Ideals

Definition and examples

Definition (Left, right and two-sided \(R\)-ideal)

Let \(R\) be a ring and \(I \subseteq R\) with \(I \neq \emptyset\). We say that \(I\) is a left \(R\)-ideal (respectively, a right ideal) if:

1. \(\forall x, y \in I\ (x + y \in I)\);
2. \(\forall r \in R\ \forall x \in I\ (r \cdot x \in I)\) (respectively, \(x \cdot r \in I\)); this property is called absorption.

If \(I\) is both a right and a left \(R\)-ideal, we say that \(I\) is an \(R\)-ideal.

Note (First properties of ideals)

Let \(I\) be an \(R\)-ideal (left or right).

1. \((I, +)\) is a group. Note that \(0 \in I\): indeed, since \(0 \in R\) and the proposition on arithmetic rules gives \(0 \cdot x = 0\) for every \(x \in R\), absorption applied to any \(x \in I\) shows that \(0 = 0 \cdot x \in I\).
2. Let \(x \in I\). Then \((-1) \cdot x = -x \in I\).
3. If \(1 \in I\), then \(I = R\), since if \(a \in R\), absorption gives \(a \cdot 1 = a \in I\).

Example (Trivial ideals and the ideals of a field)

The first examples of ideals:

1. If \(R\) is a ring, \(R\) and \(\{0\}\) are ideals; they are called the trivial ideals.

2. Let \(F\) be a field and \(I\) an ideal in \(F\) such that \(I \neq \{0\}\). Let \(a \in I\) with \(a \neq 0\). Then

   \[a^{-1} \in F \implies a^{-1} \cdot a \in I \implies 1 \in I \implies I = F.\]

   We conclude that a field possesses only the two trivial ideals.

Example (Ideals in polynomial rings)

Let \(R\) be a commutative ring.

1. Let \(I \subseteq R[x]\) be such that

   \[I = \{ p \in R[x] :\ \text{the constant term of } p \text{ is zero} \}.\]

   Then \(I\) is an ideal.

2. If \(I\) is an ideal of \(R\), define

   \[I[x] = \{ p \in R[x] :\ p(x) \text{ has all its coefficients in } I \}.\]

   Then \(I[x]\) is an ideal in \(R[x]\).

Example (Ideals in \(\mathbb{Z}\) and in \(\mathbb{Z}_{n}\))

Two families of ideals in the rings of integers:

1. \(n\mathbb{Z}\) is an ideal of \(\mathbb{Z}\).
2. Let \(n > 1\) and consider \(\mathbb{Z}_{n}\). Let \(d \in \mathbb{Z}_{n}\) be such that \(d \mid n\). Then \(\langle d \rangle\) (the additive group generated by \(d\)) is an ideal.

Example (A left ideal that is not a right ideal)

Let \(R = M_{2 \times 2}(\mathbb{R})\) and define

\[I := \left\{ \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix} :\ a, b \in \mathbb{R} \right\}.\]

Note that

\[\begin{aligned} \begin{pmatrix} x & y \\ z & w \end{pmatrix} \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix} &= \begin{pmatrix} xa + yb & 0 \\ za + wb & 0 \end{pmatrix} \in I, \\ \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix} \begin{pmatrix} x & y \\ z & w \end{pmatrix} &= \begin{pmatrix} ax & ay \\ bx & by \end{pmatrix} \notin I \ \text{in general}, \end{aligned}\]

so \(I\) is a left ideal but not a right ideal.

Definition (Nilpotent element and the nilradical)

If \(R\) is a ring, \(a \in R\) is nilpotent if there exists \(n \in \mathbb{N}\) such that \(a^{n} = 0\). The least such \(n\) is the index of nilpotency of \(a\). Moreover,

\[\operatorname{Nil}(R) = \{ x \in R :\ x \text{ is nilpotent} \}\]

is an ideal, called the nilradical of \(R\) (exercise).

