What are the prime and maximal ideals of the power set of $\Bbb N$, i.e. $\mathcal P(\Bbb N)$

Question

We Know that the power set of $\Bbb N$ with respect to the symmetric difference and intersection, i.e. $\left(\mathcal P(\Bbb N),\triangle, \cap \right)$ is a commutative ring with identity.

My question is:

What are the prime and maximal ideals of this ring? Can they be characterised?

Answer 1

Every prime ideal in this ring is maximal. There is not any nice explicit description of all the maximal ideals. The only ones that you can explicitly describe are the principal maximal ideals: for any $x\in\mathbb{N}$ determines a maximal ideal consisting of all $A\subseteq\mathbb{N}$ such that $x\not\in A$ (this ideal is principal, generated by $\mathbb{N}\setminus\{x\}$). There are many other maximal ideals that are not principal, but none of them can be proved to exist without the axiom of choice. A maximal ideal is nonprincipal iff it contains all finite subsets of $\mathbb{N}$.

Here is a characterization of the maximal ideals in terms of familiar algebra of sets. A subset $I\subseteq\mathcal{P}(\mathbb{N})$ is a maximal ideal iff it has the following three properties:

• If $A\in I$ and $B\subseteq I$, then $B\in I$.
• If $A,B\in I$, then $A\cup B\in I$.
• For any $A\subseteq\mathbb{N}$, exactly one of $A$ and $\mathbb{N}\setminus A$ is an element of $I$.

(All of these statements apply equally well to the ring $\mathcal{P}(X)$ for any set $X$, except that if $X$ is finite, there are no nonprincipal maximal ideals.)

Answer 2

Hint: One way of looking at $\mathcal{P}(\Bbb{N})$ is to regard it as a set of tuples $(x_1,x_2,\cdots)$, where $A\subseteq\Bbb{N}$ is represented by the tuple $(\cdots,x_i,\cdots)$ with $x_i=1$ if $i\in A$ and $0$ otherwise. This way, $\Delta$ translates into coordinatewise addition mod $2$ and $\cap$ translates into coordinatewise multiplication mod $2$. It's not difficult to see that $\mathcal{P}(\Bbb{N})\cong\Bbb{F}_2^{\Bbb{N}}$, the countably infinite direct product of fields of order $2$. Now, for any ring $R$ and ideal $I$, $I$ is prime $\Leftrightarrow R/I$ is an integral domain and $I$ is maximal $\Leftrightarrow R/I$ is a field. Can you take it from here?

Answer 3

If $S$ is any set, its power set $\mathcal{P}(S)$ is a Boolean ring, with symmetric difference and intersection as operations. Quoting this wikipedia entry, every prime ideal $P$ in a Boolean ring $R$ is maximal: the quotient ring $R/P$ is isomorphic to the field $\mathbb{F}_2$, which shows the maximality of $P$. Since maximal ideals are always prime, prime ideals and maximal ideals coincide in Boolean rings.