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Two conditions on an ideal, each a statement about the quotient: prime gives an integral domain, maximal gives a field, and in the integers both mean the generator is prime.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to test an ideal for being prime and for being maximal, state the two theorems that turn each condition into a statement about the quotient, say why every maximal ideal is prime and give the counterexample to the converse, count the maximal ideals containing a given one in the integers, and prove that the integers modulo a prime form a field.
Quotienting by an ideal always produces a ring, and rings sit on a ladder from has zero divisors up to field. So the question becomes: which ideals give which rung? Two conditions on the ideal answer it, and both were named for what they do in the integers.
A proper ideal $P$ is prime when $ab \in P$ implies $a \in P$ or $b \in P$. A proper ideal $M$ is maximal when no ideal lies strictly between $M$ and the whole ring. Proper excludes the whole ring in both definitions. In the integers, the ideal generated by $n$ is written $(n)$.
Two definitions, and two theorems that make them worth stating.
Prime. A proper ideal $P$ is prime when $ab \in P$ forces $a \in P$ or $b \in P$.
> $R/P$ is an integral domain exactly when $P$ is prime.
The proof is a translation: $R/P$ has no zero divisors means $(a + P)(b + P) = P$ forces one factor to be $P$, which is the definition of prime written with cosets.
Maximal. A proper ideal $M$ is maximal when no ideal sits strictly between it and $R$.
> $R/M$ is a field exactly when $M$ is maximal (for a commutative ring with $1$).
Again a translation: the ideals of $R/M$ correspond to the ideals of $R$ containing $M$, and a ring is a field exactly when it has only the two trivial ideals.
Maximal implies prime, since every field is a domain. The converse is false in general.
In the integers. $(n)$ is prime exactly when $n$ is prime — which is why the word prime was borrowed. And $(p)$ is then maximal too, because $(p) \subseteq (m)$ forces $m \mid p$, so $m$ is $1$ or $p$. The one exception is the zero ideal: prime, because a product of non-zero integers is non-zero, and not maximal, because $(0) \subset (2) \subset \mathbb{Z}$. Its quotient is $\mathbb{Z}$, a domain and not a field, exactly as the two theorems predict.
Containment reverses divisibility: $(a) \subseteq (b)$ means $b \mid a$. So a bigger ideal has a smaller generator, and the maximal ideals are generated by the primes — the smallest numbers above $1$.
Another way: picture
Picture the ideals of the integers as a family tree with the whole ring at the top and the zero ideal at the bottom, and each ideal joined to those containing it. Because containment reverses divisibility, the tree is the divisibility lattice upside down: $(12)$ sits below $(6)$, which sits below $(2)$ and $(3)$, which sit just below the top. The ideals just below the top are the prime ones, and just below the top is precisely what maximal means.
Another way: steps
To classify an ideal of the integers: 1. Find its generator $n$. 2. If $n = 0$: prime, not maximal; the quotient is the integers. 3. If $n$ is prime: both prime and maximal; the quotient is a field with $n$ elements. 4. If $n$ is composite: neither; the quotient has zero divisors. 5. If $n = 1$: the whole ring, which is excluded from both definitions.
In the integers, and in any principal ideal domain, every non-zero prime ideal is maximal. So the distinction looks like pedantry. It is not, and two examples show why.
The zero ideal. In any integral domain $(0)$ is prime — that is literally what no zero divisors says — and it is maximal only when the ring is already a field. So the distinction exists in the integers, at exactly one ideal.
Polynomials in two variables. In $\mathbb{Q}[x, y]$, the ideal generated by $x$ is prime: the quotient is $\mathbb{Q}[y]$, a domain. It is not maximal, because it sits inside the ideal generated by $x$ and $y$, whose quotient is $\mathbb{Q}$, a field. Here the gap is not an edge case but the general situation, and the chains of prime ideals inside a ring are what measure its dimension. That is the beginning of algebraic geometry, where a prime ideal is a point of a space and a maximal ideal is a point that cannot be refined.
