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Let $\phi:R\to S$ be an onto ring homomorphism. Prove that $\phi(\langle a \rangle) = \langle \phi(a)\rangle$.

This seems like an easy problem, but I can't quite figure out how to put it on paper. Could someone please help me?

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Please use MathJax to format your post; as it stands, it's impossible to tell what's being asked. –  user61527 Nov 11 '13 at 2:11
    
@Dillon: Also, please add a more descriptive title. Regards –  Amzoti Nov 11 '13 at 2:13
    
I don't know how to use MathJax, I am new to this. How do I? –  Brian Nov 11 '13 at 2:14
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@Dillon See my comment on your previous question. –  user61527 Nov 11 '13 at 2:15

1 Answer 1

First of all, definitions: we have $$ \langle \phi(a) \rangle = \{s\cdot\phi(a):s\in S\}\\ \langle a \rangle = \{r\cdot a:r\in R\}\\ \phi(\langle a \rangle) = \{\phi(r'):r' \in \langle a \rangle\} $$ It is fairly easy to show $\phi(\langle a \rangle) \subseteq \langle \phi(a) \rangle$. The challenge is going to be showing that $\langle \phi(a) \rangle \subseteq \phi(\langle a \rangle)$. Once you've shown both of these to be true, you may conclude that the two sets are equal.

Showing $\phi(\langle a \rangle) \subseteq \langle \phi(a) \rangle$:

We note that any $s \in \phi(\langle a \rangle)$ can be written as $\phi(ra)$ for some $r \in R$. Because $\phi$ is a homomorphism, we can write $\phi(ra)=\phi(r)\phi(a)$. This is a member of $\langle \phi(a) \rangle$ (why?). Thus, the above holds.

showing that $\langle \phi(a) \rangle \subseteq \phi(\langle a \rangle)$:

We note that any $s \in \langle \phi(a) \rangle$ can be written as $s'\phi(a)$ for some $s' \in S$. Since $\phi$ is onto, there is a $r \in R$ for which $\phi(r) = s'$. Thus, $s = \phi(r)\phi(a) = \phi(ra)$. This is a member of $\phi(\langle a \rangle)$ (why?). Thus the above holds.

Since the sets are subsets of one other, they are equal.

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Does ϕ(a^n)=ϕ(a)^n because it is homomorphic? –  Brian Nov 11 '13 at 2:17
    
This is generally true when $\phi$ is a homomorphism, if that's what you mean. –  Omnomnomnom Nov 11 '13 at 2:19
    
So how would that help me with this problem? I am really confused on this whole section. Please help –  Brian Nov 11 '13 at 2:23
    
Do you have a definition written somewhere for $\langle a \rangle$? Do you understand this definition? Do you understand what the statement $\phi(\langle a \rangle) = \langle \phi(a) \rangle$ means? –  Omnomnomnom Nov 11 '13 at 2:47
    
⟨a⟩={ra : r ∈ ℝ}. I know what that means. I don't necessarily know what ϕ(⟨a⟩)=⟨ϕ(a)⟩ means. Could you explain it to me please? –  Brian Nov 11 '13 at 2:50

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