TANT 11 - Applications of completeness

Hello there! These are notes for the eleventh 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 past lectures we have seen a lot of complete fields. In particular, we have seen in the fourth lecture that every complete Archimedean field is isomorphic to a finite extension of $\mathbb{R}$, and we have seen in the ninth lecture that every non-Archimedean complete field which is discretely valued and whose residue field is finite is either a finite extension of $\mathbb{Q}_p$ or a finite extension of $\mathbb{F}_p((T))$. We have seen moreover in the tenth lecture that we can describe these fields as fields of Laurent series in $p$ and in $T$ respectively (with the important difference that in $\mathbb{Q}_p$ we have to sum with carrying) or as fraction fields of inverse limits of finite rings.


 Aside 1  This analogy between $\mathbb{Q}_p$ and $\mathbb{F}_p((T))$ as fields of Laurent series in $p$ and $T$ is the starting point of the incredibly beautiful and useful theory of perfectoid spaces by Scholze, which allows us to prove statements in characteristic zero by moving to positive characteristic. See this note by Bhatt (from the "What is...?" column in the Notices of the AMS) to get a feeling of what perfectoid spaces are!

In this lecture we will use completeness to prove some facts that are incredibly useful when dealing with complete fields. The first one is Hensel's lemma, which deals with finding solutions of equations over complete fields.

Henselian rings and Hensel's lemma

We have already seen in the first lecture that solving Diophantine equations is difficult. In particular we have seen that to find integral zeros of a polynomial $f(x_1,\dots,x_n) \in \mathbb{Z}[x_1,\dots,x_n]$ it is not sufficient to find zeros of $f$ in $(\mathbb{Z}/m \mathbb{Z})^n$ for all $m \in \mathbb{N}_{\geq 1}$. Suppose nevertheless that we know for some particular polynomial $f(x_1,\dots,x_n) \in \mathbb{Z}[x_1,\dots,x_n]$ that finding solutions of $f$ with coordinates in all the finite quotients of $\mathbb{Z}$ is sufficient to find an integral solution of $f$. Then, how do we find these solutions in $(\mathbb{Z}/m \mathbb{Z})^n$?

It is easy to see using the Chinese remainder theorem that this amounts to find solutions in $(\mathbb{Z}/p^r \mathbb{Z})^n$ for all primes $p \in \mathbb{N}$ and all $r \in \mathbb{N}$. Let now $\pi_r \colon \mathbb{Z}/p^r \mathbb{Z} \twoheadrightarrow \mathbb{Z}/p^{r - 1} \mathbb{Z}$ be the projection map and observe that if $\mathbf{x} \in (\mathbb{Z}/p^r \mathbb{Z})^n$ is a solution of $f$ then $\pi_r(\mathbf{x})$ is also a solution of $f$. This implies that if we can find solutions $\mathbf{x}_r \in (\mathbb{Z}/p^r \mathbb{Z})^n$ for all $r \in \mathbb{N}_{\geq 1}$ we can also find a tower of compatible solutions, i.e. we can assume that $\pi_r(\mathbf{x}_r) = \mathbf{x}_{r - 1}$ for all $r \in \mathbb{N}_{\geq 1}$. But then we see from the description of $\mathbb{Z}_p$ as an inverse limit that we gave in the previous lecture that every tower of compatible solutions of $f(x_1,\dots,x_n)$ corresponds to a solution in $\mathbb{Z}_p^n$. So, how can we determine if $f$ has a solution in $\mathbb{Z}_p^n$?

