1.1 Recap on rings and fields
We begin by recalling some basic facts in commutative algebra. Specifically, some ring theory and field theory.
Throughout, we will not differentiate between \(\subset \) and \(\subseteq \). If such a distinction needs to be made we will state it or use \(\subsetneqq \).
A ring \(R\) is a set with two binary operations called addition \('+'\) and multiplication \('\cdot '\), such that:
\(R\) 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 \(-r \in R\).
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 \(R^\times \).
Throughout this whole course, our rings will be assumed to be unital (i.e. have a \(1\)) and unless otherwise stated, will be commutative.
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.
The map \(\phi : \mathbb {Z}\to \mathbb {Z}/p\mathbb {Z}\) defined by \(x \mapsto x \bmod p\) is a ring homomorphism.
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 \(\operatorname{Im}(\phi )\)
The set of integers \(\mathbb {Z}\) is a commutative ring.
The set \(\mathbb {Z}[x]\) of polynomials with integer coefficients is a ring. In general, if \(R\) is a ring, then \(R[x]\) is also a ring.
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
The rational numbers, \(\mathbb {Q}\), are a field. As well as the Reals \(\mathbb {R}\) and the complex numbers \(\mathbb {C}\).
If \(p\) is a prime number, then \(\mathbb {F}_p:=\mathbb {Z}/p\mathbb {Z}\) (the field of integers modulo p) is a field.
The following are not fields: The integers \(\mathbb {Z}\), the polynomial ring \(\mathbb {Z}[x]\).
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 \(ra \in \mathfrak {a}\).
Let \(R\) be a ring and \(r \in R\), then we let
This is an ideal in \(R\) and we call it the principal ideal generated by \(r\). Similarly, if we take \(r_1,\dots ,r_n\) then we can from the ideal \((r_1,\dots ,r_n):=\{ \sum _i r_i x_i : x_i \in R\} \). Note that \((0)=0\) and \((1)=R\).
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.
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.
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
Check that this ring structure is well-defined.
The kernel of a ring homomorphism is an ideal.
The image of a ring homomorphism is a subring.
If \(\phi \) is a ring homomorphism, then there is a ring isomorphism
\[ R / \ker (\phi ) \cong \operatorname{Im}(\phi ). \]
Let \(R\) be a ring and \(\mathfrak {p},\mathfrak {m}\) an ideals with neither equal to \((1)\).
The \(\mathfrak {p}\) is called prime if whenever \(xy \in \mathfrak {p}\) we have \(x \in \mathfrak {p}\) or \(y \in \mathfrak {p}\).
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 \(p\) be a prime number, then \((p) \subset \mathbb {Z}\) is both a prime ideal as well as maximal.
Every maximal ideal is prime.
Let \(R\) be a ring. Then \(\mathfrak {p}\) is a prime ideal if and only if \(R/\mathfrak {p}\) is an integral domain. Similarly \(\mathfrak {m}\) is a maximal ideal if and only if \(R/\mathfrak {m}\) is a field.
The only prime ideal in a field is \((0)\). 2
If \(\phi \) is a non-zero ring homomorphism, then \(\ker (\phi )\) is a proper ideal.
Every proper ideal (meaning one which isnt the whole ring) is contained in a maximal ideal.
Let \(\mathfrak {a},\mathfrak {b}\) be ideals in a ring \(R\). Then we let
\(\mathfrak {a}\mathfrak {b}\) be the ideal generated by the product of elements of \(\mathfrak {a}\) and \(\mathfrak {b}\).
Similarly, \(\mathfrak {a}+\mathfrak {b}\) denotes the ideal generated by sums of elements in \(\mathfrak {a},\mathfrak {b}\).
If \(\mathfrak {a}+\mathfrak {b}=R=(1)\) we say \(\mathfrak {a},\mathfrak {b}\) are coprime.
If \(\mathfrak {a},\mathfrak {b}\) are ideals we write \(\mathfrak {a}\mid \mathfrak {b}\) if there exists an ideal \(\mathfrak {c}\) such that \(\mathfrak {b}=\mathfrak {a}\mathfrak {c}\).