Question

In: Advanced Math

Let X be a non-empty set and P(X) its power set. Then (P(x), symetric difference, intersection)...

Let X be a non-empty set and P(X) its power set.

Then (P(x), symetric difference, intersection) is a ring.

Find a non-trivial ideal of P(X).

Expert Solution

Let's find a non-trivial ideal of P(X).

Let A be a non-empty proper subset of X. Then we define $I_A=\{B\in P(X):B\subseteq A}\}$ . Then it's a non-empty subset of P(X) and $\phi, A\in I_A$ . First we show that this a subring of P(X). Let $B,C\in I_A$ . Then we have $B\subseteq A$ and $C\subseteq A$ . Now we know that, in the additive group of P(X) with symmetric difference, each element is its own additive inverse [ as $Y\Delta Y=(Y\setminus Y)\cup (Y\setminus Y)=\phi$ ] . So additive inverse of C is C and we observe that $B\setminus C\subseteq B\subseteq A$ and $C\setminus B\subseteq C\subseteq A$ which implies $B\Delta C=(B\setminus C)\cup (C\setminus B)\subseteq A$ . Also we have $B\cap C\subseteq B\subseteq A$ . Hence both $B\Delta C,(B\cap C)\in I_A$ . Therefore $I_A$ is a subring of P(X).

Now to prove it's an ideal , let $R\in P(X)~and~B\in I_A$ . Then we have $B\subseteq A$ . So, $R\cap B\subseteq B\subseteq A$ which implies $R\cap B\in I_A$ . Hence $I_A$ is an ideal of P(X). Also it's a proper ideal of P(X) since A is a proper subset of X so $X\not\in I_A$ .

Note:- It may seem that we have proved it's left ideal but remember that P(X) is a commutative ring so $I_A$ is an ideal.

Colby Messinger answered 3 years ago

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