I've been pondering the parallels between rings and groups and the commonalities shared by ideals and normal subgroups. There's a theorem stating that if $R$ is a ring and it happens to be a field, then all its ideals are trivial, i.e., $\{0\}$ and $R$ itself.
So, my question is this: Suppose $F$ is a field, and we consider the set $(F, 0_{F}, +)$ solely as a group (without any ring structure). Are all of its normal subgroups trivial subgroups?
I'd appreciate your insights on this matter. Thanks!
Looking forward to your insights and explanations.
You ask if the additive group of a field only trivial normal subgroups. But every subgroup is normal (the group is commutative), so you ask if the additive group has only trivial subgroups.
Well, a group has only the trivial subgroups iff it is trivial or isomorphic to $\mathbb{Z}/p\mathbb{Z}$ for a prime number $p$ (SE/1516797).
So the answer to your question is "No". Take any infinite field, for example.
But there is a similarity between ideals and normal subgroups. Namely, ideals of a ring $R$ classify the surjective ring homomorphisms $R \to S$ (take the kernel of a ring homomorphism). And normal subgroups of a group $G$ classify the surjective group homomorphisms $G \to H$ (take the kernel of a group homomorphism).
More generally, if $\mathcal{T}$ is any algebraic theory (this won't make much sense to you right now, but maybe later), then the congruence relations on a $\mathcal{T}$-algebra $A$ classify the surjective $\mathcal{T}$-algebra homomorphisms $A \to B$. The congruence relations on a ring correspond to the ideals, the congruence relations on a group correspond to the normal subgroups. Therefore, from this perspective of universal algebra, "normal subgroups" and "ideals" are the same concept.