Maximal and Prime Ideals

*Note this article assumes that the reader is somewhat familiar with rings and ideals. If you aren’t, feel free to continue reading. I hope that it will be mostly understandable!

Today, we will learn about two types of ideals. These will help us answer the question,

What property does an ideal $I\subseteq R$ need to have for the factor ring $R/I$ to be an integral domain or field?

Before we get into the details. First, let’s recall that an ideal, $I,$ in a ring $R$ is a subring with the property (that my professor called) the absorption property.

Absorption Property of Ideals: for all $r\in R$ and $a\in I,$ we have both $ar, ra \in I.$

We will restrict our attention to commutative rings with unity. This means that we don’t need to worry about left or right ideals, since every ideal is two-sided. If you are not sure what is meant by left ideal or right ideal, then you can safely ignore this remark and simply take the following as the definition of an ideal.

Definition: (Ideal) Let $R$ be a commutative ring with unity. We call $I\subseteq R$ an ideal of $R$ if

$I$ is a subring of $R,$ and

if $I$ has the absorbing property above.

We call $I$ a proper ideal if $I$ is a proper subset of $R.$

As we will see, there are two types of ideals, called maximal and prime ideals, that will help us determine whether $R/I$ is an integral domain or a field. At first, these concepts might feel strange and unmotivated; however, I hope that after this article and after you spend some time digesting the information here, you might gain some intuition for what we talked about.

What are Maximal Ideals

Let’s begin by defining what a maximal ideal is:

Definition: (Maximal Ideal) Let $R$ be a commutative ring with unity and $I$ be a proper ideal of $R.$ We call $I$ a maximal ideal of $R$ if $I$ is not properly contained inside of another proper ideal. I.e., $I$ is a maximal ideal of $R$ if $I$ has the property:

if $J$ is an ideal of $R$ such that $I\subseteq J,$ then $I = J$ or $R = J.$

In general, we will tend to denote a maximal ideal by $M.$

You might ask, “Why should we care about maximal ideals?” And this is a good question! The answer is found in the next theorem!

Theorem ( $M =\mathrm{Maximal}\;\iff\;R/M = \mathrm{Field}$ ): Let $R$ be a commutative ring with unity, and let $M$ be an ideal of $R.$ Then,

$M$ is a maximal ideal of $R$ if and only if the factor ring $R/M$ is a field.

Proof: (Click in the Discovery)

Let $R$ be a commutative ring with unity, and $M$ be an ideal. We have two directions to prove; the forward direction takes the most effort, so we will do that one first!

Forward: $M =\mathrm{Maximal}\;\implies\;R/M = \mathrm{Field}$

Key Idea: Our goal is to prove that all nonzero elements $r_0+M\in R/M$ have an inverse. The key idea here is to properly enclose $M$ in a larger ideal $I$ where the elements of $I$ will somehow imply that each nonzero element of the factor ring $r_0+M\in R/M$ has an inverse. With this in mind, note that $r_0+M\in R/M$ is nonzero iff $r_0\notin M.$ Furthermore, if $r_0+M$ has an inverse, denoted $s_0+M\in R/M,$ then,

(r_0+M)(s_0+M) = r_0s_0+M = 1+M \;\;\;\;\implies \;\;\;\; r_0s_0 -1\in M \;\;\;\;\implies \;\;\;\; 1 = r_0s_0 +m\;\;\mathrm{for}\;m\in M.

This is our key insight.

Beginning of Forward Proof: Let $M$ be a maximal ideal of $R.$ Let $r_0+M\in R/M$ be a nonzero element of the factor ring. We will demonstrate that $r_0+M$ has an inverse in $R/M.$ Consider the set $I := \{sr_0+ m\;:\;s\in R \;\;\And \;\; m\in M \}.$ First, note that $M$ is a proper subset of $I.$ Consequently, if we can show that $I$ is an ideal of $R,$ then we can conclude that $I=R$ since $M$ is maximal.

