## Commutative Algebra 0

We’re starting a new series on commutative algebra. This has been in the works for way too long, and eventually we just decided to push ahead with it anyway. Most of the articles will be short, and we’ll try to illustrate the concepts with examples and diagrams whenever possible. Hopefully this will turn out well.

Since the notes are rather disorganized, repetitions will undoubtedly occur. For instance, we may introduce concept X in an earlier installation only to reintroduce it later as a special case of a more general machinery. However, we strive at clarity in explaining the concepts. Of course, the reader will be the best judge of how successful we are.

This is a summary of things you should already be familiar with, so we will proceed rather quickly.

# Basic Concepts

We assume you have some basic knowledge about rings and ideals. Here, all rings are assumed to be commutative with 1. For a ring $A$, any ideal $\mathfrak a\subseteq A$ gives us a ring quotient $A/\mathfrak a$. Ideals also allow us to talk about “modulo” relations.

Definition.

If $\mathfrak a\subseteq A$ is an ideal, then for $x,y\in A$, we write $x\equiv y\pmod {\mathfrak a}$ if their difference $x-y\in \mathfrak a$.

Note that this generalizes the concept of $x \equiv y \pmod n$ in modular arithmetic. Thanks to the definition of an ideal, we immediately get:

• If $x \equiv y \pmod {\mathfrak a}$ and $x' \equiv y' \pmod {\mathfrak a}$, then

$x+x' \equiv y + y' \pmod {\mathfrak a}, \quad xx' \equiv yy' \pmod {\mathfrak a}.$

• If $x\equiv y\pmod {\mathfrak a}$, then for all $n\ge 1$ we have $x^n \equiv y^n \pmod {\mathfrak a}$.

Proof : exercise.

This illustrates a general maxim here: that ideals generalize the concept of individual elements of a ring. We will see many more examples later. The most startling case of this is the example of a Dedekind domain, where we have unique factorization of ideals into prime ideals, instead of unique factorization of elements into primes.

## Subrings

Also, let us properly define a subring.

Definition.

A subring of the ring $A$ is a subset $B\subseteq A$ satisfying:

• $1 \in B$;
• $(B, +)$ is a subgroup of $(A, +)$;
• if $x, y \in B$, then $xy \in B$.

Note

The trivial ring $A = \{0\}$ cannot be a subring of any other ring except itself, because of the first condition. However, it can be a ring quotient since for any ring $A$, the ideal $\mathfrak a = A$ gives a trivial ring for the quotient $A/\mathfrak a$.

# Ideals

Recall that we have the following construction of ideals.

Definition

Let $A$ be any ring.

• If $\{\mathfrak a_i\}$ is a collection of ideals of $A$, their intersection $\cap_i \mathfrak a_i$ is also an ideal of $A$.
• If $\mathfrak a, \mathfrak b\subseteq A$ are ideals, their product $\mathfrak{ab}$ is the set of all finite sums $x_1 y_1 + \ldots + x_k y_k$, where $x_1, \ldots, x_k \in \mathfrak a$ and $y_1, \ldots, y_k \in \mathfrak b$.

Beginning students in algebra are often a bit bewildered by the presence of finite sums in the product of two ideals. But the definition is sensible, for the product satisfies

$\mathfrak {ab} = \mathfrak {ba}, \quad (\mathfrak{ab})\mathfrak c = \mathfrak a(\mathfrak {bc}), \quad A\mathfrak a = \mathfrak a.$

For this reason, we often write the ideal $A$ (whole ring) as $(1)$. Aesthetically this is pleasing for we get $\mathfrak a (1) = \mathfrak a$.

# Generating an Ideal

The fact that ideals are closed under arbitrary intersections is important, for we can take any subset $S\subseteq A$ and generate the smallest ideal containing it.

Definition

Consider the collection $\Sigma$ of all ideals $\mathfrak a\subseteq A$ containing $S$. Note that $\Sigma$ is non-empty since $A\in \Sigma$. The intersection of all $\mathfrak a\in \Sigma$ gives

$(S) := \bigcap_{\mathfrak a \in \Sigma} \mathfrak a,$

the ideal generated by set $S$. If $S = \{a_1, a_2, \ldots, a_n\}$, we also denote $(S)$ by $(a_1, a_2, \ldots, a_n)$.

Note

We will use the computer science notation `:=’ to refer to definition. Thus $A := B$ means the notation $A$ on the LHS is defined to be $B$ on the RHS.

An ideal generated by one element is easy to describe:

$(a) = \{ ab : b\in A\}.$

More generally, one sees that $(S)$ is given by the set of all finite sums of the form $a_1 b_1 + a_2 b_2 + \ldots + a_n b_n$ where $a_1, \ldots, a_n \in S$ and $b_1, \ldots, b_n \in A$. We can only restrict ourselves to finite sums since we are dealing with algebra here and infinite sums are not well-defined. [One can consider infinite sums in, say, functional analysis, but that is another story for another day.]

