• # Syllabus for the Oral Number Theory Qualifying Exam

Format: Plan for a 50 minute oral exam.

## Elementary Number Theory.

Congruences
Gauss Sums and Jacobi Sums (e.g. Chapters 6, 7 Ireland and Rosen)

## Algebraic Number Theory

Algebraic Integers
Minkowski Theory
Finiteness of the Ideal Class Group
Dirichlet’s Unit Theorem
Cyclotomic Fields

## Analytic Number Theory

Prime Number Theorem
Dirichlet L-functions (analytic continuation and functional equation)
Dirichlet’s Theorem on Primes in Arithmetic Progressions

## Class Field Theory

Ray Class Groups
Idelic Theory
Artin Reciprocity
Artin Map
Hilbert Class Fields

• QR
Gauss sums
Jacobi sums

Dirichlet’s unit theorem
Dirichlet’s theorem on primes in arithmetic progressions
Dirichlet’s class number formula
L-functions
Prime number theorem
Prime ideals
Integral basis
Discriminant

Minkowski’s bound
Cyclotomic fields
Decomposition group
Frobenius element

Cebotarev density theorem

Artin map
Frobenius element
Artin reciprocity

Ideles
Ray class group
Hilbert class fields
Class field theory inequalities
Places

Factorization of ideals
Splitting of primes

Norm of ideals
Ramification
Inertia

• # Quadratic Reciprocity

The discriminant of $\mathbb Q[\omega_p]$ is $\pm p^{p-2}$, with the plus sign holding if $p\equiv 1 \pmod 4$. (Use Vandermonde determinant). By definition, the square root of the discriminant is in the field, so $\mathbb Q(\sqrt{p^*}) \subseteq \mathbb Q[\omega_p]$.

Now $\left( \frac{q}{p }\right)=1$ iff $q$ splits completely in $F_2$ (the unique quadratic subfield), and $F_2$ must be $\mathbb Q(\sqrt{p^*})$, since it is the unique subfield with degree 2 over $\mathbb Q$.

We also know that $q$ splits completely in a quadratic field $\mathbb Q(\sqrt{m})$ when $m$ is a square mod $q$ (or for $q=2$, when $m\equiv 1 \pmod 8$).

Putting this together, we get that $\left(\frac{q}{p}\right)=\left(\frac{p^*}{q}\right)$.

In the case where $q=2$, we get that $\left (\frac{2}{q}\right)=1 \text{ when } p^*\equiv 1 \pmod 8.$

Generalizes to Artin reciprocity.

• # Gauss sums

## Quadratic Gauss sum

For a Dirichlet character modulo $p$, it’s a sum over $p$th roots of unity that has absolute value $\sqrt p$. So it gives you away to construct a quadratic element in a cyclotomic field.

If $(a,p)=1$, then $g(a,\chi) = \sum_{n=1}^p \zeta_{p-1}^{an^2} = \sum_{n=1}^p \left(\frac{n}{p}\right)\zeta_p^{an}$
These are commonly used to prove quadratic reciprocity.

## Gauss sum for Dirichlet characters

$G(\chi) = \sum_{n=1}^{N} \chi(n)\zeta_N^{n}$

If $\chi$ is primitive, $|G(\chi)|=\sqrt{N}$.

## Jacobi sum

$J(\chi, \psi) = \sum \chi(a) \psi(1 -a)$

Factors for $\chi, \psi$ non-trivial as
$J(\chi,\psi) = \frac{G(\chi)G(\psi)}{G(\chi\psi)}$.

Analogous to Beta functions factoring as Gamma functions (which I think is pretty much the prime at infinity case of all this).

• # Dirichlet’s unit theorem

We use a similar geometric approach as with Minkowski’s bound, but we take a log map of the lattice space generated by the embeddings so we can look at multiplicative structure. The kernel of this map is the roots of unity contained in the field.

Put a norm on the log space corresponding to the regular norm. The dimension is at most $r+s-1$, since units have norm 1, and 1 maps to 0.

Carefully construct a compact, convex, centrally symmetric set $E$ satisfying certain inequalities, and with volume equal the volume of the fundamental parallelopiped.
Using Minkowski’s convex body theorem, we can show $E$ is guaranteed to contain a lattice point.

We can adjust the inequalities to get lattice points with positive values for all but one dimension. Since the norm of elements of $E$ is bounded, by repeatedly applying this to a lattice point, we eventually get points with the same norm, which are different by a unit with certain properties. Doing this for each dimension, we get a suitable set of units.

These units are negative along all dimensions except for one on which they are positive. Putting them in a matrix, we can show that they have rank $r+s-1$, which proves our theorem.

Note the similarity of the proof of this theorem and the proof of the finiteness of the ideal class group. These theorems can indeed be unified into the statement that the Picard group of the field is compact. This also relates to Dirichlet’s class number formula - the volume of the Picard group is hR, where h is the class number and R is the regulator.

The Picard group is a generalization of the ideal class group.

• # Dirichlet’s theorem on primes in arithmetic progressions

First, use characters to sort the primes into $L$-functions.

Cyclotomic fields.

Relate this to Cebotarev, Artin stuff.

Hard part is to show $L(1,\chi)$ is non-zero for non-trivial characters.

• # Dirichlet’s class number formula

Relates class numbers to zeta values.

• # L-functions

Prototypical example is Riemann zeta function.

For class field theory, typically Dirichlet’s $L$-functions are used $L(s,\chi) = \sum_{n=1}^\infty \frac{\chi(n)}{n^s}.$ These are used to prove Dirichlet’s theorem on primes in arithmetic progressions, and for his class number formula.

Modularity theorem tells us that the $L$-function for an elliptic curves and the $L$-function for the associated weight 2 Hecke-eigenform are identical.

Generalize to Weber $L$-functions, where we use ray class groups.

Expected to satisfy functional equation/be identical for certain things - Langland’s program.

Expected to be zero-free in certain regions - Generalized Riemann hypothesis.

• # Prime number theorem

The asymptotic distribution of the prime numbers is $\pi(x) \sim \frac{x}{\ln x}$.

The idea is that we count the primes using a special weighting, giving us better analytic properties. For example, Chebyshev’s function $\psi(x) = \sum_{n \le x}\Lambda(n)$, where the Von-Mangoldt function $\Lambda(n)$ is $\log p$ when $n$ is a prime power, and zero otherwise.

We relate this to the Riemann zeta function by looking at the logarithmic derivative

$\displaystyle -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \Lambda(n) n^{-s}$.

This can be transformed into the explicit equation $\psi(x) = x - \sum_{\rho} \frac{x^\rho}{\rho}$, where $\rho = \sigma + \tau i$.

The crux is to show that $\zeta(\rho)$ has no zeros on the line $\tau = 1$. This can be done using a trigonometric argument - launching point to pretentious theory. Once we have this, a careful accounting of possible types of zeros gives us the essence of the prime number theorem.

The remaining gap is solved using analysis, for example a Tauberian theorem.

• # Algebraic integers form a ring

If $\alpha$ is an algebraic integer, then $\alpha A\subset A$ for some finitely generated additive subgroup $A\subset \mathbb C$, namely $\mathbb Z[\alpha]$.

