Why are ideals specifically additive subgroups?

Question

I'm trying to understand the notion of ideals better, and something I can't quite figure out is why and where the additive portion of the ideal definition comes into use. If I have an ideal $I$ of $(R,+,\cdot)$, why can't $I$ also be a multiplicative subgroup? Or if it can but is rarely used, why is this the case?

Answer 1

One way to think about ideals is as "things you can quotient by to still get a ring", or "things that can be kernels of ring homomorphisms $R \to S$". You can check that these two "definitions" are equivalent to each other, and those are what ideals are. But sometimes it's convenient to write down a more symbolic definition of an ideal in terms of the ring operations in $R$, and that's the "standard" definition you're likely referencing.

Answer 2

If it were a multiplicative subgroup it would have $1$ in it. But an ideal having $1$ as a member is the whole ring. That's sometimes called a "trivial" ideal, not much use in working with them.

Answer 3

Keep in mind that $R$ is not a group under multiplication - for one thing, $0$ is never invertible, and even if you exclude 0 it is not always the case that $R \setminus \{0\}$ is a group (such a ring is called a division ring). When $R$ has a multiplicative identity, the set of all elements of $R$ that are invertible is called the group of units, as an invertible element in a ring is called a unit. Now you can't expect an ideal to be a subgroup of the group of units, since an ideal that contains any unit at all must actually contain the entire ring! So you can't expect the ideal to be a multiplicative "subgroup" in any sense. However, it is true that an ideal is always a subring, in the sense that is an additive subgroup that is closed under multiplication.