Well, the answer is easy: just find a solution in $(\mathbb{Z}/p^r \mathbb{Z})^n$ for all $r \in \mathbb{N}_{\geq 1}$! Okay, we are chasing our tail... but the problem here is that we would have to check infinitely many finite rings to fulfill our task! Moreover the dimension of these rings increases as $r$ increases, and these rings are not fields unless $r = 1$...
Thus, it would be very nice if to check whether a polynomial $f(x_1,\dots,x_n) \in \mathbb{Z}[x_1,\dots,x_n]$ has a solution in $\mathbb{Z}_p^n$ we only had to check if it has a solution in $\mathbb{Z}/p \mathbb{Z}$, and then we could automatically "lift" this single solution to a tower of compatible solutions that would give us the desired solution in $\mathbb{Z}_p^n$!
We will look first of all to the case $n = 1$. In this case finding a solution to a polynomial is equivalent to finding a linear factor to this polynomial. Thus "lifting" a solution of a monic polynomial $f(x) \in \mathbb{Z}[x]$ in $\mathbb{Z}/p \mathbb{Z}$ to a solution of $f(x)$ in $\mathbb{Z}_p$ is equivalent to lifting a factorization $\overline{f}(x) = (x - a_0) g_0(x) \in \mathbb{Z}/p \mathbb{Z}[x]$ to a factorization $f(x) = (x - a) g(x) \in \mathbb{Z}_p[x]$ with $\overline{a} = a_0$ and $\overline{g} = g_0$. We can give a more general definition along these lines for every ring $R$.

 Definition 2  Let $R$ be a ring and $I \subseteq R$ be an ideal. We say that $R$ is Henselian with respect to $I$ (or that $(R,I)$ is a Henselian pair) if:

1. $I \subseteq \mathfrak{m}$ for all maximal ideals $\mathfrak{m} \subseteq R$;
2. for every polynomial $f(x) \in R[x]$ such that $\overline{f} = g_0 h_0 \in (R/I)[x]$ with $(g_0,h_0) = R/I$ there exist $g, h \in R[x]$ such that $f = g h \in R[x]$ with $\overline{g} = g_0$ and  $\overline{h} = h_0$.

This definition is rather technical, and we will never use it in this generality (if you are curious about henselian pairs you can read this section of the Stacks Project after having finished this lecture). We use this definition only to state the following result, which is one of the most general versions of Hensel's lemma.

 Theorem 3  Let $A$ be a ring which is complete with respect to an ideal $I \subseteq A$. Then $(A,I)$ is a Henselian pair.

 Proof  See Lemma 15.10.4 of the Stacks Project

We will content ourselves with considering a local ring $R$ and its (unique) maximal ideal $\mathfrak{m} \subseteq R$. In this case the first condition is automatically satisfied, and the second condition is indeed equivalent to "lifting roots" from $R/\mathfrak{m}$ to $R$.

 Exercise 4  Let $R$ be a local ring with maximal ideal $\mathfrak{m} \subseteq R$. Prove that $(R,\mathfrak{m})$ is a Henselian pair if and only if for every monic polynomial $f(x) \in R[x]$ and every $a_0 \in R/\mathfrak{m}$ with $\overline{f}(a_0) = 0$ and $\overline{f}'(a_0) \neq 0$ there exists $a \in R$ with $\overline{a} = a_0$ and $f(a) = 0$.

In this context of our interest, we can prove our less general version of Hensel's lemma.

 Theorem 5  Let $(K, \phi)$ be a complete, non-Archimedean field. Let $f(x) \in A_{\phi}[x]$ and suppose that there exists $a_0 \in A_{\phi}$ satisfying $\phi(f(a_0)) < \phi(f'(a_0))^2$. Then there is a unique $\alpha \in A_{\phi}$ such that:

• $f(\alpha) = 0$
• $\phi(\alpha - a_0) < \phi(f'(a_0))$

 Thus, $(A_{\phi},\mathfrak{m}_{\phi} )$ is a Henselian pair.

 Proof  Observe first of all that $a_0$ is an "approximate root" of $f(x)$. Thus we can try to construct $\alpha$ as a limit of a sequence $\{ a_n \}$ of approximate roots of $f(x)$, defined recursively as $$a_{n+1} := a_n - \frac{f(a_n)}{f'(a_n)}$$ which is exactly what goes under the name of Newton's method in real analysis. We need to prove that this sequence is well defined, and that it is a Cauchy sequence. To do so, we will prove that:

1. $a_n \in A_{\phi}$ for all $n \in \mathbb{N}$;
2. $\phi(f(a_n)) \leq \phi(f'(a_0))^2 \phi(f(a_0)/f'(a_0)^2)^{2^{n}}$;
3. $\phi(f'(a_n)) = \phi(f'(a_0))$.

