Tag Archives: ideal

Characterize the left ideal of a semigroup generated by a subset

Let S be a semigroup and let A \subseteq S be a nonempty subset. Recall that the left ideal of S generated by A is the intersection L(A) of all the ideals which contain A, and similarly for right- and two-sided- ideals.

Prove that the left ideal generated by A is A \cup SA and that the two-sided ideal generated by A is A \cup SA \cup AS \cup SAS.

Note that if L is a left ideal containing A, then SA \subseteq L. In particular, A \cup SA is contained in every left ideal which also contains A, and thus is contained in the left ideal generated by A. On the other hand, S(A \cup SA) = SA \cup SSA \subseteq SA \subseteq A \cup SA, and thus A \cup SA is a left ideal of S which certainly contains A. So the left ideal generated by A is contained in A \cup SA, so that L(A) = A \cup SA.

Similarly, if I is an ideal of S containing A, then A \cup SA \cup AS \cup SAS \subseteq I. So A \cup SA \cup AS \cup SAS is contained in the ideal \langle A \rangle generated by A. And on the other hand, since S(A \cup SA \cup AS \cup SAS) \subseteq A \cup SA \cup AS \cup SAS, this set is a left ideal, and likewise a right ideal. So A \cup SA \cup AS \cup SAS is a two sided ideal containing A. Hence \langle A \rangle = A \cup SA \cup AS \cup SAS.

The semigroup of nonempty subsets of a semigroup

Let S be a semigroup, and let P(S) denote the set of all nonempty subsets of S. Show that P(S) is a semigroup under the operation AB = \{ab \ |\ a \in A, b \in B\}. Which of the basic properties of S are preserved by P(S)?

We need only show that this operator is associative. Indeed, (AB)C = \{ab\ |\ a \in A, b \in B\}C = \{(ab)c\ |\ a \in A, b \in B, c \in C\} = \{a(bc)\ |\ a \in A, b \in B, c \in C\} = A\{bc\ |\ b \in B, c \in C\} = A(BC).

Suppose S has a left identity e. Then for all A \in P(S), we have \{e\}A = \{ea \ |\ a \in A\} = \{a \ |\ a \in A\} = A. That is, \{e\} is a left identity in P(S). Similarly if e is a right identity in S, then \{e\} is a right identity in P(S). Then if e is an identity in S, \{e\} is an identity in P(S).

Suppose S has a left zero z. Then for all A \in P(S), we have \{z\}A = \{za\ |\ a \in A\} = \{z\ |\ a \in A\} = \{z\}. Thus \{z\} is a left zero in P(S). Similarly, if z is a right zero in S then \{z\} is a right zero in P(S), and if z is a zero in S then \{z\} is a zero in P(S).

If S is commutative, then for all A,B \in P(S) we have AB = \{ab \ |\ a \in A, b \in B\} = \{ba \ |\ b \in B, a \in A\} = BA. Thus P(S) is also commutative.

Suppose e \in S is an identity element and u \in S is a unit, with uv = vu = e. Then \{u\}\{v\} = \{v\}\{u\} = \{e\}, so that \{u\} is a unit in P(S).

Suppose I \subseteq S is a left ideal. Now let A \subseteq S and B \subseteq I be nonempty subsets. Now AB = \{ab \ |\ a \in A, b \in B\} \subseteq I, since B \subseteq I, and moreover this set is nonempty since A and B are nonempty. That is to say, AB \in P(I). So P(I) \subseteq P(S) is a left ideal. Likewise if I is a right ideal in S, then P(I) is a right ideal in P(S), and if I is a two-sided ideal in S, then P(I) is a two-sided ideal in P(S). More generally, if T \subseteq S is a subsemigroup, then P(T) \subseteq P(S) is a subsemigroup.

Suppose S is a left zero semigroup. That is, ab = a for all a,b \in S. If A,B \in P(S), then AB = \{ab \ |\ a \in A, b \in B\} = \{a \ |\ a \in A, b \in B\} = A. That is, P(S) is also a left zero semigroup. Similarly, if S is a right zero semigroup then so is P(S). Suppose now that S has a zero element 0, and that S is a zero semigroup. Now \{0\} is a zero in P(S), and for all A,B \in P(S), we have AB = \{ab \ |\ a \in A, b \in B\} = \{0 \ |\ a \in A, b \in B\} = \{0\}. So P(S) is also a zero semigroup.

If S is simple, P(S) need not be simple, as we show. Consider the group G = Z_2 = \{1,x\}. We can easily verify that the only two-sided ideal in G is G itself, so that G is simple as a semigroup. Now P(Z_2) = \{\{1\}, \{x\}, Z_2\}, and evidently, \{Z_2\} is an ideal in P(Z_2), so that P(Z_2) is not simple.

