# <p> is a prime ideal iff p is a prime (integral domain)

I was asked to prove the following:

Let $$R$$ be an integral domain and let $$p\ne0$$. Then $$

$$

is a prime ideal $$\iff$$ $$p$$ is a prime.

I worked using the definitions and was able to prove it, but I can't see why I need the assumption that $$R$$ is an integral domain.

Here's one side of the proof:

$$p|ab\implies p|b$$ or $$p|a$$: There exists a $$r\in R$$ such that $$pr=ab$$. Then $$ab\in

$$

. $$

$$

is a prime ideal, so $$a\in

$$

or $$b\in

$$

. if $$a\in

$$

then there exists a $$x\in R$$ such that $$px=a$$, implying that $$p|a$$.

The other side of the proof is similar. Is the proof wrong?

The statement is valid for arbitrary commutative rings with identity, and your proof does not involve the cancellation property of an integral domain.

I think we conventionally require $$R$$ to be an integral domain because of the 'up to unit' identification. In an integral domain, two nonzero elements are associates if and only if one element is a unit multiple of the other.

Let $$R$$ be an integral domain and $$a, b \in R$$. The followings are equivalent:

(1) $$(a)=(b)$$

(2) $$a|b$$ and $$b|a$$

(3) $$\exists$$ $$r \in R^{\times}$$ such that $$a=rb$$

We define early the notions of the greatest common divisor up to unit. e.g. $$\text{gcd}(6, 10)=2$$ in $$\mathbb{Z}$$, as the positive integers are usually selected for representatives. (Note that $$\mathbb{Z}^{\times}=\{1,-1\}.)$$

This might be not satisfying answer since the relation "$$a$$ is also an associate of $$b$$" is an equivalence relation on a commutative unital ring.

If $$R$$ is UFD, we can choose representatives(with respect to unit multiplication) $$\{ p_i\}_{i \in I}$$ for irreducible numbers and give a proposition like this:

Every nonzero nonunit element $$x$$ of $$R$$ can be written $$x=up_{i_1}...p_{i_n}$$ with $$u$$ unit and $$p_{i_j}$$ irreducible. The factorization is unique up to rearrangement.

This statement holds since irreducibles in same equivalent class are related by unit. Imagine how would it be if we have only a 'up to associate' identification. Corresponding statement is for $$u \in R-\{0\}$$ and it seems that we cannot get further information of $$u$$ easily.

The assumption that $$R$$ is an integral domain gives more information about how generators of a principal ideal are related. One aspect of ideal theory is to generalize the notions of prime numbers, irreducible numbers, and related other concepts. They appear as generators of some ideals. Thus we get some advantages by letting $$R$$ to be integral domain.