A User's Guide to Serre's Arithmetic/Finite Fields

Generalities
This section sets up many of the basic notions used in this book.

Finite Fields
This chapter starts out with a discussion of the structure of finite fields. Given a field $$K$$ its characteristic is defined as the smallest number $$n$$ such that $$n\cdot 1$$ is congruent to zero in $$K$$. If this number $$n$$ is unbounded, then we say $$K$$ is of characteristic 0. This is well-defined because every ring has a unique morphism $$\mathbb{Z}\to K$$.

For a field of positive characteristic, denoted $$\mathbb{F}_q$$ where $$q=|\mathbb{F}_q|$$, he goes on to show that $$q = p^f$$ for some prime $$p$$ and some integer $$f \geq 1$$ and that the characteristic of such a field is $$p$$.

Before stating this theorem he proves a lemma showing that the frobenius
 * $$\sigma: K \to K$$ given by $$\sigma(x) = x^p$$ is an injective morphism onto a subfield of $$K^p$$, (FIX: the subfield of numbers in $$K$$ invariant under...). This can be used to show that for an algebraic closure $$\Omega$$ of $$K$$, $$\sigma$$ is an automorphism.

The theorem also states that all finite fields of order $$q = p^f$$ are isomorphic to $$\mathbb{F}_q$$. It's worth noting that the technique of looking at a polynomial and its derivative is a common and useful technical tool.

The question you should be asking yourself is:
 * How can I construct finite fields of order greater than $$p$$?

We can use the case of the real numbers to get a hint: we should look at quadratic polynomial $$x^2 -a$$ and see if
 * $$\frac{\mathbb{F}_p[x]}{(x^2 - a)}$$

is a field or not. For example, $$x^2 - 2$$ has no solutions in $$\mathbb{F}_5$$ while $$x^2 + 1$$ has $$[2],[3]$$ as solutions. This implies that


 * $$\frac{\mathbb{F}_5[x]}{(x^2 - 2)} \cong \mathbb{F}_{25}$$ while $$ \frac{\mathbb{F}_5[x]}{(x^2 + 1)} \cong \mathbb{F}_5 \times \mathbb{F}_5 $$

The rest of the chapter is dedicated to building tools for determining if a quadratic function determines a field extension of a finite field. Note that this will give us a recursive method for finding any $$\mathbb{F}_q$$. In addition, we will construct a tool and a theorem, called the Legendre Symbol and Gauss' reciprocity theorem, for efficiently figuring out if $$x^2 - a$$ determines a field extension or product of fields.

Multiplicative Group of a Finite Field
This section is dedicating to show that the multiplicative group $$\mathbb{F}_q^*$$ is cyclic of order $$q-1$$. He does this through proving a stronger result that all subgroups of $$\mathbb{F}_q^*$$ are cyclic.

In addition, while proving the theorem, he shows a generalization of Fermat's Little Theorem which states
 * $$x^{q-1} \equiv 1 \text{ } (\text{mod } p) $$

Note Fermat's original theorem proved the case $$f=1$$.

The most useful techniques used in this section are the applications of the Euler $$\phi$$-function.

Equations Over a Finite Field
This section studies sets of the form
 * $$\{p \in \mathbb{F}_q^n : f_1(p) = \cdots = f_k(p) = 0 \}$$

where $$f_i \in \mathbb{F}_q[x_1,\ldots,x_n]$$. If you are used to scheme theory, Serre studies schemes of the form
 * $$X = \text{Spec}\left(\frac{\mathbb{Z}[x_1,\ldots,x_n]}{(f_1,\ldots,f_k)}\right)$$

by looking at the sets
 * $$X(\mathbb{F}_q)$$

Power Sums
This section introduces a technical tool for proving the Chevalley-Warning theorem. It relies on the following
 * 1) $$ 0 + 1 + 2 + \cdots + p = (0 + p) + (1 + (p-1)) + (2 + (p-2)) + \cdots = k\cdot p$$

Chevalley Theorem
The Chevallay-Warning theorem gives a useful criterion for determining the number of solutions to a set of polynomials over a finite field. I will restate it here for convenience


 * Given polynomials $$f_\alpha\in K[x_1,\ldots,x_n]$$ such that $$ \sum \text{deg}(f_\alpha) < n$$. The cardinality of $$V = \{p \in K^n: f_\alpha(p) = 0\}$$ is congruent to $$0 (\text{mod } p)$$

The most interesting technical tool used in the proof of this theorem is the indicator function $$P(x) = \prod_\alpha (1 - f_\alpha^{q-1}): K^n \to \mathbb{Z}/2$$ which could be equivalently described as the function

P(x) = \begin{cases} 1 & \text{ if } x \in V \\ 0 & \text{ otherwise} \end{cases} $$ Note that the $$q-1$$-power is an application of the generalization of Fermat's little theorem proved in the last section.

