Yes/No If $R$ be a ring with unit such that $a^2=a$ for $a \in R.$ Then$ R=\{0,1\}$

Question

Is the following statement true/false ?

If $R$ be a ring with unit such that $a^2=a$ for $a \in R.$ Then $ R=\{0,1\}$

My attempt : I think this statement is true

$(a+a)^2=a +a$

$(a+a)(a+a)=a+a \implies a^2+a^2+a^2+a^2=a+a \implies a+a+a+a=a+a$

$\implies a+a=0$ by cancellation laws

Therefore $a=-a \implies 2a=0$ and $R=\mathbb{Z_2}$ i,e $R=\{0,1\}$

No jasmine, https://en.wikipedia.org/wiki/Boolean_ring#Examples — Anne Bauval, Sep 27 '23 at 10:51
okay @AnneBauval In other words, $R=(P(S);\triangle)$ where where $P(S)$ is the power set of $S$.Am i right ? — jasmine, Sep 27 '23 at 11:02
@AnneBauval where $A \triangle B =(A\cup B) \setminus (A\cap B)$ — jasmine, Sep 27 '23 at 11:05
Come on jasmine, again: read the link to Wikipedia more carefully, especially that paragraph "Examples" and in that paragraph, the first example (duplicated in the accepted answer below). Note that you need two operations for a ring, not just one. — Anne Bauval, Sep 27 '23 at 12:23
In addition to the first comments, which already completely answered your question(s), please look at Understanding boolean rings and/or Ring structure on power set of a set.. I think your post should be closed as a duplicate. — Anne Bauval, Sep 27 '23 at 12:46

Xavier Xanthopoulos · Accepted Answer · 2023-09-27T11:01:25.867

2

Consider a set X. Now consider its power set.

Now consider the operation on the power set to be:

Symmetric Difference as "addition"
Intersection as "multiplication"

In this structure, the "square" of any element is itself.

Check that all conditions for a ring hold.

edited Sep 27 '23 at 11:01

answered Sep 27 '23 at 11:00

Xavier Xanthopoulos

44
4

This answer is a duplicate (see links above). – Anne Bauval Sep 27 '23 at 12:38

score 0 · Answer 2 · answered Sep 27 '23 at 11:22

0

Posting this Answer because the other Answer ( which is valid ! ) & the Online Examples & the Comments are mostly about Power Sets & Sub-Sets , while I wanted to give a much simpler Example.

Let the ring have all n-bit Binary Strings Eg when $n=8$ , the Elements are $\{00000000,00000001,00000010,\cdots,11111110,11111111\}$ ( 256 Elements ! )

Let ADDITION $+$ be bit-wise "EXCLUSIVE-OR".
Let MULTIPLICATION $\times$ be bit-wise "AND".

We can see that $a + a = 2a = 00000000$ ( always , no matter which Binary String $a$ is )
We can also see $a \times a = a^2 = a$ ( always , no matter which Binary String $a$ is )

Hence , we can have a "large ring" with arbitrary size , not just 2 Elements $\{0,1\}$.
More-over , there is no necessity to consider just Power Sets & Sub-Sets.
There are other "types" of Examples too.

answered Sep 27 '23 at 11:22

Prem

9,669

1

This is like the other answer, in a different guise, i.e., you example of a ring of $n$-element bits is an encoding of a power-set on an $n$-element set $X={x_1,\dots,x_n}$ where each subset in encoded as a sequence of bits having $1$ in the $k$-th position iff the $x_k$ belongs to that subset. The operations coincide too. (+1) since this might be easier to understand to some. – amrsa Sep 27 '23 at 11:41
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However, this answer is a multiduplicate. See this post, and all those linked or related to it. – Anne Bauval Sep 27 '23 at 12:39
Yes , I realized that Connection , when I was typing $256=2^8$ , @amrsa , & I thought about eliminating a few elements to make it not Isomorphic to the Power Set , though I eventually decided that it would then become unintuitive & cumbersome , hence I left it as-it-was !! Mean-while , I think I have cooked up a new valid Example which may not be easily Connected to the Power Set Way , but I think I should not update my Answer , this Question being a Multi-Duplicate !! – Prem Sep 27 '23 at 16:32

Yes/No If $R$ be a ring with unit such that $a^2=a$ for $a \in R.$ Then$ R=\{0,1\}$

2 Answers2