$ we have that $\phi(x) = \phi(\pi)^k$ for some $k \in \mathbb{Z}$, which implies that $x = u \, \pi^k$ for some $u \in A_{\phi}^{\times}$. To conclude we only need to observe that if $x \in \mathfrak{m}_{\phi}$ then clearly $k \in \mathbb{N}_{\geq 1}$, and thus $\mathfrak{m}_{\phi} = \pi \, A_{\phi}$. Q.E.D.

 Exercise 3  Prove that all every discrete subgroup of $(\mathbb{R},+)$ is cyclic. Deduce as a corollary that every discrete subgroup of $(\mathbb{R}_{>0},\cdot)$ is cyclic. (Hint: for the first part, use an argument similar to the Euclidean algorithm which shows that every ideal of $\mathbb{Z}$ is principal. For the second part, think about a way of relating the two groups...)

 Example 4  Let $R$ be a Dedekind domain with field of fractions $K := \operatorname{Frac}(R)$ and let $\mathfrak{p} \subseteq R$ be a prime ideal. As we have seen in the previous lecture we have an absolute value $\lvert \cdot \rvert_{\mathfrak{p}}$ associated to $\mathfrak{p}$. It is not difficult to prove that in this case $A_{\lvert \cdot \rvert_{\mathfrak{p}}}$ is the localization of $R$ at $\mathfrak{p}$, i.e. $$A_{\lvert \cdot \rvert_{\mathfrak{p}}} = R_{\mathfrak{p}} := \left\{ \frac{x}{y} \in K \, \mid \, x \in R \ \text{and} \ y \in R \setminus \mathfrak{p} \right\}.$$ Indeed if $x \in R$ then $\lvert x \rvert_{\mathfrak{p}} \leq 1$ and if $y \in R \setminus \mathfrak{p}$ then $\lvert y \rvert_{\mathfrak{p}} = 1$, which implies that $\lvert x/y \rvert_{\mathfrak{p}} \leq 1$, i.e. that $R_{\mathfrak{p}} \subseteq A_{\lvert \cdot \rvert_{\mathfrak{p}}}$.
On the other hand let $\alpha \in A_{\lvert \cdot \rvert_{\mathfrak{p}}}$ and write $\alpha \, R = \mathfrak{a} \mathfrak{b}^{-1}$ with $\mathfrak{a}, \mathfrak{b} \subseteq R$ coprime ideals. Observe that $\operatorname{ord}_{\mathfrak{p}}(\alpha) \geq 0$ and thus we have that $\mathfrak{b}$ is coprime to $\mathfrak{p}$. Since $\mathfrak{a} = (\alpha) \, \mathfrak{b}$ we have that $\mathfrak{a}$ and $\mathfrak{b}$ define the same ideal class in $\mathfrak{Cl}(R)$. We want now to "get rid of" $\mathfrak{p}$ in the denominator of $\alpha$, and to do so we use the following Exercise 5 to take an ideal $\mathfrak{c} \subseteq R$ such that $\mathfrak{c}$ is in the class of $[\mathfrak{a}]^{-1} \in \mathfrak{Cl}(R)$ and is coprime to $\mathfrak{p}$. Since $[\mathfrak{a}] = [\mathfrak{b}] \in \mathfrak{Cl}(R)$ and $\mathfrak{c} \in [\mathfrak{a}]$ we get that $\mathfrak{a} \mathfrak{c}$ and $\mathfrak{b} \mathfrak{c}$ are principal ideals, generated by $x$ and $y$ respectively. To conclude we obtain that $$\alpha \, R = \mathfrak{a} \mathfrak{b}^{-1} = (\mathfrak{a} \mathfrak{c}) (\mathfrak{b} \mathfrak{c})^{-1} = \frac{x}{y} \, R$$ which implies (after modifying $x$ by a unit if necessary) that $\alpha = x/y$ for $x \in R$ and $y \in R \setminus \mathfrak{p}$, i.e. that $\alpha \in R_{\mathfrak{p}}$.
Observe finally that the maximal ideal $\mathfrak{m}_{\lvert \cdot \rvert_{\mathfrak{p}}}$ is equal to $\mathfrak{p} \, R_{\mathfrak{p}}$.

 Exercise 5  Prove that for every Dedekind domain $R$, every ideal $\mathfrak{a} \subseteq R$ and every prime ideal $\mathfrak{p} \subseteq R$ there exists an ideal $\mathfrak{b} \subseteq R$ such that $\mathfrak{p} \nmid \mathfrak{b}$ and $\mathfrak{b} \mathfrak{a}^{-1}$ is principal. (Hint: use the unique factorization of ideals and the Chinese remainder theorem for $R$).

We are now ready to introduce the two main theorems of this lecture. We will start with the "function field" case.