This theorem has numerous applications. First, it solves many arithmetic questions about the existence of solutions of polynomials over finite fields. This is stated in corollary 1. Also, he shows that a for a quadratic form $$Q = \sum a_{ij}x_ix_j$$ (meaning the $$a_{ij}$$ give a symmetric matrix) has a non-zero solution over every finite field.

Quadratic Reciprocity Law
This section gives us the construction of the Legendre symbol and Gauss' reciprocity theorem.

Squares in $$\mathbb{F}_q$$
The theorem is the setup for the definition of the Legendre symbol, which is defined as the second map in the short exact sequence
 * $$ 1 \to \mathbb{F}_q^{*2} \to \mathbb{F}_q^* \to \{\pm 1\} \to 1$$

This morphism is defined by using the generalization of Fermat's little theorem. Since $$x^{q-1} \equiv 1 \text{ } (\text{mod } q)$$ we have that $$x^{(q-1)/2} \equiv \pm 1 \text{ } (\text{mod } q)$$. For an application of this sequence recall that $$\mathbb{F}_9 \cong \mathbb{F}_3[i] \cong \mathbb{F}_3[x]/(x^2+1)$$. We can calculate that
 * $$(\mathbb{F}_9^*)^2 = \{1,2,i,2i\}$$

hence
 * $$\mathbb{F}_{3^3} = \mathbb{F}_{27} \cong \mathbb{F}_9[y]/(y^2 + 2i + 2) \cong F_3[x,y]/(x^2 + 1,y^2 + x + 1)$$

Legendre Symbol (Elementary Case)
Here Serre restricts to the classical case of the sequence
 * $$ 1 \to \mathbb{F}_p^{*2} \to \mathbb{F}_p^* \to \{\pm 1\} \to 1$$

and defines the second map as the Legendre symbol
 * $$\left(\frac{\cdot}{p}\right):\mathbb{F}_p^* \to \mathbb{Z}/2$$

If you embed $$\mathbb{Z}/2 \to U(1)$$, then the Legendre symbol is an example of a character (a group morphism $$\chi:G\to U(1)$$). This means that
 * $$\left( \frac{ab}{p} \right) = \left( \frac{a}{p} \right)\cdot \left( \frac{b}{p} \right)$$

In addition, the Legendre symbol can be extended to
 * $$\mathbb{F}_p$$ by setting $$\frac{0}{p}=0$$

Notice that we can lift the Legendre symbol to $$\mathbb{Z}$$ using the composition of the quotient map $$\mathbb{Z}\to \mathbb{Z}/p$$ with the Legendre symbol.

Finally, he finds a method for computing the Legendre symbol of $$[1],[-1],[2]$$. The first case is easy since $$[-1]^2 = [1] = [1]^2$$. For the last two cases he introduces a couple auxillary functions $$\varepsilon,\omega$$ from the odd integers to $$\mathbb{Z}/2$$:

\varepsilon(n) \equiv \frac{n-1}{2} \text{ } (\text{mod } 4) = \begin{cases} 0 & \text{ if } n \equiv 1 \text{ } (\text{mod } 4) \\ 1 & \text{ if } n \equiv -1 \text{ } (\text{mod } 4) \end{cases} $$

\omega(n) \equiv \frac{n^2-1}{8}\text{ } (\text{mod } 8) = \begin{cases} 0 & \text{ if } n \equiv \pm 1 \text{ } (\text{mod } 8) \\ 1 & \text{ if } n \equiv \pm 5 \text{ } (\text{mod } 8) \end{cases} $$ Recall from elementary number theory that every odd number greater than $$2$$ is of the for $$4k+1$$ or $$4k+3$$ (they can't be of the form $$4k$$ or $$4k+2$$ since those are even). Then, $$\varepsilon$$ acts as a function partitioning off the two sets of odd numbers. In addition, there are infinitely many prime numbers in both forms. Serre claims that

\left( \frac{-1}{p} \right) = \left( \frac{p-1}{p} \right) = (-1)^{\varepsilon(p)} $$ If we split $$p$$ into the two cases of odd numbers, then

p - 1 = \begin{cases} (4k + 1) - 1 = 4k \\ (4k + 3) - 1 = 4k + 2 \end{cases} $$ Then, using the definition of the Legendre symbol, we find that
 * $$\left(\frac{-1}{p}\right) = (-1)^{\frac{p-1}{2}} = \begin{cases}

(-1)^{\frac{4k}{2}} = (-1)^{2k} = 1 & \text{if } p \equiv 1 \text{ } (\text{mod } 4)\\ (-1)^{\frac{4k+2}{2}} = (-1)^{2k+1} = -1 & \text{if } p \equiv 3 \text{ } (\text{mod } 4) \end{cases}$$ as desired. In the last case, Serre again uses a function which partitions off the odd numbers. Notice that every odd number (hence ever prime greater then $$2$$) is of one of the forms
 * $$8k + 1, 8k + 3,8k + 5, 8k + 7$$

