Dependencies

A ring \(R\) is a set with two binary operations called addition \('+'\) and multiplication \('\cdot '\), such that:

1. \(R\) is an abelian group with respect to \(+\). Note this means \(R\) contains a zero element denoted \(0\) and every \(r \in R\) has an additive inverse \(-r \in R\).

2. Multiplication is associative and distributive, i.e,

   \[ (xy)z=x(yz) \qquad x(y+z)=xy+xz \qquad (y+z)x=yx+zx \]

A ring is called commutative if \(xy=yx\) and contains an identity element, denoted \(1\). Having a \(1\) is sometimes called being unital. Lastly, the subset of elements of \(R\) which have a multiplicative inverse are denoted \(R^\times \).

We say a ring \(R\) is an integral domain, if whenever \(xy=0\) then either \(x=0\) or \(y=0\), for \(x,y \in R\).

Let \(R,S\) be rings, then a ring homomorphism \(\phi :R \to S\) is a map such that

A ring homomorphism is called an isomorphism if it is bijective.

Let \(\phi : R \to S\) be a ring homomorphism. The kernel of \(\phi \) is the set of all elements \(r \in R\) such that \(\phi (r)=0\), this is denoted \(\ker (\phi )\). The image of \(\phi \) is the set \(\{ \phi (r): r \in R\} \), this is denoted \(\operatorname{Im}(\phi )\)

A field \(F\) is a commutative ring in which every non-zero element has an inverse. Equivalently, the set \(F^{\times }:=F \backslash \{ 0\} \). ¹

Let \(R\) be a ring, then an ideal \(\mathfrak {a}\) is a subset of \(R\) which is an additive subgroup of \(R\) and such that for any \(r \in R, a \in \mathfrak {a}\) we have \(ra \in \mathfrak {a}\).

If for every ideal \(\mathfrak {a}\) in an integral domain \(R\) we can find \(a \in R\) such that \(\mathfrak {a}=(a)\) then we call \(R\) an Principal ideal domain, or PID for short.

An integral domain in which every element can be written uniquely as a product of irreducible elements is called a unique factorization domain, or UFD for short.

Let \(R\) be a ring and let \(\mathfrak {a}\) be an ideal, then the quotient ring \(R/\mathfrak {a}\) is the ring whose elements are of the form \(r+\mathfrak {a}\) for \(r \in R\), with addition and multiplication given by

Let \(R\) be a ring and \(\mathfrak {p},\mathfrak {m}\) an ideals with neither equal to \((1)\).

1. The \(\mathfrak {p}\) is called prime if whenever \(xy \in \mathfrak {p}\) we have \(x \in \mathfrak {p}\) or \(y \in \mathfrak {p}\).

2. The ideal \(\mathfrak {m}\) is called maximal if there does not exist an ideal \(\mathfrak {a}\ne (1)\) such that \(\mathfrak {m}\) is properly contained in \(\mathfrak {a}\).

Let \(\mathfrak {a},\mathfrak {b}\) be ideals in a ring \(R\). Then we let

1. \(\mathfrak {a}\mathfrak {b}\) be the ideal generated by the product of elements of \(\mathfrak {a}\) and \(\mathfrak {b}\).

2. Similarly, \(\mathfrak {a}+\mathfrak {b}\) denotes the ideal generated by sums of elements in \(\mathfrak {a},\mathfrak {b}\).

3. If \(\mathfrak {a}+\mathfrak {b}=R=(1)\) we say \(\mathfrak {a},\mathfrak {b}\) are coprime.

4. If \(\mathfrak {a},\mathfrak {b}\) are ideals we write \(\mathfrak {a}\mid \mathfrak {b}\) if there exists an ideal \(\mathfrak {c}\) such that \(\mathfrak {b}=\mathfrak {a}\mathfrak {c}\).

Let \(R\) be an integral domain. A non-zero polynomial \(p(x)\) of degree at least \(1\) in \(R[x]\) is said to be irreducible if whenever \(p(x)=f(x)g(x)\) with \(f(x),g(x) \in R[x]\) then one of \(f(x)\) or \(g(x)\) is in \(R\). Note this is slightly different to Definition 1.1.14.

A polynomial is irreducible in \(\mathbb {Z}[x]\) if and only if it is irreducible in \(\mathbb {Q}[x]\).

Let \(K\) be a field containing a field \(F\). Then we call \(K\) a field extension of \(F\) (and \(F\) a subfield of \(K\)). This is denoted by \(K/F\). The degree of a field extension \(K/F\), denoted \([K:F]\) is the dimension of \(K\) as a vector space over \(F\). A field extension is said to be finite if \([K:F]\) is finite, otherwise we say its infinite.

