Basic Commutative Algebra II

Theorem. If $f:A\to B$ is a ring homomorphism, then the kernel of $f$ ($:=f^{-1}(0)$) is an ideal of $A$, since $f$ preserves multiplication. The image of $f$ is a subring of $B$. Since $\text{ker}(f)$ is a ring of $A$, we can construct a quotient ring $A/\text{ker}(f)$, which is isomorphic to $\text{img}f$.

In the following we define some important concepts.

A zero-divisor is an element in ring $A$ that in a sense “divides zero”, i.e. $x\in A$ is a zero divisor if there exists non zero element $y\in A$ such that $xy=0$. A trivial zero-divisor is zero itself.

Definition. A domain is a nonzero ring in which ab = 0 implies a = 0 or b = 0. (Sometimes such a ring is said to “have the zero-product property “.) Equivalently, a domain is a ring in which 0 is the only left zero divisor (or equivalently, the only right zero divisor). A commutative domain is called an integral domain.

Integers $\mathbb{Z}$ form an integral domain (hence the name), another important example of integral domain is the ring of polynomials with coefficients taken from field $k$, denoted by $k[x_1,\dots ,x_n]$.

An element $x\in A$ is said to be nilpotent if $x^n=0$ for some integer $n$. a nilpotent element is a zero-divisor, but the converse is not necessarily true.

A unit in $A$ is an element $x$ that “divides” 1, namely there exists elements $y\in A$ such that $xy=1$. The element $y$ is then uniquely determined and written as $x^{-1}$. The units of $a$ forms an abelian group.

Definition. A principal ideal is an ideal form by multiples of an element $x\in A$, denoted by $(x)$ or $Ax$.

If $x$ is a unit, then

$$\begin{array}{}\text{(1)}& (x)=(x{x}^{-1})=(1)=A.\end{array}$$

In other words, the principal ideal of a unit is the ring itself.

Definition. A field is a ring where $1=0$ and every non-zero element is a unit, i.e., every non-zero element has an inverse.

For example, the set of rational numbers $\mathbb{Q}$ is a field since every quotient $\frac{x}{y}$ has an inverse $\frac{y}{x}$.

Every field is also an integral domain since there can’t be zero divisors, but the converse is not necessarily true, for instance $\mathbb{Z}$ is not a field but a integral domain.

Proposition. Let $A$ be a non-zero ring. Then the following are equivalent:

1. $A$ is a field,
2. the only ideals are $(0),(1)=A$,
3. every homomorphism from $A$ to a non-zero ring $B$ is injective.

Proof. $(2)⟹(3)$. Given a homomorphism $f:A\to B$, the kernel of $f$ is an ideal of $A$, it can only be either $0$ or $A$ itself. If $\text{ker}(f)=A$ then the image is 0, which disagrees with our assumption that B is non-zero. Thus $\text{ker}(f)=0$. Then the map is injective.

Definition. An ideal $\mathfrak{p}$ in $A$ is prime if $\mathfrak{p}\ne \mathfrak{(}\mathfrak{1}\mathfrak{)}$ and $xy\in \mathfrak{p}⟹x\in \mathfrak{p}\text{ or }y\in \mathfrak{p}$. An ideal $\mathfrak{m}$ is maximal if $\mathfrak{m}\ne A$ and there is no ideal $\mathfrak{a}$ such that $\mathfrak{m}\subset \mathfrak{a}\subset A$ (strict inclusion).

Notice the definition of prime ideal is different from that of prime numbers, a number $p$ is said to be prime if $p\mid xy⟹p\mid x\text{ or }p\mid y$. For example, 6 is not prime since $6\mid 2\times 3$ but $6\nmid 2$ and $6\nmid 3$.

There might be more than one maximal ideals in $A$. $\mathfrak{m}$ is a maximal ideal only means that there is no bigger ideals between $\mathfrak{m}$ and $A$, and there could be more than one ideals satisfy this property. For example, in the ring of integers, all ideals generated by prime numbers $(p)$ are maximal.

Theorem.

• $\mathfrak{p}$ is prime $\iff$ $A/\mathfrak{p}$ is an integral domain,
• $\mathfrak{m}$ is maximal $\iff$ $A/\mathfrak{m}$ is a field.

