# Every fractional ideal of $\mathcal{O}$ is generated by two elements

I want to prove the following theorem:

Theorem. Let $I$ be a fractional ideal of $\mathcal{O}, \alpha \in I$. Then there exist $\beta \in \mathcal{O}$ such that $I=(\alpha, \beta)$.

This theorem is already asked here for Dedekind domains but I want to use "Chinese Remainder Theorem" for ideals instead of valuation. The notes I follow starts the proof as:

Let $\mathfrak{p_1},\dots,\mathfrak{p_m}$ be the prime factors of $(\alpha)$. Then, $I = \displaystyle\prod _{i=1}^{m}\mathfrak{p_i}^{k_i}$.

My first question is, why there is not any prime factors in $I$ except $\mathfrak{p_1},\dots,\mathfrak{p_m}$?

Then how can I finish the proof?

• What has $\alpha$ to do with the ideal $I$? I mean, you wrote $\alpha \in \mathcal{O}$. – Paul K Jun 17 '17 at 9:01
• I have edited the question. – Ninja Jun 17 '17 at 10:12
• You are assuming $\mathcal{O}$ to be the ring of integers of a number field? – Hurkyl Jun 17 '17 at 20:36
• Yes I am assuming it to be the ring of integers of some number field. – Ninja Jun 17 '17 at 20:43

Since $\alpha \in I$, we have $(\alpha) \subseteq I$. Thus, every prime dividing $I$ divides $\alpha$ (recall $\mathfrak p \mid I$ iff $\mathfrak p \supseteq I$). This shows that $I = \prod_{i = 1}^m \mathfrak p_i^{k_i}$.

To finish the proof:

1. Show that for a prime ideal $\mathfrak p$ the quotient $\mathcal O / \mathfrak p^k$ is a principal ideal ring.
2. Use the chinese reminder theorem to conclude that for every ideal $I$ the quotient $\mathcal O / I$ is a principal ideal ring.
3. Show the result using that $\mathcal O / (\alpha)$ is a principal ideal ring.
• Do we say that $\mathcal{p}|I$ if $\mathcal{p}$ exists in the decomposition of $I$ like it is an integer? – Ninja Jun 17 '17 at 20:35
• Usually one defines $\mathfrak p \mid I$ if and only if $\mathfrak p \supseteq I$. – Paul K Jun 17 '17 at 20:36
• I want to use CRT somehow but how can I assure that for $i \neq j, \mathcal{p}_i + \mathcal{p}_j = (1)$? Naturally I want to say that since they are "different primes", "they are coprime " but I do not know how to prove it. – Ninja Jun 17 '17 at 20:44
• Prime ideals are maximal in dedekind domains. – Paul K Jun 17 '17 at 20:45
• @Ninja Since $\mathfrak p_i$ and $\mathfrak p_j$ are different prime ideals, we have $\mathfrak p_i \subsetneq \mathfrak p_i + \mathfrak p_j$. Since $\mathfrak p_i$ is maximal, we obtain $\mathfrak p_i + \mathfrak p_j = \mathcal O$. – Paul K Jun 18 '17 at 8:34

Here is a proof using only the CRT for ideals in a Dedekind domain $R$. If $\alpha \in I = \prod P_i^{k_i}$, it suffices to find $\beta \in R$ s.t. $I=gcd (\alpha R, \beta R)$. A priori $\beta \in R$, but the previous equality ensures that $\beta \in I$. Write $\alpha R = \prod P_i^{k_i}. \prod Q_j^{h_j}$. We look for a $\beta$ s.t. no $Q_j$ divides $\beta R$ and every $P_i^{k_i}$ divides exactly $\beta R$. To this end, pick $\beta_i \in P_i^{k_i} - P_i^{{k_i}+1}$ for each $i$. Since all the powers of the $P_i$'s and of the $Q_j$'s are pairwise co-maximal, i.e. $P_i^{a_i} + Q_j^{b_j} = gcd (P_i^{a_i},Q_j^{b_j})= R$, the CRT shows the existence of $\beta$ s.t. $\beta \equiv \beta_i$ mod $P_i^{{k_i}+1}$ for all $i$ and $\beta \equiv 1$ mod $Q_j$ for all $j$. QED

But note that this is essentially the same proof as when using the language of valuations.