Operations with ideals and the generated ideal

Definition (Sum, intersection and product of ideals)

Let \(R\) be a commutative ring and let \(I, J\) be ideals. Define:

\[\begin{aligned} I + J &:= \{ a + b :\ a \in I,\ b \in J \}, \\ I \cap J &:= \{ x :\ x \in I \wedge x \in J \}, \\ I \cdot J &:= \left\{ \sum_{i=1}^{n} a_{i} \cdot b_{i} :\ a_{i} \in I,\ b_{i} \in J,\ n \in \mathbb{N} \right\}. \end{aligned}\]

Theorem (The operations on ideals produce ideals)

\(I + J\), \(I \cap J\) and \(I \cdot J\) are ideals.

Proof: Exercise.

Corollary (The intersection of ideals is an ideal)

The intersection of any family of ideals of \(R\) is an ideal of \(R\).

Proof: Let \(\{I_{\alpha}\}_{\alpha \in \Delta}\) be a family of ideals and \(I = \bigcap_{\alpha \in \Delta} I_{\alpha}\). Since \(0 \in I_{\alpha}\) for every \(\alpha\), we have that \(0 \in I\), and in particular \(I \neq \emptyset\). If \(x, y \in I\), then \(x, y \in I_{\alpha}\) for every \(\alpha\), and since each \(I_{\alpha}\) is closed under sums, \(x + y \in I_{\alpha}\) for every \(\alpha\), that is, \(x + y \in I\). Finally, if \(r \in R\) and \(x \in I\), absorption in each \(I_{\alpha}\) gives \(r x \in I_{\alpha}\) for every \(\alpha\), so that \(r x \in I\) (and analogously \(x r \in I\)). We conclude that \(I\) is an ideal.

Definition (Generated, finitely generated and principal ideal)

Let \(R\) be a commutative ring and \(U \subseteq R\).

1. \((U) := \bigcap_{U \subseteq I,\ I \text{ ideal}} I\) is the ideal generated by \(U\). Note that \((U)\) is an ideal, since the intersection of ideals is an ideal. If \(U\) is finite and \(U = \{u_{1}, \dots, u_{n}\}\), we write \((u_{1}, \dots, u_{n}) = (U)\).
2. We say that an ideal \(I\) is finitely generated if there exists a finite \(U \subseteq R\) such that \(I = (U)\).
3. We say that \(I\) is a principal ideal if there exists \(a \in R\) such that \(I = (a)\). Note that in this case \(I = (a) = \{ r \cdot a :\ r \in R \}\) (exercise).

Theorem (Characterisation of the generated ideal as combinations)

Let \(R\) be a commutative ring and \(\emptyset \neq U \subseteq R\). Then

\[(U) = \underbrace{\left\{ \sum_{i=1}^{n} a_{i} u_{i} :\ n \in \mathbb{N},\ a_{i} \in R,\ u_{i} \in U \right\}}_{J}.\]

Proof: Let us first see that \(J\) is an ideal: the sum of two combinations of the form \(\sum a_{i} u_{i}\) is again a combination of that type, and if \(r \in R\), then \(r \sum_{i=1}^{n} a_{i} u_{i} = \sum_{i=1}^{n} (r a_{i}) u_{i} \in J\), which gives absorption. Moreover \(U \subseteq J\), since \(u = 1 \cdot u\) for each \(u \in U\). Then \(J\) is one of the ideals we intersect in order to define \((U)\), and therefore \((U) \subseteq J\).

Conversely, let \(w \in J\). Then there exist \(n \in \mathbb{N}\), \(a_{i} \in R\) and \(u_{i} \in U\) such that \(w = \sum_{i=1}^{n} a_{i} u_{i}\). Let \(I\) be any ideal such that \(U \subseteq I\). Then, by absorption, \(a_{i} u_{i} \in I\) for every \(i\), and by closure of the sum \(\sum_{i=1}^{n} a_{i} u_{i} \in I\), so that \(w \in I\). Since \(I\) is arbitrary, we have that

\[w \in \bigcap_{U \subseteq I,\ I \text{ ideal}} I = (U),\]

that is, \(J \subseteq (U)\). We conclude the equality.

Example (Computations with generated ideals)

Let \(R\) be a commutative ring.