With this goal in mind, we will show that (i) $I \neq \emptyset,$ (ii) $I$ is closed under subtraction, and (iii) $I$ has the absorbing property. These three conditions will prove that $I$ is an ideal of $R.$ This is because we need to show that $I$ is a subring of $R$ and has the absorbing property.

(i) The set $I$ is nonempty since $0 = 0r_0+0 \in I.$

(ii) Let $sr_0+m\in I$ and $qr_0+n\in I.$ Then, $(sr_0+m) – (qr_0+n) = (s-q)r_0 +(m-n)\in I$ since $s-q\in R$ and $m-n\in I.$

(iii) Let $sr_0+m\in I$ and $r\in R.$ It follows, $r(sr_0+m) = (rs)r_0 + (rm)\in I$ since $rs\in R$ and $rm\in M$ because $M$ is an ideal with the absorbing property.

We have shown that $I$ is an ideal that properly contains $M.$ It follows from the maximality of $M$ that $I = R.$ Therefore, $1\in I$ and we deduce that $1 = s_0r_0 + m$ for some $s_0\in R$ and $m\in M.$ Thus, $s_0+M\in R/M$ is the inverse of $r_0+M\in R/M$ by our key idea.

$\square_{\mathrm{Forward}}$

Backward: $R/M = \mathrm{Field}\;\implies\;M =\mathrm{Maximal}$

Beginning of Backward Proof: Let $R/M$ be a field.

We aim to show that $M$ is maximal. To this end, we must show that any ideal that properly contains $M$ is in fact $R$ itself. We begin by showing that $M$ is a proper ideal of $R.$

Since $R/M$ is a field, it contains at least two distinct elements: $0+M$ and $1+M.$ Therefore, $1\notin M$ and thus $M$ is a proper subset of $R.$ Next, let $I$ be an ideal of $R$ that properly contains $M.$ It follows that there is some $i\in I$ such that $i\notin M.$ We will show that $I = R.$

Since $i\notin M,$ it follows that $i+M \neq 0+M$ and therefore has an inverse in $R/M.$ Let the inverse of $i+M$ be denoted $j+M.$ We therefore have,

(i+M)(j+M) = ij +M = 1+M \;\;\;\;\implies\;\;\;\; \exists m\in M \;\;\mathrm{such}\;\;\mathrm{that}\;\; ij-1 = m.

In particular, we have $ij+m = 1$ and thus, $1\in I$ since $ij ,m\in M \subset I.$ But, since $I$ is ideal containing 1, we have $I = R.$ This concludes the proof.

$\square_{\mathrm{Backward}}$

This theorem is quite nice! Or, at least I think it is. Let’s look at an example to help out. I will put the examples in some green boxes so that they don’t clutter this article. Also, a forewarning: there will be some facts that we motivate but do not prove; this is not to divert attention from what the example is meant to highlight.

Example: (Click in the Discovery)

Let’s work with the commutative ring with unity $\Z.$ First, we remark that $\Z$ is a principal ideal domain. That is, every ideal in $\Z$ is a principal ideal. And, a principal ideal of a ring $R$ is an ideal of the form $\langle a\rangle = \{ar\;:\;r\in R\}.$ In our situation, this means that every ideal of $\Z$ will be of the form, $n\Z\subset \Z$ where $n\in \Z.$ I.e., the only ideals of $\Z$ are of the form,

n\Z =\{0,\;n,\;-n,\;2n,\;-2n,\;3n,\;-3n,\cdots\}.

I encourage you to try to prove this on your own. As a hint, you might need the division algorithm.

We must find a maximal ideal. How might we go about this? Well, let’s just take a random integer $6\in \Z$ and consider the ideal $6\Z.$ Is $6\Z$ a maximal ideal of $\Z?$ Well, if $6\Z\subset n\Z$ for some other $n\in \Z,$ then all multiples of $6$ will be contained in $n\Z.$ In particular, multiples of $n$ contain the multiples of $6.$ For example, $6\Z\subset 2\Z.$

6\Z =\{0,\;6,\;-6,\;12,\;-12,\cdots\} \subset 2\Z =\{0,\;2,\;-2,\;4,\;-4,\;6,\;-6,\;\cdots\} .

