User talk:David815

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Axiom redundancy

It's been an interesting (and educational for several, including me) playing with redundancy of axioms on algebraic structures. However, it is not clear where this belongs in WP, even though it seems to have value; also some editors are quite definite about what they feel, even though I can't always figure what that is. —Quondum 16:20, 23 April 2015 (UTC)[reply]

Yeah, I liked your edit on the Ring page adding the section about redundancy. Your proof was better than the one I found (which I didn't discover by the way). OR or not, it's a valid point that ought to be mentioned. I'll go argue with Slawekb some more. See you. David815 (talk) 20:14, 23 April 2015 (UTC)[reply]

He restored the bare bones (since someone found a reference), so I think we have an adequate compromise. I do not see the need for a proof with a referenced statement. My proof was also something I found in a discussion thread; I liked it because the steps are pretty clear and it is brief. At Vector space there still seems to be resistance. The preferable (and probably easiest!) approach is finding a notable source. —Quondum 21:36, 23 April 2015 (UTC)[reply]

Wedderburn's little theorem

About the induction. I don't think we need an induction. For the rest of the proof to go through, we only need to know about the cardinality of the centerizer Z_x. For that, we only need to know it is a vector space over Z(A); since any f-dim such vector space has the cardinality the power of q. I suppose you can do induction, but that seems not needed. -- Taku (talk)

I agree that all we need for the rest of the proof to work is a restriction on the cardinality of the centralizer, namely that it is $q^{d}$ where $d<n$ and $d$ divides $n$ . But without the induction, although we can easily get that it is $q^{d}$ where $d<n$ , we can't say that $d$ divides $n$ . What the induction does is allow us to conclude that the centralizer is a field so that we can consider $A$ as a vector space over the centralizer, which means that the order of $A$ , $q^{n}$ , is a power of the order of the centralizer, $q^{d}$ ; thus, for some positive integer $k$ , $q^{n}=q^{dk}$ , which means that $d$ divides $n$ . Does this make sense? Or is there another way to get that $d$ divides $n$ without induction? David815 (talk) 20:49, 22 April 2017 (UTC)[reply]

Hmm... I realized there is another way: we can show that the only way that ${\frac {q^{n}-1}{q^{d}-1}}$ can be an integer (and we know it is an integer, because it is the cardinality of a conjugacy class) is if $d$ divides $n$ by using some simple algebra, but I'm not sure whether that's any nicer. David815 (talk) 21:55, 22 April 2017 (UTC)[reply]

Ok, you're right and I missed the reason for the induction (d divides n). I now agree that the induction is a good way to see the centerizer is a field. Thank you and good work for spotting the gap! -- Taku (talk) 23:25, 23 April 2017 (UTC)[reply]

Great! Thanks for pointing out the need for clarification in the cyclotomic polynomial part of the argument. David815 (talk) 02:26, 24 April 2017 (UTC)[reply]