Let \(K/F\) be a field extension. Then we say \(K\) is algebraic over \(F\) if every element of \(K\) is algebraic over \(F\).

We say a field extension \(K/F\) is finitely generated, if there are finitely many elements \(\alpha _1,\dots ,\alpha _n\) such that \(K=F(\alpha _1,\dots ,\alpha _n)\).

Let \(\alpha ,\beta \in K\) be non-zero and algebraic over \(F\). Then \(\alpha \pm \beta \), \(\alpha \beta \), \(\alpha /\beta \), \(\alpha ^{-1}\), \(\beta ^{-1}\),etc are also algebraic over \(F\).

An algebraic number is a complex number which is algebraic over \(\mathbb {Q}\). Meaning, it is a root of a polynomial \(f(x) \in \mathbb {Q}[x]\).

If \(\alpha \) is an algebraic number and \(m_\alpha (x)\) is its minimal polynomial, then the set of roots of \(m_\alpha (x)\) in \(\mathbb {C}\) are called the conjugates of \(\alpha \).

A Number field is a subfield of \(\mathbb {C}\) of finite degree over \(\mathbb {Q}\).

Let \(K\) be a number field. Then an embedding of \(K\) is a non-zero ring homomorphism \(\sigma :K \hookrightarrow \mathbb {C}\).

We say an embedding is real if its image is \(\mathbb {R}\subset \mathbb {C}\). Otherwise we say the embedding is complex. Note that if \(\sigma \) is a complex embedding, then so is its complex conjugate \(\overline{\sigma }\) (i.e. this is the embedding given by applying \(\sigma \) and then doing complex conjugation.) We call \(\sigma ,\overline{\sigma }\) a pair of complex conjugate embeddings.

Let \(K\) be a number field and let \(\alpha \in K\). Then we can associate to \(\alpha \) a linear operator

This can be written as a matrix over \(\mathbb {Q}\) by picking a basis of \(K\) over \(\mathbb {Q}\). The map \(\alpha \to A_\alpha \) is called the standard representation.

For \(K\) a number field and \(\alpha \in K\), then the characteristic polynomial of \(A_\alpha \) is known as the field polynomial of \(\alpha \). We shall denote it by \(C_\alpha \), but note that it depends on the field we are working with. In particular, if we write a basis for \(K\) over \(\mathbb {Q}\) or a basis for \(K\) over \(F\) (for \(F\) some subfield), then the field polynomial can be different in each case. Therefore, unless otherwise stated we will assume we mean a basis of \(K/\mathbb {Q}\).

Let \(K\) be a number field. We have a pairing

given by

Let \(K\) be a field extension of \(\mathbb {Q}\) (not necessarily finite). An element \(\alpha \) is called an algebraic integer if it is a root of a monic polynomial with coefficients in \(\mathbb {Z}\).

Let \(K\) be a number fields and \(\alpha \in K\). Then there exists a \(n \in \mathbb {Z}\backslash \{ 0\} \) such that \(n \alpha \) is an algebraic integer.

Let \(\alpha \in \mathbb {C}\). We say \(\mathbb {Z}[\alpha ]\) is finitely generated as an abelian group if there exits a finite set \(\{ b_1,\dots ,b_n\} \) of elements \(b_i \in \mathbb {Z}[\alpha ]\) such that every element of \(\mathbb {Z}[\alpha ]\) may be written in the form

with \(x_i \in \mathbb {Z}\).

If \(K\) is a number field, then \(\mathcal{O}_K\) is a ring.

Let \(K\) be a number field and let \(B=\{ b_1,\dots ,b_n\} \) be a set of elements in \(K\). The discriminant of \(B\) is defined as

If needed we will denote the matrix

by \(T_B\).

If \(K=\mathbb {Q}(\sqrt{-d})\) with \(d \in \mathbb {Z}_{{\gt} 0}\) a square-free integer, then \(\mathcal{O}_K^\times \) is \(\mu _K\).

Let \(K=\mathbb {Q}(\sqrt{d})\) with \(d\) a positive square-free integer. Then

In this case there is a unique unit in \(u \in \mathcal{O}_K^{\times }\) which generates \(\mathcal{O}_K^\times / \{ \pm 1\} \) and under the standard embedding we have \(u \geq 1\). This unit \(u\) is called the fundamental unit.

We say a ring has unique factorization if whenever

for \(p_i,q_i\) irreducible elements, then \(n=m\) and after possibly reordering we can find units \(u_i\) such that \(p_i=u_iq_i\) for all \(i\).

Let \(R,A\) be rings with \(A \subset R\) a subring and \(x \in R\). We say \(x\) is integral over \(A\) if there exist \(a_i \in A\) such that

for some \(n\).

