Homomorphism

Group

Let $G$ be a group with elements ${g}$, and $H$ is a homomorphism of $G$. The set of elements mapped by $H$ to the identity is the kernel $K$ of $H$. $K$ is a normal subgroup of G, and the cosets of $K$ form another group ${G \over K}$, called a factor group.

Ring

Let $R$ be a ring, and $H$ is a homomorphism of $R$. The kernel $K$ of $H$ is an ideal of $R$, somewhat analogous to the normal subgroup discussed above. The cosets of the ideal $K$ form a factor ring (sometimes: quotient ring) ${R \over K}$. We would like to know how the properties of an ideal $K$ determine the properties of the associated factor ring. We will let our rings be commutative and have identity.

Elements that multiply to zero are called zero divisors. In $Z_6$ (the integers modulo 6), 2 and 3 multiply to zero, so they are zero divisors. If $n$ is not prime, $Z_n$ will have zero-divisors. A ring with no zero divisors is an integral domain. In an integral domain, $ab=0 \implies a=0$ or $b=0$

The elements that have inverses are called units.  The units form a group.  The group of units in $Z_6$ are {1,5}.

A ring in which every element has a multiplicative inverse is a field. A field is necessarily an integral domain. If $ab =0$ , and $a \neq 0$, then multiply by $a^{-1}$ and conclude $b=0$. That the integers $Z$ is not a field shows the reverse is not true. It is true, however, that every finite integral domain is a field.

Principal Ideal

The ideal $(b)$ consisting of all multiples of an element $b$ of $R$ is a principal ideal. This is the smallest ideal containing $b$. If it's unclear how an ideal might *not* be principal, consider a polynomial in two variables over the complex numbers. The ideal generated by x and y is not principal, because if it's principal there's a generator, and if there's a generator, say $p$, this divides every element of the ideal, but that must be a constant (non-zero). Yet there are no constants in the ideal, contradiction.

We can have an ideal generated by a subset, but this is a principal ideal only if the subset is a single element.

Prime Ideal

An ideal $K$ is prime if and only if $ab$ in $K$ implies $a$ or $b$ is in $K$. If $K$ is a prime ideal, the factor ring ${R \over K}$ is an integral domain.

Maximal Ideal

An ideal $K$ is maximal if the only ideal containing $K$ is $R$. If $K$ is a maximal ideal, ${R \over K}$ is a field. A maximal ideal must be prime, but the reverse is not true: a prime ideal might not be maximal.