Let $B$ be a commutative ring with unity. We say that $B$ is a Boolean ring if $x^2=x$ for all $x \in B$. The name “Boolean” certainly rings a bell of the idea of bool values in programming, or in general, the Boolean algebra that is frequently used in logic, digital electronics and computer science.
In this post, we will examine Boolean rings on a level of commutative algebra, followed by an explicit example in algebraic number theory.
Throughout, let $B$ be a Boolean ring.
Proposition 1. In the Boolean ring $B$, we have
- $2x=0$ for all $x \in B$.
- Every prime ideal $\mathfrak{p} \subset B$ is maximal, and $B/\mathfrak{p}$ is a field with two elements.
- Every finitely generated ideal in $A$ is principal.
Proof. For 1, notice that
For 2, it suffices to show that for every prime ideal $\mathfrak{p} \subset B$, we have $B/\mathfrak{p} \cong \mathbb{Z}/2\mathbb{Z}$.
Pick $x \in B \setminus \mathfrak{p}$. Then in $B/\mathfrak{p}$ we have $\overline{x}^2=\overline{x}$, where $\overline{x}=x+\mathfrak{p} \in B/\mathfrak{p}$. Therefore $\overline{x}(\overline{x}-\overline{1})=0$. However, since $B/\mathfrak{p}$ is entire, we see that we must have $\overline{x}=\overline{1}$ since $x \not\in \mathfrak{p}$. Therefore there are exactly two elements in $B/\mathfrak{p}$, namely $\overline{0}$ and $\overline{1}$.
For 3, we use the induction. If $\mathfrak{a}$ is generated by one element, there is nothing to prove. If $\mathfrak{a}=(x,y)$, then we set
This element is interesting because
Therefore for all elements $a=rx+sy$, we have
Therefore we have $\mathfrak{a}=(u)=(x+y+xy)$.
Suppose now we have proved that all ideals generated by $n$ elements are principal. Then for an ideal generated by $n+1$ elements, let’s say $\mathfrak{a}=(x_1,\dots,x_n,x_{n+1})$, for an element
there is an element $y_{n+1} \in \mathfrak{a}$ such that $a_1x_1+\dots+a_nx_n=b_{n+1}y_{n+1}$, and if we set $u_{n+1}=x_{n+1}+y_{n+1}+x_{n+1}y_{n+1}$, then
and therefore $\mathfrak{a}=(u_{n+1})$ as expected. $\square$
Indeed, if $B$ is noetherian, then we see immediately that $\dim B = 0$, where $\dim$ denotes the Krull dimension. Besides, in this case, $B$ is automatically a PID. We should notice however a Boolean ring $B$ is not necessarily noetherian. For example
is Boolean but not noetherian because we can consider the chain of ideals
and $I_1 \subset I_2 \subset \cdots$ is not a stationary chain.
Next we investigate the topology of $\operatorname{Spec}B$. It is required to have the basic knowledge of the Zariski topology.
Proposition 2. Let $X=\operatorname{Spec}B$ and $X_f=X \setminus V(f)$, where $V(f)=\{\mathfrak{p} \in X:f \in \mathfrak{p}\}$ be the basic open sets of $X$ [recall that an open subset of $X$ is quasi-compact if and only if it is a finite union of sets $X_f$. Then
- For each $f \in B$, the set $X_f$ is both open and closed in $X$.
- Let $f_1,\dots,f_n \in B$, then $X_{f_1} \cup \cdots \cup X_{f_n}=X_f$ for some $f \in B$.
- The sets $X_f$ are the only subsets of $X$ which are both open and closed.
- $X$ is a compact Hausdorff space.
Proof. By definition $X_f$ is indeed open. To show that $X_f$ is closed, it suffices to show that $V(f)$ is always open. To do this, we use the fact that $B$ is Boolean, i.e. $f^2=f$ for all $f \in B$. We see immediately that
and on the other hand,
This is to say we have $X_f = X \setminus V(f) = V(1-f)$ to be closed all the time.
