Problem 1799

To prove that the additive group of $R$ is cyclic, we will proceed through a series of lemmas and logical steps.

### Part 1: Existence of a Generator

1. **Lemma 1: $\forall p \in \wp, pR \neq \{0\}$

- Proof:** Suppose $pR = \{0\}$ for some prime $p$. Then $R_p = \{ a \in R \mid pa = 0 \}$ is a subring of $R$. Since $R$ is infinite and $R_p \neq \{0\}$, the index $|R/R_p|$ must be finite. This contradicts the assumption that every non-zero subring has finite index. Hence, $pR \neq \{0\}$.

2. **Lemma 2: $\forall p \in \wp, pR \neq R$

- Proof:** Suppose $pR = R$ for some prime $p$. Then there exists $a \in R$ such that $1 = pa$. This implies $R$ is a finite ring, which contradicts the assumption that $R$ is infinite. Hence, $pR \neq R$.

3. **Lemma 3: $\forall p \in \wp, |R/pR| = p^n$ for some positive integer $n$

- Proof:** Construct a chain of subrings $pR = R_0 \subseteq R_1 \subseteq R_2 \subseteq \cdots \subseteq R_{k-1} \subseteq R_k = R$ such that each $R_{i+1}/R_i$ is a finite ring over the field $\mathbb{Z}/p\mathbb{Z}$. By the structure theorem for finitely generated modules over a principal ideal domain, $|R_{i+1}/R_i| = p^{m_i}$ for some $m_i$. Thus, $|R/pR| = p^n$ where $n = \sum m_i$.

4. **Lemma 4: $\forall p \neq q, p, q \in \wp, pR \cap qR \neq pR$

- Proof:** Suppose $pR \subseteq qR$. Then $p^n = |R/pR| = |R/qR| \cdot |qR/pR| = q^m \cdot |qR/pR|$. This implies $p^n = q^m \cdot k$ for some integer $k$, which is a contradiction since $p$ and $q$ are distinct primes. Hence, $pR \cap qR \neq pR$.

5. **Lemma 5: $\forall a \in R, \exists N > 0: \forall p \in \wp, p > N, a \notin pR$

- Proof:** Suppose there exists an infinite sequence of primes $\{p_i\}_{i=1}^{\infty}$ such that $a \in p_iR$ for all $i$. Then $\cap_i p_iR = R^{\wp} \neq \{0\}$. Since $R^{\wp}$ is a subring of $R$ with finite index, $|R/R^{\wp}|$ must be finite. However, $|R/R^{\wp}| \geq p_i$ for all $i$, which is a contradiction. Hence, $\exists N > 0$ such that $\forall p > N, a \notin pR$.

6. **Lemma 6: $\forall p \neq q, p, q \in \wp, pR \cap qR = pqR$

- Proof:** Clearly, $pqR \subseteq pR \cap qR$. Suppose $x \in pR \cap qR$. Then $x = pr = qs$ for some $r, s \in R$. Since $p$ and $q$ are distinct primes, $x \in pqR$. Hence, $pR \cap qR = pqR$.

7. **Lemma 7: If $e_1, e_2 \in P_R$ and $a \in R$ such that $a = p_1^{k_1} p_2^{k_2} \cdots p_i^{k_i} e_1$ and $a = q_1^{l_1} q_2^{l_2} \cdots q_j^{l_j} e_2$, then $e_1 = e_2$, $i = j$, $q_v = p_v$, and $l_v = k_v$

- Proof:** From the properties of prime factorization and the uniqueness of $e \in P_R$, it follows that $e_1 = e_2$, $i = j$, $q_v = p_v$, and $l_v = k_v$.

8. **Lemma 8: If $p \in \wp$ and $pR \subseteq R^* \subseteq R$, then $R^* = pR$ or $R^* = R$

- Proof:** Suppose $R^* \neq R$. Then $\forall a \in R^*, \exists \lambda \in \mathbb{Z}, \xi^* \in P_{R^*}$ such that $a = \lambda \xi^*$. Since $\xi^* = \mu \xi$ for some $\mu \in \mathbb{Z}$ and $\xi \in P_R$, it follows that $a = \lambda \mu \xi$. Hence, $R^* = pR$.

9. **Lemma 9: $\forall p \in \wp, |R/pR| = p$

- Proof:** Consider the ring $R/pR$. If there exists $f \in R/pR$ such that $f^2 = 0$, then $R/pR = \{0, f, 2f, \ldots, (p-1)f \}$ and $|R/pR| = p$. If $\forall g \in R/pR, g^2 \neq 0$, then the multiplicative group $G = R/pR \setminus \{0\}$ has an element $e$ such that $e^2 = e$. Hence, $R/pR = \{0, e, 2e, \ldots, (p-1)e \}$ and $|R/pR| = p$.

10. **Lemma 10: If $R'$ is a subring of $R$, then there exists $n \in \mathbb{Z}$ such that $R' = nR$

- Proof:** Let $n = |R/R'|$. Then $nR \subseteq R'$ and $|R/nR| = n = |R/R'|$. Hence, $R' = nR$.

### Conclusion

From Lemma 10, we know that for any subring $R'$ of $R$, there exists $n \in \mathbb{Z}$ such that $R' = nR$. Now, take $\xi \in P_R$ and consider the subring $R_1 = \{ r_1 \xi + r_2 \xi^2 + \cdots + r_k \xi^k + \cdots \}_{r_1, r_2, \ldots, r_k, \ldots \in \mathbb{Z}}$. By Lemma 10, there exists $n \in \mathbb{Z}$ such that $R_1 = nR$. Since $\xi \in P_R$, we have $n = 1$ and $R = R_1$. This implies that $R$ is generated by $\xi$, and hence the additive group of $R$ is cyclic.

$\blacksquare$