1. \((0) = \{0\}\) and \((1) = R\). If \(u \in R^{\times}\), then \((u) = R\). In general, if \(u_{1}, \dots, u_{n} \in I\) with \(I\) an ideal, then \((u_{1}, \dots, u_{n}) \subseteq I\).
2. If \(R = \mathbb{Z}\), then \((6, 15) = (3)\): since \(6 = 3 \cdot 2\) and \(15 = 3 \cdot 5\), we have that \(6, 15 \in (3)\) and therefore \((6, 15) \subseteq (3)\). On the other hand, \(3 = 6 \cdot (-2) + 15\), so \(3 \in (6, 15)\) and therefore \((3) \subseteq (6, 15)\).

3. Consider the polynomial ring \(R[x]\). Note that

   \[(x) = \{ x \cdot p(x) :\ p(x) \in R[x] \} = \{ r(x) :\ \text{the constant coefficient of } r(x) \text{ is zero} \}.\]

Definition (Principal ideal ring and PID)

We collect two notions:

1. A principal ideal ring is a ring in which every ideal is principal.
2. A principal ideal domain (PID) is a principal ideal ring that is an integral domain.

Example (\(\mathbb{Z}[x]\) is not a principal ideal ring)

Let \(R = \mathbb{Z}[x]\). One can prove that \((2, x)\) is not principal (exercise). Then \(\mathbb{Z}[x]\) is not a principal ideal ring.

Theorem (\(\mathbb{Z}\) is a PID, and \(K[x]\) is a PID if \(K\) is a field)

\(\mathbb{Z}\) is a principal ideal domain. If \(K\) is a field, then \(K[x]\) is a principal ideal domain.

Proof: We already know that \(\mathbb{Z}\) is an integral domain. Clearly \(\{0\} = (0)\) is principal. Let \(I \neq \{0\}\) be an ideal of \(\mathbb{Z}\). Then there exists \(a \in I\) with \(a \neq 0\); since \(-a \in I\), there is at least one element of \(\mathbb{Z}^{+}\) in \(I\). Let \(b \in I\) be the smallest positive element of \(I\). We shall prove that \(I = (b)\). Clearly \((b) \subseteq I\), by absorption. Let \(a \in I\). By the division algorithm in \(\mathbb{Z}\), there exist \(q, r \in \mathbb{Z}\) such that

\[a = q b + r, \quad \text{with } 0 \leq r < b.\]

Then \(a - q b = r \in I\), since \(a \in I\) and \(q b \in I\). By the minimality of \(b\), necessarily \(r = 0\); then \(a = q b\), so that \(a \in (b)\) and therefore \(I \subseteq (b)\). We conclude that \(I = (b)\).

The proof that \(K[x]\) is a PID if \(K\) is a field is left as an exercise.

Definition (Relatively prime ideals)

Let \(R\) be a commutative ring and let \(I, J\) be ideals. We say that \(I\) and \(J\) are relatively prime if \(I + J = R\).

Proposition (Properties of the sum and the product of ideals)

Let \(R\) be a commutative ring and let \(I, J, K\) be ideals. Then:

1. \(I + J = J + I\) and \(I J = J I\);
2. \(I + (J + K) = (I + J) + K\) and \((I J) K = I (J K)\);
3. \(I (J + K) = I J + I K\);
4. \((0) + I = I\), \(R \cdot I = I\) and \(I + R = R\);
5. \((0) \cdot I = (0)\);
6. \(I J \subseteq I \cap J\).

Proof: For (1): \(I + J = J + I\) is immediate from the commutativity of the sum in \(R\), and \(I J = J I\) because each generator satisfies \(a_{i} b_{i} = b_{i} a_{i}\), as \(R\) is commutative.

For (2): both sides of \(I + (J + K) = (I + J) + K\) are \(\{ a + b + c :\ a \in I,\ b \in J,\ c \in K \}\), by the associativity of the sum. For the product, note that every element of \((I J) K\) is a finite sum \(\sum_{i} w_{i} c_{i}\) with \(w_{i} \in I J\) and \(c_{i} \in K\); writing each \(w_{i}\) as a finite sum of products \(a b\), and distributing, \((I J) K\) is exactly the set of finite sums of triple products \((a b) c\) with \(a \in I\), \(b \in J\), \(c \in K\). The same argument describes \(I (J K)\) as the finite sums of products \(a (b c)\), and the associativity of \(R\) gives the equality.

