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-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.

(Optional) Advanced Remarks

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
3. http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf for hensel's lemma
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.

p-adic Equations

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 $p^{n}\cdot u<math>where<math>n\in \mathbb {Z}$ 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 $\mathbb {Q} _{p}^{*}$

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