Let \(A'\) be the set of all elements of \(R\) that are integral over \(A\). Then similarly to how we prove Corollary 2.1.16, \(A'\) is a ring which we call the integral closure of \(A\) in \(R\).

Moreover, if \(R\) is an integral domain and we let

be its fields of fractions (this can alternatively be defined as smallest field containing \(R\)), then we say \(R\) is integrally closed, if it is integrally closed in its field of fractions.

Let \(K\) be a number field with ring of integers \(\mathcal{O}_K\). Then \(\mathcal{O}_K\) is a Dedekind domain.

Let \(R\) be a Dedekind domain and \(K\) its field of fractions. A fractional ideal is a subset \(\mathfrak {a}\subset K\) such that

1. \(\mathfrak {a}\) is an abelian group under addition.

2. \(x \mathfrak {a}\subset \mathfrak {a}\) for every \(x \in R\).

3. There exists some \(x \in R\) such that \(x\mathfrak {a}\subset R\).

Let \(K\) be a number field. Then every ideal in \(\mathcal{O}_K\) can factored uniquely into a product of prime ideals.

Let \(R\) be a Dedekind domain and let \(\mathfrak {a}\) be an ideal. Then for any \(\alpha \in \mathfrak {a}\) with \(\alpha \) non-zero we can find \(\beta \in \mathfrak {a}\) such that \(\mathfrak {a}=(\alpha ,\beta )\).

A Dedekind domain is a UFD if and only if it is a PID.

Let \(K\) be a number field and \(\mathfrak {a}\subset \mathcal{O}_K\) an ideal. We define the norm of \(\mathfrak {a}\) to be

Let \(K\) be a number field, \(\mathfrak {a}\subset \mathcal{O}_K\) an ideal and \(\alpha \in \mathfrak {a}\). Then \(\mathfrak {a}=(\alpha )\) if and only if \(|N_{K/\mathbb {Q}}(\alpha )|=N(\mathfrak {a})\).

Let \(K/F\) be a finite extension of number fields with rings of integers \(\mathcal{O}_K\) and \(\mathcal{O}_F\). If \(\mathfrak {P}\) is a non-zero prime ideal in \(\mathcal{O}_K\) and \(\mathfrak {p}\) is a non-zero prime ideal in \(\mathcal{O}_F\) then we say \(\mathfrak {P}\) lies over \(\mathfrak {p}\) (or equivalently \(\mathfrak {p}\) lies under \(\mathfrak {P}\)), if \(\mathfrak {P} \mid \mathfrak {p}\mathcal{O}_K\). Here \(\mathfrak {p}\mathcal{O}_K\) denotes the ideal generated by \(\mathfrak {p}\) in \(\mathcal{O}_K\).

Let \(K/F\) be a finite extension of number fields with rings of integers \(\mathcal{O}_K\) and \(\mathcal{O}_F\). Then every non-zero prime ideal \(\mathfrak {P}\subset \mathcal{O}_K\) lies over a unique non-zero prime ideal \(\mathfrak {p}\subset \mathcal{O}_F\).

Conversely, every non-zero prime ideal \(\mathfrak {p}\subset \mathcal{O}_F\) lies under at least one non-zero prime ideal \(\mathfrak {P}\subset \mathcal{O}_K\).

Let \(K\) be a number field and \(p\) a prime number. We say

1. \(p\) is ramified in \(K\), if there exists a \(\mathfrak {p}|p\) such that \(e_{\mathfrak {p}|p}{\gt}1\). Otherwise we say \(p\) is unramified.

2. \(p\) is totally ramified if there is a \(\mathfrak {p}|p\) such that \(e_{\mathfrak {p}|p}=[K:\mathbb {Q}]\).

3. \(p\) is inert if \(p\) is unramified and there exists a unique prime \(\mathfrak {p}\) lying above \(p\) (which will have \(f_{\mathfrak {p}|p}=[K:\mathbb {Q}]\)).

4. \(p\) is split if it is unramified and for some \(\mathfrak {p}|p\) we have \(f_{\mathfrak {p}|p}=1\). If it is unramified and for all \(\mathfrak {p}|p\) we we have \(f_{\mathfrak {p}|p}=1\), we say \(p\) is totally split.

Let \(K/F\) be an extension of number fields. We say \(K\) is normal over \(F\) every embedding \(\sigma :K \to \mathbb {C}\) which fixes \(F\) has image again in \(K\). In other words the embedding \(\sigma \) is an automorphism \(\sigma :K \to K\) which fixes the elements of \(F\).

In particular, if \(K=F(\alpha )\) and \(K\) contains all of the conjugates of \(\alpha \), then \(K\) is normal.

