0

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.

1 Answers1

1

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.