A quantale $Q$ is a set with three binary operations on it: $\wedge,\vee$, and $\cdot$, such that

1. 1.

   $(Q,\wedge,\vee)$ is a complete lattice (with $0$ as the bottom and $1$ as the top), and

2. 2.

   $(Q,\cdot)$ is a monoid (with $1^{{\prime}}$ as the identity with respect to $\cdot$), such that

3. 3.

   $\cdot$ distributes over arbitrary joins; that is, for any $a\in Q$ and any subset $S\subseteq Q$,

   $$a\cdot\big(\bigvee S\big)=\bigvee\{a\cdot s\mid s\in S\}\quad\mbox{ and }\quad\big(\bigvee S\big)\cdot a=\bigvee\{s\cdot a\mid s\in S\}.$$

It is sometimes convenient to drop the multiplication symbol, when there is no confusion. So instead of writing $a\cdot b$, we write $ab$.

The most obvious example of a quantale comes from ring theory. Let $R$ be a commutative ring with $1$. Then $L(R)$, the lattice of ideals of $R$, is a quantale.

Remark. In the above example, notice that $IJ\leq I$ and $IJ\leq J$, and we actually have $IJ\leq I\wedge J$. In particular, $I^{2}\leq I$. With an added condition, this fact can be characterized in an arbitrary quantale (see below).

Properties. Let $Q$ be a quantale.

1. 1.

   Multiplication is monotone in each argument. This means that if $a,b\in Q$, then $a\leq b$ implies that $ac\leq bc$ and $ca\leq cb$ for all $c\in Q$. This is easily verified. For example, if $a\leq b$, then $ac\vee bc=(a\vee b)c=bc$, so $ac\leq bc$. So a quantale is a partially ordered semigroup, and in fact, an l-monoid (an l-semigroup and a monoid at the same time).

2. 2.

   If $1=1^{{\prime}}$, then $ab\leq a\wedge b$: since $a\leq 1$, then $ab\leq a1=a1^{{\prime}}=a$; similarly, $b\leq ab$. In particular, the bottom $0$ is also the multiplicative zero: $a0\leq a\wedge 0=0$, and $0a=0$ similarly.

3. 3.

   Actually, $a0=0a=0$ is true even without $1=1^{{\prime}}$: since $a\varnothing=\{ab\mid b\in\varnothing\}=\varnothing$ and $0:=\bigvee\varnothing$, we have $a0=a\bigvee\varnothing=\bigvee a\varnothing=\bigvee\varnothing=0$. Similarly $0a=0$. So a quantale is a semiring, if $\vee$ is identified as $+$ (with $0$ as the additive identity), and $\cdot$ is again $\cdot$ (with $1^{{\prime}}$ the multiplicative identity).

4. 4.

   Viewing quantale $Q$ now as a semiring, we see in fact that $Q$ is an idempotent semiring, since $a+a=a\vee a=a$.

5. 5.

   Now, view $Q$ as an i-semiring. For each $a\in Q$, let $S=\{1^{{\prime}},a,a^{2},\ldots\}$ and define $a^{*}=\bigvee S$. We observe some basic properties

   • $1^{{\prime}}+aa^{*}=a^{*}$: since $1^{{\prime}}\vee(a\bigvee S)=1^{{\prime}}\vee(\bigvee\{a1^{{\prime}},aa,aa^{2},\ldots\})=\bigvee\{1^{{\prime}},a,a^{2},\ldots\}=\bigvee S=a^{*}$

   • $1^{{\prime}}+a^{*}a=a^{*}$ as well

   • if $ab\leq b$, then $a^{*}b\leq b$: by induction on $n$, we have $a^{n}b\leq b$ whenever $a\leq b$, so that $a^{*}b=\bigvee\{a^{n}b\mid n\in\mathbb{N}\cup\{0\}\}\leq b$.

   • similarly, if $ba\leq b$, then $ba^{*}\leq b$

   All of the above properties satisfy the conditions for an i-semiring to be a Kleene algebra. For this reason, a quantale is sometimes called a standard Kleene algebra.

6. 6.

   Call the multiplication idempotent if each element is an idempotent with respect to the multiplication: $aa=a$ for any $a\in Q$. If $\cdot$ is idempotent and $1=1^{{\prime}}$, then $\cdot=\wedge$. In other words, $ab=a\wedge b$.

   Proof.

   As we have seen, $ab\leq a\wedge b$ in the 2 above. Now, suppose $c\leq a\wedge b$. Then $c\leq a$ and $c\leq b$, so $c=c^{2}\leq cb\leq ab$. So $ab$ is the greatest lower bound of $a$ and $b$, i.e., $ab=a\wedge b$. This also means that $ba=b\wedge a=a\wedge b=ab$. ∎

7. 7.

   In fact, a locale is a quantale if we define $\cdot:=\wedge$. Conversely, a quantale where $\cdot$ is idempotent and $1=1^{{\prime}}$ is a locale.

   Proof.

   If $Q$ is a locale with $\cdot=\wedge$, then $aa=a\wedge a=a$ and $a1=a\wedge 1=a=1\wedge a=1a$, implying $1=1^{{\prime}}$. The infinite distributivity of $\cdot$ over $\vee$ is just a restatement of the infinite distributivity of $\wedge$ over $\vee$ in a locale. Conversely, if $\cdot$ is idempotent and $1=1^{{\prime}}$, then $\cdot=\wedge$ as shown previously, so $a\wedge(\bigvee S)=a(\bigvee S)=\bigvee\{as\mid s\in S\}=\bigvee\{a\wedge s\mid s\in S\}$. Similarly $(\bigvee S)\wedge a=\bigvee\{s\wedge a\mid s\in S\}$. Therefore, $Q$ is a locale. ∎

Remark. A quantale homomorphism between two quantales is a complete lattice homomorphism and a monoid homomorphism at the same time.