Which quotient rings are integral domains, and which are fields? The answer depends only on the ideal, and it is the bridge from polynomials to field extensions.
Definitions. In a commutative ring $R$ with identity, a proper ideal $P \ne R$ is prime if $ab \in P$ implies $a \in P$ or $b \in P$. A proper ideal $M$ is maximal if the only ideals containing $M$ are $M$ and $R$. In $\mathbb{Z}$, $(p)$ is prime exactly when $p$ is prime (this is "a prime dividing a product divides a factor"), and $(0)$ is prime too.
Why it works
Proof of (1). $(a + P)(b + P) = 0 + P$ means $ab \in P$; "no zero divisors in $R/P$" says this forces $a \in P$ or $b \in P$, which is the definition of prime.
Proof of (2). Suppose $M$ is maximal and $a \notin M$. The ideal $M + (a) = \{m + ra\}$ is bigger than $M$, so it is $R$, and $1 = m + ra$ for some $m \in M$, $r \in R$. Then $(r + M)(a + M) = 1 + M$: every nonzero coset has an inverse. Conversely, if $R/M$ is a field and $J \supsetneq M$ is an ideal, pick $a \in J \setminus M$; then $a + M$ has an inverse $r + M$, so $1 - ra \in M \subseteq J$, hence $1 \in J$ and $J = R$. $\square$
| Ring | Ideal | Quotient | Prime? | Maximal? |
|---|---|---|---|---|
| $\mathbb{Z}$ | $(5)$ | $\mathbb{Z}_{5}$ | yes | yes |
| $\mathbb{Z}$ | $(6)$ | $\mathbb{Z}_{6}$ | no | no |
| $\mathbb{Z}$ | $(0)$ | $\mathbb{Z}$ | yes | no |
| $\mathbb{Z}[x]$ | $(x)$ | $\mathbb{Z}$ | yes | no |
| $\mathbb{Z}[x]$ | $(2, x)$ | $\mathbb{Z}_{2}$ | yes | yes |
| $\mathbb{R}[x]$ | $(x^{2} + 1)$ | $\mathbb{C}$ | yes | yes |
In $F[x]$. Every ideal is $(f)$ for one polynomial (the division algorithm gives this, as for $\mathbb{Z}$), and $(f)$ is maximal exactly when $f$ is irreducible. So $F[x]/(f)$ is a field exactly when $f$ is irreducible: this is how we will build new fields.
Worked example: Maximal ideals of $\mathbb{Z}_{n}$
- The ideals are $(d)$ for $d \mid 12$, with $\mathbb{Z}_{12}/(d) \cong \mathbb{Z}_{d}$.
- $\mathbb{Z}_{d}$ is a field exactly when $d$ is prime: $d = 2, 3$.
Answer: $(2)$ and $(3)$.
Worked example: A polynomial ideal
- Over $\mathbb{Z}_{3}$: $0^{2} + 1 = 1$, $1^{2} + 1 = 2$, $2^{2} + 1 = 5 \equiv 2$: no root, and a quadratic with no root is irreducible. Maximal.
- Over $\mathbb{Z}_{5}$: $2^{2} + 1 = 5 \equiv 0$, so $x - 2$ divides it. Not maximal (not even prime).
Answer: Yes in $\mathbb{Z}_{3}[x]$; no in $\mathbb{Z}_{5}[x]$.
Common mistake
Key idea