Why can there be more than one maximal ideal of a ring?

Question

I just started learning about ideals and I'm really confused about this.

I read that a ring can have two maximal ideals, the example that was given was $(2)$,$(3)$ for the ring $\mathbb Z$.

But on the other hand, it says that the sum of ideals is also an ideal? Wouldn't then the set $(2)+(3)$ also be an ideal of the ring $\mathbb Z$ that contains both $(2)$ and $(3)$? Why are then $(2)$ and $(3)$ maximal?

Answer 1

It's because $(2)+(3)=\Bbb Z$, which is not an proper ideal. And this is true because $1=3-2$, and therefore, for each $n\in\Bbb Z$, $n=3n-2n$.

Answer 2

Let $A$ be an arbitrary ring, and $\mathfrak{A}=\{ \mathfrak{a}\subset A : \mathfrak{a} \text{ is ideal and } \mathfrak{a} \neq A\}$ be the set of all proper ideals of $A$. One can see quite easily that $\subset$ defines a partial-order on $\mathfrak{A}$. A maximal ideal is now a maximal element of the partially ordered set ($\mathfrak{A}$,$\subset$). I.e. an ideal $\mathfrak{m}\subset A$ is maximal iff $\mathfrak{m}\neq A$ and if for all ideals $\mathfrak{n}\subset $A with $\mathfrak{n}\neq A$ it holds that $\mathfrak{n}\supset\mathfrak{m}$ implies that $\mathfrak{n}=\mathfrak{m}$. Now let $\mathfrak{m},\mathfrak{n}\subset A$ be two distinct maximal ideals then $\mathfrak{m}+\mathfrak{n}=A$, because evidently $\mathfrak{m}+\mathfrak{n}\supset \mathfrak{m}$ hence if $\mathfrak{m}+\mathfrak{n}\neq A$ then $\mathfrak{m}+\mathfrak{n}=\mathfrak{m}$. This means that for all $n\in \mathfrak{n}$ there exist $m,m'\in \mathfrak{m}$ s.t. $m+n=m'$ thus $n=m'-m\in \mathfrak{m}$ thus $\mathfrak{m}\supset \mathfrak{n}$ a contradiction. Thus, we must have $\mathfrak{m}+\mathfrak{n}=A$.