Problem 1632

To prove that the commutative ring $R$ with $|R| = p^3$ has exactly one maximal ideal if the number of elements of $\text{zd}(R)$ is of the form $p^n$ where $n \in \mathbb{N^*}$, we will use the fact that a ring $R$ has exactly one maximal ideal if and only if it has no non-trivial idempotent elements.

1. **Case Analysis on the Order of $R$**:
Since $|R| = p^3$, we consider the possible structures of $R$:
- $R$ could be a field.
- $R$ could be a direct product of fields.
- $R$ could be a local ring with a unique maximal ideal.

2. **Case 1: $R$ is a Field**:
If $R$ is a field, then $R$ has exactly one maximal ideal, which is $\{0\}$. In this case, $\text{zd}(R) = \{0\}$, and the number of elements in $\text{zd}(R)$ is $p^0 = 1$, which is of the form $p^n$ with $n = 0$.

3. **Case 2: $R \cong F_{p^3}$**:
If $R$ is isomorphic to the finite field $F_{p^3}$, then $R$ is a field, and hence has exactly one maximal ideal, $\{0\}$. The number of elements in $\text{zd}(R)$ is $p^0 = 1$.

4. **Case 3: $R \cong F_p \times F_p \times F_p$**:
If $R$ is isomorphic to $F_p \times F_p \times F_p$, then $R$ has three maximal ideals, each corresponding to one of the components being zero. The zero divisors in this ring are all elements where at least one component is zero. The number of zero divisors is $p^2 + p^2 + p^2 - 2p + 1 = 3p^2 - 2p + 1$, which is not of the form $p^n$.

5. **Case 4: $R \cong F_{p^2} \times F_p$**:
If $R$ is isomorphic to $F_{p^2} \times F_p$, then $R$ has two maximal ideals. The zero divisors in this ring are all elements where at least one component is zero. The number of zero divisors is $p^2 + p - 1 = p^2 + p - 1$, which is not of the form $p^n$.

6. **Case 5: $R \cong F_p[x]/(x^3)$**:
If $R$ is isomorphic to $F_p[x]/(x^3)$, then $R$ is a local ring with a unique maximal ideal $(x)$. The zero divisors in this ring are all elements of the form $ax + bx^2$ where $a, b \in F_p$. The number of zero divisors is $p^2$, which is of the form $p^n$ with $n = 2$.

7. Conclusion:
From the above cases, we see that the only structure of $R$ that fits the given conditions (i.e., the number of elements in $\text{zd}(R)$ being of the form $p^n$) and has exactly one maximal ideal is when $R$ is a field or a local ring with a unique maximal ideal.

Therefore, $R$ has exactly one maximal ideal.

$\blacksquare$