Conversely, let $\alpha A \subset A$ for some $\alpha \in \mathbb C$, where $A$ is a finitely generated subgroup $A\subset \mathbb C$. Let $a_i$ be a set of generators. Then $\begin{pmatrix} \alpha a_1 \\ \alpha a_n \end{pmatrix}= M \begin{pmatrix} a_1 \\ a_n\end{pmatrix}$, where $M$ is an $n\times n$ matrix over $\mathbb Z$. This means $\alpha$ is an eigenvalue of $M$, so $\det (\alpha I -M)=0$, which is a monic polynomial with coefficients in $\mathbb Z$. This $\alpha$ is an algebraic integer.

Using the converse with the finitely generated subgroup $\mathbb Z[\alpha, \beta]$, we get that the algebraic integers form a ring.

• # Integral basis

## Integral basis theorem

For a deg $n$ number field $K$, $\mathcal O_K$ is a degree $n$ $\mathbb Z$-module.

Start w a $\mathbb Q$-basis for $K$. Clear denominators so that you have a $\mathbb Q$-basis inside $\mathcal O_K$. Call the discriminant of this basis $d$, and the basis itself $\{ \alpha_i\}$.

We can write $s = \sum a_i \alpha_i$, with $a_i$ rational.

Consider the image of this under all the embeddings $\sigma$. If we solve for $a_j$ using Cramer’s rule, we get that $a_J = \gamma_j/\delta$, where $\delta = |\sigma_i(\alpha_j)|$, and $\gamma_j$ is constructed like $\delta$ except we replace the $j$th column by $\sigma_i(s)$.

Thus, $\delta$ and $\gamma_j$ are both algebraic integers, with $\delta^2 = d$. So $da_j = \delta \gamma_j$, which implies $da_j$ is an integer. This shows that $\mathcal O_K$ is contained in a free abelian group of rank $n$.

## Discriminant

Can be thought of as the square of the volume of the integral basis, int the lattice spaced generated from all the embeddings.

Is an invariant of a number field.

• # Minkowski’s bound

The ideal class group of a ring is finite.

The basic idea is to prove every ideal $I$ contains an element $\alpha$ that is ‘small enough’ so something like $|N_{\mathbb Q}^K(\alpha)| \le \lambda ||I||$ where $\lambda$ is independent of $I$.

Here’s how we use this: take an ideal class $C$, and consider an ideal $I\in C^{-1}$. Use the above to find such an $\alpha \in I$. So $I$ contains $(\alpha)$, therefore $(\alpha) = IJ$ for some ideal $J\in C$.

Since $|N_{\mathbb Q}^K(\alpha)| = ||(\alpha)|| = ||I|| ||J||$, the ideal $J$ has ideal norm less than $\lambda$. Since $\lambda$ is fixed, this bounds the possible divisors of $J$, and so there are only a finite number of such $J$‘s. Since we can find such a $J$ for each ideal class, the ideal class group is finite!

Minkowski used a geometric argument to get a really nice $\lambda$. Every ideal class contains an ideal $J$ such that
$||J||\le \frac{n!}{n^n}\left(\frac{4}{\pi}\right)^s\sqrt{|\textsf{disc}(R)|}.$

To show this, we embed our ring of integers $R$ into $\mathbb R^r \times \mathbb C^s$ as a lattice, using the different embeddings. The discriminant is the square of the volume of the fundamental parallelopiped.

An ideal will map to a sublattice, and the fundamental volume of this will be equal to $||I||$ times $\sqrt{|\textsf{disc}(R)|}$.

We can define a norm on this space that matches the standard norm. Now carefully construct a space $A$ where each element has norm has absolute value bounded by 1, which is compact, convex, and centrally symmetric, and has volume $\frac{n^n}{n!}2^r \left(\frac{\pi}{2}\right)^s.$

Applying Minkowski’s convex body theorem to $E$, which is $A$ scaled to have volume equal to $||I||\sqrt{|\textsf{disc(R)}|}$ (the volume of the fundamental parallelopiped for the lattice generated by $I$), we are guaranteed that there is a non-zero lattice point for $I$ inside $E$.

This point gives us the $\alpha$ that we need for the theorem.

• # Cyclotomic fields

Discriminant for prime case is $\pm p^{p-2}$.

Extension has degree $\varphi(m)$ over $\mathbb Q$.

The Galois group is isomorphic to $\left(\mathbb Z/m\mathbb Z\right)^\times.$

Important special case for building up larger theorems in class field theory.

## Primes $q$ that split completely in degree $d$ subfields of cyclotomic extensions are $d$th powers mod $p$.

For a cyclotomic field $\mathbb Q[\omega_p]$, the Galois group over $\mathbb Q$ is cyclic of order $p-1$. Let $F_d\subset \mathbb Q[\omega_p]$ be the unique subfield of degree $d$ over $\mathbb Q$, when $d|p-1$.

$q$ splits into $r$ primes in $\mathbb Q[\omega_p]$, and $f=(p-1)/r$ is the order of $q$ mod $p$. Since this is a cyclic group, the $d$th powers form the unique subgroup of order $(p-1)/d$.

Thus, TFAE

1. $q$ is a $d$th power mod $p$
2. $f|(p-1)/d$
3. $d|r$
4. $F_d\subset F_r$

Now observe that $F_r$ is the decomposition field for any prime $Q$ over $q$, since this is the only field over $\mathbb Q$ of degree $r$.

Therefore, $F_d \subset F_r$ is equivalent to $q$ splitting completely in $F_d$.

• # Galois theory of number fields

## Inertia

The inertial degree of $Q$ over $P$ is defined as the degree of the extension $\mathcal O_L/Q$ over $\mathcal O_K/P$.

## Decomposition group

The decomposition group of $Q$ over $P$ is $D = \{\sigma \in G : \sigma Q = Q\}.$ The inertia group is $E = \{\sigma \in G: \sigma(\alpha) \equiv \alpha \pmod Q\}$, which is a normal subgroup of $D$.

The quotient $D/E$ is a cyclic group of order $f$ - the inertial degree.

## Decomposition/ramification fields

By Galois theory we can look at the fixed fields of $D$ and $E$. If $D$ is a normal subgroup of $G$, it turns out that $L_D$ has degree $r$ over $K$, and $P$ splits into $r$ primes here. $L_E$ has degree $f$ over $L_D$, and the primes of $L_D$ over $P$ have inertial degree $f$ here. Finally, $L$ has degree $e$ over $L_E$, and the primes of $L_E$ over $P$ all have ramification index $e$ in $L$.

• # Cebotarev density theorem

This theorem gives a density for the prime ideals that split in a certain way (or alternatively, have an Artin symbol in a specific conjugacy class).

For an extension $L/K$ with Galois group $G$, then the set of primes of $\mathcal O_K$, $\mathcal S = \{\mathfrak p : \mathfrak p \text{ is unramified in } L \text{ and } \left(\frac{L/K}{\mathfrak p}\right) = \langle \sigma \rangle \}$, and where $\sigma \in \text{Gal(L/K)}$, has Dirichlet density
$\delta(S) = \frac{|\langle \sigma \rangle |}{[L:K]}$.

The case when $\sigma = 1$ gives the density of the primes that split completely.

