I understand that similar questions have been asked before, but I am looking for an explanation of certain steps in Marcus' proof of the same, in Theorem $17$, Chapter $3$ of Number Fields. I shall paraphrase the proof slightly for convenience.
Theorem. Let $I$ be an ideal in a Dedekind domain $R$, and let $\alpha$ be any nonzero element of $I$. Then there exists $β ∈ I$ such that $I = (α, β)$.
Proof. It is sufficient to construct $\beta\in R$ such that $I = \gcd((\alpha),(\beta))$.
- I know that $\gcd(J,J') = J+J'$, so we expect $I = (\alpha) + (\beta)$ to imply $I = (\alpha,\beta)$. Indeed, $(\alpha) + (\beta) = (\alpha,\beta)$ follows from definitions, as $(\alpha) + (\beta) = R\alpha + R\beta = (\alpha,\beta)$.
Let $P_1^{n_1}P_2^{n_2}\ldots P_r^{n_r}$ be the prime decomposition of $I$, where $P_i$ are distinct. Then, $(\alpha)$ is divisible by all $P_i^{n_i}$. Let $Q_1,\ldots, Q_s$ denote the other primes (if any) which divide $(\alpha)$. We must construct $\beta$ such that none of the $Q_j$ divide $(β)$, and for each $i$, $P_i^{n_i}$ is the exact power of $P_i$ dividing $(β)$. Equivalently, $$\beta\in \bigcap_{i=1}^r (P_i^{n_i} - P_i^{n_i+1}) \cap \bigcap_{j=1}^s (R - Q_j).$$
- Why must we construct such a $\beta$? I know that $A|B$ if and only if $B\subset A$, but I do not see an immediate contradiction.
This can be accomplished via the Chinese Remainder Theorem. Fix $β_i ∈ P_i^{n_i} − P_{i}^{n_i+1}$ (which is necessarily nonempty by unique factorization) and let $β$ satisfy the congruences $$\beta\equiv \beta_i \bmod P_i^{n_i+1}, \quad i = 1, \ldots, r$$ $$\beta\equiv 1 \bmod Q_j, \quad j= 1,\ldots,s$$ (To show that such a $β$ exists we have to show the powers of the $P_i$ and the $Q_j$ are pairwise co-maximal: that the sum of any two is $R$. This is easy to verify if one interprets the sum as the greatest common divisor.)
- I understand the Chinese Remainder Theorem in the following form: Let $R$ be a commutative ring with unity, and $I_1, \ldots, I_n$ be pairwise relatively prime ideals in $R$. Then, $$R\left/\bigcap_{i=1}^n I_i\right. \cong R/I_1 \times \ldots \times R/I_n$$ and the canonical map $r + \bigcap_{i=1}^n I_i \mapsto (r+ I_1, \ldots, r+I_n)$ is a ring isomorphism. How does this help us construct the required $\beta$? Moreover, what does congruence modulo an ideal mean? My guess is that $a \equiv b \bmod I$ if and only if $a-b \in I$. Even if this is true, I have trouble seeing where the congruences in Marcus' proof come from.
I would appreciate any details and explanations. Thank you very much!