For (3): “\(\subseteq\)” because each generator satisfies \(a (b + c) = a b + a c \in I J + I K\), and the finite sums of these lie in the ideal \(I J + I K\). “\(\supseteq\)” because \(I J \subseteq I (J + K)\) (every \(b \in J\) is written as \(b = b + 0 \in J + K\)) and \(I K \subseteq I (J + K)\), and the ideal \(I (J + K)\) is closed under sums.

For (4): \((0) + I = I\) since \(0 + a = a\); \(R \cdot I = I\) since “\(\subseteq\)” is absorption and “\(\supseteq\)” follows from \(a = 1 \cdot a\); and \(I + R = R\) since every \(a \in R\) is written as \(a = 0 + a\).

For (5): each generator of \((0) \cdot I\) is of the form \(0 \cdot b = 0\), so \((0) \cdot I = \{0\} = (0)\).

For (6): each generator \(a b\) with \(a \in I\), \(b \in J\) belongs to \(I\) (absorption of \(I\), with \(b \in R\)) and to \(J\) (absorption of \(J\), with \(a \in R\)); since \(I \cap J\) is an ideal, the finite sums of such products lie in \(I \cap J\), that is, \(I J \subseteq I \cap J\).

Maximal ideals and Krull’s theorem

Definition (Maximal ideal and the maximal spectrum)

Let \(R\) be a commutative ring. An ideal \(\mathfrak{m} \neq R\) is maximal if whenever \(J\) is an ideal such that \(\mathfrak{m} \subseteq J\), then \(J = R\) or \(J = \mathfrak{m}\). We denote by \(\operatorname{mSpec}(R)\) the set of maximal ideals of \(R\).

Example (Known maximal spectra)

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

1. If \(R = \{0\}\), then \(\operatorname{mSpec}(R) = \emptyset\).
2. If \(K\) is a field, then \(\operatorname{mSpec}(K) = \{ \{0\} \}\).
3. Let \(R = \mathbb{Z}_{4}\). The ideals are \((0)\), \((2) = \{0, 2\}\) and \((1) = \mathbb{Z}_{4} = (3)\). Note that \(\operatorname{mSpec}(\mathbb{Z}_{4}) = \{(2)\}\).
4. In \(\mathbb{Z}\) all the ideals are of the form \(n\mathbb{Z}\) with \(n \in \mathbb{N}\). Show that \(n\mathbb{Z}\) is maximal if and only if \(n\) is prime.

Axiom (Zorn’s lemma)

Let \((X, \leq)\) be a partially ordered set. If

1. \(X \neq \emptyset\), and
2. every totally ordered subset of \(X\) possesses an upper bound in \(X\),

then \(X\) possesses a maximal element.

Theorem (Krull: every proper ideal is contained in a maximal one)

Let \(R \neq \{0\}\) be a commutative ring and let \(I\) be a proper ideal. Then there exists a maximal ideal \(\mathfrak{m}\) such that \(I \subseteq \mathfrak{m}\).

Proof: Let

\[\Omega := \{ J :\ J \text{ ideal},\ J \neq R,\ I \subseteq J \},\]

which is a set partially ordered by inclusion. Note that \(\Omega \neq \emptyset\), since \(I \in \Omega\). In order to apply Zorn’s lemma, let \((C_{\alpha})_{\alpha \in \Delta}\) be a chain in \(\Omega\), which we may suppose non-empty (the empty chain has \(I\) as an upper bound), and let \(N = \bigcup_{\alpha \in \Delta} C_{\alpha}\). We shall prove that \(N \in \Omega\).