A prime ideal can’t be written as the section or two ideals. Conversely, if an ideal is the section of two other ideals, then it is not prime. A maximal ideal is always prime, since $xy\in \mathfrak{m}⟹x\in \mathfrak{m}$ or $y\in \mathfrak{m}$. If not, we can always combine these two ideals and make a bigger one, then $\mathfrak{m}$ wouldn’t be the maximal idea.

If the zero ideal is prime, then $xy=0$ implies $x=0$ or $y=0$, then $A$ has no zero-divisors and it is an integral domain.

In general, homomorphic maps will preserve the properties defined by multiplication and addition. If $f:A\to B$ is a homomorphism and $\mathfrak{p}$ is a prime ideal of $B$, then $f^{-1}\mathfrak{p}$ is a prime ideal in $A$. However, maximal ideals are not defined by multiplication or addition, so homomorphism in general does not preserve it, let $\mathfrak{m}$ be the maximal ideal in $B$, $f^{-1}(\mathfrak{m})$ may not be the maximal ideal in $A$, there could be a bigger ideal in $A$. For example, we can construct a homomorphism from $\mathbb{Z}$ to $\mathbb{Q}$, something like $g:z\mapsto \frac{z}{p}$ where $p$ is some prime number, then this map preserves addition and multiplication, the maximal ideal of $\mathbb{Q}$ is zero since $\mathbb{Q}$ is a field, but $g^{-1}(0)=0$ is clearly not a maximal ideal in $\mathbb{Z}$.

Prime ideals are most important to the whole of commutative algebra.

Theorem. Every ring (commutative) $A\ne 0$ has at least one maximal ideal.

Corollary.

1. If $\mathfrak{a}\ne (1)$ is an ideal of $A$, then there is a maximal ideal containing $\mathfrak{a}$ (could be $\mathfrak{a}$ itself).
2. Every non-unit of $A$ is contained in a maximal ideal.

Definition. There exists rings with exactly one maximal ideal, for example fields. A ring $A$ with exactly one maximal ideal $\mathfrak{m}$ is called a local ring. The field $k=A/\mathfrak{m}$ is called the residue field of $A$.

The set of all non-units of ring $A$ does not form an ideal, since it might not be closed under addition, the sum of two non-units is not necessarily a unit. But there is a theorem connecting maximal rings with non-units:

Theorem Let $A$ be a ring and $\mathfrak{m}\ne (1)$ an ideal such that every $x\in A-\mathfrak{m}$ is a unit in $A$. Then A is a local ring and $\mathfrak{m}$ its maximal ideal.

To prove this, recall that all every ideal $\mathfrak{a}\ne (1)$ consists of non-units, hence is contained in $\mathfrak{m}$. Hence $\mathfrak{m}$ is the only maximal ideal of $A$. An example is ${\mathbb{Z}}_3$.

Let $A,\mathfrak{m}$ be a local ring and the maximal ideal of that ring, if $1+\mathfrak{m}$ is a unit in $A$, then $A$ is a local ring. To prove this, let $x\in A-\mathfrak{m}$. Since $\mathfrak{m}$ is the maximal ideal, the union of $(x)$ and $\mathfrak{m}$ is $(1)$. Let $m\in \mathfrak{m}$, then every element in $A$ can be written in the form $xa+mb$, where $a,b\in A$ . Since $mb\in \mathfrak{m}$ by the definition of ideal, every element of $A$ can be expressed in the form $ma+\mathfrak{m}$, including $1$, hence there exists $y\in A$ such that $xy+\mathfrak{m}=1$, which implies $xy=-\mathfrak{m}+1=\mathfrak{m}+1$. According to the assumption, $xy$ is a unit, then $x,y$ are both units. Since now $A-\mathfrak{m}$ are units, we conclude that $A$ is a local ring.

For the ring of polynomials $k[x_1,\dots ,x_n]$, the maximal ideals are all the polynomials with zero constant term, these polynomials are the kernel of homomorphism: evaluation at $x_i=0$.

Definition. A principal ideal domain, (PID) is an integral domain in which every ideal is principal. Recall that a principal ideal is an ideal generated by a single element.

We claim without proof that, in PID every non-zero prime ideal is maximal.