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Suppose $R$ is a Boolean ring. Prove that $a+a=0$ for all $a\in R$. Also prove that $R$ is commutative. Give an example (with explanation) of a Boolean ring.

From what I know, a Boolean ring is a ring for which $a^2=a$ for all $a\in R$.

Under addition a ring is a commutative group.

$a + b = b + a$ (commutative)

$(a + b) + c = a + (b + c)$ (associative)

$a + (-a) = 0$ (inverse exists for every element)

$a + 0 = a$ (identity exists)

Where $a,b,c \in R$

But I'm not really sure how to proceed with the proof from here. Any idea?

5 Answers5

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Hint Applying the definition of Boolean gives $$a + a = (a + a)^2 = a^2 + a^2 + a^2 + a^2 .$$ What does applying it again give?

Travis Willse
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  • So $a+a=(a+a)^2=(a+a)(a+a) = (a+a)a+(a+a)a= a^2+a^2+a^2+a^2$ on one side and $a+a=a^2+a^2$ on the other. Then can I directly write $a^2+a^2=0$? What property are we using? –  May 23 '18 at 15:22
  • Also, I'm not sure what property we're using when we write $(a+a)(a+a)=(a+a)a+(a+a)a$ –  May 23 '18 at 15:23
  • @user48929: That's just the distributive law, also one of the ring axioms. – hmakholm left over Monica May 23 '18 at 15:24
  • @HenningMakholm Oh, nice. Got that part. But how are we concluding $a^2+a^2=0$ from the first part? –  May 23 '18 at 15:26
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    I don't think starting out with $a+a=a^2+a^2$ here is particularly helpful -- just $$ a+a=(a+a)^2=aa+aa+aa+aa=a+a+a+a$$ and cancel $a+a$. – hmakholm left over Monica May 23 '18 at 15:26
  • @HenningMakholm Great, yes! That would be better. By the way, in this method I'm not sure how the existence of $1$ is needed. –  May 23 '18 at 15:28
  • @HenningMakholm I had in mind that canceling in $a^2 + a^2 = a^2 + a^2 + a^2 + a^2$ implies that $a^2 + a^2 = 0$ and thus $a + a = 0$, but you're right that that approach is more parsimonious---I've tweaked the answer to reflect this. – Travis Willse May 23 '18 at 15:35
  • @user48929 NB this solution does use distributivity, which is one of the standard ring axioms, but which was not listed in the question. – Travis Willse May 23 '18 at 15:39
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Hint. Compute $(a+1)^2$ in two ways -- once by the Boolean ring property, another time by using the distributive law that works in all rings to multiply out $(a+1)(a+1)$. Then apply $a^2=a$ once more and cancel terms that appear in both results.

hmakholm left over Monica
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  • How are you defining $1$? –  May 23 '18 at 15:12
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    @user48929: That's the multiplicative identity which has to exist according to the ring axioms. (There are concepts of ring that don't demand this, sometimes known as "rng"s with the i removed -- but they certainly don't lead to Boolean rings). – hmakholm left over Monica May 23 '18 at 15:14
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    Are you sure about that? Afaik rings do not necessarily have a multiplicative identity. They only have an additive identity. It's nowhere mentioned in the question that it is a ring with an identity. Do you have any source for this? –  May 23 '18 at 15:15
  • @user48929: The usual concept of ring does require an identity -- see e.g. Wikipedia's definition. There are some authors who use the word "ring" to mean non-unital ring, but can you point to any of those authors who also define a "Boolean ring" to be one of their rings with just the additional axiom $x=x^2$? – hmakholm left over Monica May 23 '18 at 15:18
  • @user48929 Every Boolean ring gives rise to a Boolean algebra and each Boolean algebra gives rise to a Boolean ring. The additive identity $0$ is equivalent to the empty set while the multiplicative identity $1$ is equivalent to the whole set, the complement of the empty set. – Henry May 23 '18 at 15:19
  • See also https://en.wikipedia.org/wiki/Ring_(mathematics)#Multiplicative_identity:_mandatory_vs._optional for more discussion. – hmakholm left over Monica May 23 '18 at 15:20
  • @HenningMakholm Interesting. But it seems Travis's method doesn't require multiplicative identity (or does it?). I'm confused now –  May 23 '18 at 15:24
  • @user48929: That is correct. But we still need to be speaking about unital rings before $a^2=a$ alone works as a definition of "Boolean ring". Otherwise, the set of all finite subsets of $\mathbb N$, with intersection as multiplication and symmetric difference as addition, would be a Boolean ring -- but that doesn't correspond to any Boolean algebra because there's no complements of anything. – hmakholm left over Monica May 23 '18 at 15:32
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$a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a\Rightarrow 0=2a$

Kempa
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Take $x \in R$, with $R$ boolean ring, in particular is a ring, therefore $(x+x) \in R$ and $(x+x)^{2}=(x+x)$. Thus \begin{align*} (x+x)^{2} &= (x+x)(x+x) \\ &=(x+x)x + (x+x)x \\ &= x^2 + x^2 + x^2 + x^2 \end{align*} since $x^2 = x$ and $(x+x)^2 = (x+x)$ because $R$ is boolean, then \begin{align*} (x+x)^{2} &= x+x+x +x\\ &= (x+x) + (x+x)\\ &= (x+x) \end{align*} and finally we add both sides the right inverse of $-(x+x)$. \begin{align*} (x+x) + (x+x) -(x+x) &= (x+x) - (x+x) \end{align*} we conclude
\begin{align*} (x+x) = 2x = 0 \end{align*}

Gustavo Andres Pava Parra
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Another way different than the others presented here.

Take $a\in R$. Then also $-a\in R$. Hence $ -a=(-a)^2=a^2=a$ and the assertion is proved.

Temoi
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