For 2, we can simply use the identity $X_f=V(1-f)$ proved above. Indeed,
where $F(f_1,\dots,f_n) \in B$ is a finite sum and product of $f_1,\dots,f_n$ and is the element $f$ that we were looking for.
For 3, we pick a open and closed set $Y \subset X$. Since $Y$ is open, we can write $Y = \bigcup_{i \in I}X_{f_i}$ for some index set $I$. Since $Y$ is closed in $X$, we see that $Y$ is quasi-compact, and therefore the index set $I$ can be chosen to be finite. By 2, there is therefore a $f \in B$ such that $Y=X_f$.
Finally, we show that $X$ is Hausdorff. Indeed, if $\mathfrak{p},\mathfrak{q} \in X$ with $\mathfrak{p} \ne \mathfrak{q}$, then without loss of generality we can assume that there exists $x \in \mathfrak{p}$ such that $x \not\in \mathfrak{q}$. We see then $\mathfrak{p} \in V(x)$ and $\mathfrak{q} \in V(1-x)$, and both $V(x)$ and $V(1-x)$ are open, while $V(x) \cap V(1-x)=\varnothing$. $\square$
Let $K$ be a number field and let $\mathcal{O}_K$ be the ring of integers of $K$. We would expect that $\mathcal{O}_K=\mathbb{Z}[x]$ for some $x \in \mathcal{O}_K$. For example, if $d \in \mathbb{Z}\setminus\{0,1\}$ is a care-free integer, and if we set $K=\mathbb{Q}(\sqrt{d})$, then
So now we pose a question : if we consider $K=\mathbb{Q}(\sqrt{-7},\sqrt{17})$, then does there exist $x \in \mathcal{O}_K$ such that $\mathcal{O}_K = \mathbb{Z}[x]$?
Instead of trying to find such a $x$ manually, we will solve this question with a general setting.
Proposition 3. Let $m,n \in \mathbb{Z}\setminus\{0,1\}$ be distinct integers, square-free, such that $m \equiv n \equiv 1 \pmod{4}$. If we put $K = \mathbb{Q}(\sqrt{m},\sqrt{n})$, then $\mathcal{O}_K = \mathbb{Z} \oplus \mathbb{Z}\alpha \oplus \mathbb{Z}\beta\oplus\mathbb{Z}\alpha\beta$, where $\alpha=\frac{1+\sqrt{n}}{2}$ and $\beta = \frac{1+\sqrt{m}}{2}$.
Proof. First of all we notice that $[K:\mathbb{Q}]=4$ and that $\{1,\alpha,\beta,\alpha\beta\}$ as well as $\{1,\sqrt{m},\sqrt{n},\sqrt{mn}\}$ are two $\mathbb{Q}$-basis of $K$. The Galois group $G(K/\mathbb{Q})$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, generated by $\sigma:\sqrt{n} \mapsto -\sqrt{n}$ and $\gamma:\sqrt{m} \mapsto -\sqrt{m}$. Therefore
We next want to show that
On one hand, $\alpha,\beta \in \mathcal{O}_K$ should be clear. Notice that
and
Therefore if we put $f(X)=X^2-X-\frac{n-1}{4}$, then $f(\alpha)=0$. Likewise, if we put $g(X)=X^2-X-\frac{m-1}{4}$, then $g(\beta)=0$. The first inclusion $\mathbb{Z} \oplus \mathbb{Z}\alpha \oplus \mathbb{Z}\beta\oplus\mathbb{Z}\alpha\beta \subset \mathcal{O}_K$ is then proved.
On the other hand, pick an arbitrary $x = a+b\sqrt{n}+c\sqrt{m}+d\sqrt{mn} \in \mathcal{O}_K \subset K$. We know on the first place that $a,b,c,d\in\mathbb{Q}$. However, we notice that
is an algebraic integer as it is the root of a monic polynomial $(X-2a)^2-4b^2n$. At the same time, we have $2a+2b\sqrt{n}$. Therefore $2a+2b\sqrt{n} \in \mathcal{O}_{\mathbb{Q}(\sqrt{n})}=\mathbb{Z}\left[\frac{1+\sqrt{n}}{2}\right]$ (since $n\equiv 1\pmod{4}$). Therefore there exists $a’,b’\in\mathbb{Z}$ such that
which implies that
Likewise, we see
and in the same way we can prove that $4c \in \mathbb{Z}$.