 For $n = 0$ these three conditions are obvious. Suppose now that these three conditions are true for $n$. Then $a_{n + 1}$ is well defined because $f'(a_n) \neq 0$, since $\phi(f'(a_n)) = \phi(f'(a_0)) \neq 0$. Moreover $$\phi(a_{n + 1}) \leq \max\left(\phi(a_n),\phi\left(\frac{f(a_n)}{f'(a_n)}\right) \right) \leq 1$$ since $\phi(a_n) \leq 1$ by 1. (for n) and $$\frac{\phi(f(a_n))}{\phi(f'(a_n))} = \frac{\phi(f(a_n))}{\phi(f'(a_0))} \leq \phi(f'(a_0)) \phi(f(a_0)/f'(a_0)^2)^{2^{n}} \leq 1$$ by 2. and 3. (for n). This proves that 1. is true for every $n \in \mathbb{N}$.

To prove 2. suppose that $f(x) = \sum_{j = 0}^n c_j x^j \in A_{\phi}[x]$ and write $$f(x+y) = \sum_{j = 0}^n c_j (x+y)^j = \sum_{j = 0}^n c_j x^j + \sum_{j = 1}^n j c_j x^{j - 1} y + g(x,y) \, y^2 = f(x) + f'(x) y + g(x,y) y^2 \in A_{\phi}[x,y]$$ for some polynomial $g(x,y) \in A_{\phi}[x,y]$ (this is a sort of Taylor series for the function $f(x+y)$ around $y = 0$). Using this we see immediately that $$f(a_{n + 1}) = f\left( a_n - \frac{f(a_n)}{f'(a_n)} \right) = f(a_n) + f'(a_n) \left( - \frac{f(a_n)}{f'(a_n)} \right) + \gamma \left( \frac{f(a_n)}{f'(a_n)} \right)^2 = \gamma \left( \frac{f(a_n)}{f'(a_n)} \right)^2$$ where $\gamma := g(a_n,- f(a_n)/f'(a_n)) \in A_{\phi}$. This implies that $$\phi(f(a_{n+1})) = \phi(\gamma) \, \left(\frac{\phi(f(a_n))}{\phi(f'(a_{n}))}\right)^2 \leq \left(\frac{\phi(f(a_n))}{\phi(f'(a_0))}\right)^2 \leq \left( \phi(f'(a_0)) \phi\left(\frac{f(a_0)}{f'(a_0)^2}\right)^{2^{n}} \right)^2 = \phi(f'(a_0))^2 \phi\left(\frac{f(a_0)}{f'(a_0)^2}\right)^{2^{n+1}}$$ which proves that 2. holds for all $n \in \mathbb{N}$.

To prove 3. we write again $f(x) = \sum_{j = 0}^n c_j x^j \in A_{\phi}[x]$ and we see that $$f'(x) - f'(y) = \sum_{j = 2}^n j c_j (x^{j - 1} - y^{j - 1}) = (x - y) h(x,y) \in A_{\phi}[x,y]$$ for some polynomial $h(x,y) \in A_{\phi}[x,y]$. Thus we can use 2. and 3. for $n$ to see that $$\phi(f'(a_{n + 1}) - f'(a_n)) = \phi(a_{n+1} - a_n) \phi(\delta) \leq \phi(a_{n+1} - a_n) = \frac{\phi(f(a_n))}{\phi(f'(a_n))} = \frac{\phi(f(a_n))}{\phi(f'(a_0))} < \phi(f'(a_0))$$ where $\delta := h(a_{n+1},a_n) \in A_{\phi}$. Since $\phi$ is non-Archimedean this implies that $\phi(f'(a_{n+1})) = \phi(f'(a_0))$ which proves that 3. holds for all $n \in \mathbb{N}$.