When $L$ is a cyclotomic extension of $\mathbb Q$, this reduces to Dirichlet’s theorem on primes in arithmetic progressions.

The proof is important because it uses the technique of reducing to relatively cyclotomic extensions - used for the proof of Artin reciprocity.

• # Artin reciprocity

For a ring $R$ with characteristic $p$, there is an homomorphism called the Frobenius homomorphism taking $x$ to $x^p$ (and for the prime at infinity, it is complex conjugation). The basic question that reciprocity laws answer is ‘which ring homomorphism is it?’

For a prime field, every homomorphism is the identity - including the Frobeinus homomorphism. This is Fermat’s Little Theorem.

For a quadratic extension $\mathbb F(\sqrt{d})$ of a prime field $\mathbb F$ with char $p$, we could either get the identity or conjugation as the Frobenius homomorphism. We can look at this for all the primes (except for the ones dividing $4d$) at once by studying $\mathbb Q[\sqrt{d}]$.

In this case, Artin’s reciprocity law is that there exists a group homomorphism $(\mathbb Z / 4d\mathbb Z)^\times \rightarrow \{\pm 1\}$, which takes $(p \bmod 4d) \mapsto \left(\frac{d}{p}\right)$ for any prime $p$ not dividing $4d$. This is equivalent to quadratic reciprocity.

## Artin map

More generally, suppose we have an abelian Galois extension $\mathbb Q[\alpha]$, with $f$ the minimal polynomial of $\alpha$, and Galois group $G$. Then there exists a group homomorphism $(\mathbb Z / \Delta(f) \mathbb Z)^\times \rightarrow G$ which takes $(p \bmod \Delta(f)) \mapsto \varphi_p$ for primes $p$ not dividing $\Delta(f)$.

This map is called the Artin map. It identifies primes $p$ with $\varphi_p$, which is the unique element of $G$ that extends the Frobenius map of $\mathbb F_p[\alpha]$. This also applies to the prime at infinity.

Even more generally, for $K/F$, and $\mathfrak P$ a prime of $K$ over $\mathfrak p$ a prime of $F$, then we define the Artin symbol $\left(\frac{K/F}{\mathfrak P}\right)$ to be the $N(\mathfrak p)$-power map modulo $\mathfrak P$. This can be extended by multiplicativity to a homomorphism $I_F(\mathfrak m) \rightarrow \text{Gal}(K/F)$, which is the Artin map.

Another important fact is that the order of the Artin symbol is the inertial degree of $\mathfrak P$ over $\mathfrak p$ when $\mathfrak p$ is unramified. This implies that $\left(\frac{K/F}{\mathfrak P}\right) = 1$ iff $\mathfrak p$ splits completely in $K$. More generally, the Artin map tells you how $\mathfrak p$ factors in $K$.

## Artin reciprocity

Let $K/F$ be an abelian extension, and let $\mathfrak m$ be a modulus divisible by all the primes of $F$ which ramify in $k$ (including infinite). Then

1. The Artin map $\Phi_\mathfrak m$ is surjective
2. The modulus $\mathfrak m$ can be chosen (with high enough powers of primes dividing it) so that we have an isomorphism $I_F(\mathfrak m)/\text{ker}(\Phi_\mathfrak m) \simeq \text{Gal}(K/F)$

This gives a nice relation between generalized class groups and Galois groups, which can be used in particular for Hilbert class fields.

• # Ideles

It’s more convenient to work with completions for all the primes at once, including the primes at infinity. Accounting for all the primes reveals a nice symmetry which can be used to state and prove the theorems of class field theory in a natural way.

## Places

We can define different absolute values on number fields, which must be multiplicative homomorphisms mapping 0 to 0, and satisfying a weakend form of the triangle inequality. We can arrange these into different equivalence classes based on the induced topology. These are called the palces. It turns out by Ostrowski’s theorem that each place corresponds to

1. $\mathfrak p$-adic absolute value, or a finite place
2. an absolute value for a real embedding, infinite real place
3. or an absolute value from an imaginary embedding (which is the same for the conjugate embedding), the infinite imaginary place.

For each place $v$ of a field $F$ we can construct the completion with respect to that place (analogously to the construction of $\mathbb R$ from $\mathbb Q$ - which is a special case). We call this $F_v$.

## Takagi’s formulation

Let $F_v$ be the completion of $F$ at the place $v$. We define $J_F$ as $\prod_v F_v^\times$, where this is the restricted topological product. Also, $\mathcal E_F = \prod_v \mathcal U_v$, where $\mathcal U_v$ are the elements with norm 1.

The major theorems of class field theory can be compactly stated as the following. There is an order reversing, bijective correspondence between the set of all finite abelian extensions $K/F$, and the set of all open subgroups $\mathcal H$ of $J_F$ which contain $F^\times$. Furthermore, $\text{Gal}(K/F)\simeq J_F/\mathcal H$.

Note that $J_F/F^\times \mathcal E_{F,\mathfrak m}^+ \simeq \mathcal R_{F,\mathfrak m}^+$, the strict ray class group.

## Other applications

For another example, the functional equation for certain $L$-functions can be elegantly proved using idelic theory, à la Tate’s thesis.

• # Ray class group

Generalization of ideal class group relative to a modulus $\mathfrak m$.

Principal ideals relative to $\mathfrak m$: $\mathcal P_{F,\mathfrak m} =\{\langle\alpha\rangle : \alpha \in \mathcal O_F, \alpha\equiv 1 \pmod {\mathfrak m} \}$

Ideal relative to $\mathfrak m$: $\mathcal I_F(\mathfrak m) = \{\mathfrak a : \text{ord}_{\mathfrak p}\mathfrak a = 0 \text{ for all } \mathfrak p | \mathfrak m \}$.

Ray class group is $\mathcal R_{F,\mathfrak m} = \mathcal I_F(\mathfrak m)/\mathcal P_{F,\mathfrak m}$.

Strict if $\mathfrak m$ includes prime at infinity

Used to generalize Dirichlet’s Theorem on Primes in Arithmetic Progressions - Are there infinitely many prime ideals in each ray class? Weber $L$-functions are defined for characters over Ray class groups.

• # Hilbert class fields

The maximal unramified abelian extension of a number field $K$ (unramified at the infinite places also). It always exists.

A particularly important property is that its
Galois group is isomorphic to the ideal class group of $K$ (uniquely!). This gives it the ‘right’ to be called a class field, and allows us to use Galois theory and theorems from class field theory to prove results about the class group.

For example, we use this to prove Cebotarev’s density theorem.

For imaginary quadratic fields, the $j$-invariant can be used to give you the Hilbert class field.

• # Class field theory inequalities

The inequalities tell us about the relation between $J_F/\mathcal H$ and $K/F$, where $\mathcal H = F^\times N_{K/F} J_K$.

## Universal Norm Index Inequality

$\mathcal H$ is an open subset of $J_F$, so if $\mathfrak m$ is chosen so that $\mathcal E_{F,\mathfrak m}^+ \subseteq \mathcal H$, then $[J_F : \mathcal H] \le [K : F]$

This generalizes the fact that $L_\mathfrak m (1,\chi) \neq 0$ for non-trivial characters which are trivial on $\mathcal H$.