 Theorem 6  Let $F$ be a field and let $R := F[t]$ and $K := \operatorname{Frac}(R) = F(t)$. Then every absolute value on $K$ which is trivial on $F$ is either trivial tout court or equivalent to $\lvert \cdot \rvert_{\infty}$ or to an absolute value $\lvert \cdot \rvert_{\mathfrak{p}}$ for some prime ideal $\mathfrak{p} \subseteq F[t]$.

 Proof  Let $\phi \colon K = F(t) \to \mathbb{R}_{\geq 0}$ be a non-trivial absolute value which is trivial on $F$. Then since $\{ n \cdot 1_{K} \}_{n \in \mathbb{Z}} \subseteq F \subseteq F(t)$ we have that $\phi$ is non-Archimedean by applying the result proved in Exercise 6 of the previous lecture.

Suppose now that $\phi(t) \leq 1$. In this case for every polynomial $P(t) = \sum_{j = 0}^n a_j t^j \in F[t]$ we have that $$\phi(P(t)) \leq \max_j\{\phi(a_j) \phi(t)\} \leq 1$$ because $\phi(a_j) = 1$ by assumption and $\phi(t) \leq 1$ as we have supposed at the beginning of this paragraph. Thus $F[t] \subseteq A_{\phi}$ and $\mathfrak{p} := F[t] \cap \mathfrak{m}_{\phi}$ is a prime ideal of $F[t]$. To conclude we want to show that $\phi$ is equivalent to $\lvert \cdot \rvert_{\mathfrak{p}}$.

To do so observe first of all that $F[t]$ is a principal ideal domain and thus in particular that $\mathfrak{p} = p(t) \, F[t]$ for some irreducible $p(t) \in F[t]$. Thus we have that $\mathfrak{m}_{\lvert \cdot \rvert_{\mathfrak{p}}} = p(t) \, R_{\mathfrak{p}}$. So, if we proceed as in the proof of Lemma 2 and we observe that $K = \operatorname{Frac}(R_{\mathfrak{p}})$, we obtain that we can write every $\alpha \in K^{\times}$ as $\alpha = u \, p(t)^k$ for some $u \in R_{\mathfrak{p}}^{\times}$ and $k \in \mathbb{Z}$. In particular $k = \operatorname{ord}_{\mathfrak{p}}(\alpha)$.
Observe now that we can write $u = x/y$ with $x,y \in R \setminus \mathfrak{p}$, which implies that $\phi(u) = \phi(x)/\phi(y) = 1/1 = 1$. Thus we have that $$\phi(\alpha) = \phi(p(t))^{\operatorname{ord}_{\mathfrak{p}}(\alpha)} = \left(\frac{1}{\phi(p(t))}\right)^{- \operatorname{ord}_{\mathfrak{p}}(\alpha)}$$ and thus we get that $\phi$ is equivalent to $\lvert \cdot \rvert_{\mathfrak{p}}$, because $\phi(p(t)) < 1$.

Suppose finally that $\phi(t) > 1$. Then we have that $F[t^{-1}] \subseteq A_{\phi}$ and we can define again $\mathfrak{p} := F[t^{-1}] \cap \mathfrak{m}_{\phi}$. Observe that $t^{-1} \in \mathfrak{m}_{\phi}$ and thus $\mathfrak{p} = t^{-1} \, F[t^{-1}]$. Since we have that $F(t) = \operatorname{Frac}(F[t^{-1}])$ we can proceed as before and prove that $\phi$ is equivalent to $\lvert \cdot \rvert_{\mathfrak{p}}$. To conclude we only need to observe that for every polynomial $f(t) = \sum_{j = 0}^d a_j t^j \in F[t]$ we have that $f(t) = t^{-d} \, \sum_{j = 0}^d a_{d - j} t^{- j}$ which implies that $\operatorname{ord}_{\mathfrak{p}}(f(t)) = -d = -\deg(f(t))$ and so $\lvert \cdot \rvert_{\mathfrak{p}} = \lvert \cdot \rvert_{\infty}$. Q.E.D.

The main ingredient of the previous proof was the fact that the ring $R = F[t]$ is a principal ideal domain, which allowed us to relate the absolute value $\phi$ that we started with and the absolute value $\lvert \cdot \rvert_{\mathfrak{p}}$.

We want now to prove a similar statement when $R = \mathcal{O}_K$ is the ring of integers of a number fields. If we try to adapt the previous proof to this case, we run into two obstacles:

• first of all, it is not true anymore that every absolute value $\phi$ on $K = \operatorname{Frac}(R)$ is non-Archimedean. Indeed we have already seen in the previous lecture that every embedding $K \hookrightarrow \mathbb{C}$ gives rise to an Archimedean absolute value on $K$;
• secondly, it is not true anymore that $R = \mathcal{O}_K$ is a principal ideal domain: take for example $K = \mathbb{Q}(\sqrt{-5})$. More specifically, we can measure how "far" the ring $\mathcal{O}_K$ is from being a principal ideal domain by means of the class group $\mathfrak{Cl}(\mathcal{O}_K)$, which is trivial precisely when $\mathcal{O}_K$ is a principal ideal domain. It is not known at present whether there exist infinitely many number fields $K$ such that $\mathfrak{Cl}(\mathcal{O}_K)$ is trivial, and more generally there is a conjecture of Cohen and Lenstra which tries to explain the statistical behavior of these groups in families of number fields. If you are curious about this you can have a look at this paper by Goldfeld (concerning the characterization of imaginary quadratic fields of class number one), this master thesis of Nevin and this recent preprint of Bartel and Lenstra.

