# A User's Guide to Serre's Arithmetic/p-adic Fields

## The ring $\mathbb {Z} _{p}$ and the field $\mathbb {Q} _{p}$ The section introduces one of the main players in arithmetic geometry: the p-adics. This chapter studies a few basic properties of the p-adics including their topological structure, multiplicative structure, and solutions of affine polynomials in them.

For example, if you have an arithmetic scheme $X\in {\textbf {Sch}}/\mathbb {Z}$  (such as ${\text{Spec}}(\mathbb {Z} [x_{1},\ldots ,x_{n}]/(f_{1},\ldots ,f_{k}))$  or ${\text{Proj}}(\mathbb {Z} [x_{1},\ldots ,x_{n}]/(f_{1},\ldots ,f_{k}))$ ) then you can consider the base change to $X_{p}\in {\textbf {Sch}}/(\mathbb {Z} /p)$ . From the inverse system

$\cdots \to \mathbb {Z} /(p^{3})\to \mathbb {Z} /(p^{2})\to \mathbb {Z} /(p)$

there is an associated direct system of schemes

$X_{p}\to X_{p^{2}}\to X_{p^{3}}\to \cdots$

which gives ${\mathfrak {X}}_{p}={\text{Spf}}(X_{p^{k}})$ . Another example of a system of schemes is in deformation theory. For example, consider a scheme

${\begin{matrix}X_{1}\\\downarrow \\\mathbb {Z} /(p)\end{matrix}}$

Deformation theory can be used to ask if there is a scheme $X_{2}\in {\textbf {Sch}}/(\mathbb {Z} /(p^{2}))$  which fits into a cartesian square

${\begin{matrix}X_{1}&\to &X_{2}\\\downarrow &&\downarrow \\{\text{Spec}}(\mathbb {Z} /p)&\to &{\text{Spec}}(\mathbb {Z} /p^{2})\end{matrix}}$

This question can be repeatedly asked to get a directed system of schemes

${\begin{matrix}X_{1}&\to &X_{2}&\to &X_{3}&\to \cdots \\\downarrow &&\downarrow &&\downarrow &\\{\text{Spec}}(\mathbb {Z} /p)&\to &{\text{Spec}}(\mathbb {Z} /p^{2})&\to &{\text{Spec}}(\mathbb {Z} /p^{3})&\to \cdots \end{matrix}}$

where each square is cartesian. It turns out these questions are cohomological. All deformations depend on the cohomology group $H^{1}(X_{1},T_{X_{1}})$  and all "obstructions" to a deformation live in a group depending on $H^{2}(X_{1},T_{X_{1}})$ . If we have an algebraic curve $X/\mathbb {F} _{p}$  then $H^{2}(X,T_{X})=0$  because of dimension reasons. This implies that we can always deform and get a direct system of schemes as above. We can make a minor generalization of this case by considering an arithmetic surface $X/\mathbb {Z}$  which is an algebraic curve over each point $(p)$ . Then, the surface can be deformed into such a system. Deformations then give us another example of constructing a formal scheme ${\mathfrak {X}}={\text{Spf}}(X_{i})$ .

### Definitions

Set $A_{k}=\mathbb {Z} /p^{k}$ . You should think of elements in $A_{k}$  as finite sums

$a_{0}+a_{1}p+a_{2}p^{2}+\cdots +a_{k-1}p^{k-1}+a_{k}p^{k}$  where each $a_{i}\in \{0,1,2,\ldots ,p-1\}$

There is an obvious morphism $\phi _{k}:A_{k}\to A_{k-1}$  with kernel $p^{k-1}A_{k}$  sending

$\phi _{k}(a_{0}+a_{1}p+a_{2}p^{2}+\cdots +a_{k-1}p^{k-1}+a_{k}p^{k})=a_{0}+a_{1}p+a_{2}p^{2}+\cdots +a_{k-1}p^{k-1}$

We can use these morphisms to construct an inverse system

$\cdots \to A_{3}\to A_{2}\to A_{1}$

whose inverse limit is defined as the p-adic integers $\mathbb {Z} _{p}=\lim _{\leftarrow }A_{k}$ . Elements in $\mathbb {Z} _{p}$  should be thought of infinite sums