Characterize the ideals and congruences on the semigroup of natural numbers under max

Characterize the ideals and congruences on the semigroup \mathbb{N} of natural numbers under the \mathsf{max} operator. (I.e., a\ \mathsf{max}\ b = a if a \geq b, b otherwise.)

First we will make some definitions.

Definition: A subset I \subseteq \mathbb{N} is called an interval if, whenever a \leq c \leq b and a,b \in I, c \in I.

Lemma 1: Every nonempty interval in \mathbb{N} has the form [a,b] = \{c \ |\ a \leq c \leq b\} for some a,b or [a,\infty) = \{c \ |\ a \leq c\}. Proof: Every nonempty interval on \mathbb{N} has a least element by the well-ordering property of natural numbers. If a nonempty interval I has a greatest element b, then c \in I for all a \leq c \leq b, and so we have I = [a,b]. If I does not have a greatest element, then for every c \geq a, there exists b \in I with a \leq c \leq b, and thus c \in I. So we have I = [a,\infty). \square

Lemma 2: If I and J are disjoint intervals and there exist a \in I and b \in J such that a \leq b, then for all x \in I and y \in J we have x \leq y. Proof: Suppose to the contrary that there exist x \in I and y \in J with y \leq x. There are six possible orderings for \{a,b,x,y\} which preserve a \leq b and y \leq x. Specifically, these are a \leq b \leq y \leq x, a \leq y \leq b \leq x, a \leq y \leq x \leq b, y \leq a \leq b \leq x, y \leq a \leq x \leq b, and y \leq x \leq a \leq b. In each case, I \cap J is not empty. \square

Next we consider the ideals on \mathbb{N}. To that end, let I \subseteq \mathbb{N} be a subset.

Suppose that I is an ideal under the \mathsf{max} operator. If I is nonempty (as we insist our ideals to be), then it has a least element a by the well-ordering property of the natural numbers. If b \geq a, then \mathsf{max}(a,b) = b \in I, and if b < a, then b \notin I since a is minimal. Thus I = [a,\infty).

Conversely, suppose I = [a,\infty) for some a \in \mathbb{N}. If t \in I and s \in \mathbb{N}, then we have t \geq a. So \mathsf{max}(s,t) \geq a, and thus \mathsf{max}(s,t) \in I. So I is an ideal.

Hence the ideals of (\mathbb{N}, \mathsf{max}) are precisely the intervals [a,\infty) where a \in \mathbb{N}.

Now we will consider the congruences on \mathbb{N}. To that end, let \sigma \subseteq \mathbb{N} \times \mathbb{N} be an equivalence relation.

Suppose \sigma is a congruence, and let A \subseteq \mathbb{N} be a \sigma-class. We claim that A is an interval. To that end, let a,b \in A and suppose a \leq c \leq b. Now a \sigma b and c \sigma c, so that (since \sigma is a congruence) ac \sigma bc, and thus c \sigma b. So c \in A. Thus A is an interval (by our definition above). Thus if \sigma is a congruence on \mathbb{N} under \mathsf{max}, then the classes of \sigma are pairwise disjoint intervals.

Conversely, suppose the classes of the equivalence relation \sigma are pairwise disjoint intervals. We claim that \sigma is a congruence. To that end, suppose a \sigma b and c \sigma d, and suppose without loss of generality that a \leq c. By Lemma 2, we have b \leq d. So ac = c \sigma d = bd. Thus \sigma is a congruence.

Hence the congruences on (\mathbb{N}, \mathsf{max}) are precisely the paritions of \mathbb{N} which consist of intervals.

A characterization of principal ideals in a semigroup

Let S be a semigroup and let a \in S. Let L(a), R(a), and J(a) be the left, right, and two-sided ideals generated by a, respectively. Prove that L(a) = a \cup Sa, R(a) = a \cup aS, and J(a) = a \cup Sa \cup aS \cup SaS.

Recall that by definition, L(a) is the intersection over the class of left ideals which contain a. Suppose L is a left ideal containing a. Then for all s \in S, sa \in L, and thus a \cup Sa \subseteq L. So a \cup Sa \subseteq L(a). On the other hand, we have S(a \cup Sa) = Sa \cup SSa \subseteq Sa \subseteq a \cup Sa. So a \cup Sa is a left ideal containing a, and thus L(a) \subseteq a \cup Sa. So L(a) = a \cup Sa.

The proofs for R(a) and J(a) are similar; show that every (right,two sided) ideal contains the set in question, and that the set in question is a (right, two-sided) ideal.

The classes of left, right, and two-sided ideals of a semigroup are closed under finite intersection and arbitrary union

Let S be a semigroup, and let L(S), R(S), and I(S) denote the sets of left- right- and two-sided ideals of S. Let X \in \{L(S), R(S), I(S)\} and let Y \subseteq X be nonempty. Show that \bigcup Y \in X and that if \bigcap Y \neq \emptyset, then \bigcap Y \in X. Further, show that if Y \subseteq I(S) is finite, then \bigcap Y is not empty (and so is an ideal).

