I am reading a proof of this result that uses the Chinese Remainder Theorem on (the finite number of) prime ideals $P_i$. In order to apply CRT we should assume that the prime ideals are coprime, i.e. the ring is equal to $P_h + P_k$ for $h \neq k$, but I can't see it. How does it follow?
[Math] Dedekind domain with a finite number of prime ideals is principal
abstract-algebracommutative-algebradedekind-domainprincipal-ideal-domainsring-theory
Related Solutions
Background
Here is the problem:
and the two problems it refers to:
Scanning through the book, the author uses $j(R)$ to refer to the Jacobson radical and $rad(R)$ for nilradicals, so the OP's interpretation is accurate according to the text.
What happened?
zcn's adept example of the semilocalization of $\Bbb Z$ at the complemement of $(2)\cup (3)$ readily produces a domain with just three prime ideals (generated by $0$, $2$ and $3$), and of course a domain can't factor into a direct product of rings since it lacks nontrivial idempotent elements, and yet it's not local. The nilradical is zero, but the Jacobson radical is the intersection of the ideals generated by $2$ and $3$, and it is not nilpotent.
We can speculate that the author, at some point in writing or memory, thought that this would work with finitely many prime ideals and their intersection, when actually it should be finitely many maximal ideals and their intersection. In fact, the problem will go through if we assume a nilpotent Jacobson radical and finitely many maximal ideals.
Let's see how things will work out.
but where are the comaximal ideals?
Good question indeed! The first obvious candidate is the set of maximal ideals of $R$, which are guaranteed to be comaximal. However, applying the Chinese remainder theorem then only yields an isomorphism of $R/J(R)$ with a product of fields, which isn't exactly what we hoped for. But we are talking about powers of the radical: what about powers of maximal ideals?
Then one gets the idea to check if $\{m_i^n\mid i=1\ldots k\}$ is a set of comaximal ideals, where $n$ is the smallest $n$ such that $J(R)^n=\{0\}$ and the $m_i$ are a complete irredundant set of maximal ideals of $R$. This is indeed the case, and is now your first assigment.$(\ast)$
Once this is done, you can apply the Chinese remainder theorem and conclude that $R/\cap m_i^n\cong \prod R/m_i^n$, and you have two further justifications to convince yourself of $(\ast)$:
- $\cap m_i^n=\{0\}$
- $R/m_i^n$ is local for each $i$.
Superfluous bonus stuff
The best you can say is this:
A commutative ring is a finite direct product of local rings iff it is a semiperfect ring. (Here semiperfect means "has finitely many maximal ideals and idempotents lift mod $J(R)$.")
This encompasses more direct products of local rings because there are semiperfect rings whose radicals are not nilpotent (any nonfield local domain, for example.) However, if $J(R)$ is nilpotent, then idempotents lift mod $J(R)$, so this does cover the hypotheses above.
I don't think the CRT can prove this general case. Instead, one can show that having finitely many maximal ideals means you can find a maximal set of finitely many mutually orthogonal idempotents $e_1,\ldots e_k$, such that $e_iRe_i$ is a ring in $R$ with only trivial idempotents. Combined with the lifting property, one can then show that $e_iRe_i$ is local for each $i$. Thus $R\cong \prod e_iRe_i$ yields the decomposition.
Don't forget your exercises marked by $(\ast)$s above!
In fact, the inclusion $IJ \subseteq I \cap J$ is always true. Given ideals $I$ and $J$, an element of $IJ$ is of the form $x = a_1 b_1 + \cdots + a_t b_t$ for some $t \in \mathbb{Z}_{\geq 0}$ and $a_k \in I$ and $b_k \in J$. Since ideals are closed under arbitrary multiplication, then each term of the above sum is in both $I$ and $J$, and since ideals are closed under addition, then the sum is in $I$ and $J$, so $x \in I \cap J$.
Now, let's consider the reverse inclusion. For ease of notation, let's just consider two primary ideals $Q_1$ and $Q_2$. By part (b) we have $Q_1 = (p_1^m)$ and $Q_2 = (p_2^n)$ for some primes $p_1, p_2$ and, as per my comment, assume $p_1$ and $p_2$ are non-associate. Given $x \in Q_1 \cap Q_2$, then $x = p_1^m a$ and $x = p_2^n b$ for some $a,b \in R$. Then $p_1 \mid p_2^nb$, so $p_1 \mid p_2^n$ or $p_1 \mid b$. If $p_1 \mid p_2^n$, then by repeated using the fact that $p_1$ is prime we find that $p_1 \mid p_2$. Then $p_2 = p_1 c$ for some $c \in R$. But since primes are irreducible, then $c$ must be a unit, which contradicts that $p_1$ and $p_2$ are non-associate. Thus $p_1 \mid b$, so $b = p_1 b_1$ for some $b_1 \in R$.
Now we have $$ p_1^m a = x = p_2^n b = p_2^n p_1 b_1 \, . $$ Since $R$ is a domain, then we cancel the factor of $p_1$, which yields $p_1^{m-1} a = p_2^n b_1$. Repeating the same argument as above (or by induction, if you want to be rigorous), then $p_1 \mid b_1$ so $p_1^2 \mid b$, and so on and so forth until we get $p_1^m \mid b$. Then $b = p_1^m b_m$ for some $b_m \in R$, so $$ x = p_2^n b = p_2^n p_1^m b_m \in (p_1^m)(p_2^n) = Q_1 Q_2 \, . $$
We can use the above in the induction step to prove this result for any finite number of primary ideals. I'll leave the details to you.
Suppose we know the result for $k$ primary ideals. Then \begin{align*} Q_1 \cap \cdots \cap Q_k \cap Q_{k+1} &= (Q_1 \cap \cdots \cap Q_k) \cap Q_{k+1} = (Q_1 \cdots Q_k) \cap Q_{k+1} \, . \end{align*} Given $x \in \bigcap_{i=1}^{k+1} Q_i = (Q_1 \cdots Q_k) \cap Q_{k+1}$, then $$x = p_1^{n_1} \cdots p_k^{n_k} a \qquad \text{and } \qquad x = p_{k+1}^{n_{k+1}} b $$ for some $a,b \in R$. You can use the same proof as above to show that $p_{k+1}^{n_{k+1}} \mid a$ which will finish the induction.
Best Answer
Here's one proof. Let $R$ be a Dedekind ring and assume that the prime ideals are $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$. Then $\mathfrak{p}_1^2,\mathfrak{p}_2,\ldots,\mathfrak{p}_n$ are coprime. Pick an element $\pi \in \mathfrak{p}_1\setminus \mathfrak{p}_1^2$ and by CRT you can find an $x\in R$ s.t.
$$ x\equiv \pi\,(\textrm{mod } \mathfrak{p}_1^2),\;\; x\equiv 1\,(\textrm{mod } \mathfrak{p}_k),\; k=2,\ldots,n $$
Factoring we must have $(x)=\mathfrak{p}_1$. It follows that all prime ideals are principal, so all ideals are principal and $R$ is a PID.
EDIT: The definition of a Dedekind domain is a Noetherian integrally closed, integral domain of dimension 1. The last condition means precisely that every nonzero prime ideal is maximal, so maximality of nonzero primes is tautological. Maximal ideals are always coprime.