## Tag Archives: principal ideal

### Exhibit the idempotents and principal ideals in a given semigroup

Denote the function $\mathsf{max}$ on $\mathbb{N}$ by $\wedge$. Let $S = \{0,1\} \times \mathbb{N}$, and define a binary operator $\star$ on $S$ by $(s,a) \star (t,b) = (0, a \wedge b)$ if $s = t$ and $(1, a \wedge b)$ if $s \neq t$. Show that $(S,\star)$ is a semigroup and exhibit all of its idempotents and principal ideals. Does $S$ have a kernel?

First, we argue that $\star$ is associative. To show this, we refer to the following tree diagram.

Associativity diagram

This diagram is to be read from left to right. Labels on an edge indicate an assumption that holds in all subsequent branches. Each path from the root to a leaf corresponds to a string of equalities, and together these imply that $\star$ is associative.

So $(S,\star)$ is a semigroup.

Suppose $(s,a)$ is idempotent. Then $(s,a) = (s,a) \star (s,a)$ $= (0, a)$, and we have $s = 0$. Conversely, $(0,a)$ is clearly idempotent for all $a \in \mathbb{N}$. So the idempotents in $S$ are precisely elements of the form $(0,a)$ with $a \in \mathbb{N}$.

Next we claim that $S$ is commutative. Indeed, if $s = t$, then $(s,a) \star (t,b) = (0,a \wedge b) = (0,b \wedge a)$ $= (t,b) \star (s,a)$ and if $s \neq t$ then $(s,a) \star (t,b) = (1, a \wedge b)$latex = (1, b \wedge a)\$ $= (t,b) \star (s,b)$.

We claim that the principal left ideal $L(s,a) = (s,a) \cup S(s,a)$ is $\{(t,b)\ |\ b \geq a\}$. Indeed, if $(t,b) \in L(s,a)$, then either $(t,b) = (s,a)$ or $(t,b) = (u,c)(s,a)$ for some $(u,a)$. But then $b = c \wedge a \geq a$. Conversely, consider $(t,b)$ with $b \geq a$, and let $u$ be 0 if $s =1$ and 1 otherwise. Now if $t = 0$, then $(t,b) = (s,b) \star (s,a)$ and if $t = 1$ then $(t,b) = (u,b) \star (s,a)$. So $L(s,a) = \{(t,b)\ |\ b \geq a\}$.

In particular, for every element $(s,a) \in S$, there is an ideal of $S$ not containing $(s,a)$ (for instance, $(s,a+1)$.) So $S$ has no kernel.

### Exhibit all of the principal ideals of a given semigroup

Let $S = [1,2] \cup [3,4] \cup [5,\infty) \subseteq \mathbb{R}$, and define a binary operator $\star$ on $S$ by $x \star y = x+y$ if $x,y \geq 5$ and $\mathsf{max}(x,y)$ otherwise. Show that $S$ is a semigroup and exhibit all of its principal ideals. Does $S$ have a kernel? Compute the orders of the elements of $S$.

Note first that $\star$ is commutative.

Given a triple $(a,b,c)$ of elements in $S$, each is either less than 5 or not. This leads us to consider 8 cases. Let $\alpha,\beta, \gamma \in S$ be at least 5, and let $a,b,c \in S$ be less than 5.

• $(\alpha \star \beta) \star \gamma = (\alpha + \beta) \star \gamma$ $= (\alpha + \beta) + \gamma$ $= \alpha + (\beta + \gamma)$ $= \alpha \star (\beta + \gamma)$ $= \alpha \star (\beta \star \gamma)$
• $(a \star \beta) \star \gamma = \beta \star \gamma$ $= \beta + \gamma$ $= a \star (\beta + \gamma)$ $= a \star (\beta \star \gamma)$
• $\alpha \star b) \star \gamma = \alpha \star \gamma$ $= \alpha \star (b \star \gamma)$
• $(\alpha \star \beta) \star c = (\alpha + \beta) \star c$ $= \alpha + \beta$ $= \alpha \star \beta$ $= \alpha \star (\beta \star c)$
• $(a \star b) \star \gamma = a \star \gamma$ $= \gamma$ $= a \star \gamma$ $= a \star (b \star \gamma)$ (Without loss of generality, $a \geq b$.)
• $(a \star \beta) \star c = \beta \star c$ $= \beta$ $= a \star \beta$ $= a \star (\beta \star c)$
• $(\alpha \star b) \star c = \alpha \star c$ $= \alpha$ $= \alpha \star (b \star c)$
• $(a \star b) \star c = \mathsf{max}(\mathsf{max}(a,b),c)$ $= \mathsf{max}(a,\mathsf{max}(b,c))$ $= a \star (b \star c)$.

