Let $R=\mathbb{Z}$, since $\mathbb{Z}$ is an integer domain the ideal $(0)$ is prime. I must prove that

The prime ideals of $\mathbb{Z}$ are precisely the ideals $(n)$ where $n$ is a prime.

On the implication: $p$ is a prime then $(p)$ is a prime ideal there are no problems.

Now, let $I=(n)$ a prime ideal of $\mathbb{Z}$.

First proof

We suppose to be absurd that $n$ is not prime, then $n=n_1n_2$ where $n_1\ne\pm 1$ and $n_2\ne\pm 1$. Then $n_1\notin (n)$, in fact if it were $n_1\in(n)$, $\exists k\in\mathbb{Z}$ such that $n_1=kn$, therefore $n=(kn)n_2\ne n$, absurd. So, $n_1\notin (n)$ and $n_2\notin (n)$. However $n_1n_2=n\in (n)$.

Summing up if $n$ is not prime, then $$n_1n_2\in(n)\Rightarrow n_1\notin (n)\quad\text{and}\quad n_2\notin (n),$$ therefore $(n)$ is not prime ideal, absurd.

Second proof

If we suppose to know that every ideal $I$ of $\mathbb{Z}$ is of the form $I=(n)$ where $n$ is the least integer not negative which belongs to $I$, then when we suppose that $n$ is not prime, then $n=n_1n_2$, where $1<n_1,n_2<n$. At this point if it were $n_1\in (n)$ $\exists k\in\mathbb{Z}_+$ such that $n_1=kn$, then $n_1>n$ absurd. Then, also in this case we have that $$n_1n_2\in(n)\Rightarrow n_1\notin (n)\quad\text{and}\quad n_2\notin (n),$$ therefore $(n)$ is not prime ideal, absurd.

In the same hypothesis we could conclude by saying that: $n=n_1n_2\in (n)$, since $(n)$ is prime ideal then $n_1\in (n)$ or $n_2\in (n)$, but if were $n_1\in (n)$, then $\exists k\in\mathbb{Z}_+$ such that $n_1=kn$ therefore $n_1>n$, therefore we would contradict the condition $1<n_1,n_2<n$.

It's correct?



We prove that for $a \neq 0$ the following statements are equivalent:

$(a)$ is a prime ideal $\iff$ $a$ is prime in $\mathbb{Z}$

$"\Rightarrow"$ If $a$ is not prime, then $a = bc$ where $b,c \notin \{-1,1\}$. Then $bc \in (a)$, and because it is a prime ideal we have $b \in (a)$ or $c \in (a)$. Assume the latter, that is $c = ka$ for some $k \in \mathbb{Z}$. Then $a = bc = abk$ meaning that $bk = 1$, thus $b$ is invertible. Contradiction.

$"\Leftarrow"$ Assume $a$ is prime. If $bc \in (a)$, then $a|(bc)$ and then $a|b$ or $a|c$ by elementary number theory. I.e. $b \in (a)$ or $c \in (a)$ and we conclude that $(a)$ is prime. $\square$

  • $\begingroup$ Thanks for your ansewer, but at this level of the theory, in general, irreducible elements are not introduced, therefore in the proofs I have to use instruments that I possess at the moment. $\endgroup$ – Jack J. Dec 3 '18 at 11:26
  • 1
    $\begingroup$ I edited my answer. Let me know if something is unclear. $\endgroup$ – user370967 Dec 3 '18 at 12:32

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