Primary ideals of $\mathbb{Z}$

Question

Let $R=\mathbb{Z}$ I must prove that

In $\mathbb{Z}$ the primary ideals s are precisely the ideals $(p^k)$ where $p$ is a prime number and $n\ge 1$.

$$$$ The definition of primary ideal.

Definition. An ideal $I$ of the ring $R$ is called primary if for all $a,b\in R$ $$(ab\in I\quad\text{and}\quad a\notin I)\Rightarrow b^n\in I, \text{for some positive integer $n$}.$$

Therefore I must prove that

$$(n)\;\text{is primary in}\;\mathbb{Z}\iff n=p^k, k\ge1,p\ge2,\;\text{where}\;p\;\text{is a prime}.$$

Proof ($\Leftarrow$) We suppose that $n=p^k$ for same prime $p$ and $k\ge 1$. Let $a,b\in\mathbb{Z}$ such that $ab\in(p^k)$ and $a\notin (p^k)$. Then $p^k\nmid a$, but $p^k \mid ab$ I no longer know how to proceed.

$(\Rightarrow)$ Let $(n)$ a primary ideal and we suppose for absurd that $n$ is not a power of a prime, then? how could I proceed?

Now, let $ab\in (p^k)$ with $a\notin(p^k)$, then $p^k\mid ab.$ Can I conclude that $p^k\mid b$?

How can I set up the other implication? Thanks

Answer 1

No. $2 \cdot 2^5 \in (2^6)$ and $2 \not\in (2^6)$, but $2^6 \not \mid 2^5$. What you do know is there are $6$ powers of two among $2$ and $2^5$.

If $a b \in (p^k)$, then $a b = m p^k$ for some $m \in \mathbb{Z}$. Since $a \not \in (p^k)$, $a = m' p^\ell$ for $m,\ell \in \mathbb{Z}$ where $m'|m$, $(m',p) = 1$, and $0 \leq \ell < p$. Then $b = \frac{m}{m'}p^{k-\ell}$. Since $k-\ell \geq 1$, $b^k \in (p^k)$.

For the other direction, you should see how to use a composite, as indicated above, to show that a non-prime-power generator of the ideal can be split into parts preventing the ideal being primary.

Answer 2

A primary ideal is a proper ideal by its definition. Any proper nonzero ideal of $\mathbb{Z}$ is of the form $k\mathbb{Z}$ with $k>1$.

⇒ Suppose $k\mathbb{Z}$ is a primary ideal of $\mathbb{Z}$. We need to show that $k\mathbb{Z}=p^n\mathbb{Z}=(p^n)$.
Since $\mathbb{Z}$ is a primary ideal, for any $ab∈\mathbb{Z}$ if $a∉\mathbb{Z}$ then $b^m∈k\mathbb{Z}$ for some positive integer $m$.
Since $k>1$, the prime decomposition of $k$ contains at least one factor, say $p^n$, where $p$ is prime not equal to $1$ and $n≥1$.
Suppose the prime decomposition of $k$ contains two factors, that is, $k=q^α p^n$, where $q,p$ are distinct primes and $α,n∈\mathbb{Z}^+$. Let $a=q^α p^{n-1}$ and $b=p$. Then $ab=q^αp^{n-1}p=q^α p^n=k∈k\mathbb{Z}$ and $a∉k\mathbb{Z}$ (since $k=q^α p^n∤q^α p^{n-1}=a$). And $b^m=p^m$ hence $k=q^α p^n∤p^m=b^m$, therefore $b^m∉k\mathbb{Z}$ for any positive integer $m$. So we constructed an element $ab∈k\mathbb{Z}$ of a primary ideal $k\mathbb{Z}$ such that $a∉k\mathbb{Z}$ and also $b^m∉k\mathbb{Z}$ for any positive integer $m$, a contradiction to the definition of a primary ideal.
(If the prime decomposition of $k$ contains more than two factors, then $k=(\prod_{i}q_i^{α_i})p^n$ and letting $a=(\prod_{i}q_i^{α_i})p^{n-1}$ and $b=p$ we reach the same contradiction just substituting "$q^α$" with "$(\prod_{i}q_i^{α_i})$" throughout the argument).
⦁ Thus, the prime decomposition of $k$ cannot contain more than one factor. And, at the same time, it contains the factor $p^n$. Hence it can only be that $k=p^n$, so that $k\mathbb{Z}=p^n \mathbb{Z}=(p^n)$.

⇐ Suppose $k\mathbb{Z}=p^n\mathbb{Z}=(p^n)$. We need to show that $(p^n)$ is a primary ideal of $\mathbb{Z}$.
Take any $ab∈p^n\mathbb{Z}$ such that $a∉p^n\mathbb{Z}$. It suffices to show that there is some positive integer $m$ such that $b^m∈p^n\mathbb{Z}$. Since $a∉p^n\mathbb{Z}$, it follows that $p^n∤a$, hence the prime decomposition of $a$ either doesn’t contain this prime $p$ or contains a factor $p^{l_1}$ with $1≤l_1<n$, that is, the prime decomposition of $a$ contains a factor $p^{l_1}$ with $0≤l_1<n$. Then, since $k∣ab$, the prime decomposition of $b$ contains a factor $p^{l_2}$ with $l_2≥n-l_1$, so that $n≥l_1+l_2$ and hence $p^n∣p^{l_1+l_2}$. We definitely can take some $m∈\mathbb{Z}^+$ such that $ml_2≥n$. Then, raising $b$ to this power $m$, we get $b^m$, whose prime decomposition contains a factor $p^{ml_2}$, so that $p^n∣b^m$ and hence $b^m∈p^n\mathbb{Z}$.
⦁ Thus, $(p^n)$ is a primary ideal of $\mathbb{Z}$.