So $\star$ is associative, and $(S,\star)$ is a semigroup. Since $S$ is commutative, the left, right, and two-sided ideals of $S$ coincide. Recall that the principal left ideal generated by $a \in S$ is $L(a) = a \cup Sa$.

If $a < 5$, then $L(a) = a \cup Sa$ $= a \cup \{sa \ |\ s \in S, s < a\} \cup \{sa\ |\ s \in S, a \leq s\}$ $= a \cup ([a,\infty) \cap S)$.

If $a \geq 5$, then $L(a) = a \cup Sa$ $= a \cup \{sa\ |\ s \in S, s < 5\} \cup \{sa\ |\ s \in S, s \geq 5\}$ $= a \cup [a+5,\infty)$

In particular, for every element $a$ in $S$, there is a principal ideal which does not contain $a$. (For example, $L(a+5)$.) So the kernel of $S$ is empty.

If $a < 5$, then $a^2 = a$, so that $a$ has order 1. If $a \geq 5$, then $a$ has infinite order.

### 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.

### Exhibit an extended principal generator for an ideal in an algebraic integer ring

Let $A = (3,1+2\sqrt{-5})$ be an ideal in the ring of integers of $K = \mathbb{Q}(\sqrt{-5})$. Find an algebraic integer $\kappa$ such that $A = (\kappa) \cap \mathcal{O}_E$, where $E = K(\kappa)$.

We saw in the text that the class number of $K$ is 2. Note that $A^2 = (3)$, as indeed $3 = (1+2\sqrt{-5})^2 - 2 \cdot 3^2$. Let $\kappa = \sqrt{3}$. By our proof of Theorem 10.6, $A = \{ \tau \in \mathcal{O}_K \ |\ \tau/\kappa \in \mathcal{O}_K \}$.

### 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.

### In an algebraic integer ring, the norm of an ideal principally generated by a rational integer is the absolute value of the norm of its generator

Let $K$ be an algebraic extension of $\mathbb{Q}$ of degree $n$, and let $\mathcal{O}$ be the ring of integers in $K$. Let $k \in \mathbb{Z}$. Show that the norm $N((k))$ is precisely $|N(k)|$. (The first norm is the norm of $(k)$ as an ideal, while the second norm is the norm of $k$ as an element.)

We can assume without loss of generality that $k$ is positive.

Recall that the conjugates of $k$ over $K$ are all the same (Theorem 5.10 in TAN), so that $N(k) = k^n$.

Let $\{\omega_i\}$ be an integral basis for $K$. Suppose $A$ is an ideal containing $k$. By Theorem 9.3 in TAN, there exists an element $\alpha = \sum c_i\omega_i$ such that $A = (k,\alpha)$. Note that we may assume that each $c_i$ is in $[0,k)$. In particular, $N((k)) \leq k^n$. We claim that this $\alpha$ is in fact unique. To that end, suppose $\sum c_i\omega_i \equiv \sum d_i\omega_i$ mod $(k)$ for some $0 \leq c_i, d_i < k$. Then we have $\sum c_i\omega_i - \sum d_i \omega_i = \sum ke_i\omega_i$ for some $e_i$. Since $\{\omega_i\}$ is an integral basis, we have $c_i \equiv d_i$ mod $k$ for each $i$, and thus $c_i = d_i$. So in fact $N((k)) = k^n = N(k)$, as desired.