$a_{0}+a_{1}p+a_{2}p^{2}+\cdots$  such that $a_{i}\in \{0,1,\ldots ,p-1\}$

It is sometimes convenient to write these infinite sums as infinite tuples

$(a_{0},a_{1},a_{2},\ldots )$

Let's play around with $\mathbb {Z} _{5}$  to try and get a feel for what the $p$ -adics are about. Since there is a unique morphism $\mathbb {Z} \to \mathbb {Z} _{p}$  we can ask what the image of elements in $\mathbb {Z}$  look like. If we consider $89$ , then

$89=4+2\cdot 5+3\cdot 5^{2}=(4,2,3,0,0,\ldots )$

So all we did was find the decomposition of the integer in terms of base-$5$ . Negative numbers are a little more tricky since we need to figure out what $-1$  "means" in $\mathbb {Z} _{p}$ . Notice if we take the sum

{\begin{aligned}(1,0,0,0,\ldots )+(4,4,4,4,\ldots )&=(4+1,4,4,4,4,\ldots )\\&=(0,1,0,0,\ldots )+(0,4,4,\ldots )&{\text{ since we carried the }}1\\&=(0,4+1,4,4,\ldots )&\\&=(0,0,4+1,4,\ldots )&\\&=(0,0,0,0,\ldots )&\end{aligned}}

Then, in $\mathbb {Z} _{p}$  we can see that

$-1=(p-1,p-1,p-1,\ldots )$

In $\mathbb {Z} _{5}$  we can find that $-7$  is

$(0,3,4,4,4,\ldots )$

An interesting set of numbers to look at are the $-5^{k}$ 's. For example,

$-125=(0,0,0,1,4,4,4,\ldots )$

We can then look to see what the units in $\mathbb {Z} _{p}$  are like. Observe that for $a,b\in \mathbb {Z} _{p}$

{\begin{aligned}a\cdot p&=(a_{0}+a_{1}\cdot p+a_{2}\cdot p^{2}+\cdots )\cdot (b_{0}+b_{1}\cdot p+b_{2}\cdot p^{2}+\cdots )\\&=(a_{0}b_{0})+(a_{0}b_{1}+a_{1}b_{0})\cdot p+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0})\cdot p^{2}+\cdots \end{aligned}}

If we have $a\cdot b=1$  then

{\begin{aligned}a_{0}b_{0}&=1\\a_{0}b_{1}+a_{1}b_{0}&=0\\a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0}&=0\\\cdots \end{aligned}}

From this we see a $p$ -adic integer $a$  is invertible if and only if the $a_{0}\neq 0$ .

### Properties of $\mathbb {Z} _{p}$ The previous observations/computations should make the first two propositions easy to parse.

The last part of this section shows how to topologize the $p$ -adics. From proposition 2 we know that any $p$ -adic integer is of the form $p^{k}u$  where $u$  is a unit. We define the $p$ -adic valuation of this integer as

$v_{p}:\mathbb {Z} _{p}\to \mathbb {Z}$  by $v_{p}(p^{k}u)=k$  and $v_{p}(0)=+\infty$

For example

$v_{p}(-25)=v_{p}(0,0,1,4,4,\ldots )=2$  and $v_{p}(-1)=v_{p}(4,4,4,4,4,\ldots )=0$

Notice that

$v_{p}(xy)=v_{p}(x)+v_{p}(y)$  and $v_{p}(x+y)\geq {\text{inf}}(v_{p}(x),v_{p}(y))$

In particular

$v_{p}(-x)=v_{p}((-1)\cdot x)=v_{p}(-1)+v_{p}(x)=v_{p}(x)$

The $p$ -adic valuation can be used to topologize $\mathbb {Z} _{p}$  by defining the metric

$d(x,y)=e^{-v_{p}(x,y)}$

From the definition of the $p$ -adic valuation and it's properties with respect to negatives we can see that

$d(x,x)=e^{-\infty }=0$  and $d(x,y)=d(y,x)$

Since

$v_{p}(x-z)=v_{p}((x-y)+(y-z))\geq {\text{inf}}(v_{p}(x-y),v_{p}(y-z))$

and

$-v_{p}(x-z)\leq -{\text{inf}}(v_{p}(x-y),v_{p}(y-z))$

we can see that the triangle inequality holds

$d(x,z)\leq d(x,y)+d(y,z)$

We could have also taken the algebraic approach of defining the topology in terms of the neighborhoods $p^{k}\mathbb {Z} _{p}$  of $0$ . There are equal to the set

$\{a\in \mathbb {Z} _{p}:v_{p}(a)\geq k\}$

Finally, we could have given it the topology from the product of the $\prod A_{k}$  where each $A_{k}$  is equipped with the discrete topology. From Tynchenoff's theorem, we know that this is a compact space. And since $\mathbb {Z} _{p}\subset \prod A_{k}$  is closed it is also compact.

