View source for Abstract Algebra/Rings, ideals, ring homomorphisms

== Basic definitions ==

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 10.1''':

A '''ring''' is a set <math>R</math> together with two binary operations <math>+: R \times R \to R</math> and <math>\cdot: R \times R \to R</math> and two special elements, the unit <math>1</math> and the zero <math>0</math>, such that:
#<math>R</math> is an abelian group with respect to <math>+</math> with neutral element <math>0</math>.
#<math>R</math> is a [[w:monoid|monoid]] (that is, a group without inversion) with respect to <math>\cdot</math> with neutral element <math>1</math>.
#The distributive laws hold: <math>a \cdot (b + c) = a \cdot b + a \cdot c</math>, <math>(b + c) \cdot a = b \cdot a + c \cdot a</math>.}}

'''Examples 10.2''':

* The whole numbers <math>\mathbb Z</math> with respect to usual addition and multiplication are a ring.
* Every field is a ring.
* If <math>R</math> is a ring, then all polynomials over <math>R</math> form a ring. This example will be explained later in the section on polynomial rings.

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 10.3''':

Let <math>R</math> be a ring. A '''left ideal''' of <math>R</math> is a subset <math>I \subseteq R</math> such that the following two things hold:
#<math>(I, +)</math> is a subgroup of <math>(R, +)</math>.
#<math>\forall r \in R: rI \subseteq I</math>, where <math>rI = \{ri| i \in I\}</math> (closedness by left multiplication).
Replacing closedness by left multiplication by closedness by right multiplication, we can define right ideals, and then both-sided ideals. If <math>I \subseteq R</math> is a both-sided ideal of <math>R</math>, we write <math>I \le R</math>.}}

We'll now show an important property of the set of all ideals of a given ring, namely that it's inductive. This means:

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 10.4''':

Let <math>(S, \le)</math> be a partially ordered set (that is, the usual conditions transitivity, reflexivity and anti-symmetry are satisfied). <math>S</math> is called ''inductive'' if and only if every ascending chain of elements of <math>S</math> (that is, a sequence <math>(s_n)_{n \in \mathbb N}</math> in <math>S</math> such that <math>s_1 \le s_2 \le s_3 \le \cdots \le s_k \le \cdots</math>) has an upper bound (that is, an element <math>b \in S</math> such that <math>\forall n \in \mathbb N: b \ge s_n</math>).}}

With this definition, we observe:

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Theorem 10.5''':

If a commutative ring <math>R</math> is given, the set of all ideals <math>I</math> of <math>R</math>, partially ordered by inclusion (i.e. <math>S = \{I \subset R | I \le R\} \subset 2^R</math>, where we use the convention of [[w:Donald Knuth|Donald Knuth]] and denote the power set of a set <math>T</math> by <math>2^T</math>) is inductive.}}

'''Proof''':

If
:<math>I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots \subseteq I_k \subseteq \cdots</math>
is an ascending chain of ideals, we set
:<math>J := \bigcup_{n \in \mathbb N} I_n</math>
and claim that <math>J \le R</math>. Indeed, if <math>a, b \in J</math>, find <math>m, n \in \mathbb N</math> such that <math>a \in I_n</math> and <math>b \in I_m</math>. Then set <math>N := \max\{m,n\}</math>, so that <math>a, b, a+b \in I_N \subseteq J</math> since <math>I_N \le R</math>. Similarly, if <math>a \in J</math> and <math>r \in R</math>, pick <math>n \in \mathbb N</math> such that <math>a \in I_n</math>, whence <math>ra \in I_n \subseteq J</math> since <math>I_n \le R</math>.{{QED}}

== Residue class rings ==

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition and theorem 10.4''':

Let <math>R</math> be a ring, and <math>I \le R</math>. Then we define a relation <math>\sim_I</math> on <math>R</math> as follows:
:<math>a \sim_I b : \Leftrightarrow a - b \in I</math>.
This relation is an equivalence relation, and an equivalence class <math>[a]</math> shall be denoted by <math>a + I</math> for <math>a \in R</math>. If we define an addition
:<math>(a + I) + (b + I) := (a + b) + I</math>
and a multiplication
:<math>(a + I) \cdot (b + I) := a \cdot b + I</math>,
then these two are well-defined (i. e. independent of the choice of the representatives <math>a</math> and <math>b</math>) and turn <math>R/I</math> into a ring, called the '''residue class ring''' with respect to the ideal <math>I</math>.}}

'''Proof''':

First, we check that <math>\sim_I</math> is an equivalence relation.
#Reflexiveness: <math>a - a = 0 \in I</math> since <math>I</math> is an additive subgroup.
#Symmetry: <math>a - b \in I \Leftrightarrow -(a - b) \in I</math> since inverses are in the subgroup.
#Transitivity: Let <math>a - b \in I</math> and <math>b - c \in I</math>. Then <math>a - c = a - b + (b - c) \in I</math>, since a subgroup is closed under the group operation.

Then we check that addition and multiplication are well-defined. Let <math>a + I = a' + I</math> and <math>b + I = b' + I</math>. Then
:<math>a + b - (a' + b') = a + b - (a + i + b + j) = -i -j \in I</math> for certain <math>i, j \in I</math>.
Furthermore,
:<math>a \cdot b - a' \cdot b' = a \cdot b - a \cdot b - a \cdot j - i \cdot b - i \cdot b</math>
for these same <math>i, j \in I</math>; this is in <math>I</math> by closedness by left and right multiplication.

The ring axioms directly carry over from the old ring <math>R</math>.{{QED}}

== Ring homomorphisms ==

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 10.5''':

Let <math>R, S</math> be rings. A '''ring homomorphism''' between the two is a map
:<math>\varphi: R \to S</math>
such that:
#For all <math>a, b \in R</math> <math>\varphi(a+b) = \varphi(a)+ \varphi(b)</math> and <math>\varphi(a \cdot b) = \varphi(a) \cdot \varphi(b)</math>.
#<math>\varphi(1_R) = 1_S</math> (<math>1_R</math> is the unit of <math>R</math> and <math>1_S</math> of <math>S</math>).}}

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