If $A$ is a Dedekind domain, and $\{0\} \ne I \vartriangleleft A$ is an ideal, then $A/I$ is a principal ideal ring

Question

If $A$ is a Dedekind domain, and $\{0\} \ne I \vartriangleleft A$ is an ideal, then $A/I$ is a principal ideal ring.

I have questions about a proof of the said result here. I am reproducing the proof from the source (by @jspecter) below:

Factor $I = \displaystyle\prod_{i =1}^n \mathfrak{p}_i^{e^i}.$ Then by the Chinese Remainder Theorem $A/I \cong \displaystyle\bigoplus_{i =1}^n A/\mathfrak{p}_i^{e^i}.$ So it is enough to show each factor $A/\mathfrak{p}_i^{e^i}$ is principal. The ideals of $A/\mathfrak{p}_i^{e^i}$ are exactly the images of the ideals of $A$ containing $\mathfrak{p}_i^{e^i},$ i.e., $\mathfrak{p}_i^{n}$ for $1\le n \leq e_i,$ under the projection map $\pi:A \rightarrow A/\mathfrak{p}_i^{e^i}.$ If $\pi(\mathfrak{p}_i) = \pi(\mathfrak{p}_i^2)$ then $\pi( \mathfrak{p}_i) = 0$ and $A/\mathfrak{p}_i^{e_i}$ is a field. Otherwise, let $\alpha \in\pi( \mathfrak{p}_i) \setminus \pi(\mathfrak{p}_i^2).$ Then $(\alpha)$ is a proper ideal such that $(\alpha) \not\subset \pi(\mathfrak{p}_i^n) $ for any $n\ge 2.$ It follows $(\alpha) = \pi(\mathfrak{p}_i).$ We conclude $\pi(\mathfrak{p}_i^n) = (\alpha^n)$ and hence $A/\mathfrak{p}_i^{e_i}$ is principal.

1. If $\pi(\mathfrak{p}_i) = \pi(\mathfrak{p}_i^2)$, why is $\pi(\mathfrak p_i) = \{0\}$? If this holds, I understand why $A/\mathfrak{p}_i^{e_i}$ is a field.

2. Why is $(\alpha)$ a proper ideal? If $(\alpha)$ is not proper, then $\alpha \in \pi(\mathfrak{p}_i) \setminus \pi(\mathfrak{p}_i^2)$ is a unit in $A/\mathfrak{p}_i^{e_i}$, i.e., there exists $\beta \in A/\mathfrak{p}_i^{e_i}$ such that $\alpha\beta = 1_A + \mathfrak{p}_i^{e_i}$. I don't see an immediate contradiction. We also have $(\alpha)\subset \pi(\mathfrak{p}_i)$, since $\pi(\mathfrak{p}_i)$ is an ideal.

3. I understand $(\alpha) = \pi(\mathfrak{p}_i)$ (assuming previous steps.) How do we get $\pi(\mathfrak{p}_i^n) = (\alpha^n)$?

4. Why is it enough to show that each "factor" $A/\mathfrak p_i^{e_i}$ is a principal ideal ring? Suppose $I$ is a non-zero ideal in $A$ with the factorization $I = \prod_{i=1}^n\mathfrak{p}_i^{e_i}$ as above, and that every $A/\mathfrak{p}_i^{e_i}$ is a principal ideal ring. Let $\{0\} \ne J \vartriangleleft A/I$ be an ideal, and $\phi$ denote the ring isomorphism $A/I \to \displaystyle\bigoplus_{i =1}^n A/\mathfrak{p}_i^{e^i}$. Then, $\phi(J)$ is an ideal in $\displaystyle\bigoplus_{i =1}^n A/\mathfrak{p}_i^{e^i}$. By this lemma, $J = J_1 \times \ldots \times J_n$ where $J_i \vartriangleleft A/\mathfrak{p}_i^{e_i}$. As every $J_i = (j_i)$ is principal, we have $J = (j_1,\ldots,j_n)$. Is this correct?

Thanks a ton!

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P.S. The notation $A/I \cong \displaystyle\bigoplus_{i =1}^n A/\mathfrak{p}_i^{e^i}$ is terribly confusing. Isn't this a direct product instead of a direct sum? I do not wish to complicate things unnecessarily, but the direct sum doesn't function as the coproduct in the category of rings.

Answer 1

About your PS: there is no difference between a direct product and direct sum of finitely many groups or rings or modules in terms of its algebraic structure. I disagree that the use of the direct sum is terribly confusing or an unnecessary complication.

The proof at your source by Bill Dubuque is simpler than the proof by jspecter. It is based on the fact that every nonzero ideal $I$ in $A$ has two generators, where one of them can be an arbitrary nonzero element of $I$. That fact is proved with the Chinese remainder theorem (as is the argument you ask about) and does not require knowing that $A/I$ is a principal ideal ring, so this won't be a circular argument.

Here is a proof of the result you ask about by using the same argument that Dubuque wrote in your source. Each ideal in $A/I$ has the form $J/I$ for an ideal $J$ of $A$ that contains $I$. Pick a nonzero $\alpha \in I$, so $\alpha \in J$ and thus $J = (\alpha,\beta)$ for some $\beta \in J$. Then in $A/I$, its ideal $J/I$ is generated by $\overline{\alpha}$ and $\overline{\beta}$. Since $\alpha \in I$, we have $\overline{\alpha} = \overline{0}$ in $J/I$, so $J/I = (\overline{\beta})$ in $A/I$. Hence each ideal in $A/I$ is principal.

As for the proof you are asking about, in the ring $R = A/\mathfrak p^e$ each proper ideal has the form $\mathfrak p^i/\mathfrak p^e$ where $1 \leq i \leq e$, and this ideal a power (more precisely, the $i$th power) of the ideal $P = \mathfrak p/\mathfrak p^e$, so it suffices to show $P$ is principal in $R$ (a product of principal ideals is principal: $(x)(y) = (xy)$).

Pick $\alpha \in \mathfrak p - \mathfrak p^2$, so $(\alpha)$ is divisible by $\mathfrak p$ exactly once: $(\alpha) = \mathfrak p\mathfrak a$ where $\mathfrak p \nmid \mathfrak a$. We'll show that in $R$, $\overline{\alpha}$ generates $\mathfrak p/\mathfrak p^e$. Since $\mathfrak p$ divides $(\alpha)$ just once, $\gcd((\alpha),\mathfrak p^e) = \mathfrak p$, so $(\alpha)+ \mathfrak p^e = \mathfrak p$ in $A$. Reducing that equation modulo $\mathfrak p^e$, we get $(\overline{\alpha}) = \mathfrak p/\mathfrak p^e$ in $R = A/\mathfrak p^e$.