Ideals, quotient rings, prime & maximal ideals, CRT

In one sentence

An ideal is prime when the quotient is a domain and maximal when the quotient is a field, so maximal implies prime and the converse fails exactly where a domain falls short of being a field.

Why the exam asks it

Because the definitions of prime and maximal look unrelated — one is about products of elements, the other about there being nothing in between — and the quotient reformulation makes both mechanical. Once you compute $R/I$, the answer is visible.

The idea

An ideal $I \subseteq R$ absorbs multiplication: $rI \subseteq I$ for every $r \in R$. Ideals are exactly the kernels of ring homomorphisms, and $R/I$ is a ring.

Prime: $ab \in P \Rightarrow a \in P$ or $b \in P$, with $P \ne R$.

Maximal: no ideal strictly between $M$ and $R$.

$P$ is prime $\iff$ $R/P$ is an integral domain. $M$ is maximal $\iff$ $R/M$ is a field.

Since every field is a domain, maximal $\Rightarrow$ prime. The converse fails whenever the quotient is a domain but not a field, and the two standard witnesses are worth memorising:

• $(0) \subset \mathbb{Z}$: quotient $\mathbb{Z}$, a domain, not a field. Prime, not maximal.
• $(x) \subset \mathbb{Z}[x]$: quotient $\mathbb{Z}$ again. Prime, not maximal. The maximal ideal above it is $(2, x)$, with quotient $\mathbb{F}_2$.

That last example also shows $\mathbb{Z}[x]$ is not a PID: $(2,x)$ needs two generators.

In a PID, every nonzero prime ideal is maximal. So in $\mathbb{Z}$ or $K[x]$ the distinction collapses except at $(0)$, which is why it is easy to forget it exists.

Computing quotients

The technique that answers most questions: $R/I$ is "$R$ with the elements of $I$ set to zero".

• $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$ — kill $x$.
• $\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$ — make $x$ a square root of $-1$.
• $\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$, a field, so $(x^2+1)$ is maximal in $\mathbb{R}[x]$.
• $\mathbb{F}_p[x]/(f)$ is a field of $p^{\deg f}$ elements exactly when $f$ is irreducible.

For $K[x]$ with $K$ a field: $(f)$ is maximal $\iff$ $(f)$ is prime $\iff$ $f$ is irreducible.

Chinese remainder theorem

If $I + J = R$ (the ideals are comaximal), then

$$R/(I \cap J) \cong R/I \times R/J .$$

In $\mathbb{Z}$, $(m)$ and $(n)$ are comaximal exactly when $\gcd(m,n) = 1$, giving $\mathbb{Z}/mn \cong \mathbb{Z}/m \times \mathbb{Z}/n$. Comaximality is the hypothesis: $\mathbb{Z}/8 \not\cong \mathbb{Z}/2 \times \mathbb{Z}/4$.

Counting units on both sides gives $\varphi(mn) = \varphi(m)\varphi(n)$ for coprime $m,n$, which is where multiplicativity of Euler's function comes from.

Where intuition breaks

"Prime implies maximal." $(0)$ in $\mathbb{Z}$, and $(x)$ in $\mathbb{Z}[x]$. True in a PID for nonzero ideals, and the qualifier is what options remove.

"$(0)$ is not an ideal / not prime." It is an ideal, and it is prime exactly when $R$ is a domain. Recognising $(0)$ as prime in $\mathbb{Z}$ is often the whole question.

"Every ideal is principal." $(2,x)$ in $\mathbb{Z}[x]$. Only PIDs have that property, and $\mathbb{Z}[x]$ is not one even though $\mathbb{Z}$ is.

"$R/I$ is a field, so $I$ is prime but maybe not maximal." A field quotient is exactly maximality. The implication is an equivalence in both rows.

"CRT applies to any two ideals." Comaximality is required. Without it, $R/(I\cap J)$ can be much bigger than the product.

"$\mathbb{Z}/n$ is a field for every $n$." Only for prime $n$. For composite $n$ there are zero divisors, so it is not even a domain.

"An ideal containing a unit is still a proper ideal." It is all of $R$. That is why maximal ideals never contain units, and why "proper" is in the definition of prime.

The exam's angle

1. Compute the quotient. Domain means prime, field means maximal; nothing else is needed.
2. In $K[x]$, irreducibility is the whole story — $(f)$ maximal $\iff$ $f$ irreducible.
3. Keep $(0)$ and $(x) \subset \mathbb{Z}[x]$ ready as the prime-not-maximal witnesses.
4. For CRT, check comaximality first, which in $\mathbb{Z}$ means checking a gcd.
5. To show an ideal is not principal, count generators needed — $(2,x)$ is the model argument.

The night before

• Prime $\iff$ $R/P$ a domain. Maximal $\iff$ $R/M$ a field. Maximal $\Rightarrow$ prime.
• Converse fails: $(0)\subset\mathbb{Z}$ and $(x)\subset\mathbb{Z}[x]$, both prime, neither maximal.
• In a PID, nonzero prime $\iff$ maximal.
• $(2,x)$ is maximal in $\mathbb{Z}[x]$ with quotient $\mathbb{F}_2$, and is not principal — so $\mathbb{Z}[x]$ is not a PID.
• $K[x]$: $(f)$ maximal $\iff$ $f$ irreducible. $\mathbb{F}_p[x]/(f)$ has $p^{\deg f}$ elements.
• CRT needs $I + J = R$. In $\mathbb{Z}$: $\gcd(m,n)=1$. Hence $\varphi(mn) = \varphi(m)\varphi(n)$.
• $\mathbb{Z}/n$ is a field $\iff$ $n$ prime; a domain under the same condition.