Suppose Y \subseteq L(S) is a nonempty collection of left ideals of S. Certainly \bigcup Y \neq \emptyset. If a \in \bigcup Y, then we have a \in T for some T \in Y. Now if s \in S, then sa \in T \in \bigcup Y. So S(\bigcup Y) \subseteq \bigcup Y, and thus \bigcup Y is a left ideal of S. Now suppose \bigcup Y \neq \emptyset, and say a \in \bigcap Y. So a \in T for all T \in Y. If s \in S, then sa \in T for all T \in Y, and so sa \in \bigcap Y. Thus \bigcap Y is a left ideal.

Likewise, the results hold for R(S). Since every two-sided ideal is also a left and a right ideal, the results also follow for I(S).

Now suppose Y = \{T_i\}_{i=1}^n is a finite collection of ideals of S. Now \prod T_i \subseteq T_k for each k, so that \prod T_i \subseteq \bigcap Y. Since each T_i is nonempty, \prod T_i is nonempty, and so \bigcap Y is nonempty.

If A² is principal over an algebraic number field, then A² is principal over any extension

Let K_1 and K_2 be algebraic number fields with K_1 \subseteq K_2, and having integer rings \mathcal{O}_1 and \mathcal{O}_2, respectively. Suppose (A)_{\mathcal{O}_1} \subseteq \mathcal{O}_1 is an ideal such that (A)^2 is principal. Show that (A)_{\mathcal{O}_2}^2 is principal in \mathcal{O}_2.

We will let G_1 and G_2 denote the ideal class groups of K_1 and K_2, respectively. For brevity, if S \subseteq \mathcal{O}_1 is a subset, then we will denote by (S)_1 the ideal generated by S in \mathcal{O}_1 and by (S)_2 the ideal generated by S in \mathcal{O}_2.

We claim that if (A)_1 \sim (B)_1, then (A)_2 \sim (B)_2. To see this, suppose we have \alpha and \beta such that (\alpha)_1(A)_1 = (\beta)_1(B)_1. Then (\alpha A)_1 = (\beta B)_1. Now if x = \sum r_i \alpha a_i \in (\alpha A)_2, then each \alpha a_i is in (\beta B)_1, and so has the form \sum s_{i,j} \beta b_j. So x \in (\beta B)_2. Conversely, (\beta B)_2 \subseteq (\alpha A)_2, and so (A)_2 \sim (B)_2.

The mapping A \mapsto (A)_2 then induces a well-defined mapping \Psi : G_1 \rightarrow G_2 given by [A] \mapsto [(A)_2]. Moreover, since \Psi([A][B]) = \Psi([AB]) = [(AB)_2] = [(A)_2][(B)_2] = \Psi(A) \Psi(B), \Psi is a group homomorphism.

In particular, if A^2 \sim (1) in \mathcal{O}_1, then \Psi(A) \sim (1) in \mathcal{O}_2, as desired.

The exponent of the smallest power of an ideal which is principal divides the class number

Let \mathbb{O} be the ring of integers in an algebraic number field K of class number k. Let A be an ideal. Show that if k is minimal such that A^k is principal, then k|h.

This k is precisely the order of [A] in the class group of K. The result then follows by Lagrange’s Theorem.

Detect equivalences among ideals of an algebraic integer ring

Determine which of the following ideals of \mathbb{Z}[\sqrt{-6}] are equivalent: A = (2,\sqrt{-6}), B = 5-2\sqrt{-6},2+2\sqrt{-6}), C = (2+\sqrt{-6}), and D = (5,1+2\sqrt{-6}).

First we argue that B is principal. Indeed, B = (5-2\sqrt{-6}, 2+2\sqrt{-6}) = (7,2+2\sqrt{-6}) = (1+\sqrt{-6})(1-\sqrt{-6},2). Now (1-\sqrt{-6})(1+\sqrt{-6}) - 3 \cdot 2 = 1 \in (1-\sqrt{-6},2), so that B = (1+\sqrt{-6})(1) = (1+\sqrt{-6}). Indeed, 5-2\sqrt{-6} = (1+\sqrt{-6})(-1-\sqrt{-6}) and 2+2\sqrt{-6} = 2(1+\sqrt{-6}), and 1+\sqrt{-6} = (1-\sqrt{-6})(5-21\sqrt{-6}) + 4(2+2\sqrt{-6}).

In particular, B \sim C \sim (1).

