Prime Ideals

Definition and characterisations

Definition (Prime ideal)

Let \(R\) be a ring and \(I\) an \(R\)-ideal. We say that \(I\) is prime if for all \(x, y \in R\), if \(x y \in I\) then \(x \in I\) or \(y \in I\).

Notation (The prime spectrum)

We denote by \(\operatorname{Spec}(R)\) the set of prime ideals of \(R\).

Proposition (Characterisation of primality via products of ideals)

Let \(R\) be a ring and let \(I\) be an \(R\)-ideal. The following are equivalent:

1. \(I\) is prime;
2. for all ideals \(J_{1}, J_{2}\) of \(R\): if \(J_{1} J_{2} \subseteq I\), then \(J_{1} \subseteq I\) or \(J_{2} \subseteq I\).

Proof: Exercise.

Proposition (The zero ideal is prime if and only if the ring is an integral domain)

Let \(R\) be a commutative ring with \(R \neq \{0\}\). Then \(R\) is an integral domain if and only if \(\{0\}\) is prime.

Proof: “\(\Rightarrow\)”: Let \(x, y \in R\) be such that \(x \cdot y \in \{0\}\), that is, \(x \cdot y = 0\). Since \(R\) is an integral domain, it has no zero divisors, and hence \(x = 0\) or \(y = 0\); that is, \(x \in \{0\}\) or \(y \in \{0\}\). We conclude that \(\{0\}\) is prime.

“\(\Leftarrow\)”: We argue by contraposition. If \(R\) is not an integral domain, then, since \(R\) is commutative, the only possibility is that it has zero divisors: there exist \(x, y \in R\) such that \(x \neq 0\), \(y \neq 0\), but \(x \cdot y = 0\). Then \(x \cdot y \in \{0\}\), but \(x \notin \{0\}\) and \(y \notin \{0\}\), so that \(\{0\}\) is not prime.

Maximal ideals, primes and the nilradical

Proposition (Every maximal ideal is prime)

Let \(R\) be a commutative ring with \(R \neq \{0\}\). Then every maximal ideal is prime. In particular, \(\operatorname{Spec}(R) \neq \emptyset\).

Proof: Let \(\mathfrak{m}\) be a maximal ideal, and let \(x, y \in R\) be such that \(x \cdot y \in \mathfrak{m}\). Suppose, for a contradiction, that \(x \notin \mathfrak{m}\) and \(y \notin \mathfrak{m}\). Consider

\[(x) + \mathfrak{m} = \{ r \cdot x + m :\ r \in R,\ m \in \mathfrak{m} \} \subseteq R,\]

which is an ideal, being a sum of ideals. Note that \(\mathfrak{m} \subseteq (x) + \mathfrak{m}\), and the inclusion is strict: \(x = 1 \cdot x + 0 \in (x) + \mathfrak{m}\), but \(x \notin \mathfrak{m}\). That is, \(\mathfrak{m} \subsetneq (x) + \mathfrak{m}\), and analogously \(\mathfrak{m} \subsetneq (y) + \mathfrak{m}\).

Since \(\mathfrak{m}\) is maximal and both are ideals properly containing \(\mathfrak{m}\), we have that

\[(x) + \mathfrak{m} = R = (y) + \mathfrak{m}.\]

Since \(1 \in R\), there exist \(m_{1}, m_{2} \in \mathfrak{m}\) and \(r_{1}, r_{2} \in R\) such that

\[1 = r_{1} x + m_{1}, \qquad 1 = r_{2} y + m_{2}.\]

Multiplying the two equalities,

\[1 = (r_{1} x + m_{1})(r_{2} y + m_{2}) = \underbrace{r_{1} r_{2}\, x y}_{\in\, \mathfrak{m}} + \underbrace{r_{1} x\, m_{2}}_{\in\, \mathfrak{m}} + \underbrace{m_{1}\, r_{2} y}_{\in\, \mathfrak{m}} + \underbrace{m_{1} m_{2}}_{\in\, \mathfrak{m}} \in \mathfrak{m},\]

where the first summand lies in \(\mathfrak{m}\) because \(x y \in \mathfrak{m}\) and \(\mathfrak{m}\) absorbs products, and the remaining ones by direct absorption. Then \(1 \in \mathfrak{m}\), whence \(\mathfrak{m} = R\), which contradicts the definition of a maximal ideal. We conclude that \(x \in \mathfrak{m}\) or \(y \in \mathfrak{m}\), that is, \(\mathfrak{m}\) is prime.

Finally, by the corollary to Krull’s theorem, \(\operatorname{mSpec}(R) \neq \emptyset\), and since every maximal ideal is prime, \(\operatorname{Spec}(R) \supseteq \operatorname{mSpec}(R) \neq \emptyset\).

Example (Well-known prime spectra)

We compute \(\operatorname{Spec}\) in the basic examples:

1. Let \(K\) be a field. Then \(\operatorname{mSpec}(K) = \operatorname{Spec}(K) = \{ \{0\} \}\).

2. Let \(R = \mathbb{Z}\). Then

   \[\operatorname{mSpec}(\mathbb{Z}) = \{ p\mathbb{Z} :\ p \text{ prime} \}, \qquad \operatorname{Spec}(\mathbb{Z}) = \operatorname{mSpec}(\mathbb{Z}) \cup \{ \{0\} \}.\]

3. Let \(R = \mathbb{R}[x]\). The ideal \(\langle x^{2} - 1 \rangle\) is not prime:

   \[x^{2} - 1 = (x - 1)(x + 1) \in \langle x^{2} - 1 \rangle,\]

   but (exercise) \((x - 1), (x + 1) \notin \langle x^{2} - 1 \rangle\).

Theorem (The nilradical is the intersection of the prime ideals)

Let \(R\) be a commutative ring. Then

\[\bigcap_{I \text{ prime}} I = \operatorname{Nil}(R) = \{ x \in R :\ \exists n \in \mathbb{N}\ (x^{n} = 0) \}.\]

Proof: Exercise.