 Nevertheless, we will manage to overcome the second obstacle using our very own Lemma 2 that we proved before. To overcome the first obstacle we will need to wait until the following lecture!

 Theorem 7  Let $K$ be an algebraic number field and let $\phi \colon K \to \mathbb{R}_{\geq 0}$ be a non-Archimedean absolute value on $K$. Then $\phi$ is equivalent to $\lvert \cdot \rvert_{\mathfrak{p}}$ for some prime ideal $\mathfrak{p} \subseteq \mathcal{O}_K$.

 Proof  Observe first of all that $\phi(n) = \phi(1+\cdots+1) \leq \phi(1) = 1$ for every $n \in \mathbb{N}$ (and thus for every $n \in \mathbb{Z}$) because $\phi$ is non-Archimedean (this was the easy part of Exercise 6 of the previous lecture).

We want now to prove that $\mathcal{O}_K \subseteq A_{\phi}$. To do so observe that every $x \in \mathcal{O}_K$ satisfies an equation of the form $x^n = \sum_{j = 0}^{n - 1} a_j \, x^j$ for $a_0,\dots,a_{n - 1} \in \mathbb{Z}$. Thus we get that $$\phi(x)^n = \phi(x^n) = \phi\left( \sum_{j = 0}^{n - 1} a_j \, x^j \right) \leq \max_{j = 0, \dots,n-1}(\phi(a_j) \, \phi(x)^j) \leq \max_{j = 0, \dots,n-1}(\phi(x)^j)$$ which clearly implies that $\phi(x) \leq 1$ and thus that $\mathcal{O}_K \subseteq A_{\phi}$.

Now as before we set $\mathfrak{p} := \mathcal{O}_K \cap \mathfrak{m}_{\phi}$ and we want to prove that $\phi$ is equivalent to $\lvert \cdot \rvert_{\mathfrak{p}}$. Note first of all that by definition $\lvert \cdot \rvert_{\mathfrak{p}}$ is discrete and so by Lemma 2 we have that $A_{\lvert \cdot \rvert_{\mathfrak{p}}} = (\mathcal{O}_K)_{\mathfrak{p}}$ is a discrete valuation ring with a principal maximal ideal $\mathfrak{m}_{\lvert \cdot \rvert_{\mathfrak{p}}} = \mathfrak{p} (\mathcal{O}_K)_{\mathfrak{p}} = \pi (\mathcal{O}_K)_{\mathfrak{p}}$ for some element $\pi \in (\mathcal{O}_K)_{\mathfrak{p}}$.

Now we proceed exactly as before! Every element $x \in K^{\times}$ can be written as $x = u \, \pi^k$ for some $k \in \mathbb{Z}$ and $u \in (\mathcal{O}_K)_{\mathfrak{p}}^{\times}$ and we have that $\phi(v) = 1$ for every $v \in (\mathcal{O}_K)_{\mathfrak{p}}^{\times}$. Thus we have that $\phi(x) = (1/\phi(\pi))^{-\operatorname{ord}_{\mathfrak{p}}(x)}$ for all $x \in K^{\times}$. Observe finally that $\phi(\pi) < 1$ because $\pi \in \mathfrak{p} \subseteq \mathfrak{m}_{\phi}$ and thus $\phi$ is equivalent to $\lvert \cdot \rvert_{\mathfrak{p}}$. Q.E.D.

We are thus left with the analysis of the set of Archimedean places $\Sigma_{K,\infty}$ of a number field $K$. We already know from Example 10 of the previous lecture that every embedding $K \hookrightarrow \mathbb{C}$ induces an abolute value on $K$. We will prove in the following lecture that every possible Archimedean absolute value of $K$ arises in this way!

Conclusions and references

In this lecture we managed to:

• describe a way to associate a local ring $A_{\phi}$ to every absolute value $\phi$;
• characterize up to equivalence the set of all absolute values on a function field $F(t)$ which are trivial when restricted to $F$;
• completely characterize the set $\Sigma_K^{\infty}$ of non-Archimedean places of a number field $K$.

The main reference is again Section 1 of these notes of Peter Stevenhagen. Moreover, you can have a look at:

• Keith Conrad, "Ostrowski for $F(T)$".
• Keith Conrad, "Ostrowski for number fields".