### If A is an ideal in an algebraic integer ring and a an element of A, then there is an ideal B such that AB = (a)

Let $\mathcal{O}$ be the ring of integers in an algebraic number field $K$. Suppose $A \subseteq \mathcal{O}$ is an ideal and $\alpha \in A$ an element. Show that there exists an ideal $B \subseteq \mathcal{O}$ such that $AB = (\alpha)$.

Write $A = \prod_{i=1}^n P_i$ as a product of maximal ideals, and let $A^{-1} = \prod_{i=n}^1 P_i$. Note that as subsets of $\mathcal{O}$, we have $(A^{-1}(\alpha))A = A^{-1}A(\alpha) = \mathcal{O}(\alpha) = (\alpha)$. (Using Lemma 8.23 in TAN.) On the other hand, since $(\alpha) \subseteq A$, $A^{-1}(\alpha)$ is an ideal of $\mathcal{O}$, as we argue. By definition, $A^{-1}(\alpha) \subseteq \mathcal{O}$. If $x,y \in A^{-1}(\alpha)$ and $r \in \mathcal{O}$, where $x = \sum h_ia_i\alpha$ and $y = \sum k_ib_i\alpha$, then $x+ry = \sum h_ia_i\alpha + \sum k_irb_i\alpha \in A^{-1}(\alpha)$. Since $0 \in A^{-1}(\alpha)$, by the submodule criterion $A^{-1}(\alpha)$ is an ideal of $\mathcal{O}$.

### Principal ideals generated by irreducible elements need not be prime

Suppose $\pi$ is an irreducible element in an algebraic integer ring $\mathcal{O}$. Must $(\pi)$ be prime as an ideal in $\mathcal{O}$?

Recall that $1+\sqrt{-5}$ is irreducible in $\mathcal{O}_K$, where $K = \mathbb{Q}(\sqrt{-5})$, since no element in this ring has norm 3. Moreover, $1 - \sqrt{-5} \notin (1+\sqrt{-5})$, as we show. If to the contrary we have $(a+b\sqrt{-5})(1+\sqrt{-5}) = 1 - \sqrt{-5}$, then comparing coefficients we have $a-5b = 1$ and $a+b = -1$, so that $3b = -2$ for some $b \in \mathbb{Z}$, a contradiction. Using this previous exercise, we have $(1+\sqrt{-5}) \subsetneq (1+\sqrt{-5}, 1-\sqrt{-5}) \subsetneq \mathcal{O}$. In particular, $1+\sqrt{-5}$ is irreducible, but $(1+\sqrt{-5})$ is not maximal and thus not prime.

So it need not be the case that $(\pi)$ is prime if $\pi$ is irreducible.

### Over an algebraic integer ring, if (a)|(b) then a|b

Let $\mathcal{O}$ be the ring of integers in an algebraic number field $K$, and let $\alpha,\beta \in \mathcal{O}$. Prove that if $(\alpha)|(\beta)$, then $\alpha|\beta$ in $\mathcal{O}$.

Recall that $(\alpha)|(\beta)$ means that $(\beta) = C(\alpha)$ for some ideal $C$. Now $(\beta) \subseteq (\alpha)$, so that $\beta \in (\alpha)$, and thus $\beta = \alpha\gamma$ for some $\gamma$. Thus $\alpha|\beta$ in $\mathcal{O}$.

### Characterize the ideals in F[x] principally generated by monic irreducibles

Let $F$ be a field, let $E$ be an extension of $F$, and let $\alpha \in E$ be algebraic over $F$ with minimal polynomial $p(x)$. Prove that the ideal $(p(x))$ in $F[x]$ is equal to $\{ q(x) \in F[x] \ |\ q(\alpha) = 0 \}$.

If $t(x) \in (p(x))$, then $t(x) = p(x)s(x)$ for some $s(x) \in F[x]$. Thus $t(\alpha) = p(\alpha)s(\alpha) = 0$.

Conversely, if $t(\alpha) = 0$, then the minimal polynomial $p(x)$ divides $t(x)$. So $t(x) \in (p(x))$.