We can use now properties 2. and 3. to show that the sequence $\{ a_n \}$ is a Cauchy sequence. Indeed we have that $$\phi(a_{n+1} - a_n) = \frac{\phi(f(a_n))}{\phi(f'(a_n))} = \frac{\phi(f(a_n))}{\phi(f'(a_0))} \leq \phi(f'(a_0)) \phi\left(\frac{f(a_0)}{f'(a_0)^2}\right)^{2^{n}}$$ which shows that the sequence is Cauchy using Exercise 6 since $\phi(f(a_0)/f'(a_0)^2) < 1$ and thus the series $\sum_{n = 0}^{+\infty} \phi(f(a_0)/f'(a_0)^2)^{2^{n}}$ converges.

Since $(K,\phi)$ is complete we can take $\alpha := \lim_{n \to +\infty} a_n$. Observe first of all that $\phi(\alpha) = \lim_n \phi(a_n) \leq 1$, so $\alpha \in A_{\phi}$. Observe moreover that $$\phi(f(\alpha)) =  \lim_{n \to +\infty} \phi(f(a_n)) \leq \lim_{n \to +\infty} \phi(f'(a_0))^2 \phi\left(\frac{f(a_0)}{f'(a_0)^2}\right)^{2^{n}} = 0$$ which implies that $f(\alpha) = 0$.
To prove now that $\phi(\alpha - a_0) < \phi(f'(a_0))$ we observe that $$\phi(a_1 - a_0) = \phi\left(\frac{f(a_0)}{f'(a_0)}\right) < \phi(f'(a_0))$$ and we use this as the first step to prove by induction that $\phi(a_{n} - a_0) = \phi(f(a_0)/f'(a_0)) < \phi(f'(a_0))$ for every $n \in \mathbb{N}_{\geq 1}$. Indeed we can use what we have proved in the previous paragraph to show that $$\phi(a_{n+1} - a_n) \leq \phi(f'(a_0)) \phi\left(\frac{f(a_0)}{f'(a_0)^2}\right)^{2^{n}} < \phi(f'(a_0)) \phi\left(\frac{f(a_0)}{f'(a_0)^2}\right) = \phi\left(\frac{f(a_0)}{f'(a_0)}\right)$$ for every $n \in \mathbb{N}$. Thus if we assume by induction that $\phi(a_{n} - a_0) = \phi(f(a_0)/f'(a_0))$ we get that $\phi(a_{n+1} - a_n) < \phi(a_{n} - a_0)$ and thus that $$\phi(a_{n+1} - a_0) = \phi(a_{n+1} - a_n + a_n - a_0) = \phi(a_{n} - a_0) = \phi(f(a_0)/f'(a_0))$$ because $\phi$ is non-Archimedean.

To conclude we need to show that $\alpha$ is unique. Suppose that $\beta \in A_{\phi}$ is such that $f(\beta) = 0$ and $\phi(\beta - a_0) < \phi(f'(a_0))$. Let $h := \beta - \alpha$ and observe that $$0 = f(\beta) = f(\alpha + h) = f(\alpha) + f'(\alpha) h + \mu h^2 = f'(\alpha) h + \mu h^2$$ where $\mu = g(\alpha,h) \in A_{\phi}$. If $h \neq 0$ this implies that $f'(\alpha) = -\mu h$ and thus that $$\phi(f'(\alpha)) = \phi(\mu) \phi(h) \leq \phi(h) = \phi(\beta - a_0 + a_0 - \alpha) < \phi(f'(a_0)) = \phi(f'(\alpha))$$ which is a contradiction. Thus $h = 0$ and $\beta = \alpha$. Q.E.D.

 Exercise 6  Let $(K,\phi)$ be a valued field. Prove that a sequence $\{ \alpha_n \} \subseteq K$ is Cauchy if and only if the series $\sum_{n = 0}^{+\infty} \phi(\alpha_{n+1} - \alpha_n)$ converges.