Let \(n\) be an integer and \(\zeta _n\) an \(n\)-th root of unity. Let \(K=\mathbb {Q}(\zeta _n)\) and \(p\) a prime number. Then

where \(\mathfrak {p}_i\) are the primes of \(\mathcal{O}_K\) over \(p\), which moreover all have the same inertial degree.

There are only finitely many primes of  \(\mathbb {Z}\) which ramify in a number field \(K\).

Let \(n\) be a positive integer and \(\zeta _n\) an \(n\)-th root of unity. Let \(K=\mathbb {Q}(\zeta _n)\) and \(p\) a prime number. Write \(n=p^km\) with \(p \nmid m\) and set \(e=\varphi (p^k)\) (where \(\varphi \) is Euler’s Totient function). Lastly, let \(f\) be the (multiplicative) order of \(p\) modulo \(m\).

Then

(so \(e=e_{\mathfrak {p}_i|p}\)) and moreover \(f=f_{\mathfrak {p}_i|p}\).

The class group of \(K\) is defined to be quotient group

Its elements are called ideal classes. If \(\mathfrak {a}\) is a fractional ideal, we let \([\mathfrak {a}]\) denote its class in \(\operatorname{Cl}_K\). We let \(h_K\) denote the size of \(\operatorname{Cl}_K\) (which we will see below is finite), this is called the class number of \(K\).

The quantity \( \frac{n!}{n^n} \left( \frac{4}{\pi } \right)^{r_2} |\Delta (\mathcal{O}_K)|^{1/2}\) is known as the Minkowski bound and we will denote it by \(M_K\).

A lattice \(\Lambda \subset \mathbb {R}^n\) is a subgroup (under addition) generated by \(n\) linearly independent vectors.

If \(\Lambda \subset \mathbb {R}^n\) is a lattice generated by \(e_i\) then

is called the fundamental domain of \(\Lambda \).

Let \(\Lambda \subset \mathbb {R}^n\) be a lattice. Then the volume of \(P(\Lambda )\) does not depend on the choice of basis of \(\Lambda \). Moreover, if \(\{ e_i\} \) is the basis, then

(here the right hand side is the determinant of the matrix whose columns are given by the \(e_i\)).

A subset \(S \subset \mathbb {R}^n\) is called:

1. Convex if whenever \(x,y \in S\) then the line segment joining \(x\) and \(y\) is also contained in \(S\).

2. Centrally symmetric if whenever \(x \in S\) then \(-x \in S\).

Let \(K\) be a number field with \(r_1\) real embeddings and \(r_2\) complex conjugate pairs of embeddings, then the canonical embedding is

given by

where the first \(r_1\) of the \(\sigma _i\) are the real embeddings then rest are the complex ones and \(\Re ,\Im \) denote real and imaginary parts.

If \(\mathfrak {a},\mathfrak {b},\mathfrak {c}\) are non-zero ideals in a Dedekind domain and \(\mathfrak {a}\mathfrak {b}=\mathfrak {a}\mathfrak {c}\) then \(\mathfrak {b}=\mathfrak {c}\).

Let \(K\) be a number field of degree \(n\) over \(\mathbb {Q}\) and let \(\alpha \in \mathcal{O}_K\) be a primitive element so that \(K\cong \mathbb {Q}(\alpha )\).

Let \(p\) be a prime number not dividing \([\mathcal{O}_K:\mathbb {Z}[\alpha ]]\) and let

be the reduction modulo \(p\) of the minimal polynomial \(m_\alpha \) of \(\alpha \). Then in \(\mathcal{O}_K\) the ideal \((p)\) factorizes as

where \(\mathfrak {p}_i=(p,m_i(\alpha ))\) and \(f_{\mathfrak {p}_i|p}=\deg (\overline{m}_i)\) and \(e_i=e_{\mathfrak {p}_i|p}\).

Let \(K=\mathbb {Q}(\alpha )\) be a number field with \(m_\alpha (x)=x^3+ax+b\). Then

Let \(K\) be a number field and \(\alpha \in K\). Then \(\alpha \) is an algebraic integer if and only if \(C_\alpha \) has integer coefficients.

Let \(K\) be a number field and \(\beta \in K\), then

If \(K/F\) is a finite extension, then is it algebraic.

Let \(K\) be a number field. Then there are only finitely many ideals with a given norm.

Let \(\mathfrak {a},\mathfrak {b}\) be ideals in a Dedekind domain. Then \(\mathfrak {a}\mid \mathfrak {b}\) if and only if  \(\mathfrak {b}\subset \mathfrak {a}\).

In (3.4) and (3.6) of Theorem 3.5.17 the ideals appearing are distinct.

There exists a (finite) integral basis in \(K\).

Let \(K/F\) be a field extension and \(\alpha \in K\) algebraic over \(F\). Then

and moreover \([F(\alpha ):F]=\deg (m_{\alpha ,F}(x))\).