In order to take advantage of this partition, he embeds $$\mathbb{F}_p \to \overline{\mathbb{F}}_p$$ and claims that $$(\zeta_8 + \zeta_8^{-1})^2 = 2$$ where $$\zeta_8$$ is the primitive $$8$$-th root of unity (in the complex numbers $$\zeta_8 = e^{\frac{2\pi i}{8}})$$. This follows from the observation that
 * $$\zeta_8^4 = -1$$ hence $$\zeta_8^{-4} = -1$$ since $$\zeta_8^4\cdot\zeta_8^{-4} = \zeta_8^0 = 1$$ and $$\zeta_8^4\cdot\zeta_8^{-4} = (-1) \cdot \zeta_8^4$$

implying that
 * $$(\zeta_8^2 + \zeta_8^{-2})^2 = \zeta_8^4 + 2 + \zeta_8^{-4} = 0$$ forcing $$ \zeta_8^2 + \zeta_8^{-2} = 0$$

hence $$y = \zeta_8 + \zeta_8^{-1}$$ satisfies $$y^2 = 2$$ since
 * $$ y^2 = (\zeta_8 + \zeta_8^{-1})^2 = \zeta_8^2 + 2 + \zeta_8^{-2} = 2$$

Since the Frobenius $$\sigma: \overline{\mathbb{F}}_p \to \overline{\mathbb{F}}_p$$ is an automorphism of $$\overline{\mathbb{F}}_p$$, we have that
 * $$y^p = \zeta_8^p + \zeta_8^{-p}$$

If $$p \equiv \pm 1 \text{ } (\text{mod } 8)$$ then $$y^p = \zeta_8^p + \zeta_8^{-p} = \zeta_8 + \zeta_8^{-1} = y$$. This implies
 * $$\left(\frac{2}{p}\right) = y^{p-1} = 1 $$

Otherwise, if $$p \equiv \pm 5 \text{ } (\text{mod } 8)$$ then $$y^p = \zeta_8^5 + \zeta_8^{-5} = -(\zeta_8 + \zeta_8^{-1}) = -y$$ (draw a picture of the unit circle to check that $$-\zeta_8 = \zeta_8^5$$ and $$-\zeta_8^{-1}=\zeta_8^{-5}$$). Hence $$y^{p-1} = -1$$.

Small Remark
Furthermore, using the fact that the Legendre symbol is a group morphism, we can compute the Legendre symbol of $$[\mathbb{F}_p^*:(2)] + [\mathbb{F}_p^*:(p-1)]$$ many elements for $$p >3$$ without having to compute explicit squares.

Quadratic Reciprocity Law
This section is dedicated to proving Quadratic reciprocity. As we have said before, this is a useful computational tool for determining if
 * $$\frac{\mathbb{F}_p[x]}{(x^2 - a)}$$ is a field extension

He gives a computation of the Legendre symbol to determine that
 * $$\frac{\mathbb{F}_{43}[x]}{(x^2 - 29)} \cong \mathbb{F}_{1849}$$

I will simply state quadratic reciprocity and give references to other proofs.

Theorem: Given a pair of distinct odd prime number $$p$$,$$l$$ we have the following reciprocity law:
 * $$\left( \frac{l}{p} \right) = (-1)^{\varepsilon(p)\varepsilon(l)}\left( \frac{p}{l} \right)$$

His sample computation goes as follows:

\begin{align} \left( \frac{29}{43} \right) &= \left( \frac{43}{29} \right) & \text{ since } \varepsilon(43) = 0 \\ &= \left( \frac{14}{29} \right) & \text{ since } [43] = [14] \\ &= \left( \frac{2}{29} \right)\cdot\left( \frac{7}{29} \right) & \text{ from being a group morphism } \\ &= -\left( \frac{7}{29} \right) & \text{ since } 29 \equiv 5 \text{ } (\text{mod } 8) \\ &= -\left( \frac{29}{7} \right) & \text{ since } \varepsilon(7) = 0 \\ &= -\left( \frac{1}{7} \right) = -1 & \end{align} $$ There are nice discussions about the proofs of quadratic reciprocity on mathoverflow and here is a compilation of hundreds of proofs for quadratic reciprocity
 * https://mathoverflow.net/questions/1420/whats-the-best-proof-of-quadratic-reciprocity
 * http://www.rzuser.uni-heidelberg.de/~hb3/fchrono.html

Try reading the proof https://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity#Proof_using_algebraic_number_theory to motivate the generalization to Artin reciprocity.