 Remark 7  There is also a multivariate version of Hensel's lemma, involving a system of polynomials $f_1,\dots,f_r \in A_{\phi}[x_1,\dots,x_n]$ with $r \leq n$ and the $r \times r$-minors of the Jacobian matrix $(\partial f_i/\partial x_j)_{i,j}$. For the precise statement of this result see Lemma 1.3 of the paper "Artin approximation" by Dorin Popescu.

Vector spaces over complete fields

Let's now leave for a second Hensel's lemma to prove a new interesting consequence of completeness which concerns vector spaces. Recall first of all that in real analysis one defines for every $p \in [1,+\infty]$ a $p$-norm on $\mathbb{R}^n$ by setting $$\left\| \mathbf{x} \right\| _p := \left( \sum_{i=1}^n \left| x_i \right| ^p \right) ^{1/p} \ \text{for} \ p < +\infty \qquad \text{and} \qquad \left\| \mathbf{x} \right\| _\infty := \max_{i} \left| x_i \right|$$ for every vector $\mathbf{x} = (x_1,\dots,x_n) \in \mathbb{R}^n$. We can generalize this to every valued field $(K,\phi)$, and in order to do so we need to define what we mean by norm on a vector space.

 Definition 8  Let $(K,\phi)$ be a valued field and let $V$ be a vector space over $K$. Then a norm on $V$ is a function $\psi \colon V \to \mathbb{R}_{\geq 0}$ such that:

1. $\psi(\mathbf{x}) = 0$ if and only if $\mathbf{x} = \mathbf{0}$;
2. $\psi(\lambda \mathbf{x}) = \phi(\lambda) \psi(\mathbf{x})$ for all $\lambda \in K$ and $\mathbf{x} \in V$;
3. $\psi(\mathbf{x} + \mathbf{y}) \leq \psi(\mathbf{x}) + \psi(\mathbf{y})$ for all $\mathbf{x}, \mathbf{y} \in V$.

 Definition 9  Two norms $\psi_1$ and $\psi_2$ are said to be equivalent if there exist constants $C_1, C_2 \in \mathbb{R}_{> 0}$ such that $$C_1 \psi_1(\mathbf{x}) \leq \psi_2(\mathbf{x}) \leq C_2 \psi_1(\mathbf{x})$$ for all $\mathbf{x} \in V$.

 Exercise 10  Let $(K,\phi)$ be a valued field and let $V$ be a vector space over $K$ with a norm $\psi$ on it. Prove that the map $$\begin{align} V \times V &\to \mathbb{R}_{\geq 0} \\ (\mathbf{x},\mathbf{y}) &\mapsto \psi(\mathbf{x} - \mathbf{y}) \end{align}$$ is a metric on $V$ and that two norms are equivalent if and only if the two topologies induced by these metrics coincide.

 Definition 11  Let $(K,\phi)$ be a valued field and let $V$ be a finite dimensional vector space over $K$. Let $\mathcal{B} = \{ \mathbf{v}_1,\dots,\mathbf{v}_n \}$ be a basis of $V$ and $p \in [1,+\infty]$. We define the $p$-norm associated to $\mathcal{B}$ as $$\psi_p^{\mathcal{B}}(\mathbf{x})  := \left( \sum_{i=1}^n \phi(x_i)^p \right) ^{1/p} \ \text{for} \ p < +\infty \qquad \text{and} \qquad \psi_{\infty}^{\mathcal{B}}(\mathbf{x}) := \max_{i} \phi(x_i)$$ for every $\mathbf{x} = \sum_{j = 1}^n x_j \mathbf{v}_j \in V$.

Suppose now that $(K,\phi)$ is a complete valued field, and let $V$ be a finite dimensional vector space over $K$. The next theorem tells us that we can define only one norm over $V$, up to equivalence.

Before being able to prove it we need to observe that we can again define the notion of a Cauchy sequence and of a limit with respect to a norm $\psi$ on a vector space $V$ defined over a valued field $(K,\phi)$. In this sense we can speak about complete vector spaces over valued fields.