Let \(K\) be a number field with \(K=\mathbb {Q}(\alpha )\) for \(\alpha \) a primitive element and let \(f \in \mathbb {Z}[x]\) be any monic polynomial such that \(f(\alpha )=0\). If \(p\) is a prime number such that \(p \nmid N_{K/\mathbb {Q}}(f'(\alpha ))\) then \(p\) is unramified.

Let \(K\) be a number field. Then every non-zero prime ideal in \(\mathcal{O}_K\) is maximal.

Let \(K/F\) be a normal extension of number fields and let \(\mathfrak {P}\) and \(\mathfrak {P}'\) be two primes lying above the prime \(\mathfrak {p}\). Then

If \(B\) is a basis consisting of integral elements and \(\Delta (B)\) is square-free, then \(B\) is an integral basis.

Let \(\alpha ,\beta \) be algebraic integers, then so are \(\alpha +\beta \) and \(\alpha \beta \).

Let \(K/F\) be a field extension and let \(\alpha \in K\). Then we say \(\alpha \) is algebraic over \(F\) if there exists a polynomial \(f(x) \in F[x]\) such that \(f(\alpha )=0\). Otherwise we say \(\alpha \) is transcendental.

A Dedekind domain is an integral domain \(R\) such that:

1. \(R\) is Noetherian.

2. Every non-zero prime ideal is maximal.

3. \(R\) is integrally closed.

Let \(K\) be a number field, and let \(B\) be an integral basis. Then we define the discriminant of \(K\) as \(\Delta (B)\). Note that by Proposition 2.2.13, this definition does not depend on the choice of integral basis. So we will sometimes denote it simply by \(\Delta (\mathcal{O}_K)\).

Let \(K/F\) be a finite extension of number fields and let \(\sigma :F \hookrightarrow \mathbb {C}\) and \(\nu : K \hookrightarrow \mathbb {C}\) be embeddings. Then we say \(\nu \) extends \(\sigma \) if \(\nu \) restricted to \(F\) agrees with \(\sigma \). Symbolically, we say \(\nu {\mid _F}=\sigma \).

Let \(R\) be a Dedekind domain and let \(\mathfrak {a},\mathfrak {b}\) be ideals. Then we define the greatest common divisor \(\gcd \) and least common multiple \(\operatorname{lcm}\) as follows: Let \(\mathfrak {a}=\prod _i \mathfrak {p}_i^{n_i}\) and \(\mathfrak {b}=\prod _i \mathfrak {p}_i^{m_i}\) then

Here the \(\mathfrak {p}_i\) are all different.

Let \(K\) be a number field and let \(\{ b_1,\dots ,b_n\} \) be a basis for \(K/\mathbb {Q}\). We call this an integral basis if \(\mathcal{O}_K=\mathbb {Z}[b_1,\dots ,b_n]=\{ x_1b_1+\cdots x_nb_n \mid x_i \in \mathbb {Z}\} \) as an additive group. In practice, the set \(\{ b_i\} \) will be some thing like \(\{ 1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}\} \) in which case, as rings, we have \(\mathbb {Z}[\alpha ]=\mathbb {Z}[1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}]\)

Let \(R\) be an integral domain. We say an element \(r \in R\) is irreducible if whenever \(r=ab\) we must have exactly one of \(a,b\) being a unit.

Let \(R\) be a commutative ring. Then \(R\) is called Noetherian any of the following equivalent conditions holds:

1. Every ideal is finitely generated.

2. Every increasing chain of ideals \(I_1 \subset I_2 \subset \dots \) is eventually constant.

3. Every non-empty set \(S\) of ideals contains a (not necessarily unique) maximal member.

Let \(K/\mathbb {Q}\) be a number field and \(\alpha \in K\). Then we define the norm of \(\alpha \) by

and the trace of \(\alpha \) by

Here \(\operatorname{Trace}(A_\alpha )\) is the sum of the diagonal entries of \(A_\alpha \).

Let \(R\) be a Dedekind ring and \(K\) its field of fractions. For each \(x \in K\) we call the fractional ideal

a principal fractional ideal.

Let \(K/F\) be a finite extension of number fields with rings of integers \(\mathcal{O}_K\) and \(\mathcal{O}_F\) and let \(\mathfrak {p}\) be a non-zero prime ideal in \(\mathcal{O}_F\). Then

We call the \(e_i\) the ramification indices and if needed we will denote them by \(e_{\mathfrak {P}_i|\mathfrak {p}}\).

