# More on Ideals

In Rings, Fields, and Ideals, we talked about the concept of an ideal, which we described as a subset of a ring which is closed under addition among its own elements and under multiplication by the elements of the ring. Here is a more rigorous definition, lifted verbatim from the book Algebra by Michael Artin:

An ideal $I$of a ring $R$ is a nonempty subset of $R$ with these properties:

(i) $I$ is closed under addition.

(ii) If $s$ is in $I$ and $r$ is in $R$, then $rs$ is in $I$.

Now we show two important examples of ideals that all rings have:

The first one we will present here is called the zero ideal, which we also write as $(0)$. It is the ideal consisting only of one element, which is the zero element, which we will write as $0$ regardless of whether or not we are working in the ring of integers $\mathbb{Z}$. It is good to mention at this point two important properties of $0$. First, it is the identity element under addition and subtraction; in other words, $0$ added to or subtracted from any other element gives back that element. Second, any number multiplied to or by $0$ gives back $0$. A familiar and intuitive example of these concepts is given in the arithmetic of $\mathbb{Z}$.

Now to show that the set consisting of the zero element alone is an ideal, we check if the properties that define the concept of an ideal hold:

For property (i) above, we see from

$\displaystyle 0+0=0$

and

$\displaystyle 0-0=0$

that whenever the left hand side contains only elements from $(0)$, namely only $0$, then the right hand also only contains elements from $(0)$, which is, again, $0$.

As for property (ii), this is satisfied since any element of $R$ multiplied to or by $0$ is once again $0$, which is the one and only element of the set $(0)$.

Therefore $(0)$ is indeed an ideal.

The second example of an ideal is called the unit ideal, which we write down using the symbol $(1)$. It is none other than the entire ring $R$ itself. Obviously, since the definition of a ring requires that it be closed under addition, subtraction and multiplication, the entire ring $R$ is an ideal of itself, but in its context as an ideal we refer to it as the unit ideal.

Notice that $(0)$ can be considered to be the subset of $R$ consisting of all multiples of $0$ and that $(1)$ can be considered to be the subset of $R$ consisting of all multiples of $1$. These are special cases of a more general kind of ideal, called a principal ideal. Once again we take the rigorous definition almost verbatim from Artin’s Algebra:

In any $R$, the multiples of a particular element $a$ form an ideal called the principal ideal generated by $a$. An element $b$ of $R$ is in this ideal if and only if $b$ is a multiple of $a$, which is to say, if and only if $a$ divides $b$ in $R$.

There are several notations for this principal ideal:

$(a) = aR = Ra = \{ra|r\in R\}$.

Principal ideals are the most basic kind of ideals, but they are not the only kind. As with most of mathematics, most of the interesting stuff happens when we go beyond the basics. Examples of ideals that are not principal will be tackled in future posts. For now, we will mention and take note of the fact that in the ring $\mathbb{Z}$, every ideal is principal. We will discuss in this post two more special cases of ideals, maximal ideals and prime ideals, and we show concrete examples in the ring $\mathbb{Z}$.

Ideals are sets, and one of the most basic concepts involving sets is when one set is a subset of another. A set $B$ is said to be a subset of another set $A$ when every element of $B$ is also an element of $A$. In symbols we write $B \subseteq A$ or $A \supseteq B$. For example, we can say that the set of all reptiles are a subset of the set of all animals, since every reptile is also an animal. We have been using this concept implicitly (hoping that intuition will carry the day) in the definition of ideals of a ring.

When all elements of $A$ are also elements of $B$, and at the same time all elements of $B$ are also elements of $A$, then we may say that $A$ and $B$ have the same elements, and that $A$=$B$. In more symbolic notation, we say that whenever both $B \subseteq A$ and $A \subseteq B$, then $A=B$. Whenever we can write $B \subseteq A$ but not $A = B$, then we can write $B \subset A$ and say that $B$ is a proper subset of $A$. Intuitively we can think of $A$ as containing $B$ and being bigger than $B$.

We now show examples of ideals containing other ideals. Let us stay in the ring of integers $\mathbb{Z}$. Consider the ideal made up of integer multiples of $10$, i.e. $(10)$. Its elements include

$\displaystyle ..., -40, -30, -20, -10, 0, 10 , 20, 30, 40, ...$

Consider also the ideal $(5)$. Its elements include

$\displaystyle ...,-20, -15, -10, -5, 0, 5 , 10, 15, 20, ...$

Not all elements are displayed (since there are an infinite number of them) but one can see that every element of $(10)$ is also an element of $(5)$. We can therefore say that $(10) \subseteq (5)$. But since not all elements of $(5)$ are elements of $(10)$, we cannot also write $(10)=(5)$. Instead we say that $(10) \subset (5)$ . In other words, $(10)$  is a proper subset of $(5)$.

A maximal ideal is an ideal that is a proper subset of the unit ideal $(1)$ (the entire ring itself) but is not a proper subset of any other ideal. So the ideal $(10)$ is not maximal, because it is a proper subset of the ideal $(5)$$(5)$, however, is maximal. And so are all the principal ideals of $\mathbb{Z}$ of the form $(p)$ where $p$ is a prime number. Any other ideal that is not of this form, for example $(10)$, $(24)$, or $(25)$ is a proper subset of an ideal of this form. One may now see some sort of an analogy with prime and composite numbers, since any natural number greater than $1$ is either a prime or a product of primes.

Speaking of primes, a prime ideal is an ideal, once again not the unit ideal $(1)$ (therefore also a proper subset of the unit ideal), which we will now define. Let $a$ and $b$ be elements of the ring $R$. Let their product, the element $ab$, be one of the elements of the ideal $I$. If this fact ensures that either $a$ or $b$ is also an element of the ideal the ideal $I$, then we say that $I$ is a prime ideal.

Still staying in the ring $\mathbb{Z}$, the ideal $(10)$ is not a prime ideal since $10$ is the product of $2$ and $5$ and yet neither $2$ nor $5$ is an element of $(10)$. Once again, principal ideals of the form $(p)$ where $p$ is a prime number, are prime ideals, because the only factors of $p$ are $1$ and itself, and while $1$ cannot be in $(p)$ (or else it would be the unit ideal), certainly $p$ is in $(p)$. Therefore all maximal ideals in $\mathbb{Z}$ are also prime ideals.

There is, however, one prime ideal in $\mathbb{Z}$ which is not a maximal ideal. It is the zero ideal $(0)$. It is a prime ideal since the only factor of $0$ is itself, and it is certainly an element (in fact the only element) of $(0)$. It is not, however, a maximal ideal since its only element, the integer $0$, is also an element of every other ideal in $\mathbb{Z}$; therefore, the ideal $(0)$ is a proper subset of every other ideal and is not maximal. The prime ideals in $\mathbb{Z}$ are those of the form $(p)$ where $p$ is a prime number, i.e. the maximal ideals, as well as the zero ideal $(0)$.

References:

Ideal in Wikipedia

Maximal Ideal in Wikipedia

Prime Ideal in Wikipedia

Algebra by Michael Artin