## Global Cyclic Index Inequality

If $K/F$ is a cyclic extension, and $\mathfrak m$ is divisible by a sufficiently high power of every ramified prime in $K/F$, then $[J_F : \mathcal H] \ge [K:F]$.

Combining this with the Universal Norm Index inequality gives us equality.

• # Factorization of ideals

Ideals in number fields have unique factorization into prime ideals. In algebraic number theory, the main focus is on how a prime ideal factors once it has been lifted up in a field extension.

## Splitting of primes

In an extension of a number field, there are three different ways that primes in the base field factor.

They can split, ramify, or have inertia.

## Split primes via factoring polynomials

One way to see how a prime splits is to see how the minimal polynomial for the extension factors modulo that prime.
Say $L = K (\alpha)$, and let $f$ be the minimal polynomial of $\alpha$ over $K$. Let $p$ be a prime not dividing $[\mathcal O_L:K[\alpha]]$, and let $\mathfrak p$ be a prime ideal of $K$ over $p$. Then if $f$ factors mod $\mathfrak p$ as $f(x) \equiv \pi_1(x)^{e_1}\cdots \pi_g(x)^{e_g} \bmod p$, we have that $\mathfrak p$ factors as $\mathfrak p = \mathfrak q_1^{e_1}\cdots \mathfrak q_g^{e_g}$. Furthermore, $\mathfrak q_i = \mathfrak p + (\pi_i(\alpha))$.

## Split primes via Artin symbol

We can also use the Artin symbol to determine how an unramified prime splits. Consider an extension $K/F$, along with a Galois extension $M/F$ containing $K$, with Galois group $G$. Call the Galois group of $M/K$, $H$. Fix a prime $\mathfrak U$ of $M$ lying over $\mathfrak P$ of $K$, lying over $\mathfrak p$ a prime of $F$. Then for each $\sigma \in G$, consider the coset $H\sigma$, and look at the orbits from right-action by the Artin symbol of $\mathfrak U$ over $\mathfrak p$. The size of the orbit is the inertial degree of $\sigma \mathfrak P$ over $\mathfrak p$, and thus the Artin symbol determines the splitting of $\mathfrak p$ in $K$.

## Galois group permutes primes transitively

Take $L/K$ to be a normal extension.

Suppose that $\sigma(Q)\neq Q'$ for all $\sigma\in G$. Then by the C.R.T., there is a solution to the system of congruences $x\equiv 0 \pmod {Q'}$ and $x\equiv 1 \pmod {\sigma(Q)}$.

Let $\alpha\in \mathcal O_L$. Then $N(\alpha) \in \mathcal O_K \cup Q' = P$, since $\alpha \in Q'$.

On the other hand, $\alpha \not\in \sigma(Q)$, so $\sigma^{-1}(\alpha)\not\in Q$. We can also write $N(\alpha)$ as the product of all $\sigma^{-1}(\alpha)$, so this implies that $N(\alpha)\not\in Q$, which is a contradiction.

• # Norm of ideals

The norm of an ideal $I$ of $\mathcal O_K$ is $|\mathcal O_K/I|$. This is multiplicative, and the norm a the principal ideal $(\alpha)$ is $|N^K(\alpha)|$.

The norm of a prime $Q$ above a prime $P$ is $|P|^f$, where $f$ is the inertial degree.

## Degree of extension is $\small \sum_i e_if_i$

Take a prime $p\in\mathbb Z$.
$p\mathcal O_k = \prod_i Q_i^{e_i}$

Now take ideal norms of everything. Since these are multiplicative, and since $||Q_i||=p^{f_i}$, the theorem follows.

• # Ramification

When looking at an extension of number fields $K/F$, a prime $P$ ramifies if $P\mathcal O_K = \prod P_i^{e_i}$ for $e_i > 1$ for some $i$.

## Ramified primes divide the discriminant

Let $P$ be a prime of $R$ lying over $p$ which ramifies. Then $pR=PI$, with $I$ divisible by all the primes of $R$ lying over $p$.

Take any integral basis. Replace one of the elements by $\alpha\in I \setminus pR$. It is sufficient to show that $p$ divides the discriminant of this basis.

Assume the field is Galois (if not, use the Galois ext). For each automorphism, notice that $\sigma^{-1}(Q)$ is a prime lying over $p$, so it contains $\alpha$. Thus, $\sigma_i(\alpha)\in Q$ for all $i$.

This means that $Q$ contains the discriminant of this basis, and since the discriminant is an integer, it has to be in $p\mathbb Z$.