Now, let \(\mathfrak {P}\) lie over \(\mathfrak {p}\). Recall that \(k_{\mathfrak {P}}:=\mathcal{O}_K/\mathfrak {P}\) and \(k_\mathfrak {p}:=\mathcal{O}_F/\mathfrak {p}\) are both finite fields, since \(\mathfrak {P},\mathfrak {p}\) are both maximal ideals and by Proposition 3.2.4 we know the quotient ring is always finite. These are called the residual fields attached to \(\mathfrak {P}\) and \(\mathfrak {p}\) respectively. Moreover, \(k_\mathfrak {p}\) is naturally a subfield of \(k_{\mathfrak {P}}\) (convince yourself of this). Therefore, \(k_{\mathfrak {P}}\) is a finite extension of \(k_{\mathfrak {p}}\) and \([k_{\mathfrak {P}}:k_{\mathfrak {p}}]\) is called the inertial degree or residue degree of \(\mathfrak {P}\) over \(\mathfrak {p}\), which will be denoted \(f_{\mathfrak {P}|\mathfrak {p}}\).

Let \(K/F\) be a field extension and let \(\alpha _1,\dots ,\alpha _n \in K\). Let \(F(\alpha _1,\dots ,\alpha _n)\) denote the smallest subfield of \(K\) containing \(F\) and \(\alpha _1,\dots ,\alpha _n\). We call it the field generated by \(\alpha _1,\dots ,\alpha _n\).

A field generated by a single element is called a simple extension. In other words, \(F(\alpha )\) is a simple extension of \(F\). We call \(\alpha \in K\) the primitive element for the extension.

Let \(\alpha \in \mathbb {C}\). We let \(\mathbb {Z}[\alpha ]=\{ f(\alpha ) \mid f \in \mathbb {Z}[x]\} \), in other words its the smallest ring containing both \(\mathbb {Z}\) and \(\alpha \).

Let

be a polynomial in \(\mathbb {Z}[x]\) (with \(n \geq 1)\) and let \(p\) be a prime number. Suppose \(a_n\) is not divisible by \(p\), \(a_{n-1},\dots ,a_0\) are all divisible by \(p\) and \(a_0\) is not divisible by \(p^2\). Then \(f(x)\) is irreducible.

Let \(K\) be a number field and \(\mathfrak {a}\subset \mathfrak {b}\subset \mathcal{O}_K\) non-zero ideals. Then \(\mathfrak {a}=\mathfrak {b}\) if and only if \(N(\mathfrak {a})=N(\mathfrak {b})\)

For \(n\) any integer, \(\Phi _n\) is an irreducible polynomial of degree \(\varphi (n)\) (where \(\varphi \) is Euler’s Totient function).

Let \(R\) be a Dedekind domain. Then every ideal contains a product of prime ideals.

Let \(f\) be a monic irreducible polynomial over a number field \(K\) and let \(\alpha \) be one of its roots in \(\mathbb {C}\). Then

where the product is over the roots of \(f\) different from \(\alpha \).

Let \(K\) be a number field and \(B=\{ b_1,\dots ,b_n\} \) be a basis for \(K/\mathbb {Q}\) consisting of algebraic integers. If \(B\) is not an integral basis then there exists an algebraic integer of the form

where \(p\) is a prime and \(x_i \in \{ 0,\dots ,p-1\} \) with not all \(x_i\) zero. Moreover, if \(x_i \neq 0\) and we let \(B'\) be the basis obtained by replacing \(b_i\) with \(\alpha \), then

In particular \(p^2 \mid \Delta (B)\).

Let \(K=\mathbb {Q}(\alpha )\)and \(\alpha \) be an algebraic integer such that \(m_\alpha \) satisfies Eisensteins Criterion 1.2.15 for a prime \(p\). Then none of the elements

is an algebraic integer, where \(n=\deg (m_\alpha )\) and \(x_i \in \{ 0,\dots ,p-1\} .\)

A finite integral domain is a field.

Let \(K\) be a number field and \(\mathfrak {a}\subset \mathcal{O}_K\) a non-zero ideal. Let \(a=N(\mathfrak {a})\). Then \(a \in \mathfrak {a}\).

Let \(R\) be a Dedekind domain and \(\mathfrak {a},\mathfrak {b},\mathfrak {c}\) ideals such that

and suppose \(\mathfrak {a},\mathfrak {b}\) are coprime. Then there exist ideals \(\mathfrak {e},\mathfrak {d}\) such that

With the above notation. In \(\mathbb {Q}(\alpha )\) we have

with \((1-\alpha )\) a prime ideal.

In \(\mathbb {Q}(\beta )\), \(p\) is unramified and for each \(\mathfrak {p}\) over \(p\), we have \(f_{\mathfrak {p}|p}=f\).