Also, $6\Z \subset 3\Z,$

6\Z =\{0,\;6,\;-6,\;12,\;-12,\cdots\} \subset 3\Z =\{0,\;3,\;-3,\;6,\;-6,\;9,\;-9,\;\cdots\} .

So $6\Z$ is not a maximal ideal of $\Z$ since $6\Z\subset 2\Z \subset \Z.$ But we might notice/conjecture that $2\Z$ or $3\Z$ are maximal ideals, since 2 and 3 are prime numbers and therefore there are no smaller numbers (other than 1) whose multiples will contain the multiples of 2 or 3. More generally, we conjecture $n\Z\subset m\Z$ iff $m\mid n.$ Consequently, we conjecture $n\Z$ is a maximal ideal of $\Z$ iff $n$ is prime. Indeed, this is the case! I also encourage you to prove these facts as well! You’re welcome for all these fun facts for you to try and prove.

Thus, we conclude that $n\Z$ is maximal if and only if $n$ is prime. Thus, let $p$ be some prime number and let $p\Z$ be a maximal ideal of $\Z.$ Our theorem implies that $\Z/p\Z$ is a field. Indeed! As we might know $\Z/p\Z$ contains all the residue classes modulo $p,$ and we might have known that every nonzero residue class has an additive inverse, multiplicative inverse, and there is an additive identity, multiplicative identity, etc., since $p$ is a prime. Making $\Z/p\Z$ a field. How amazing!

$\blacksquare$

What are Prime Ideals

I absolutely love the concept of a prime ideal! I find it to be an absolutely beautiful construction when you generalize the concept of prime numbers using prime ideals. Here, however, we will not discuss that. We will discuss a nice property that prime ideals do have, as a hint, it has to do with their factor rings!

Definition (Prime Ideals): Let $R$ be a commutative ring with unity and $I$ be a proper ideal of $R.$ We call $I$ a prime ideal of $R$ if for all $ab\in I$ we have either $a\in I$ or $b\in I.$

i.e., $ab\in I \implies a\in I \;\mathrm{or}\;b\in I.$

As a brief and perhaps confusing remark, the reason why we call this type of ideal prime follows from the following property that prime numbers have: If $p\in \N$ is prime, then

$p\mid ab\implies p\mid a \;\mathrm{or}\;p\mid b.$

This is known as Euclid’s Lemma. We will use this fact in our example later.

The main reason we care about prime ideals, at least for the purposes of this article, follows from the following theorem. We will use $P$ to denote a prime ideal.

Theorem ( $P = \mathrm{Prime} \;\iff\; R/P = \mathrm{Integral}\;\mathrm{Domain}$ ): Let $R$ be a commutative ring with unity and $P$ be a proper ideal of $R.$ Then,

$P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.

Proof: (Click in the Discovery)

Let $R$ be a commutative ring with unity, and $P$ be a proper ideal of $R.$

Forward: $P = \mathrm{Prime} \;\implies\; R/P = \mathrm{Integral}\;\mathrm{Domain}$

Let $P$ be a prime ideal. We are aiming to prove that $R/P$ is an integral domain, that is, if $(r+P)(s+P) = 0+P,$ then $r+P = 0+P$ or $s+P = 0+P.$ I.e., that either $r\in P$ or $s\in P .$ To this end, suppose that $(r+P)(s+P) = 0+P,$ it follows, $rs+P = 0+P,$ and thus $rs\in P.$ But, since $P$ is prime, it follows that $r\in P$ or $s\in P .$ Consquently, $r+P = 0+P$ or $s+P = 0+P.$

Backward: $R/P = \mathrm{Integral}\;\mathrm{Domain} \;\implies\;P = \mathrm{Prime}$

Let $R/P$ be an integral domain. Since we are aiming to show that $P$ is prime, consider some $ab\in P.$ It follows that $ab+P = (a+P)(b+P) = 0+P.$ And, since $R/P$ is an integral domain, we either have $a+P = 0+P$ or $b+P = 0+P.$ But, this is equivalent to the statement $a\in I$ or $b\in I.$ Therefore, $P$ is prime.

