First, let me explicitly assume $R$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$.
Proposition. The Krull dimension of $R$ is at most $1$.
Proof. Let $\mathfrak{m} = (t)$, and let $\mathfrak{p}$ be prime. We know $\mathfrak{p} \subseteq \mathfrak{m}$, so it is enough to show that either $\mathfrak{p} = \mathfrak{m}$ or $\mathfrak{p} = (0)$. Suppose $\mathfrak{p} \ne \mathfrak{m}$: then $t \notin \mathfrak{p}$. Let $a_0 \in \mathfrak{p}$. For each $a_n$, because $\mathfrak{p}$ is prime, there exists $a_{n+1}$ in $\mathfrak{p}$ such that $a_n = a_{n+1} t$. By the axiom of dependent choice, this yields an infinite ascending sequence of principal ideals:
$$(a_0) \subseteq (a_1) \subseteq (a_2) \subseteq \cdots$$
Since $R$ is noetherian, for $n \gg 0$, we must have $(a_n) = (a_{n+1}) = (a_{n+2}) = \cdots$. Suppose, for a contradiction, that $a_0 \ne 0$. Then, $a_n \ne 0$ and $a_{n+1} \ne 0$, and there is $u \ne 0$ such that $a_{n+1} = a_n u$. But then $a_n = a_{n+1} t = a_n u t$, so cancelling $a_n$ (which we can do because $R$ is an integral domain), we get $1 = u t$, i.e. $t$ is a unit. But then $\mathfrak{m} = R$ – a contradiction. So $a_n = 0$. $\qquad \blacksquare$
Here's an elementary proof which shows why we can reduce to the case where $R$ is an integral domain.
Proposition. Any non-trivial ring $A$ has a minimal prime.
Proof. By Krull's theorem, $A$ has a maximal ideal, which is prime. Let $\Sigma$ be the set of all prime ideals of $A$, partially ordered by inclusion. The intersection of a decereasing chain of prime ideals is a prime ideal, so by Zorn's lemma, $\Sigma$ has a minimal element. $\qquad \blacksquare$
Thus, we can always assume that a maximal chain of prime ideals starts at a minimal prime and ends at a maximal ideal. But if $R$ is a noetherian local ring with principal maximal ideal $\mathfrak{m}$ and $\mathfrak{p}$ is a minimal prime of $R$, then $R / \mathfrak{p}$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$, and $\dim R = \sup_\mathfrak{p} \dim R / \mathfrak{p}$, as $\mathfrak{p}$ varies over the minimal primes.
Update. Georges Elencwajg pointed out in a comment that the first proof actually works without the assumption that $R$ is a domain, because $(1 - u t)$ is always invertible.
Your responses for the first three are exactly correct. To see why $\langle 0 \rangle$ is not maximal in $\mathbb{Z}$, notice that $\langle 0 \rangle \subsetneq \langle x \rangle$ for any nonzero $x \in \mathbb{Z}$.
For the fourth, the definition of a maximal ideal $I$ is that there are no other ideals $J$ such that $I \subsetneq J \subsetneq R$. For example, $\langle 4 \rangle$ is not maximal since $\langle 4 \rangle \subsetneq \langle 2 \rangle \subsetneq R$. If you can convince yourself that this is true, then you should be able to proceed with your problem.
As an aside, there are some very useful facts you should be aware of (and prove, if possible):
- An ideal $I$ of a commutative ring $R$ is maximal $\iff$ $R/I$ is a field.
- An ideal $I$ of a commutative ring $R$ is prime $\iff$ $R/I$ is an integral domain.
- All maximal ideals are prime.
- If $R$ is a principal ideal domain ($\mathbb{Z}$, e.g.), then all nonzero prime ideals are maximal.
- If $R$ is a field, then $\langle 0 \rangle$ is the only maximal ideal.
Best Answer
Basically you need to verify that Zorn's lemma applies. If you have a chain of ideals that all contain $p$ and all don't contain $1$ then their union contains $p$ and doesn't contain $1$ (every chain has an upper bound). So the set of proper ideals containing $p$ contains a maximal element and that element is a maximal ideal.
Which should be familiar.