Let \(\alpha \) be an algebraic number with minimal polynomial \(m_\alpha \) and let \(\beta \in \mathbb {Q}(\alpha )\) be such that \(\alpha =\frac{a}{b\beta +c}\) with \(a,b,c \in \mathbb {Q}\) and \(b \neq 0\). Then \(\deg (m_\alpha )=\deg (m_\beta )\).

Let \(S\) be a compact, convex and centrally symmetric subset of \(\mathbb {R}^n\) and \(\Lambda \) a lattice. If

then \(S\) contains a point of \(\Lambda \).

Let \(K\) be a number field and \(\alpha ,\beta \in \mathcal{O}_K\). If \((\alpha )=(\beta )\) then there is \(u \in \mathcal{O}_K^\times \) such that \(\alpha =u\beta \).

Let \(R\) be a Dedekind domain with field of fractions \(K\) and \(\mathfrak {a}\) a proper ideal . Then there is an element \(x \in K \backslash R\) such that \(x \mathfrak {a}\subset R\).

If \(A\) is a matrix with integer coefficients, then its eigenvalues are algebraic integers.

(Separability Lemma) Let \(K\) be a number field and let \(f(x) \in K[x]\) be an irreducible polynomial of degree \(n \geq 1\). Then \(f(x)\) has \(n\) distinct roots.

Moreover, if \(\sigma :K \hookrightarrow \mathbb {C}\) is any embedding and \(f^\sigma (x)\) denotes the polynomial obtained by applying \(\sigma \) to each coefficient, then \(f^\sigma \) also has \(n\) distinct roots.

Let \(S \subset \mathbb {R}^n\) be a measurable set (i.e. \(\operatorname{Vol}(S)=|\idotsint _S dx_1\cdots dx_n|\) exists) and \(\Lambda \) is a lattice. Then if  \(\operatorname{Vol}(S) {\gt} \operatorname{Vol}(P(\Lambda ))\) then there exist \(x,y \in S\) with \(x \neq y\) such that \(x-y \in \Lambda \).

Furthermore, if \(S\) is compact, then the same conclusion holds if \(\operatorname{Vol}(S) \geq \operatorname{Vol}(P(\Lambda ))\).

Let \(S_t \subset \mathbb {R}^{r_1} \times \mathbb {C}^{r_2} \cong \mathbb {R}^n\) be a subset given by points \((y_i,z_i) \in \mathbb {R}^{r_1} \times \mathbb {C}^{r_2}\) such that

Then \(S\) is compact, convex and centrally symmetric and moreover

Let \(f \in \mathbb {Z}[x]\) be monic and assume that \(f=gh\) with \(g,h \in \mathbb {Q}[x]\) which are also monic. Then \(g,h \in \mathbb {Z}[x]\).

Let \(\alpha \in \mathbb {C}\). The following are equivalent:

1. \(\alpha \) is an algebraic integer.

2. As an additive group, \(\mathbb {Z}[\alpha ]\) is finitely generated.

3. \(\alpha \) is an element of some subring of \(\mathbb {C}\) having a finitely generated additive group.

4. \(\alpha A \subset A\) with \(A \subset \mathbb {C}\) some finitely generated additive subgroup. Here \(\alpha A\) is the set of elements of the form \(\alpha x\) for \(x \in A\).

Let \(R\) be a commutative ring and let \(\mathfrak {a},\mathfrak {b}\) be coprime ideals (i.e., \(\mathfrak {a}+\mathfrak {b}=(1)\)) then

Let \(K\) be a number field. Then \(\operatorname{Cl}_K\) is finite.

Let \(R\) be a Dedekind domain, then every ideal \(\mathfrak {a}\) can be written uniquely as a product of prime ideals, i.e,

with \(\mathfrak {p}_i\) prime ideals (not necessarily distinct).

Let \(K/F\) be an extension of number fields and \(\alpha \in \mathcal{O}_K\) a primitive element so that \(K \cong F(\alpha )\).

Let \(\mathfrak {p}\subset \mathcal{O}_F\) be a prime ideal and \(p\) the prime number such that \((p)=\mathfrak {p}\cap \mathbb {Z}\). Assume that \(p\) does not divide the index ¹

Let \(m_{\alpha ,F}\) be the minimal polynomial of \(\alpha \) over \(F\) and let

be its reduction modulo \(\mathfrak {p}\). Now, write \(\overline{m}_{\alpha ,F}\) as a product of powers of irreducible polynomials (i.e factorize it)

Next, let \(m_i \in \mathcal{O}_F[x]\) be a polynomial whose reduction modulo \(\mathfrak {p}\) is \(\overline{m}_i\) and let

Then:

1. The ideals \(\mathfrak {P}_i\) are independent of choice of \(m_i\). (This is by construction)

2. The \(\mathfrak {P}_i\) are distinct prime ideals and they are precisely the prime ideals of \(\mathcal{O}_K\) lying over \(\mathfrak {p}\). Therefore

   \[ \mathfrak {p}\mathcal{O}_K=\mathfrak {P}_1^{e_1}\dots \mathfrak {P}_r^{e_r} \]

3. \(f_{\mathfrak {P}_i|\mathfrak {p}}=\deg (\overline{m}_i)\) and \(e_i=e_{\mathfrak {P}_i|\mathfrak {p}}\)

Let \(K\) be a number field and let

This is the set of roots of unity in \(K\). Let \(r_1\) denote the number of real embeddings of \(K\) and \(r_2\) the number of complex conjugate pairs of embeddings. Then

Let \(K=\mathbb {Q}(\alpha )\) a number field with \(m_\alpha (x)=x^n+ax+b\). Then

Let \(F\) be a field and \(p(x)\) an irreducible polynomial in \(F[x]\). Moreover, let \(K/F\) be a field extension containing a root \(\alpha \) of \(p(x)\). Then there is an isomorphism

given by sending \(f(x)\) to \(f(\alpha )\). ¹

A field extension \(K/F\) is finite if and only if \(K\) is generated by a finite number of algebraic elements over \(F\).

Let \(K/F\) be a normal extension of number fields. Let \(\mathfrak {p}\) be a prime ideal in \(\mathcal{O}_F\) and \(\mathfrak {P},\mathfrak {P}'\) be two prime ideals of \(\mathcal{O}_K\) above \(\mathfrak {p}\). Then \(\sigma (\mathfrak {P})\) is again a prime ideal lying over \(\mathfrak {p}\), moreover there is some element \(\sigma \in \operatorname{Gal}(K/F)\), such that \(\sigma (\mathfrak {P})=\mathfrak {P}'\).

Let \(R\) be a Dedekind domain and \(\mathfrak {a}\) an ideal in \(R\). Then there is an ideal \(\mathfrak {b}\) such that \(\mathfrak {a}\mathfrak {b}\) is principal.

Let \(d\) be a square free integer and let \(K=\mathbb {Q}(\sqrt{d})\) then

Let \(K\) be a number field with \(r_1\) real embeddings and \(r_2\) conjugate pairs of complex embeddings. Let \([K:\mathbb {Q}]=n\) and let \(\mathfrak {a}\) be an ideal of \(\mathcal{O}_K\). Then there is an element \(a \in \mathfrak {a}\) such that

Let \(K/F\) be a finite extension of number fields. Then there exists \(\alpha \in K\) such that \(K \cong F(\alpha )\).

Let \(p\) be a prime number and \(K\) a number field. If \(p\) is ramified in \(\mathcal{O}_K\) then \(p \mid \Delta (\mathcal{O}_K)\).

Let \(K\) be a number field with \([K:\mathbb {Q}]=n\) and \(p\) a prime number. If \(\mathfrak {p}_i\) are the prime ideals in \(\mathcal{O}_K\) dividing \(p\mathcal{O}_K\), then

with notation as in Definition 3.5.5.

Let \(\zeta _p\) be a \(p\)-th root of unity for \(p\) an odd prime, let \(\lambda _p=1-\zeta _p\) and \(K=\mathbb {Q}(\zeta _p)\). Then \(\mathcal{O}_K=\mathbb {Z}[\zeta _p]=\mathbb {Z}[\lambda _p]\) moreover

Let \(n\) be a positive integer and \(\zeta _n\) a root of unity. If \(K=\mathbb {Q}(\zeta _n)\) then

Let \(n\) be a square-free integer, \(K:=\mathbb {Q}(\sqrt{n})\) and \(p\) a prime number.

1. If \(p\mid n\) then \((p)=(p,\sqrt{n})^2\).

2. If \(n\) is odd then

   \[ (2)= \begin{cases} (2,1+\sqrt{n})^2, & \text{if } n \equiv 3 \pmod4 \hspace{0.5cm} (3.3) \\ \left(2,\frac{1+\sqrt{n}}{2}\right)\left(2,\frac{1-\sqrt{n}}{2}\right) & \text{if } n \equiv 1 \pmod8 \hspace{0.5cm} (3.4)\\ (2) & \text{if } n \equiv 5 \pmod8 \hspace{0.5cm} (3.5) \end{cases} \]

3. If \(p\) is odd and \(p \nmid n\), then

   \[ (p)=\begin{cases} (p,a+\sqrt{n}) (p,a-\sqrt{n}) & \text{if } n \equiv a^2 \pmod p \hspace{0.5cm} (3.6)\\ (p) & \text{if } \left( \frac{n}{p}\right)=-1 \end{cases} \]

Moreover, in (3.4) and (3.6) the ideals appearing are distinct.