1. edit/reorganize
2. show density is obvious
4. Completeness of compact metric space - https://math.stackexchange.com/questions/627667/every-compact-metric-space-is-complete

### The field $\mathbb {Q} _{p}$ From the computation earlier, if we wanted to invert an element $p^{k}u$  we would have to find $v=u^{-1}$  but also invert the $p^{k}$ . This should give us the hint that the fraction field $\mathbb {Q} _{p}$  of $\mathbb {Z} _{p}$  is isomorphic to

$\mathbb {Z} _{p}\left[{\frac {1}{p}}\right]$

This is called the field of $p$ -adic numbers. A $p$ -adic number should be thought of as an infinite sum of the form

${\frac {a_{-k}}{p^{k}}}+{\frac {a_{-k+1}}{p^{k-1}}}+\cdots +{\frac {a_{-1}}{p}}+a_{0}+a_{1}\cdot p+a_{2}\cdot p^{2}+\cdots$

A useful tool for computing inverses is the formal power series

${\frac {1}{1+x}}=1-x+x^{2}-x^{3}+x^{4}-\cdots$

For example, setting $x=p$  we find that the inverse of $1+p$  in $\mathbb {Z} _{5}$  is

$(1,4,1,4,1,4,1,4,\ldots )$

and the inverse of $1-p$  is

$(1,1,1,1,1,1)$

In general, you have to use iterated long division to find the $p$ -adic expansion of a rational number.

We can extend the $p$ -adic valuation to $v_{p}:\mathbb {Q} _{p}\to \mathbb {Z} \cup \{\infty \}$  by

$v_{p}(p^{k}u)=k$  and $v_{p}(0)=+\infty$

The metric constructed previously on $\mathbb {Z} _{p}$  extends to $\mathbb {Q} _{p}$  and defines a locally compact topology. In addition, $\mathbb {Q}$  is dense in $\mathbb {Q} _{p}$  using a similar kind of argument as before.

#### Absolute Values (Extra)

There is an alternative construction of the p-adic numbers using a valuation on $\mathbb {Q}$ . Given a rational number $a/b\in \mathbb {Q}$  such that $gcd(a,b)=1$  we can construct the $p$ -adic absolute value

$|\cdot |_{p}:\mathbb {Q} \to \mathbb {R} _{\geq 0}$  defined by $|a/b|_{p}=p^{-v_{p}(a/b)}=p^{v_{p}(b)-v_{p}(a)}$

using the $p$ -adic valuation on $\mathbb {Q} _{p}$ . This absolute value satisfies the following axioms

1. $|x|_{p}=0$  if and only if $x=0$
2. $|xy|_{p}=|x|_{p}\cdot |y|_{p}$
3. $|x+y|_{p}\leq |x|_{p}+|y|_{p}$

In addition, it satisfies a stronger version of 3. called the non-archimedian property

1. $|x+y|_{p}\leq {\text{max}}\{|x|_{p},|y|_{p}\}$

A natural question to ask then is if there exists a classification scheme for absolute values on $\mathbb {Q}$ . This turns out to be true and is called Ostrowski's Theorem. These notes give an introduction and proof to this theorem. In addition, there is a generalization to a number field $K/\mathbb {Q}$  (meaning it is a finite field extension of $\mathbb {Q}$ ) which shows that the isomorphism classes of absolute values on $K$  are classified by the closed points of ${\text{Spec}}({\mathcal {O}}_{K})$ . This is discussed in these notes by Keith Conrad.

This section gives us the criterion for finding $p$ -adic varieties, or even better, schemes in ${\textbf {Sch}}/\mathbb {Z} _{p}$ .

1. add section with discussion of Hensel's lemma in both the simple and general cases
2. given $a\in \mathbb {Z}$  which is square free, we can show that the vanishing locus of $ay^{2}=f(x)$  has no rational points.