Finally,
from which it follows that $4d \in \mathbb{Z}$. We have therefore proved that
Finally, since
we have
If we consider the discriminant of $\mathcal{O}_K$, noted by $\Delta_K$, then
and at the same time,
However, since $m^2n^2$ is impair (as $m\equiv n \equiv 1 \pmod{4}$), we can only have
which forces $\{1,\alpha,\beta,\alpha\beta\}$ to be a $\mathbb{Z}$-basis of $\mathcal{O}_K$. $\square$
To answer our question, we restrict ourselves to the case $m \equiv n \equiv 1 \pmod{8}$. In this question we will see that the Boolean ring arises naturally.
Proposition 4. Let $m,n \in \mathbb{Z}\setminus\{0,1\}$ be distinct integers, square-free, such that $m \equiv n \equiv 1 \pmod{8}$. If we put $K = \mathbb{Q}(\sqrt{m},\sqrt{n})$, then there does not exist $t \in \mathcal{O}_K$ such that $\mathcal{O}_K = \mathbb{Z}[t]$.
Proof. The proposition invites us to try to write $\mathcal{O}_K$ as a polynomial ring over $\mathbb{Z}$. As one can see easily,
where $\alpha=\frac{1+\sqrt{n}}{2}$ and $\beta=\frac{1+\sqrt{m}}{2}$ as above with the isomorphism induced by the map
Since in our question, $m\equiv n \equiv 1 \pmod{8}$, we see that $\frac{1-n}{4},\, \frac{1-m}{4} \in 2\mathbb{Z}$. Therefore by a modulo of $2$, we obtain
where $\mathbf{F}_2$ is the finite field of $2$ elements. Here, the ring $B=\mathcal{O}_K/2\mathcal{O}_K$ is a Boolean ring. Indeed, we can now even explicitly write down $\mathcal{O}_K/2\mathcal{O}_K$ as $\mathbf{F}_2[x,y]$ with $x^2=x$ and $y^2=y$. All elements of $\mathcal{O}_K/2\mathcal{O}_K$ can be identified as $a+bx+cy+dxy$ with $a,b,c,d\in\mathbf{F}_2$. There are $2^4=16$ elements in total, and it can be easily seen that $(a+bx+cy+dxy)^2=a+bx+cy+dxy$.
Since $B\cong \mathbf{F}_2[x,y]$ is Boolean, all prime ideals are maximal. There are exactly $4$ maximal ideals:
For a homomorphism $\varphi:B \to \mathbf{F}_2$, we have $\varphi(0)=0$, $\varphi(1)=1$ so $\varphi$ is surjective. The kernel $\ker\varphi$ is therefore a maximal ideal. There are thus exactly $4$ homomorphisms $B \to \mathbf{F}_2$, which correspond to, sending $x$ to $0$ and $y$ to $1$, sending $x$ to $1$ and $y$ to $0$, sending $x$ and $y$ to $1$ and finally sending $x$ and $y$ to $0$, respectively.
Now we show that we cannot pick $t \in \mathcal{O}_K$ such that $\mathcal{O}_K=\mathbb{Z}[t]$. To reach a contradiction, we suppose that such a $t$ exist. It follows that
where $P$ is a polynomial of degree $4$. However this is absurd because for a homomorphism
we can only have two possibilities: $\psi(X)=1$ or $\psi(X)=0$. However we have shown that $\mathcal{O}_K/2\mathcal{O}_K$ can be mapped onto $\mathcal{F}_2$ in $4$ ways. A contradiction. $\square$
Therefore unfortunately, for the number field $K=\mathbb{Q}(\sqrt{-7},\sqrt{17})$, we cannot find $x \in \mathcal{O}_K$ such that $\mathcal{O}_K=\mathbb{Z}[x]$.
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