Now we claim that AD and A^2 are both principal. Indeed, AD = (2,\sqrt{-6})(5,1+2\sqrt{-6}) = (10,2+4\sqrt{-6},5\sqrt{-6},-12+\sqrt{-6}) = (10,2+4\sqrt{-6},5\sqrt{-6},-10) = (10,2-\sqrt{-6},5\sqrt{-6}). Since N(2-\sqrt{-6}) = 10 and (2-\sqrt{-6})(-3+\sqrt{-6}) = 5\sqrt{-6}, we have AD = (2-\sqrt{-6}). So AD \sim (1). Similarly, A^2 = (s,\sqrt{-6})^2 = (4,2\sqrt{-6},-6) = (2), so that A^2 \sim (1). Now AD \sim A^2, so that A \sim D.

Finally, we claim that A is not principal. If (2,\sqrt{-6}) = (\alpha), then \alpha|2 and \alpha|\sqrt{-6}, so that N(\alpha) divides both 4 and 6. So N(\alpha) \in \{1,2\}. Note, however, that no element of \mathbb{Z}[\sqrt{-6}] has norm 2 since the equation a^2 + 6b^2 = 2 has no solutions in \mathbb{Z}. If N(\alpha) = 1, then we have 1 \in D. But then 1 = 2(h+k\sqrt{-6}) + \sqrt{-6}(u+v\sqrt{-6}), so that 1=2h-6v, which is impossible mod 2. So D is not principal.

In summary, we have B \sim C \sim (1), A \sim D, and (1) \not\sim D.

In an algebraic integer ring, the smallest positive integer contained in a given ideal divides the norm of the ideal

Let K be an algebraic number field with ring of integers \mathcal{O}, and let A \subseteq \mathcal{O} be and ideal. Suppose m is the smallest positive integer contained in A. Show that m|N(A). Suppose now that N(A) is a rational prime; show that \mathcal{O}/A = \{\overline{t}\ |\ 0 \leq t < p\}.

By Theorem 9.16, N(A) \in A. Now A \cap \mathbb{Z} = (m) is an ideal of \mathbb{Z}, which is principal since \mathbb{Z} is a principal ideal domain. Certainly then m is the smallest positive integer in A. Thus m|N(A).

Now suppose N(A) is a rational prime. Suppose there is a (minimal) integer t \in A such that 0 < t < N(A). By part (1), t|N(A), a contradiction. So none of the integers in (0,N(A)) are in A, and thus t \not\equiv 0 mod A for all t \in (0,N(A)). Moreover, we have t \not\equiv s mod A for all t,s \in (0,N(A)) with s \neq t, since otherwise (assuming s > t) we have s-t \equiv 0 mod A. Since N(A) = p, the cosets \{\overline{t}\ |\ t \in [0,N(A))\} exhaust \mathcal{O} and are mutually exclusive.

Factor a given ideal in an algebraic integer ring

Let K = \mathbb{Q}(\sqrt{-6}) and let \mathcal{O} = \mathbb{Z}[\sqrt{-6}] be the ring of integers in K. Factor the ideals (2) and (5) in \mathcal{O}.

We claim that (2) = (2,\sqrt{-6})^2. Indeed, 2 = -\sqrt{-6}^2 - 2^2, so that (2) \subseteq (2,\sqrt{-6})^2. The reverse inclusion is clear.

Now we claim that (2,\sqrt{-6}) is maximal. By Corollary 9.11, N((2)) = 4, and by Theorem 9.14, we have N((2,\sqrt{-6}))^2 = 4, so that N((2,\sqrt{-6})) = 2. By Corollary 9.15, (2,\sqrt{-6}) is a prime ideal.

Thus (2) = (2,\sqrt{-6})^2 is the prime factorization of (2) in \mathcal{O}.

Now we claim that (5) = (5,1+2\sqrt{-6})(5,1-2\sqrt{-6}). Indeed, we have (5,1+2\sqrt{-6})(5,1-2\sqrt{-6}) = (25,5+10\sqrt{-6}, 5-10\sqrt{-6}, 25) = (25, 5+10\sqrt{-6}, 10) = (5).

Next we claim that Q_1 = (5,1+2\sqrt{-6}) and Q_2 = (5,1-2\sqrt{-6}) are proper. Suppose to the contrary that 1 \in Q_1; then we have 1 = 5(a+b\sqrt{-6}) + (1+2\sqrt{-6})(h+k\sqrt{-6}) for some a,b,h,k \in \mathbb{Z}. Comparing coefficients, 5a+h-12k = 1 and 5b+k+2h = 0, which yields a contradiction mod 5. So Q_1 is proper. Likewise, Q_2 is proper. In particular, neither Q_1 nor Q_2 have norm 1 as ideals. Now 25 = N((5)) = N(Q_1)N(Q_2), and neither factor is 1, so that N(Q_1) = N(Q_2) = 5. By Corollary 9.15, Q_1 and Q_2 are prime ideals.

Thus (5) = (5,1+2\sqrt{-6})(5,1-2\sqrt{-6}) is the prime factorization of (5) in \mathcal{O}.