Abstract Algebra/Rings, ideals, ring homomorphisms

Basic definitions

Examples 10.2:

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

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

With this definition, we observe:

Proof:

If

$I_1\subseteq I_2\subseteq I_3\subseteq \cdots \subseteq I_k\subseteq \cdots$

is an ascending chain of ideals, we set

$J:=\bigcup _{n\in \mathbb{N}}{I}_n$

and claim that $J\le R$. Indeed, if $a,b\in J$, find $m,n\in \mathbb{N}$ such that $a\in I_n$ and $b\in I_m$. Then set $N:=\max \{m,n\}$, so that $a,b,a+b\in I_N\subseteq J$ since $I_N\le R$. Similarly, if $a\in J$ and $r\in R$, pick $n\in \mathbb{N}$ such that $a\in I_n$, whence $ra\in I_n\subseteq J$ since $I_n\le R$.$◻$

Residue class rings

Proof:

First, we check that ${\sim }_I$ is an equivalence relation.

1. Reflexiveness: $a-a=0\in I$ since $I$ is an additive subgroup.
2. Symmetry: $a-b\in I\iff -(a-b)\in I$ since inverses are in the subgroup.
3. Transitivity: Let $a-b\in I$ and $b-c\in I$. Then $a-c=a-b+(b-c)\in I$, since a subgroup is closed under the group operation.

Then we check that addition and multiplication are well-defined. Let $a+I=a^{\prime }+I$ and $b+I=b^{\prime }+I$. Then

$a+b-(a^{\prime }+b^{\prime })=a+b-(a+i+b+j)=-i-j\in I$ for certain $i,j\in I$.

Furthermore,

$a\cdot b-a^{\prime }\cdot b^{\prime }=a\cdot b-a\cdot b-a\cdot j-i\cdot b-i\cdot b$

for these same $i,j\in I$; this is in $I$ by closedness by left and right multiplication.

The ring axioms directly carry over from the old ring $R$.$◻$