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(Use Vandermonde determinant). By definition, the square root of the discriminant is in the field, so $\\mathbb Q(\\sqrt{p^*}) \\subseteq \\mathbb Q[\\omega_p]$. \n\n\nNow $\\left( \\frac{q}{p }\\right)=1$ iff $q$ splits completely in $F_2$ (the unique quadratic subfield), and $F_2$ must be $\\mathbb Q(\\sqrt{p^*})$, since it is the unique subfield with degree 2 over $\\mathbb Q$. \n\nWe also know that $q$ splits completely in a quadratic field $\\mathbb Q(\\sqrt{m})$ when $m$ is a square mod $q$ (or for $q=2$, when $m\\equiv 1 \\pmod 8$). \n\nPutting this together, we get that $\\left(\\frac{q}{p}\\right)=\\left(\\frac{p^*}{q}\\right)$. \n\nIn the case where $q=2$, we get that $\\left (\\frac{2}{q}\\right)=1 \\text{ when } p^*\\equiv 1 \\pmod 8.$\n\nGeneralizes to Artin reciprocity. \n","deleted":false},{"_id":"4718acf3ad2a45ed2a000030","treeId":"384915e096bc4c7cd0000017","seq":7922908,"position":2,"parentId":"4718aafdad2a45ed2a00002e","content":"# Gauss sums\n\n## Quadratic Gauss sum\nFor a Dirichlet character modulo $p$, it's a sum over $p$th roots of unity that has absolute value $\\sqrt p$. So it gives you away to construct a quadratic element in a cyclotomic field. \n\nIf $(a,p)=1$, then $g(a,\\chi) = \\sum_{n=1}^p \\zeta_{p-1}^{an^2} = \\sum_{n=1}^p \\left(\\frac{n}{p}\\right)\\zeta_p^{an}$\nThese are commonly used to prove quadratic reciprocity. \n\n## Gauss sum for Dirichlet characters\n\n$G(\\chi) = \\sum_{n=1}^{N} \\chi(n)\\zeta_N^{n}$\n\nIf $\\chi$ is primitive, $|G(\\chi)|=\\sqrt{N}$. \n\n## Jacobi sum\n$J(\\chi, \\psi) = \\sum \\chi(a) \\psi(1 -a)$\n\nFactors for $\\chi, \\psi$ non-trivial as\n$J(\\chi,\\psi) = \\frac{G(\\chi)G(\\psi)}{G(\\chi\\psi)}$. \n\nAnalogous to Beta functions factoring as Gamma functions (which I think is pretty much the prime at infinity case of all this). "},{"_id":"4718ad52ad2a45ed2a000031","treeId":"384915e096bc4c7cd0000017","seq":7922885,"position":3,"parentId":"4718aafdad2a45ed2a00002e","content":"# Dirichlet's unit theorem\n\nWe use a similar geometric approach as with Minkowski's bound, but we take a log map of the lattice space generated by the embeddings so we can look at multiplicative structure. The kernel of this map is the roots of unity contained in the field. \n\nPut a norm on the log space corresponding to the regular norm. The dimension is at most $r+s-1$, since units have norm 1, and 1 maps to 0.\n\nCarefully construct a compact, convex, centrally symmetric set $E$ satisfying certain inequalities, and with volume equal the volume of the fundamental parallelopiped. \nUsing Minkowski's convex body theorem, we can show $E$ is guaranteed to contain a lattice point. \n\nWe can adjust the inequalities to get lattice points with positive values for all but one dimension. Since the norm of elements of $E$ is bounded, by repeatedly applying this to a lattice point, we eventually get points with the same norm, which are different by a unit with certain properties. Doing this for each dimension, we get a suitable set of units. \n\nThese units are negative along all dimensions except for one on which they are positive. Putting them in a matrix, we can show that they have rank $r+s-1$, which proves our theorem. \n\n\n****\nNote the similarity of the proof of this theorem and the proof of the finiteness of the ideal class group. These theorems can indeed be unified into the statement that the Picard group of the field is compact. This also relates to Dirichlet's class number formula - the volume of the Picard group is hR, where h is the class number and R is the regulator. \n\nThe Picard group is a generalization of the ideal class group. \n"},{"_id":"4718ada1ad2a45ed2a000032","treeId":"384915e096bc4c7cd0000017","seq":7922891,"position":4,"parentId":"4718aafdad2a45ed2a00002e","content":"# Dirichlet's theorem on primes in arithmetic progressions\n\nFirst, use characters to sort the primes into $L$-functions. \n\n\nCyclotomic fields.\n\nRelate this to Cebotarev, Artin stuff.\n\nHard part is to show $L(1,\\chi)$ is non-zero for non-trivial characters. "},{"_id":"4753f35e42a10fe4b7000033","treeId":"384915e096bc4c7cd0000017","seq":7922888,"position":4.5,"parentId":"4718aafdad2a45ed2a00002e","content":"# Dirichlet's class number formula\n\nRelates class numbers to zeta values. \n\n##$\\displaystyle \\lim_{s\\to 1} (s-1)\\zeta_K(s) = \\frac{ 2^r (2\\pi)^s h_K \\textsf{Reg}_K}{w_K \\sqrt {|\\textsf{disc}(K)|}}$"},{"_id":"4718ae2ead2a45ed2a000033","treeId":"384915e096bc4c7cd0000017","seq":7922886,"position":5,"parentId":"4718aafdad2a45ed2a00002e","content":"# L-functions\n\nPrototypical example is Riemann zeta function. \n\nFor class field theory, typically Dirichlet's $L$-functions are used $L(s,\\chi) = \\sum_{n=1}^\\infty \\frac{\\chi(n)}{n^s}.$ These are used to prove Dirichlet's theorem on primes in arithmetic progressions, and for his class number formula. \n\nModularity theorem tells us that the $L$-function for an elliptic curves and the $L$-function for the associated weight 2 Hecke-eigenform are identical.\n\nGeneralize to Weber $L$-functions, where we use ray class groups.\n\nExpected to satisfy functional equation/be identical for certain things - Langland's program. \n\nExpected to be zero-free in certain regions - Generalized Riemann hypothesis. "},{"_id":"4718ae67ad2a45ed2a000034","treeId":"384915e096bc4c7cd0000017","seq":7922918,"position":6,"parentId":"4718aafdad2a45ed2a00002e","content":"# Prime number theorem\n\nThe asymptotic distribution of the prime numbers is $\\pi(x) \\sim \\frac{x}{\\ln x}$.\n\n*****\n\nThe idea is that we count the primes using a special weighting, giving us better analytic properties. For example, Chebyshev's function $\\psi(x) = \\sum_{n \\le x}\\Lambda(n)$, where the Von-Mangoldt function $\\Lambda(n)$ is $\\log p$ when $n$ is a prime power, and zero otherwise. \n\nWe relate this to the Riemann zeta function by looking at the logarithmic derivative\n\n $\\displaystyle -\\frac{\\zeta'(s)}{\\zeta(s)} = \\sum_{n=1}^\\infty \\Lambda(n) n^{-s}$. \n\nThis can be transformed into the explicit equation $\\psi(x) = x - \\sum_{\\rho} \\frac{x^\\rho}{\\rho}$, where $\\rho = \\sigma + \\tau i$. \n\nThe crux is to show that $\\zeta(\\rho)$ has no zeros on the line $\\tau = 1$. This can be done using a trigonometric argument - launching point to pretentious theory. Once we have this, a careful accounting of possible types of zeros gives us the essence of the prime number theorem. \n\nThe remaining gap is solved using analysis, for example a Tauberian theorem. \n\n<maybe talk about contour integrals>"},{"_id":"48b16361ad3b3301d2000033","treeId":"384915e096bc4c7cd0000017","seq":7922892,"position":7.5,"parentId":"4718aafdad2a45ed2a00002e","content":"# Algebraic