# Equivalent properties of a proper ideal of a generalized boolean algebra

I do not understand the item c) of the following question, the exercise 9 from section 1.2 from the book "Lattice-ordered Rings and Modules" from Stuart A. Steinberg:

A generalized boolean algebra is a distributive lattice $$X$$ with least element $$0$$ which is relatively complemented; this means that complements exist in each closed interval $$[a,b]=\{x\in X:a\leq x\leq b\}$$.

An ideal of a lattice $$X$$ is a nonempty subset $$I$$ such that if $$a,b\in I$$ and $$c\in X$$ with $$c\leq a$$, then $$a\sqcup b\in I$$ and $$c\in I$$. $$I$$ is a prime ideal of $$X$$ if and only if $$I$$ is proper ideal and whenever $$a,b\in X$$ with $$a\sqcap y\in I$$, then $$a\in I$$ or $$b\in I$$. Let $$I$$ be a subset of the generalized boolean algebra $$X$$. Show that $$I$$ is an ideal of $$X$$ if and only if $$I$$ is an ideal of the [induced boolean] ring $$X$$. Moreover, the following are equivalent for the proper ideal $$I$$:

a) $$I$$ is a prime ideal.

b) $$I$$ is a maximal ideal.

c) If $$x\in X$$ and $$y$$ is its complement in some interval $$[a,b]$$, then $$x\in I$$ or $$y\in I$$.

d) $$|X/I|=2$$.

Just help me to interpret correctly the item (c) (I was able to prove the equivalence of the other items).

• Condition $c)$ sounds wrong. It implies that $I=X$. Indeed, we can take $x=y=a=b$ and so $a \in I$ for all $a \in X$. Try using this condition instead: c') If $x\in X$ and $y$ is its complement in some interval $[0,b]$, then $x\in I$ or $y\in I$. Dec 22 '18 at 8:46
• Thanks, I also used this condition you suggested (c'). Dec 22 '18 at 16:14
• Let me post the proof as an answer for the people who are gonna stumble upon this question. Dec 23 '18 at 7:23

I think there might be a typo on the book. Indeed, condition $$c)$$ is not equivalent to the others. It implies that $$I=X$$. Indeed, we can take $$x=y=a=b$$ and so $$a \in I$$ for all $$a \in X$$. Thus there is no proper ideal satisfying $$c)$$. Whereas there are clearly proper ideals satisfying $$a),b)$$ and $$d)$$.
$$c')$$ If $$x \in X$$ and $$y$$ is its complement in some interval $$[0,b]$$, then $$x \in I$$ or $$y \in I$$.
Indeed, we can prove it is equivalent to $$a)$$:
$$a) \Rightarrow c').$$ If $$x \in X$$ and $$y$$ is its complement in some interval $$[0,b]$$, then $$x \wedge y=0 \in I$$. Since $$I$$ is prime, $$x \in I$$ or $$y \in I$$.
$$c') \Rightarrow a).$$ Let $$x,y \in X$$ such that $$x \wedge y \in I$$. Let $$z$$ be the complement of $$y$$ in $$[0, x \vee y]$$. Hence, $$z$$ is the difference "$$x-y$$". By $$c')$$ we have $$y \in I$$ or $$z \in I$$. If $$y \in I$$, we are done. If $$z \in I$$, then $$z \vee (x \wedge y) \in I$$ because $$I$$ is an ideal. But $$z \vee (x \wedge y)= (z \vee x) \wedge (z \vee y)=x \wedge (x \vee y)=x$$ We used $$z \vee x=x$$. That follows from $$z= z \wedge (x \vee y)=(z \wedge x ) \vee (z \wedge y)= (z \wedge x ) \vee 0= z \wedge x$$, which means $$z \leq x$$.