 Exercise 12  Let $(K,\phi)$ be a valued field and let $V$ be a finite dimensional vector space over $K$ with a basis $\mathcal{B} = \{ \mathbf{v}_1,\dots,\mathbf{v}_n \}$. Prove that a sequence $\{ \mathbf{x}_k = \sum_{j = 1}^n x_k^j \mathbf{v}_j \} \subseteq V$ is Cauchy for $\psi_{\infty}^{\mathcal{B}}$ if and only if all the sequences $\{ x_k^1 \}_k, \dots, \{ x_k^n \}_k$ are Cauchy with respect to $\phi$. Prove moreover that the sequence $\{ \mathbf{x}_k \}$ converges to $\mathbf{x}_{\infty} = \sum_{j = 1}^n x_{\infty}^j \mathbf{v}_j$ if and only if $\{ x_k^j \}_k$ converges to $x_{\infty}^j$ for all $j \in \{ 1, \dots, n \}$.

 Exercise 13  Let $V$ be a vector space defined over a valued field $(K,\phi)$ and let $\psi_1$ and $\psi_2$ be two equivalent norms on $V$. Prove that $V$ is complete with respect to $\psi_1$ if and only if it is complete with respect to $\psi_2$.

 Theorem 14  Let $(K,\phi)$ be a complete valued field, and let $V$ be a finite dimensional vector space over $K$. Then all the norms over $V$ are equivalent. In particular, $V$ is complete with respect to any norm.

 Proof  It is sufficient to show that any norm $\psi$ on $V$ is equivalent to $\psi_{\infty}^{\mathcal{B}}$, where $\mathcal{B} = \{ \mathbf{v}_1,\dots,\mathbf{v}_n \}$ is any basis of $V$. Using the triangle inequality we can observe that $$\psi(\mathbf{x}) = \psi\left( \sum_{j = 1}^n x_j \mathbf{v}_j \right) \leq n \, \max_j (\phi(x_j) \psi(\mathbf{v}_j)) \leq C_2 \psi_{\infty}^{\mathcal{B}}(\mathbf{x})$$ where $C_2 := \max_j (\psi(\mathbf{v}_j)) \in \mathbb{R}_{> 0}$.

We want now to prove the other inequality, i.e. that there exists a constant $C_1 \in \mathbb{R}_{> 0}$ such that $C_1 \psi_{\infty}^{\mathcal{B}}(\mathbf{x}) \leq \psi(\mathbf{x})$ for all $\mathbf{x} \in V$. We will proceed by induction on $n$, the dimension of $V$. If $n = 1$ there is nothing to prove, because $\psi_{\infty}^{\mathcal{B}} = \phi$ and $\psi = C \, \phi$, where $C = \psi(1)$.

Suppose now that the theorem is true for vector spaces of dimension up to $n - 1$, and suppose by contradiction that this is not true for $V$. This means that the quotient $\psi(\mathbf{x})/\psi_{\infty}^{\mathcal{B}}(\mathbf{x})$ is arbitrarily small as $\mathbf{x}$ ranges over all the possible vectors of $V$, i.e. that we can find a sequence $\{ \mathbf{x}_k \} \subseteq V$ such that $\psi_{\infty}^{\mathcal{B}}(\mathbf{x}_k) > k \psi(\mathbf{x}_k)$ for all $k \in \mathbb{N}$. Write now $\mathbf{x}_k = \sum_{j = 1}^n x_k^j \mathbf{v}_j$ and define $f \colon \mathbb{N} \to \{ 1,\dots,n \}$ as $f(k) := \{ j_0 \mid \phi(x_k^{j_0}) = \max_j(\phi(x_k^j)) \}$. Thus we have that $\psi_{\infty}^{\mathcal{B}}(\mathbf{x}_k) = \phi(x_k^{f(k)})$ and thus $\phi(x_k^{f(k)}) > k \psi(\mathbf{x}_k)$ for all $k \in \mathbb{N}$. Since $\{ 1,\dots,n \}$ is finite and $\mathbb{N}$ is infinite there exists $i \in \{ 1,\dots,n \}$ such that $f(k) = i$ for an infinite number of $k \in \mathbb{N}$. If we change the order of vectors in the basis we can assume that $i = n$ and if we restrict to a sub-sequence $\{ \mathbf{x}_{m_k} \}$ of $\{ \mathbf{x}_k \}$ we can assume that $\psi_{\infty}^{\mathcal{B}}(\mathbf{x}_{m_k}) = \phi(y_{m_k}^n)$ and thus that $\phi(y_{m_k}^{n}) > k \, \psi(\mathbf{y}_{m_k})$ for all $k \in \mathbb{N}$.