$\square$

Again, how remarkable! Let’s go through an example.

Example: (Click in the Discovery)

Our example will again use the ring of integers $\Z.$ However, we already know that all ideals of $\Z$ are of the form $n\Z.$ So, the question is: for what $n\in\Z$ is $n\Z$ a prime ideal? As the question suggests, probably when $n$ is a prime. Indeed, this is the case. Let’s see why.

Let $p\in \Z$ be a prime and consider the ideal $p\Z.$ We want to show that for any $ab\in p\Z,$ it must be the case that $a\in p\Z$ or $b\in p\Z.$ Indeed, since $ab\in p\Z$ implies that $ab$ is a multiple of $p,$ and thus $p\mid ab$ for some $k\in\Z.$ Recall that primes have the property that $p\mid ab$ implies $p\mid a$ or $p\mid b.$ Let’s suppose that $p\mid a.$ Then, we see that $a\in p\Z.$

Great! Now we know that $p\Z$ is a prime ideal. Therefore, we see that $\Z/p\Z$ is an integral domain. However, this isn’t a surprise since we already knew that $\Z/p\Z$ is a field.

$\blacksquare$

Cooooool Corollary

As we might have noticed in our examples, once we see that a particular ideal is maximal, we know it is prime too.

Corollary ( $\mathrm{Maximal}\;\implies \mathrm{Prime}$ ): Let $R$ be a commutative ring with unity. If $I$ is a maximal ideal of $R$ then, $I$ is a prime ideal of $R.$

Proof: (Click in the Discovery)

Let $R$ be a commutative ring with unity, and $I$ be a maximal ideal. Then,

$I=\mathrm{Maximal} \;\iff\;R/I = \mathrm{Field} \implies R/I = \mathrm{Integral}\;\mathrm{Domain} \iff I = \mathrm{Prime}.$

$\square$

What is the ideal sponge? One that has the absorbing property!

I hope that this article taught you something or entertained you (somehow). Ideals are incredibly important, and knowing when an ideal is maximal, or prime, is very useful. For example, if you continue learning about rings and fields, you may come across polynomial rings. For example, let $\mathbb{F}$ be a field, then we call the following a polynomial ring,

\mathbb{F}[x] = \{a_0 + a_1x+\cdots + a_nx^n\;:\;a_i\in \mathbb{F}\}.

That is, $\mathbb{F}[x]$ is a ring whose elements are polynomials with coefficients in $\mathbb{F}.$ For example, consider the field of real numbers $\R$ and the ring of polynomials with real coefficients $\R[x].$ One aspect of algebra is to know when there is a root of a particular polynomial. For example, $2x-1 \in \R[x]$ has a real root $\frac{1}{2}\in \R.$ However, $x^2+1 \in \R[x]$ does not have a root in $\R.$ We don’t like this one bit! Wait, I hear some of you say, $x^2+1$ does have a root. In fact, it has two complex roots $i,-i\in \mathbb{C}.$ We say that the complex numbers is an extension of the real numbers in such a way as to contain roots of polynomials in $\R[x].$ We can make this more precise, but here we just want motivation.

Let $\mathbb{F}$ be any field and suppose that the polynomial $p(x)\in \mathbb{F}[x]$ doesn’t have a root in $\mathbb{F}.$ Then, as it turns out, we can extend $\mathbb{F}$ to a larger field $\mathbb{E}$ that does contain the roots of $p(x).$ But, in the proof of this fact, we need to be able to conclude that $\mathbb{F}[x]/\langle p(x)\rangle$ is a field. This is done by showing that $\langle p(x)\rangle$ is a maximal ideal of $\mathbb{F}[x].$ So, when you come to this in your studies, you can think back to this article and say that you already knew that!

As always, please let me know if you found any typos in this article or have any constructive criticism that will help me be a better math(s) communicator! Thank you in advance!

Be Kind. Be Curious. Be Compassionate. Be Creative.

And Have Fun!

A Kick in the Discovery