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I'm having a bit of trouble understanding the second half of the proof presented here:

http://crazyproject.wordpress.com/2010/08/09/the-order-of-conductor-f-in-the-ring-of-quadratic-integers-is-a-subring-and-has-index-f/#comment-2051

In particular, why is it that $f x S=S$ ? I do not understand why the index of $S$ justifies this.

Thank you!

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If I'm reading the linked proof correctly, I think there is a typo in that the expression $fxS$ should be $fx + S$. Similarly, $f(a+b\omega)S$ should be $f(a + b\omega) + S$, and so on, in that paragraph. Making this change, and carrying on as shown in the blog, the argument there seems to make a bit more sense.

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Hi, I am aware of that since the ring is not necessarily a group wrt multiplication. It only makes sense to define the cosets wrt addition. I still do not see why fx+S=S however. –  MR1992 Jan 15 '13 at 18:50
    
Since the index of $S$ is equal to $f$, therefore, $fx + S = f(x + S) = S$, for any $x$. (Think of these as elements of the additive quotient group $\mathbb{Z}[\omega]/S$, which has order $f$.) –  James Jan 15 '13 at 19:16
    
I'm having difficulty understanding the equalities you've presented. Would you mind clarifying please? –  MR1992 Jan 15 '13 at 21:29
    
The expression $fx + S$ means $x + x + \cdots + x + S$, where the $x$ appears $f$ times (remember $f$ is a positive integer). This can be rewritten as $(x+S) + (x+S) + \cdots + (x+S)$, with $f$ summands, so this is $f(x+S)$. But $f$ times any coset of $S$ is equal to $S$, since $S$ is the zero of the additive quotient group $\mathbb{Z}[\omega]/S$, and the order of any element of this quotient group divides the order ($f$) of the group. Does this help? –  James Jan 16 '13 at 1:08
    
Perfect! Understood! Thank you so much! My group theory from first semester is rusty. I completely forgot that given the ring structure requires an abelian additive group structure, all subgroups are normal and therefore allow for us to define quotient groups. –  MR1992 Jan 16 '13 at 3:38
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