Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Repeating for my exam in commutative algebra.

Let G be a topological abelian group, i.e. such that the mappings $+:G\times G \to G$ and $-:G\to G$ are continuous. Then we have the following Lemma:

Let H be the intersection of all neighborhoods of $0$ in $G$. Then $H$ is a subgroup.

The proof in the books is the following one-liner: "follows from continuity of the group operations". (this is from "Introduction to Commutative Algebra" by Atiyah-MacDonald)

I must admit that I don't really see how that "follows". If there is an easy explanation aimed at someone who has not encountered topological groups in any extent, I'd be happy to read it.

share|improve this question

1 Answer 1

up vote 8 down vote accepted

If $U$ is a neighbourhood of $0$ then so is $-U=\{-x:x\in U\}$. This shows that if $x\in H$ then $-x\in H$.

To show that $H$ is closed under addition, use the fact that if $U$ is a neighbourhood of $0$ then there is another neighbourhood $V$ of $0$ with $V+V\subseteq U$. The existence of $V$ follows from the continuity of addition at $(0,0)$.

share|improve this answer
    
Thanks. So since $+$ is continous, there exists a neighborhood $W \subset G\times G$ such that $+(W)\subseteq U$. Projecting $W$ down to $G$, we get our $V$. Is this correct? –  Fredrik Meyer Dec 7 '10 at 11:53
    
Roughly speaking yes, you need that $W$ contains a set $V\times V$ where $V$ is a neighbourhood of $0$ in $G$. –  Robin Chapman Dec 7 '10 at 14:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.