Given all the multiples of a prime number $p \in \mathbb{Z}$, is $p\mathbb{Z}$ an ideal of $\mathbb{Z}$?

Question

So I'm having a little trouble understanding the concept of an ideal. The book gives the "classic example" of $2\mathbb{Z}$, the even integers, saying these form an ideal. Would I be correct in thinking that given any other prime $p$, like $3$ or $-17$, $p\mathbb{Z}$ is also an ideal of $\mathbb{Z}$? or maybe this even goes for any integer other than $-1, 0, 1$?

(and yeah, I've looked at some of the questions the computer thinks "may already have your answer," and they might, but at such a level of generality that I can't see how it applies to this question without some extra help).

Answer 1

Indeed, $n\mathbb{Z}$ is an ideal of $\mathbb{Z}$ for any integer $n$. I recommend checking this for yourself: it's quite easy to show this using the definition of an ideal.

Furthermore, these are the only ideals of $\mathbb{Z}$, which is a bit more work to show. I'll provide a proof of this fact:

Let $I$ be an ideal of $\mathbb{Z}$. If $I$ contains only zero, then it equals $0\mathbb{Z}$. Otherwise, $I$ contains some nonzero number $x$. Because $I$ is an ideal, it is an additive subgroup, so it contains $-x$. One of these is positive. Therefore $I$ contains a positive integer. Let $n$ be the smallest positive integer in $I$. We claim that $I = n\mathbb{Z}$.

Proof of claim:

First, $I$ must contain every multiple of $n$, so $n\mathbb{Z} \subseteq I$. To show the reverse containment, let $x$ be any element of $I$. By the division algorithm, there exist $q$ and $r$ in $\mathbb{Z}$ such that $x = qn + r$, where $0 \leq r < n$. Note that $r = x - qn$ is in $I$, since $x$ and $n$ are in $I$. If $r$ is nonzero, then $r$ is a positive element of $I$ which is strictly smaller than $n$, contradicting the definition of $n$. Therefore $r=0$, which means $x = qn$, so $x \in n\mathbb{Z}$.

Answer 2

Yes, in fact even more generally $n\mathbb{Z}$ is an ideal for any $n$. The prime case is special because they are examples of what are called prime ideals (because in the integers the prime ideals are $p\mathbb{Z}$), and even maximal ideals. If you've seen how ideals $I$ in a ring $R$ gives a quotient ring $R/I$ then prime and maximal ideals correspond to how nice the quotient is.

Answer 3

$k\Bbb{Z}$ is indeed an ideal over $\Bbb{Z}$ for any $k \in \Bbb{Z}$, but most of the time people are interested in prime ideals.

$p\Bbb{Z}$ with $p$ a prime integer is always a prime ideal.