## 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 $\mathcal{O}$.