This was given to me by my friend.
Prove that $I=\langle \{2,3,4,...\}\rangle $ is maximal ideal of $(P(\mathbb{N}),\Delta,\cap)$
My thoughts:-
Let $A\in P(\mathbb{N})$ Then $A\cap \mathbb{N}=\mathbb{N}\cap A=A$ ,so $\mathbb{N}$ is multiplicative identity. Similarly $\phi$ is additive identity.
Now I claim that ideal $I$ contains all possible subsets of $\mathbb{N}\setminus \{1\}$.
Trial Proof of the claim by me:-
$I$ is the principal ideal generated by $\mathbb{N}\setminus \{1\}$ and it contains all elements of the form $A\cap (\mathbb{N}\setminus \{1\}$) where $A\in P(\mathbb{N})$. Since $A$ is arbitary subset of $\mathbb{N}$, the ideal $I$ contains all subsets of $\mathbb{N}\setminus \{1\}$.
Now this proof might be incomplete .If yes , please help me complete it.
To show $I$ is maximal , let us assume that it is properly contained in a maximal ideal $M$(say).
Again I claim that $M$ must contain a set (obviously as an element)larger than $\mathbb{N}\setminus \{1\} $ or otherwise if for every $X\in M$ (where $X \in P(\mathbb{N}))$ , we have $X \subseteq \mathbb{N}\setminus \{1\}$,then $M\subseteq I$, a contradiction.
Hence $M$ must contains a set larger than $\mathbb{N}\setminus \{1\}$ but the only set larger than that is $\mathbb{N}$ which is the multiplicative identity and hence $M=P(\mathbb{N})$. Thus , a contradiction .So $I$ must be maximal.
Again , I am in doubt here.I need your expertise to check my proof or if it is totally wrong, please give a proof or at least a hint.
Thanks for your valuable time!!
It’s close, but not quite right. $M$ need not contain a set larger than $\Bbb N\setminus\{1\}$: it just needs to contain a set that is not in $\Bbb N\setminus\{1\}$. Thus, $M$ must contain some set $A$ such that $1\in A$. (For instance, $\{1\}$ might be an element of $M$.) You want to use $A$ to show that $\Bbb N\in M$.
Now $\Bbb N\setminus A\subseteq\Bbb N\setminus\{1\}$, so $\Bbb N\setminus A\in I\subseteq M$; what is $A\triangle(\Bbb N\setminus A)$?