I am extremely new to abstract math. I was given the following problem and below each of the questions, I have my answer. I can't imagine it is right because I am so confused. Please point me in the right direction and help me understand this. Thank you so much, as always.

Let $U = \{0, 1\}$ and $R$ the collection of all subsets of $U$. This is called the power set of $U$, denoted $\mathcal{P}(U)$. In this case, $R = \mathcal{P}(U) = \{\varnothing, \{0\}, \{1\}, U\}$. Define two binary operations, denoted $\oplus$ and $\otimes$ on $R$ as follows:

$A \oplus B = (A \cup B) \setminus (A \cap B) = \{x \mid x\text{ is an element of }A\text{ or }B\text{, but not both}\}$

$A \otimes B = A \cap B$

With these operations $(R,{\oplus}, {\otimes} )$ is a ring.

a) What is the additive identity in this ring?

The empty set? Because $\{0\} \oplus \varnothing = \{0\}$?

b) What is the additive inverse of $\{0\}$ element in $R$? $\{0\}$ would be, right? Since: $\{0\} \oplus \{0\} = \varnothing$

c) Does this ring have a multiplicative identity? If so, what is it and it's units?

I'm really lost on this. Because, I think it has one...but it isn't unique, which is a requirement of the properties of rings, right? Example: $U \otimes U = U$; $\{0\} \otimes \{0\} = \{0\}$, etc. So, that is a multiplicative inverse, but it isn't unique because there are several, right? What does that mean?

d)Verify associativity I think I got this one, it's pretty simple...just going through the thing verifying they are equal. I guess I'm just having trouble with the above ideas behind rings.

  • $\begingroup$ I've prettified up your notation -- edit the question to see how formulas are written with MathJax, or see this tutorial for more details than you probably want. $\endgroup$ – hmakholm left over Monica Oct 26 '14 at 23:05

(a): correct, except your argument should be that $A\oplus\varnothing=A$ for every $A\in R$, not just $A=\{0\}$. Fortunately this is true.

(b): correct.

(c): No -- the requirement for a multiplicative identity is that you have one element $e$ such that $e\otimes A=A$ and $A\otimes e=A$ for every $A$. And $e=\{0\}$ doesn't work, for example because $\{0\}\otimes \{1\}=\varnothing\ne \{1\}$.

A ring can have only one multiplicative identity -- if $e$ and $f$ are both multiplicative identities then $e\otimes f$ must simultaneously be $e$ and $f$, which is only possible if they're actually the same element.

What you seem to be looking for is idempotent elements, namely $A$s that satisfy $A\otimes A=A$ -- which is indeed the case for all elements of your ring.

(d) is not quite simple -- you need to provide an argument that $(A\otimes B)\otimes C=A\otimes (B\otimes C)$ no matter which sets $A$, $B$, and $C$ are (this is fairly easy), as well as that $(A\oplus B)\oplus C=A\oplus(B\oplus C)$ (this is somewhat more involved). In both cases your attack strategy should probably be to prove that $x\in (A\odot B)\odot C$ if and only if $x\in A\odot(B\odot C)$, for all choices of $x$, $A$, $B$, and $C$.

  • $\begingroup$ Thank you so much for the clear explanation. I feel as though I understand this much more. I definitely intend on practicing more examples to fully understand. Thank you so much for your time. $\endgroup$ – Gilly Oct 26 '14 at 23:15
  • $\begingroup$ @Gilly: If the answer was very helpful, you might upvote it. $\endgroup$ – Martin Brandenburg Oct 28 '14 at 4:26
  • $\begingroup$ @MartinBrandenburg: He can't because he has less than 15 rep. $\endgroup$ – hmakholm left over Monica Oct 28 '14 at 10:13

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