Relationship Between Subgroups of Abelian Groups & Ideals/Rings.

Question

Clarification. I am currently reading from Dummit and Foote. Given $R$-module $M,$ we require $(1)$ $R$ is unital, and $(2)$ $1\cdot x=x$ for all $x\in M.$ When discussing rings $R,$ for the purposes of this question, we may assume unital.

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Question. Let's say I give you an abelian group $(A,+)$ with (non-zero proper normal) subgroup $I\triangleleft A.$ Assume this group is chosen in such a way so that it may induce ring structure (so as to avoid some trivial counter-examples). Will there always exist a nontrivial multiplicative operator so that $(A,+,\cdot\ )$ denotes a ring which $I$ will be an ideal of? And assuming there exists a multiplicative binary operator such that $I\subset A$ becomes an ideal, will this multiplicative operator be uniquely defined?

Efforts. It's clear that in the case where $(A,+)$ denotes a cyclic group, then there does exist unique ring structure on $A.$ And since all subgroups of $(A,+)$ remain ideals in this instance, all my questions are answered affirmatively. (This results seems to be confirmed with this post.) In the instance of finitely generated abelian groups, the Fundamental Theorem for Finitely Generated Abelian Groups immediately demonstrates how a component-wise multiplicative operator (defined in the natural (modular) arithmetic manner) may serve sufficient in giving us a ring where all subgroups are ideals of some subring with underlying additive group $A.$ But then the next question becomes whether this operator is the only operator which constructs a ring (which is not a field)? To which I don't have an answer. Non-finitely generated abelian groups are unfamiliar to me, so not much progress has been made with respect to this case. (Though if somebody is willing to provide an explanation which generalizes to the non-finitely generated case, I would really appreciate it nonetheless!)

I do not anticipate these statements to be true, though I'm struggling to prove they're not. Any help is appreciated. Thank you all so much for your time.

Answer 1

When the multiplication exists, it need not be unique.

Fix a prime $p$, and let $A$ be the abelian group $$A = \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}}\times \frac{\mathbb{Z}}{p\mathbb{Z}}.$$ Let $\mathbf{R}$ denote the ring (with its usual multiplication) $(\mathbb{Z}/p\mathbb{Z})\times\mathbb{Z}/p\mathbb{Z}$, and let $\mathbf{F}$ denote the field with $p^2$ elements, whose additive group structure can be identified with $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$.

Now consider the rings $R_1 = (\mathbb{Z}/p\mathbb{Z})\times \mathbf{R}$ and $R_2 = (\mathbb{Z}/p\mathbb{Z})\times\mathbf{F}$. Both rings are unital, have underlying additive structure equal to $A$, but the multiplicative structure is different.

Now consider the subgrup $I=\{(a,0,0)\in A\mid a\text{ arbitrary}\}$. This is a nontrivial proper ideal in both $R_1$ and $R_2$, since both $R_1$ and $R_2$ are direct products of two rings and $I$ is the first factor in both products. So we have two distinct multiplications defined on $(A,+)$ that turn it into a ring, and that make $I$ into an ideal. However, the two are not isomorphic, since $R_2$ has a quotient which is isomorphic to a field with $p^2$ elements, but $R_1$ does not (all its quotients are powers of the field with $p$ elements).

Answer 2

So unless I am misunderstanding your question, I believe such a multiplication is not uniquely defined. Let $G =\{0,a,b,c\}$ be the Klein-4 group (with $0$ being the identity). Then $(c) = \{0,c\}$ is a proper, non-trivial normal subgroup. Define the multiplicative operation as follows: $a$ will serve as our unit, $b^2=b$, $c^2=c$, and $bc = cb = 0$. Then $\{0, c\}$ is an ideal. However, we made the arbitrary choice that $a$ is our unit. We could have instead chosen $b$ to be our unit and gotten a different multiplication.