### Solutions

This section starts out with a useful technical lemma: a projective system

$\cdots \to D_{3}\to D_{2}\to D_{1}$

of finite non-empty sets has a non-empty inverse limit $D=\lim _{\leftarrow }D_{i}$ . This is directly applied to the case of considering a finite set of polynomials $f_1,\ldots,f_k \in \mathbb{Z}_p[x_1,\ldots,x_n]$: they have a non-empty vanishing locus in $(\mathbb {Z} _{p})^{m}$  if and only if their reductions ${\text{mod }}p^{k}$  have a solution in $(\mathbb {Z} /p^{k})^{m}$  for each $k$ . This proposition can be considered in the homogeneous case as well.

We should then be asking ourselves: how can we guarantee that there is a solution in each $(\mathbb {Z} /p^{k})^{m}$ ? This is answered in the next subsection where Serre proves Hensel's lemma.

### =Applications

In the next chapter Serre will be applying the tools here to study the polynomial

$z^{2}=ax^{2}+by^{2}$  in $\mathbb {Z} _{p}$

## The Multiplicative Group of $\mathbb {Q} _{p}$ The section studies the various multiplicative groups we have encountered so far: $\mathbb {F} _{p}^{*},\mathbb {Z} _{p}^{*},\mathbb {Q} _{p}^{*}$  and the squares of these groups. This tools in this section will be useful in the next chapter when Serre discusses the Hilbert symbol.

### The Filtration of the Group of Units

This subsection determines some of the roots of unity containted in $\mathbb {Z} _{p}$ , hence $\mathbb {Q} _{p}$ . Serre does this through a filtration on the group of units

$U=\mathbb {Z} _{p}^{*}=\{a_{0}+a_{1}\cdot p+a_{2}\cdot p^{2}+\cdots |a_{0}\neq 0\}$

given by

$U\supseteq U_{1}\supseteq U_{2}\supseteq U_{3}\supseteq \cdots$

where

$U_{n}=1+p^{n}\mathbb {Z} _{p}=\{1+0\cdot p+\cdots +0\cdot p^{n-1}+a_{n}\cdot p^{n}+\cdots \}$

Notice that each $U_{n}$  is the kernel of the morphism

$\varepsilon _{n}:\mathbb {Z} _{p}^{*}\to (\mathbb {Z} /p^{n})^{*}$  sending $a_{0}+a_{1}\cdot p+a_{2}\cdot p^{2}+\cdots \mapsto a_{0}+a_{1}\cdot p+\cdots +a_{n-1}\cdot p^{n-1}$

We can see that

$U\cong \lim _{\leftarrow }{\frac {U}{U_{n}}}$  since ${\frac {U}{U_{n}}}=\{a_{0}+a_{1}\cdot p+\cdots +a_{n-1}\cdot p^{n-1}|a_{0}\neq 0\}$

There is a short exact sequence

$1\to U_{1}\to U\to \mathbb {F} _{p}^{*}\to 1$

since $U_{1}$  contains the $p$ -adic integers of the form $1+a_{1}\cdot p+a_{2}\cdot p^{2}+\cdots$  while $U$  can have any $a_{0}\neq 0$ . Furthermore, there are short exact sequences of the form

$1\to U_{n+1}\to U_{n}\to \mathbb {Z} /p\to 0$

This is because if we take two elements $(1+p^{n}x),(1+p^{n}y)\in U_{n}$  we can multiply them together to get

$(1+p^{n}x)\cdot (1+p^{n}y)=1+p^{n}(x+y)+p^{2n}xy\equiv 1+p^{n}(x+y){\text{ }}({\text{mod }}p^{n+1})$

Serre then introduces a useful auxillary lemma to analyze the following direct system of short exact sequences

### Structure of the Group $U_{1}$ This subsection determines the structure of the group $\mathbb {Q} _{p}^{*}$ . It uses the observation that an $x\in \mathbb {Q} _{p}^{*}$  is equal to $p^{n}\cdot u[itex]where[itex]n\in \mathbb {Z}$  and $u\in U$ , hence we can decompose this group as the product $\mathbb {Z} \times U$ . Now we are reduced to determining the group structure of $U$  — this is done in proposition 8.

### Squares in $\mathbb {Q} _{p}^{*}$ ${\frac {\mathbb {Q} _{p}^{*}}{(\mathbb {Q} _{p}^{*})^{2}}}={\begin{cases}(\mathbb {Z} /2)^{3}&p=2\\(\mathbb {Z} /2)^{2}&p{\text{ odd}}\\{\text{ }}\mathbb {Z} /2&p=\infty \end{cases}}$