integers form a ring\n\nIf $\\alpha$ is an algebraic integer, then $\\alpha A\\subset A$ for some finitely generated additive subgroup $A\\subset \\mathbb C$, namely $\\mathbb Z[\\alpha]$. \n\nConversely, let $\\alpha A \\subset A$ for some $\\alpha \\in \\mathbb C$, where $A$ is a finitely generated subgroup $A\\subset \\mathbb C$. Let $a_i$ be a set of generators. Then $\\begin{pmatrix} \\alpha a_1 \\\\ \\alpha a_n \\end{pmatrix}= M \\begin{pmatrix} a_1 \\\\ a_n\\end{pmatrix}$, where $M$ is an $n\\times n$ matrix over $\\mathbb Z$. This means $\\alpha$ is an eigenvalue of $M$, so $\\det (\\alpha I -M)=0$, which is a monic polynomial with coefficients in $\\mathbb Z$. This $\\alpha$ is an algebraic integer. \n\n\nUsing the converse with the finitely generated subgroup $\\mathbb Z[\\alpha, \\beta]$, we get that the algebraic integers form a ring. \n"},{"_id":"4718aef5ad2a45ed2a000036","treeId":"384915e096bc4c7cd0000017","seq":7922893,"position":8,"parentId":"4718aafdad2a45ed2a00002e","content":"# Integral basis\n\n\n## Integral basis theorem\nFor a deg $n$ number field $K$, $\\mathcal O_K$ is a degree $n$ $\\mathbb Z$-module. \n****\nStart w a $\\mathbb Q$-basis for $K$. Clear denominators so that you have a $\\mathbb Q$-basis inside $\\mathcal O_K$. Call the discriminant of this basis $d$, and the basis itself $\\{ \\alpha_i\\}$. \n\nWe can write $s = \\sum a_i \\alpha_i$, with $a_i$ rational. \n\nConsider the image of this under all the embeddings $\\sigma$. If we solve for $a_j$ using Cramer's rule, we get that $a_J = \\gamma_j/\\delta$, where $\\delta = |\\sigma_i(\\alpha_j)|$, and $\\gamma_j$ is constructed like $\\delta$ except we replace the $j$th column by $\\sigma_i(s)$. \n\nThus, $\\delta$ and $\\gamma_j$ are both algebraic integers, with $\\delta^2 = d$. So $da_j = \\delta \\gamma_j$, which implies $da_j$ is an integer. This shows that $\\mathcal O_K$ is contained in a free abelian group of rank $n$. \n\n##Discriminant\n\nCan be thought of as the square of the volume of the integral basis, int the lattice spaced generated from all the embeddings. \n\nIs an invariant of a number field. \n\n"},{"_id":"471a5d5217b090ade1000022","treeId":"384915e096bc4c7cd0000017","seq":7922553,"position":11,"parentId":"4718aafdad2a45ed2a00002e","content":"# Minkowski's bound\n\nThe ideal class group of a ring is finite. \n\nThe basic idea is to prove every ideal $I$ contains an element $\\alpha$ that is 'small enough' so something like $|N_{\\mathbb Q}^K(\\alpha)| \\le \\lambda ||I||$ where $\\lambda$ is independent of $I$. \n\nHere's how we use this: take an ideal class $C$, and consider an ideal $I\\in C^{-1}$. Use the above to find such an $\\alpha \\in I$. So $I$ contains $(\\alpha)$, therefore $(\\alpha) = IJ$ for some ideal $J\\in C$. \n\nSince $|N_{\\mathbb Q}^K(\\alpha)| = ||(\\alpha)|| = ||I|| ||J||$, the ideal $J$ has ideal norm less than $\\lambda$. Since $\\lambda$ is fixed, this bounds the possible divisors of $J$, and so there are only a finite number of such $J$'s. Since we can find such a $J$ for each ideal class, the ideal class group is finite!\n\n*****\nMinkowski used a geometric argument to get a really nice $\\lambda$. Every ideal class contains an ideal $J$ such that \n$||J||\\le \\frac{n!}{n^n}\\left(\\frac{4}{\\pi}\\right)^s\\sqrt{|\\textsf{disc}(R)|}.$\n\n\nTo show this, we embed our ring of integers $R$ into $\\mathbb R^r \\times \\mathbb C^s$ as a lattice, using the different embeddings. The discriminant is the square of the volume of the fundamental parallelopiped. \n\nAn ideal will map to a sublattice, and the fundamental volume of this will be equal to $||I||$ times $\\sqrt{|\\textsf{disc}(R)|}$. \n\nWe can define a norm on this space that matches the standard norm. Now carefully construct a space $A$ where each element has norm has absolute value bounded by 1, which is compact, convex, and centrally symmetric, and has volume $\\frac{n^n}{n!}2^r \\left(\\frac{\\pi}{2}\\right)^s.$\n\nApplying Minkowski's convex body theorem to $E$, which is $A$ scaled to have volume equal to $||I||\\sqrt{|\\textsf{disc(R)}|}$ (the volume of the fundamental parallelopiped for the lattice generated by $I$), we are guaranteed that there is a non-zero lattice point for $I$ inside $E$. \n\nThis point gives us the $\\alpha$ that we need for the theorem. "},{"_id":"471a5d9e17b090ade1000023","treeId":"384915e096bc4c7cd0000017","seq":7922556,"position":12,"parentId":"4718aafdad2a45ed2a00002e","content":"# Cyclotomic fields\n\nDiscriminant for prime case is $\\pm p^{p-2}$. \n\nExtension has degree $\\varphi(m)$ over $\\mathbb Q$. \n\nThe Galois group is isomorphic to $\\left(\\mathbb Z/m\\mathbb Z\\right)^\\times.$\n\nImportant special case for building up larger theorems in class field theory. \n\n##Primes $q$ that split completely in degree $d$ subfields of cyclotomic extensions are $d$th powers mod $p$. \n\nFor a cyclotomic field $\\mathbb Q[\\omega_p]$, the Galois group over $\\mathbb Q$ is cyclic of order $p-1$. Let $F_d\\subset \\mathbb Q[\\omega_p]$ be the unique subfield of degree $d$ over $\\mathbb Q$, when $d|p-1$. \n\n$q$ splits into $r$ primes in $\\mathbb Q[\\omega_p]$, and $f=(p-1)/r$ is the order of $q$ mod $p$. Since this is a cyclic group, the $d$th powers form the unique subgroup of order $(p-1)/d$. \n\nThus, TFAE\n1. $q$ is a $d$th power mod $p$\n2. $f|(p-1)/d$\n3. $d|r$\n4. $F_d\\subset F_r$\n\nNow observe that $F_r$ is the decomposition field for any prime $Q$ over $q$, since this is the only field over $\\mathbb Q$ of degree $r$. \n\nTherefore, $F_d \\subset F_r$ is equivalent to $q$ splitting completely in $F_d$. \n "},{"_id":"471a5dfd17b090ade1000024","treeId":"384915e096bc4c7cd0000017","seq":7927372,"position":13,"parentId":"4718aafdad2a45ed2a00002e","content":"# Galois theory of number fields\n\n## Inertia\n\nThe inertial degree of $Q$ over $P$ is defined as the degree of the extension $\\mathcal O_L/Q$ over $\\mathcal O_K/P$. \n\n## Decomposition group\nThe decomposition group of $Q$ over $P$ is $D = \\{\\sigma \\in G : \\sigma Q = Q\\}.