Let $I$ be an ideal of $\mathbb{Z}$, the ring of integers. We need to analyze the given statements about the properties of such an ideal.

Ideal in Integers

The ring of integers, $\mathbb{Z}$, is a very important example in abstract algebra. Its structure is quite simple and well-understood. A key property of $\mathbb{Z}$ is that it is a Principal Ideal Domain (PID). A Principal Ideal Domain is an integral domain where every ideal is a principal ideal.

Principal Ideal Property

A principal ideal is an ideal $I$ that can be generated by a single element $a$ from the ring, meaning $I = \{ra \mid r \in R\}$ for some ring $R$ and element $a \in R$. In $\mathbb{Z}$, any ideal $I$ can be described as the set of all multiples of some non-negative integer $n$. That is, $I = \{kn \mid k \in \mathbb{Z}\} = n\mathbb{Z}$. This ideal is generated by the single element $n$. This property holds for every ideal in $\mathbb{Z}$. If $I$ is the zero ideal $\{0\}$, it is generated by 0 ($0\mathbb{Z} = \{0\}$). If $I$ is a non-zero ideal, let $n$ be the smallest positive integer in $I$. Then it can be shown that $I = n\mathbb{Z}$.

Therefore, any ideal $I$ of $\mathbb{Z}$ is a principal ideal.

Maximal and Prime Ideals

Let $R$ be a commutative ring with unity. An ideal $M$ of $R$ is called a maximal ideal if $M \neq R$ and there is no ideal $J$ such that $M \subsetneq J \subsetneq R$. In simpler terms, a maximal ideal is an ideal that is as large as possible without being the entire ring itself.

An ideal $P$ of $R$ is called a prime ideal if $P \neq R$ and for any $a, b \in R$, if $ab \in P$, then $a \in P$ or $b \in P$. In $\mathbb{Z}$, a non-zero ideal $p\mathbb{Z}$ is prime if and only if $p$ is a prime number. The zero ideal $(0) = \{0\}$ is also a prime ideal in $\mathbb{Z}$ because if $ab \in \{0\}$, then $ab=0$, which in the integral domain $\mathbb{Z}$ implies $a=0$ or $b=0$, so $a \in \{0\}$ or $b \in \{0\}$.

Now, let's consider the relationship between maximal and prime ideals in a commutative ring with unity. A fundamental result in ring theory states that every maximal ideal in a commutative ring with unity is a prime ideal.

Since $\mathbb{Z}$ is a commutative ring with unity (the unity element is 1), this general result applies to $\mathbb{Z}$. Therefore, if $I$ is a maximal ideal in $\mathbb{Z}$, it must also be a prime ideal of $\mathbb{Z}$.

Let's summarize the findings based on the options:

• Statement 1: I is a principal ideal. As explained, $\mathbb{Z}$ is a PID, so every ideal $I$ is principal. This statement is true.
• Statement 4: If I is a maximal ideal in $\mathbb{Z}$, then I is a prime ideal of $\mathbb{Z}$. As explained, in any commutative ring with unity, every maximal ideal is prime. This statement is true for $\mathbb{Z}$.

The other statements are not universally true for all ideals $I$ of $\mathbb{Z}$. For instance, the ideal $4\mathbb{Z}$ is principal but not prime (since $2 \times 2 = 4 \in 4\mathbb{Z}$, but $2 \notin 4\mathbb{Z}$). Also, the ideal $(0)$ is prime but not maximal in $\mathbb{Z}$ because $(0) \subsetneq 2\mathbb{Z} \subsetneq \mathbb{Z}$. Statement 3 would require non-zero prime ideals in $\mathbb{Z}$ to be maximal, which is true because $\mathbb{Z}$ is a PID (non-zero prime ideals are maximal in a PID), but the statement as written might imply it holds for all prime ideals, including the zero ideal, which is not maximal.

Based on the properties of ideals in $\mathbb{Z}$, the true statements are that any ideal is principal, and any maximal ideal is prime.