The single-variable case is the one this course needs. In $F[x]$ for a field $F$, every ideal is principal — the division algorithm again — so the non-zero primes are $(p)$ for irreducible $p$, they are all maximal, and
$$F[x]/(p) \text{ is a field exactly when } p \text{ is irreducible}.$$
That single sentence is the construction of every finite field, and unit 6 spends its last two lessons on it. The analogy is worth stating once: irreducible polynomials are to $F[x]$ what primes are to $\mathbb{Z}$, and both give fields for the same reason.
*Thinking prime means the generator has no factors in the ideal.* The condition is about products landing in the ideal: $ab \in P$ forces $a \in P$ or $b \in P$.
Thinking maximal means largest. The whole ring is larger than every proper ideal and is excluded by definition. Maximal means largest among the proper ones.
Assuming prime implies maximal. True in the integers for non-zero ideals, false in general, and false for the zero ideal even there.
Forgetting the zero ideal is prime. In any integral domain it is. It catches people because its quotient is the whole ring back again.
Expecting the theorems without commutativity and unity. The result that maximal gives a field needs a commutative ring with $1$. Every ring in this course has both.
Is $(6)$ prime? Test the definition: $2 \times 3 = 6 \in (6)$, but neither $2$ nor $3$ is a multiple of $6$.
The definition, tested directly.
So $(6)$ is not prime, and the quotient $\mathbb{Z}_6$ should have zero divisors — as it does: $2 \times 3 = 0$ there.
The theorem, confirmed.
Nor is it maximal: $(6) \subset (2) \subset \mathbb{Z}$, with $(2)$ strictly between.
Not maximal either.
Is $(0)$ prime? $ab = 0$ in the integers forces $a = 0$ or $b = 0$, which is exactly the condition. So yes.
Prime.
Is it maximal? No: $(0) \subset (2) \subset \mathbb{Z}$, and $(2)$ is a proper ideal strictly containing it.
Not maximal.
The quotient is $\mathbb{Z}/(0) \cong \mathbb{Z}$, an integral domain that is not a field — exactly what the two theorems together predict.
Both theorems, agreeing.
The quotient sends each polynomial to its constant term, so it is isomorphic to the rationals.
Identify the quotient first.
The rationals are a field, so the ideal is maximal — and every maximal ideal is prime.
Both, by the two theorems.
In one variable the two conditions agree for every non-zero ideal. In two variables they would not: the ideal generated by $x$ inside $\mathbb{Q}[x, y]$ is prime and not maximal.
Match each ideal of the integers to a description of its quotient.
| a field with five elements | a ring with zero divisors | the integers again | a ring with a single element | an infinite field | |
|---|---|---|---|---|---|
| the multiples of $5$ | |||||
| the multiples of $6$ | |||||
| the zero ideal | |||||
| the whole ring |
For each of these quotients of the integers, fill in how many elements it has and how many of them are units.
| Elements | Units | |
|---|---|---|
| by the multiples of 5 | 5 | |
| by the multiples of 6 | 6 | |
| by the multiples of 7 | 7 | |
| by the multiples of 8 | 8 |
Build the proof that the integers modulo a prime $p$ form a field.
This task has no paper form; do it on a device.
Quotient the integers by the multiples of $13$. What do you get?
In the integers, how many maximal ideals contain the ideal generated by $49$?
Answer:
Select every statement about prime and maximal ideals that is true.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Quotient the integers by the multiples of $4$. What do you get?
You can classify an ideal as prime, maximal or neither and say what its quotient is. Say in your own words why every maximal ideal is prime, and name a prime ideal that is not maximal. Next: polynomial rings, where irreducible polynomials play the part the primes played here.
9. Your turn: is the ideal generated by $x$ in the polynomials with rational coefficients prime, maximal, or neither?, step 3