- #1
Bashyboy
- 1,421
- 5
Homework Statement
Let ##R## be a principal ideal domain. (a) Every proper ideal is a product of ##P_1...P_n## of maximal ideals which are uniquely determined up to an order. (b) ##P## is a primary (not prime) ideal if and only ##P=(p^n)## for some prime ##p \in R## and ##n \in \Bbb{N}##. (c) If ##P_1,...,P_k## are primary ideals such that ##P_i = (p_i^{n_i})## and the ##p_i## are distinct primes, then ##P_1...P_k = P_1 \cap ... \cap P_k##. (d) Every proper ideal in ##R## can be expressed (uniquely up to order) as the intersection of a finite number of primary ideals.
Homework Equations
The Attempt at a Solution
I am currently working on part (c) and could use some help. Here is what I have so far. Since we are taking the ##p_i## to be distinct, I believe we can also take them not be associates of each other, but I'm not entirely certain. First note that ##(p_1^{n_1})...(p_k^{n_k}) = (p_1^{n_1}..p_k^{n_k})##, since we are working in a commutative ring. Let ##x \in (p_1^{n_1}..p_k^{n_k})##. Then ##p_1^{n_1}...p_k^{n_k} \mid x## and therefore ##p_i^{n_i} \mid x## for every ##i## and finally ##x \in (p_i^{n_i})## for every ##i##, proving that ##(p_1^{n_1})...(p_k^{n_k}) \subseteq \bigcap_{i=1}^k (p_i^{n_i})##.
Now we try to prove the other set inclusion. First, since ##\bigcap_{i=1}^k (p_i^{n_i})## is the intersection of ideals, it must also be an ideal, from which we can conclude ##\bigcap_{i=1}^k (p_i^{n_i}) = (y)## for some ##y \in R##. This ##y## must be nonzero and not a unit, which means that there exist primes/irreducibles ##c_i## and integers ##m_i \in \Bbb{N}## such that ##y = c_i^{m_1} ... c_s^{m_s}##, because ##R## is also a UFD. Since ##(y) = \bigcap_{i=1}^k (p_i^{n_i})##, ##c_i^{m_1} ... c_s^{m_s}## must be in the intersection, which implies that, for every ##i##, ##p_i^{n_i} \mid c_i^{m_1} ... c_s^{m_s}## and therefore ##p_i \mid c_i^{m_1} ... c_s^{m_s}##. Because ##p_i## is prime, it follows that ##p_i \mid c_j^{m_j}## for some ##j##, and through induction we get ##p_i \mid c_j##. This means that ##c_j = \ell_{ij} p_i## for some ##\ell_{ij} \in \Bbb{N}##, and since ##c_j## is also irreducible, ##\ell_{ij}## must be a unit, from which it follows ##c_j## and ##p_i## are associates. So, for every ##i##, there exists a ##j## such that ##p_i## and ##c_j## are associates, so ##k \le s## (can we take, WLOG, ##k=s##?) For convenience, relabel the elements so that ##p_i## and ##c_i## are associates...
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Okay. At this point, I am unsure of how to proceed. I am trying to show that ##c_i^{m_1} ... c_s^{m_s} \in (p_1^{n_1})...(p_k^{n_k})##, but the path is obscure for some reason, although I believe I am close. I could use some hints.