$ The inertia group is $E = \\{\\sigma \\in G: \\sigma(\\alpha) \\equiv \\alpha \\pmod Q\\}$, which is a normal subgroup of $D$. \n\nThe quotient $D/E$ is a cyclic group of order $f$ - the inertial degree. \n\n## Decomposition/ramification fields\nBy Galois theory we can look at the fixed fields of $D$ and $E$. If $D$ is a normal subgroup of $G$, it turns out that $L_D$ has degree $r$ over $K$, and $P$ splits into $r$ primes here. $L_E$ has degree $f$ over $L_D$, and the primes of $L_D$ over $P$ have inertial degree $f$ here. Finally, $L$ has degree $e$ over $L_E$, and the primes of $L_E$ over $P$ all have ramification index $e$ in $L$. \n\n\n"},{"_id":"471a5f0217b090ade1000027","treeId":"384915e096bc4c7cd0000017","seq":7922894,"position":16,"parentId":"4718aafdad2a45ed2a00002e","content":"# Cebotarev density theorem\n\nThis theorem gives a density for the prime ideals that split in a certain way (or alternatively, have an Artin symbol in a specific conjugacy class). \n\nFor an extension $L/K$ with Galois group $G$, then the set of primes of $\\mathcal O_K$, $\\mathcal S = \\{\\mathfrak p : \\mathfrak p \\text{ is unramified in } L \\text{ and } \\left(\\frac{L/K}{\\mathfrak p}\\right) = \\langle \\sigma \\rangle \\}$, and where $\\sigma \\in \\text{Gal(L/K)}$, has Dirichlet density\n$\\delta(S) = \\frac{|\\langle \\sigma \\rangle |}{[L:K]}$. \n\nThe case when $\\sigma = 1$ gives the density of the primes that split completely.\n\nWhen $L$ is a cyclotomic extension of $\\mathbb Q$, this reduces to Dirichlet's theorem on primes in arithmetic progressions. \n\nThe proof is important because it uses the technique of reducing to relatively cyclotomic extensions - used for the proof of Artin reciprocity. "},{"_id":"471a5f6017b090ade1000028","treeId":"384915e096bc4c7cd0000017","seq":7922873,"position":17,"parentId":"4718aafdad2a45ed2a00002e","content":"# Artin reciprocity\n\nFor a ring $R$ with characteristic $p$, there is an homomorphism called the Frobenius homomorphism taking $x$ to $x^p$ (and for the prime at infinity, it is complex conjugation). The basic question that reciprocity laws answer is 'which ring homomorphism is it?' \n\nFor a prime field, every homomorphism is the identity - including the Frobeinus homomorphism. This is Fermat's Little Theorem.\n\n\n## Quadratic reciprocity\nFor a quadratic extension $\\mathbb F(\\sqrt{d})$ of a prime field $\\mathbb F$ with char $p$, we could either get the identity or conjugation as the Frobenius homomorphism. We can look at this for all the primes (except for the ones dividing $4d$) at once by studying $\\mathbb Q[\\sqrt{d}]$. \n\nIn this case, Artin's reciprocity law is that there exists a group homomorphism $(\\mathbb Z / 4d\\mathbb Z)^\\times \\rightarrow \\{\\pm 1\\}$, which takes $(p \\bmod 4d) \\mapsto \\left(\\frac{d}{p}\\right)$ for any prime $p$ not dividing $4d$. This is equivalent to quadratic reciprocity. \n\n## Artin map\nMore generally, suppose we have an abelian Galois extension $\\mathbb Q[\\alpha]$, with $f$ the minimal polynomial of $\\alpha$, and Galois group $G$. Then there exists a group homomorphism $(\\mathbb Z / \\Delta(f) \\mathbb Z)^\\times \\rightarrow G$ which takes $(p \\bmod \\Delta(f)) \\mapsto \\varphi_p$ for primes $p$ not dividing $\\Delta(f)$. \n\nThis map is called the Artin map. It identifies primes $p$ with $\\varphi_p$, which is the unique element of $G$ that extends the Frobenius map of $\\mathbb F_p[\\alpha]$. This also applies to the prime at infinity. \n\nEven more generally, for $K/F$, and $\\mathfrak P$ a prime of $K$ over $\\mathfrak p$ a prime of $F$, then we define the Artin symbol $\\left(\\frac{K/F}{\\mathfrak P}\\right)$ to be the $N(\\mathfrak p)$-power map modulo $\\mathfrak P$. This can be extended by multiplicativity to a homomorphism $I_F(\\mathfrak m) \\rightarrow \\text{Gal}(K/F)$, which is the Artin map. \n\nAnother important fact is that the order of the Artin symbol is the inertial degree of $\\mathfrak P$ over $\\mathfrak p$ when $\\mathfrak p$ is unramified. This implies that $\\left(\\frac{K/F}{\\mathfrak P}\\right) = 1$ iff $\\mathfrak p$ splits completely in $K$. More generally, the Artin map tells you how $\\mathfrak p$ factors in $K$. \n\n\n## Artin reciprocity\nLet $K/F$ be an abelian extension, and let $\\mathfrak m$ be a modulus divisible by all the primes of $F$ which ramify in $k$ (including infinite). Then\n1. The Artin map $\\Phi_\\mathfrak m$ is surjective\n2. The modulus $\\mathfrak m$ can be chosen (with high enough powers of primes dividing it) so that we have an isomorphism $I_F(\\mathfrak m)/\\text{ker}(\\Phi_\\mathfrak m) \\simeq \\text{Gal}(K/F)$\n\nThis gives a nice relation between generalized class groups and Galois groups, which can be used in particular for Hilbert class fields. "},{"_id":"471a5fdb17b090ade1000029","treeId":"384915e096bc4c7cd0000017","seq":7922895,"position":18,"parentId":"4718aafdad2a45ed2a00002e","content":"# Ideles\n\nIt's more convenient to work with completions for *all* the primes at once, including the primes at infinity. Accounting for all the primes reveals a nice symmetry which can be used to state and prove the theorems of class field theory in a natural way. \n\n## Places\nWe can define different absolute values on number fields, which must be multiplicative homomorphisms mapping 0 to 0, and satisfying a weakend form of the triangle inequality. We can arrange these into different equivalence classes based on the induced topology. These are called the palces. It turns out by Ostrowski's theorem that each place corresponds to \n1. $\\mathfrak p$-adic absolute value, or a *finite place*\n2. an absolute value for a real embedding, *infinite real place*\n3. or an absolute value from an imaginary embedding (which is the same for the conjugate embedding), the *infinite imaginary place*.\n\nFor each place $v$ of a field $F$ we can construct the completion with respect to that place (analogously to the construction of $\\mathbb R$ from $\\mathbb Q$ - which is a special case). We call this $F_v$. \n\n## Takagi's formulation\nLet $F_v$ be the completion of $F$ at the place $v$. We define $J_F$ as $\\prod_v F_v^\\times$, where this is the restricted topological product. Also, $\\mathcal E_F = \\prod_v \\mathcal U_v$, where $\\mathcal U_v$ are the elements with norm 1. \n\nThe major theorems of class field theory can be compactly stated as the following. There is an order reversing, bijective correspondence between the set of all finite abelian extensions $K/F$, and the set of all open subgroups $\\mathcal H$ of $J_F$ which contain $F^\\times$. Furthermore, $\\text{Gal}(K/F)\\simeq J_F/\\mathcal H$. \n\nNote that $J_F/F^\\times \\mathcal E_{F,\\mathfrak m}^+ \\simeq \\mathcal R_{F,\\mathfrak m}^+$, the strict ray class group.\n\n## Other applications \nFor another example, the functional equation for certain $L$-functions can be elegantly proved using idelic theory, *à la* Tate's thesis. "},{"_id":"471a603517b090ade100002a","treeId":"384915e096bc4c7cd0000017","seq":7922897,"position":19,"parentId":"4718aafdad2a45ed2a00002e","content":"# Ray class group\n\nGeneralization of ideal class group relative to a modulus $\\mathfrak m$. \n\nPrincipal ideals relative to $\\mathfrak m$: $\\mathcal P_{F,\\mathfrak m} =\\{\\langle\\alpha\\rangle : \\alpha \\in \\mathcal O_F, \\alpha\\equiv 1 \\pmod {\\mathfrak m} \\}$ \n\nIdeal relative to $\\mathfrak m$: $\\mathcal I_F(\\mathfrak m) = \\{\\mathfrak a : \\text{ord}_{\\mathfrak p}\\mathfrak a = 0 \\text{ for all } \\mathfrak p | \\mathfrak m \\}$.