Coprime ideal definition

Question

I am learning about ideals in my algebra class. If I have a ring $R$, I know that two ideals $I$ and $J$ in $R$ are coprime if $I+J=R$. I also know that $\mathbb{Z}$ is a principal ideal domain. I was TOLD that if you have two integers that are coprime then the ideals that they generate are coprime.

However, I know that $1 \in \mathbb{Z}$ is a unit. And I know that if you have a unit in your ideal then you end up generating the whole ring.

So actually my question has to do with a general ring: how is it possible to have two ideals be coprime (and not trivial)?

Answer 1

If $R$ is a local commutative ring with $1$, then it has not coprime ideals. If $R$ has two maximal ideals $m$, $n$, then $m$ and $n$ are coprime.

---

In $\mathbb{Z}$, if different prime numbers appear in the decomposition of two integers, then the ideals generated by them are coprime.

Answer 2

Consider ideals $(2)$ and $(3)$ in $\mathbb Z$.

The ideal $(2)$ is all integers of the form $2z$ for $z \in \mathbb Z$ (that is the even integers, which can be written as $2 \mathbb Z$).

$(2)$ is an additive subgroup of $\mathbb Z$ viewed as an additive group, meaning the additive identity $0 \in (2)$, but observe the multiplicative identity $1 \not \in (2)$ (if it were, then it would be the whole ring).

Similarly, $(3)$ is the ideal that is all multiples of $3$.

By Bézout's lemma, we know there exist $x,y \in \mathbb Z$ such that $2x + 3y = 1$, therefore by definition of ideal addition, $1 \in (2) + (3)$, and since you know $1$ can generate the whole ring, you can see $(2) + (3) = (1) = \mathbb Z$.