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: $$a - b \in I \Leftrightarrow -(a - b) \in I$$ 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 $$a + I = a' + I$$ and $$b + I = b' + I$$. Then
 * $$a + b - (a' + b') = a + b - (a + i + b + j) = -i -j \in I$$ for certain $$i, j \in I$$.

Furthermore,
 * $$a \cdot b - a' \cdot b' = a \cdot b - a \cdot b - a \cdot j - i \cdot b - i \cdot b$$

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 $$R$$.