First, \(N\) is an ideal. Note that \(N \neq \emptyset\), since \(C_{\alpha} \subseteq N\) for \(\alpha \in \Delta\) and each \(C_{\alpha}\) is non-empty. Let \(x, y \in N\). Then there exist \(\alpha, \beta \in \Delta\) such that \(x \in C_{\alpha}\) and \(y \in C_{\beta}\). Since \((C_{\alpha})\) is a chain, we have that \(C_{\alpha} \subseteq C_{\beta}\) or \(C_{\beta} \subseteq C_{\alpha}\); without loss of generality, \(C_{\alpha} \subseteq C_{\beta}\). Then \(x, y \in C_{\beta}\), and since \(C_{\beta}\) is an ideal, \(x + y \in C_{\beta} \subseteq N\). In the same way, if \(x \in N\) and \(r \in R\), then \(x \in C_{\alpha}\) for some \(\alpha\), and \(r \cdot x \in C_{\alpha} \subseteq N\) by absorption.

Second, \(N \neq R\): if \(N = R\), then \(1 \in N\), so that there exists \(\alpha\) such that \(1 \in C_{\alpha}\), and therefore \(C_{\alpha} = R\), which contradicts that \(C_{\alpha} \in \Omega\). Finally, \(I \subseteq C_{\alpha} \subseteq N\) for any \(\alpha \in \Delta\), so that \(N \in \Omega\) and \(N\) is an upper bound of the chain \((C_{\alpha})_{\alpha \in \Delta}\).

Hence, by Zorn’s lemma, there exists a maximal element \(\mathfrak{m}\) in \(\Omega\). Let us see that \(\mathfrak{m}\) is a maximal ideal of \(R\): by construction \(\mathfrak{m} \neq R\) and \(I \subseteq \mathfrak{m}\). If \(J\) is an ideal such that \(\mathfrak{m} \subseteq J\) and \(J \neq R\), then \(I \subseteq \mathfrak{m} \subseteq J\), so \(J \in \Omega\); by the maximality of \(\mathfrak{m}\) in \(\Omega\), we have that \(J = \mathfrak{m}\). That is, every ideal containing \(\mathfrak{m}\) is \(R\) or is \(\mathfrak{m}\), which is the definition of a maximal ideal.

Corollary (The maximal spectrum is non-empty and detects the non-units)

If \(R \neq \{0\}\) is a commutative ring, then \(\operatorname{mSpec}(R) \neq \emptyset\). Moreover,

\[\bigcup_{\mathfrak{m} \in \operatorname{mSpec}(R)} \mathfrak{m} = R \setminus R^{\times}.\]

Proof: Since \(R \neq \{0\}\), the ideal \((0)\) is proper, and by Krull’s theorem there exists a maximal ideal containing it; then \(\operatorname{mSpec}(R) \neq \emptyset\).

“\(\supseteq\)”: If \(x \in R \setminus R^{\times}\), then \((x) \neq R\): otherwise \(1 \in (x)\), that is, \(1 = r x\) for some \(r \in R\), and \(x\) would be a unit. By Krull, there exists a maximal \(\mathfrak{m}\) such that \((x) \subseteq \mathfrak{m}\). Conclude that \(x \in \bigcup_{\mathfrak{m} \in \operatorname{mSpec}(R)} \mathfrak{m}\).

“\(\subseteq\)”: If \(x \in \mathfrak{m}\) with \(\mathfrak{m}\) maximal, then \(x \notin R^{\times}\): if \(x\) were a unit, we would have \(1 = x^{-1} x \in \mathfrak{m}\) by absorption, whence \(\mathfrak{m} = R\), which contradicts \(\mathfrak{m} \neq R\).

Definition (Local ring)

A commutative ring \(R \neq \{0\}\) is local if it possesses exactly one maximal ideal.

Corollary (Characterisation of local rings)

If \(R \neq \{0\}\) is a commutative ring, then \(R\) is local if and only if \(R \setminus R^{\times}\) is an ideal.

Proof: Exercise.

Definition (Jacobson radical)

Let \(R \neq \{0\}\) be a commutative ring. We define the Jacobson radical as

\[\operatorname{Rad}(R) = \bigcap_{\mathfrak{m} \in \operatorname{mSpec}(R)} \mathfrak{m}.\]

Exercise (Characterisation of the Jacobson radical)

Show that \(x \in \operatorname{Rad}(R)\) if and only if \(1 - x y \in R^{\times}\) for each \(y \in R\).