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

## The ring ${\displaystyle \mathbb {Z} _{p}}$ and the field ${\displaystyle \mathbb {Q} _{p}}$

The section introduces one of the main players in arithmetic geometry: the p-acids. 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 ${\displaystyle X\in {\textbf {Sch}}/\mathbb {Z} }$  (such as ${\displaystyle {\text{Spec}}(\mathbb {Z} [x_{1},\ldots ,x_{n}]/(f_{1},\ldots ,f_{k}))}$  or ${\displaystyle {\text{Proj}}(\mathbb {Z} [x_{1},\ldots ,x_{n}]/(f_{1},\ldots ,f_{k}))}$ ) then you can consider the base change to ${\displaystyle X_{p}\in {\textbf {Sch}}/(\mathbb {Z} /p)}$ . From the inverse system

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

there is an associated direct system of schemes

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

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

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

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

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

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

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

where each square is cartesian. It turns out these questions are cohomological. All deformations depend on the cohomology group ${\displaystyle H^{1}(X_{1},T_{X_{1}})}$  and all "obstructions" to a deformation live in a group depending on ${\displaystyle H^{2}(X_{1},T_{X_{1}})}$ . If we have an algebraic curve ${\displaystyle X/\mathbb {F} _{p}}$  then ${\displaystyle H^{2}(X,T_{X})=0}$  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 ${\displaystyle X/\mathbb {Z} }$  which is an algebraic curve over each point ${\displaystyle (p)}$ . Then, the surface can be deformed into such a system. Deformations then give us another example of constructing a formal scheme ${\displaystyle {\mathfrak {X}}={\text{Spf}}(X_{i})}$ .

### Definitions

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

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

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

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

We can use these morphisms to construct an inverse system

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

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

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

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

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

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

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

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

{\displaystyle {\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 ${\displaystyle \mathbb {Z} _{p}}$  we can see that

${\displaystyle -1=(p-1,p-1,p-1,\ldots )}$

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

${\displaystyle (0,3,4,4,4,\ldots )}$

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

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

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

{\displaystyle {\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 ${\displaystyle a\cdot b=1}$  then

{\displaystyle {\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 ${\displaystyle p}$ -adic integer ${\displaystyle a}$  is invertible if and only if the ${\displaystyle a_{0}\neq 0}$ .

### Properties of ${\displaystyle \mathbb {Z} _{p}}$

The previous observations/computations should make the first two propositions easy to parse.

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

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

For example

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

Notice that

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

In particular

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

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

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

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

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

Since

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

and

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

we can see that the triangle inequality holds

${\displaystyle d(x,z)\leq d(x,y)+d(y,z)}$

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

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

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

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

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

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

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

A useful tool for computing inverses is the formal power series

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

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

${\displaystyle (1,4,1,4,1,4,1,4,\ldots )}$

and the inverse of ${\displaystyle 1-p}$  is

${\displaystyle (1,1,1,1,1,1)}$

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

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

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

The metric constructed previously on ${\displaystyle \mathbb {Z} _{p}}$  extends to ${\displaystyle \mathbb {Q} _{p}}$  and defines a locally compact topology. In addition, ${\displaystyle \mathbb {Q} }$  is dense in ${\displaystyle \mathbb {Q} _{p}}$  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 ${\displaystyle \mathbb {Q} }$ . Given a rational number ${\displaystyle a/b\in \mathbb {Q} }$  such that ${\displaystyle gcd(a,b)=1}$  we can construct the ${\displaystyle p}$ -adic absolute value

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

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

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

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

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

A natural question to ask then is if there exists a classification scheme for absolute values on ${\displaystyle \mathbb {Q} }$ . 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 ${\displaystyle K/\mathbb {Q} }$  (meaning it is a finite field extension of ${\displaystyle \mathbb {Q} }$ ) which shows that the isomorphism classes of absolute values on ${\displaystyle K}$  are classified by the closed points of ${\displaystyle {\text{Spec}}({\mathcal {O}}_{K})}$ . This is discussed in these notes by Keith Conrad.

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

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

### Solutions

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

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

of finite non-empty sets has a non-empty inverse limit ${\displaystyle D=\lim _{\leftarrow }D_{i}}$ . This is directly applied to the case of considering a finite set of polynomials ${\displaystyle f_{1},\ldots ,f_{k}\in \mathbb {Z} _{p}[x_{1},\ldots ,x_{n}]}$ : they have a non-empty vanishing locus in ${\displaystyle (\mathbb {Z} _{p})^{m}}$  if and only if their reductions ${\displaystyle {\text{mod }}p^{k}}$  have a solution in ${\displaystyle (\mathbb {Z} /p^{k})^{m}}$  for each ${\displaystyle k}$ . 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 ${\displaystyle (\mathbb {Z} /p^{k})^{m}}$ ? 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

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

## The Multiplicative Group of ${\displaystyle \mathbb {Q} _{p}}$

The section studies the various multiplicative groups we have encountered so far: ${\displaystyle \mathbb {F} _{p}^{*},\mathbb {Z} _{p}^{*},\mathbb {Q} _{p}^{*}}$  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 ${\displaystyle \mathbb {Z} _{p}}$ , hence ${\displaystyle \mathbb {Q} _{p}}$ . Serre does this through a filtration on the group of units

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

given by

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

where

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

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

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

We can see that

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

There is a short exact sequence

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

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

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

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

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

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

### Structure of the Group ${\displaystyle U_{1}}$

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

### Squares in ${\displaystyle \mathbb {Q} _{p}^{*}}$

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