## Sums of Ideals

Now suppose $\mathfrak a_i$ is a collection of ideals of $A$. If $S = \cup_i \mathfrak a_i$, the ideal generated by S is called the sum of the ideals $\mathfrak a_i$. It is clear that the sum can also be defined as follows.

Definition.

The ideal $\sum_i \mathfrak a_i$ is the collection of all finite sums $x_1 + x_2 + \ldots + x_n$, where

$x_1 \in \mathfrak a_{i_1}, \ x_2 \in \mathfrak a_{i_2}, \ \ldots\ , x_n \in \mathfrak a_{i_n}.$

As expected, multiplication is distributive over addition.

# Integral Domains and Fields

Next, we have the following.

Definition.

An element $a$ of ring $A$ is a zero-divisor if there exists $b\in A$, $b\ne 0$ such that $ab=0$.

Note that the trivial ring $A = \{0\}$ has no zero-divisor by definition. Otherwise for non-trivial ring $A$, the element $0\in A$ is a zero-divisor since $0\cdot 1 = 0$ and $1\ne 0$Be forewarned that the trivial ring can be a pitfall for the unwary.

Definition.

The ring $A$ is an integral domain (or just domain) if it is non-trivial, and the only zero-divisor in $A$ is $0$.

Note that we have explicitly excluded the case of the trivial ring. Examples of integral domains include

• $\mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C$;
• $\mathbb Z[i] = \{a + bi : a, b\in\mathbb Z\}$, where $i =\sqrt{-1}$.

Clearly any subring of an integral domain is an integral domain.

Definition.

An element $a$ of a ring $A$ is a unit if there exists $b\in A$ such that $ab=1$.

Note that the set of units of a ring A forms a group under multiplication. We denote this by $U(A)$ and call it the unit group of A. For example, $U(\mathbb Z) = \{+1, -1\}$ and $U(\mathbb Z[i]) = \{+1, +i, -1, -i\}$. Finally, we define:

Definition.

The ring $A$ is said to be a field if it is non-trivial and any non-zero element of $A$ is a unit.

For example, $\mathbb Q\subset \mathbb R\subset \mathbb C$ are all fields and $\mathbb Z$ is not a field. The following is standard, whose proof we skip.

Theorem.

• Any field is an integral domain.
• Any finite integral domain is a field.

Note

A ring A is a field if and only if it has exactly two ideals: 0 and A. [The trivial ring has only one ideal.]

# Prime and Maximal Ideals

Given an ideal $\mathfrak a$ of ring $A$, one would like to know when $A/\mathfrak a$ is an integral domain or a field.

Theorem.

• $A/\mathfrak a$ is an integral domain if and only if $\mathfrak a\ne A$ and

$x, y\in A, \ xy \in \mathfrak a \implies x \in \mathfrak a \text{ or } y\in \mathfrak a.$

• $A/\mathfrak a$ is a field if and only if $\mathfrak a\ne A$ and, for any ideal $\mathfrak b\subseteq A$ containing $\mathfrak a$, we have $\mathfrak b = \mathfrak a$ or $\mathfrak b = A$.

Note

For the first case, we say that $\mathfrak a$ is a prime ideal of A. For the second case, we say that $\mathfrak a$ is a maximal ideal of A. By definition, the whole ring $A = (1)$ is neither prime nor maximal. This should not come as a surprise, since in our definition of prime numbers, we explicitly exclude 1.

We will not prove the above theorem, but the proof of the first result is immediate. For the second result, we use the fact that a field has exactly two ideals, and that the ideals of the ring quotient $A/\mathfrak a$ are in bijection with ideals $\mathfrak b$ of $A$ containing $\mathfrak a$.

Pictorially, the lattices of ideals correspond as follows (where the arrows represent inclusion):

Furthermore, the prime and maximal ideals coincide, i.e. $\mathfrak b/\mathfrak a$ is a prime (resp. maximal) ideal of the ring $A/\mathfrak a$ if and only if $\mathfrak b$ is a prime (resp. maximal) ideal of the ring $A$.

Thus, taking the quotient $A/\mathfrak a$ corresponds to “chopping off” ideals from lower branches in the lattice.

This entry was posted in Advanced Algebra and tagged , , , , . Bookmark the permalink.

### 2 Responses to Commutative Algebra 0

1. Vanya says:

While declaring the operations on modulo $\equiv$ did you mean $x \equiv x'$ instead of \$latex x \equiv y”?

Also at the end of the post : ” For a more concrete example, here is the correspondence between ideals of \mathbb Z/(30) and ideals of \mathbb Z containing (30). ” , but no description is given thereafter.

• limsup says:

Thanks fixed it. I also removed the example on $\mathbb Z/(30)$ since it’s a bit too easy.