# Prime ideals of $\mathbb{Z}$: equivalent proof?

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 $$1. At this point if it were $$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.

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. $$\square$$