Consider now the sequence $\{ \mathbf{y}_k := (x_{m_k}^n)^{-1} \mathbf{x}_{m_k} \}$. Then if $\mathbf{y}_k = \sum_{j = 1}^n y_k^j \, \mathbf{v}_j$ we have that $y_k^n = 1$ and thus the sequence of vectors $\{ \mathbf{u}_k := \mathbf{y}_k - \mathbf{v}_n \}$ lies in the $n - 1$-dimensional subspace $W \subseteq V$ generated by all the basis vectors $\{ \mathbf{v}_1,\dots, \mathbf{v}_{n - 1} \}$. Moreover the sequence $\{ \mathbf{u}_k \} \subseteq W$ is a Cauchy sequence for the restriction of $\psi$ to $W$ because $$\psi(\mathbf{u}_{k+l} - \mathbf{u}_k) \leq \psi(\mathbf{u}_{k+l} + \mathbf{v}_n) + \psi(\mathbf{u}_k + \mathbf{v}_n) = \psi(\mathbf{y}_{k+l}) + \psi(\mathbf{y}_k) = \frac{\psi(\mathbf{x}_{m_{k+l}})}{\psi_{\infty}^{\mathcal{B}}(\mathbf{x}_{m_{k+l}})} + \frac{\psi(\mathbf{x}_{m_k})}{\psi_{\infty}^{\mathcal{B}}(\mathbf{x}_{m_k})} < \frac{1}{m_{k+l}} + \frac{1}{m_{k}}$$   for all $k, l \in \mathbb{N}$.

Now we can use the induction hypothesis to see that $W$ is complete with respect to (the restriction of) $\psi$, and thus there exists $\mathbf{u} \in W$ such that $\mathbf{u}_k \to \mathbf{u}$. But we have that $\psi(\mathbf{u} + \mathbf{v}_n) = \lim_k \psi(\mathbf{u}_k + \mathbf{v}_n) = 0$ and thus that $\mathbf{v}_n = -\mathbf{u} \in W$ which is a contradiction from the definition of $W$. This means that the quotient $\psi(\mathbf{x})/\psi_{\infty}^{\mathcal{B}}(\mathbf{x})$ cannot be arbitrarily small and thus we have proved that $\psi$ is equivalent to $\psi_{\infty}^{\mathcal{B}}$. Q.E.D.

 Exercise 15  Let $(K,\phi)$ be a complete valued field and let $V$ be a vector space over $K$ with a norm $\psi$ on it. Prove that every finite dimensional subspace of $V$ is closed in the topology induced by $\psi$. (Hint: prove that the topological closure of a subset equals the set of limits of sequences in that subset and use Theorem 1)

Conclusions and references

In this lecture we managed to:

• define the concept of Henselian pair;
• prove that if $(K,\phi)$ is a complete, non-Archimedean field then \( (A_{\phi},\mathfrak{m}_{\phi}) is a Henselian pair;
• prove that all the norms defined on a finite dimensional vector space over a complete field are equivalent.

References for this lecture include:

•  The end of Section 2 and the beginning of Section 3 of these notes by Peter Stevenhagen;
• "Hensel's lemma" and "Equivalence of norms" by Keith Conrad;
• Section 3.4 and Sections 5.1-5.2 of the book "p-adic Numbers" by Fernando Q. Gouvêa;
• Section 4.4 of the book "Number theory" by Helmut Koch;
• Chapter 7 of the book "Commutative algebra" by David Eisenbud.