Krdytkyu

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 $n_1nepm 1$ and $n_2nepm 1$.
Then $n_1notin (n)$, in fact if it were $n_1in(n)$, $exists kinmathbb{Z}$ such that $n_1=kn$, therefore $n=(kn)n_2ne n$, absurd. So, $n_1notin (n)$ and $n_2notin (n)$. However $n_1n_2=nin (n)$.

Summing up if $n$ is not prime, then $$n_1n_2in(n)Rightarrow n_1notin (n)quadtext{and}quad n_2notin (n),$$
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 $n_1in (n)$ $exists kinmathbb{Z}_+$ such that $n_1=kn$, then $n_1>n$ absurd. Then, also in this case we have that $$n_1n_2in(n)Rightarrow n_1notin (n)quadtext{and}quad n_2notin (n),$$
therefore $(n)$ is not prime ideal, absurd.

In the same hypothesis we could conclude by saying that: $n=n_1n_2in (n)$, since $(n)$ is prime ideal then $n_1in (n)$ or $n_2in (n)$, but if were $n_1in (n)$, then $exists kinmathbb{Z}_+$ such that $n_1=kn$ therefore $n_1>n$, therefore we would contradict the condition $1<n_1,n_2<n$.

It's correct?

Thanks!

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. $square$

All ideals of $mathbb{Z}$ are prime and we know, since $mathbb{Z}$ is a P.I.D. (Principal Ideal Domain), that its ideals are all of the form $nmathbb{Z}=(n)$, $ngeq0$. This is regardless of the primality of $n$.

It is not hard in $mathbb{Z}$ to see that $pmathbb{Z}=(p)$ a prime ideal ideal if and only if $p$ is prime if and only if $pmathbb{Z}=(p)$ is a maximal ideal.