\n\nRay class group is $\\mathcal R_{F,\\mathfrak m} = \\mathcal I_F(\\mathfrak m)/\\mathcal P_{F,\\mathfrak m}$.\n\nStrict if $\\mathfrak m$ includes prime at infinity\n\nUsed to generalize Dirichlet's Theorem on Primes in Arithmetic Progressions - Are there infinitely many prime ideals in each ray class? Weber $L$-functions are defined for characters over Ray class groups.\n\n"},{"_id":"471a607917b090ade100002b","treeId":"384915e096bc4c7cd0000017","seq":7922896,"position":20,"parentId":"4718aafdad2a45ed2a00002e","content":"# Hilbert class fields\n\nThe *maximal* unramified abelian extension of a number field $K$ (unramified at the infinite places also). It always exists. \n\nA particularly important property is that its\nGalois group is isomorphic to the ideal class group of $K$ (uniquely!). This gives it the 'right' to be called a class field, and allows us to use Galois theory and theorems from class field theory to prove results about the class group. \n\nFor example, we use this to prove Cebotarev's density theorem. \n\nFor imaginary quadratic fields, the $j$-invariant can be used to give you the Hilbert class field. \n\n"},{"_id":"471a60c517b090ade100002c","treeId":"384915e096bc4c7cd0000017","seq":7922901,"position":21,"parentId":"4718aafdad2a45ed2a00002e","content":"# Class field theory inequalities\n\nThe inequalities tell us about the relation between $J_F/\\mathcal H$ and $K/F$, where $\\mathcal H = F^\\times N_{K/F} J_K$. \n\n## Universal Norm Index Inequality\n \n$\\mathcal H$ is an open subset of $J_F$, so if $\\mathfrak m$ is chosen so that $\\mathcal E_{F,\\mathfrak m}^+ \\subseteq \\mathcal H$, then $[J_F : \\mathcal H] \\le [K : F]$\n\nThis generalizes the fact that $L_\\mathfrak m (1,\\chi) \\neq 0$ for non-trivial characters which are trivial on $\\mathcal H$. \n## Global Cyclic Index Inequality\n\nIf $K/F$ is a cyclic extension, and $\\mathfrak m$ is divisible by a sufficiently high power of every ramified prime in $K/F$, then $[J_F : \\mathcal H] \\ge [K:F]$. \n\nCombining this with the Universal Norm Index inequality gives us equality. "},{"_id":"471a615317b090ade100002e","treeId":"384915e096bc4c7cd0000017","seq":7922898,"position":23,"parentId":"4718aafdad2a45ed2a00002e","content":"# Factorization of ideals\n\nIdeals in number fields have unique factorization into prime ideals. In algebraic number theory, the main focus is on how a prime ideal factors once it has been lifted up in a field extension. \n\n## Splitting of primes\n\nIn an extension of a number field, there are three different ways that primes in the base field factor.\n\nThey can split, ramify, or have inertia.\n\n\n## Split primes via factoring polynomials\n\nOne way to see how a prime splits is to see how the minimal polynomial for the extension factors modulo that prime. \nSay $L = K (\\alpha)$, and let $f$ be the minimal polynomial of $\\alpha$ over $K$. Let $p$ be a prime not dividing $[\\mathcal O_L:K[\\alpha]]$, and let $\\mathfrak p$ be a prime ideal of $K$ over $p$. Then if $f$ factors mod $\\mathfrak p$ as $f(x) \\equiv \\pi_1(x)^{e_1}\\cdots \\pi_g(x)^{e_g} \\bmod p$, we have that $\\mathfrak p$ factors as $\\mathfrak p = \\mathfrak q_1^{e_1}\\cdots \\mathfrak q_g^{e_g}$. Furthermore, $\\mathfrak q_i = \\mathfrak p + (\\pi_i(\\alpha))$. \n\n\n## Split primes via Artin symbol\nWe can also use the Artin symbol to determine how an unramified prime splits. Consider an extension $K/F$, along with a Galois extension $M/F$ containing $K$, with Galois group $G$. Call the Galois group of $M/K$, $H$. Fix a prime $\\mathfrak U$ of $M$ lying over $\\mathfrak P$ of $K$, lying over $\\mathfrak p$ a prime of $F$. Then for each $\\sigma \\in G$, consider the coset $H\\sigma$, and look at the orbits from right-action by the Artin symbol of $\\mathfrak U$ over $\\mathfrak p$. The size of the orbit is the inertial degree of $\\sigma \\mathfrak P$ over $\\mathfrak p$, and thus the Artin symbol determines the splitting of $\\mathfrak p$ in $K$. \n\n##Galois group permutes primes transitively\n\nTake $L/K$ to be a normal extension. \n\nSuppose that $\\sigma(Q)\\neq Q'$ for all $\\sigma\\in G$. Then by the C.R.T., there is a solution to the system of congruences $x\\equiv 0 \\pmod {Q'}$ and $x\\equiv 1 \\pmod {\\sigma(Q)}$. \n\nLet $\\alpha\\in \\mathcal O_L$. Then $N(\\alpha) \\in \\mathcal O_K \\cup Q' = P$, since $\\alpha \\in Q'$. \n\nOn the other hand, $\\alpha \\not\\in \\sigma(Q)$, so $\\sigma^{-1}(\\alpha)\\not\\in Q$. We can also write $N(\\alpha)$ as the product of all $\\sigma^{-1}(\\alpha)$, so this implies that $N(\\alpha)\\not\\in Q$, which is a contradiction."},{"_id":"471a619e17b090ade100002f","treeId":"384915e096bc4c7cd0000017","seq":7922902,"position":24,"parentId":"4718aafdad2a45ed2a00002e","content":"# Norm of ideals\n\nThe norm of an ideal $I$ of $\\mathcal O_K$ is $|\\mathcal O_K/I|$. This is multiplicative, and the norm a the principal ideal $(\\alpha)$ is $|N^K(\\alpha)|$. \n\nThe norm of a prime $Q$ above a prime $P$ is $|P|^f$, where $f$ is the inertial degree.\n\n##Degree of extension is $\\small \\sum_i e_if_i$\n\nTake a prime $p\\in\\mathbb Z$. \n$p\\mathcal O_k = \\prod_i Q_i^{e_i}$\n\nNow take ideal norms of everything. Since these are multiplicative, and since $||Q_i||=p^{f_i}$, the theorem follows. "},{"_id":"471a61e117b090ade1000030","treeId":"384915e096bc4c7cd0000017","seq":7922871,"position":25,"parentId":"4718aafdad2a45ed2a00002e","content":"# Ramification\n\nWhen looking at an extension of number fields $K/F$, a prime $P$ ramifies if $P\\mathcal O_K = \\prod P_i^{e_i}$ for $e_i > 1$ for some $i$. \n\n##Ramified primes divide the discriminant\n\nLet $P$ be a prime of $R$ lying over $p$ which ramifies. Then $pR=PI$, with $I$ divisible by all the primes of $R$ lying over $p$. \n\nTake any integral basis. Replace one of the elements by $\\alpha\\in I \\setminus pR$. It is sufficient to show that $p$ divides the discriminant of this basis. \n\nAssume the field is Galois (if not, use the Galois ext). For each automorphism, notice that $\\sigma^{-1}(Q)$ is a prime lying over $p$, so it contains $\\alpha$. Thus, $\\sigma_i(\\alpha)\\in Q$ for all $i$. \n\nThis means that $Q$ contains the discriminant of this basis, and since the discriminant is an integer, it has to be in $p\\mathbb Z$. \n\n"}],"tree":{"_id":"384915e096bc4c7cd0000017","name":"Number Theory Qual","publicUrl":"number-theory-qual","latex":true}}