Boolean ring is defined with operations of ring multiplication corresponding to conjunction or meet ∧, and ring addition to exclusive disjunction or symmetric difference (not disjunction ∨). I understand that algebraic structure involving conjunction and disjunction is lattice. Yet, I struggle to witness what ring axiom is violated if we take disjunction as ring addition.

up vote 1 down vote accepted

If we take disjunction for addition then the ring axiom "for every $x$ there exists $y$ such that $x+y=0$" is violated.

Looking at things as a lattice, if $x>0$ then $x\lor y\ge x >0$, so $x+y\ne 0$.

Or to be concrete, consider the power set of $X$. Given $A\subset X$ with $A\ne\emptyset$, there is no $B\subset X$ with $A\cup B=\emptyset$.

  • Indeed. Sloppily, I was misreading 1 instead of 0 in this axiom. – Tegiri Nenashi Jan 21 '16 at 22:51
  • I had an unfair advantage here. Having read that a Boolean algebra "is" a Boolean ring, for thirty years I just assumed that the addition was disjunction. Had a reason to read up on that stuff about a year ago, was puzzled by the same question at first. – David C. Ullrich Jan 21 '16 at 22:59
  • Without inverse for addition it is semiring. – Tegiri Nenashi Feb 1 '16 at 17:53
  • Started noticing sloppiness in the literature. arxiv.org/pdf/1602.01171.pdf page 34 "multiplication of matrices in the Boolean ring" $\langle B,∨,∧,0,1\rangle$. With conjunction and disjunction operations (not symmetric difference) they really meant "semiring". – Tegiri Nenashi Feb 6 '16 at 18:36

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