## 305 -Rings, ideals, homomorphisms (3)

March 21, 2009

In order to understand the construction of the quotient ring from last lecture, it is convenient to examine some examples in details. We are interested in ideals ${I}$ of ${{\mathbb F}[x],}$ where ${{\mathbb F}}$ is a field. We write ${{\mathbb F}[x]/I}$ for the quotient ring, i.e., the set of equivalence classes ${[a]_\sim}$ of polynomials ${a}$ in ${F[x]}$ under the equivalence relation ${a\sim b}$ iff ${a-b\in I.}$

• If ${I=\{0\},}$ then for any ${a,}$ the equivalence class ${[a]_\sim}$ is just the singleton ${\{a\}}$ and the homomorphism map ${h:{\mathbb F}[x]\rightarrow{\mathbb F}[x]/I}$ given by ${h(a)=[a]_\sim}$ is an isomorphism.

To understand general ideals better the following notions are useful; I restrict to commutative rings with identity although they make sense in other contexts as well:

Definition 1 Let ${R}$ be a commutative ring with identity. An ideal ${I}$ is principal iff it is the ideal generated by an element ${a}$ of ${R,}$ i.e., it is the set ${(a)}$ of all products ${ab}$ for ${b\in R.}$

For example, ${\{0\}=(0)}$ is principal. In ${{\mathbb Z}}$ every subring is an ideal and is principal, since all subrings of ${{\mathbb Z}}$ are of the form ${n{\mathbb Z}=(n)}$ for some integer ${n.}$

We continue from last lecture. Examination of a few particular cases finally allows us to complete Example 6. The answer to whether ${\mathbb Z}_n$ (with the operations and constants defined last time) is a field splits into three parts. The first is straightforward.
Lemma 10. For all positive integers $n>1,$ ${\mathbb Z}_n$ satisfies all the properties of fields except possible the existence of multiplicative inverses. $\Box$.
This reduces the question of whether ${\mathbb Z}_n$ is a field to the problem of finding multiplicative inverses, which turns out to be related to properties of $n.$