Say, has any one really been far even as to do look more like ?
Consider a polynom with coefficients over a non-commutative ring. How do you define roots of such a polynom ? Basicly, how do you define the induced factorisation of such a polynom ? What can you tell about the aforementionned root (numbers, relations, sauce) ?
And the least important part to us mathematicians : is this useful for anything ?
I've never understood abstract algebra and even my number theory is kinda shit.
But as to the last question, I read in this very boring book called "the mathematician's apology" that the point of higher more "useless" math is to make the processing of more applied math easier.
A (probably somewhat stupid) example: No one really cared about algorithms until computers came about. Now There's an entire field devoted to that and it's all rooted in the same stuff as ever. If/When quantum computers take off, more branches of math will need to be developed.
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Anonymous2009-12-02 2:59
>>3 I read in this very boring book called "the mathematician's apology"
Meh, is it really boring? I was going to read that one.
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Anonymous2009-12-02 3:07
>>4
I really only picked it up because of the title. Knowing a little about it might have helped.
I didn't realize, for example, that it's really about math and not a little book of anecdotes about math teachers.
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Anonymous2009-12-02 3:32
Howdy, chaps ?
I had an answer yesterday : the teacher told us that for example if you consider X²+1 (or something like this, there was some noise at that time) over the Quaternion field, there is an infinite number of roots... in any case, this polynom has at least 3 roots in H, i j and k. Nevertheless I still need a proper definition of a root in non-commutative business.
Btw, I heard that theorical physics need non-commutative algebra so it might be more useful than I thought at first sight.
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Anonymous2009-12-02 5:15
>>5
I guess you didn't realise that the title means "in defense of mathematics", not "I'm sorry for being a mathematican".
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Anonymous2009-12-02 23:29
what kind of "proper" definition are you looking for, it's the _root_ of the polynomial... since it's noncommutative, you just have left and right coefficients for your polynomial
r is a root of f(x) if and only if f(r) = 0.
its harder to factor shit since you're not used to it, but you can still do it
i.e. f(x) = ax^2b-ax-xb+1 = (ax-1)(xb-1) has roots that are the right inverse of a or left inverse of b, if they exist.
At first I did'nt think of "left" or "right" thingies. I wondered how you could obtain a root factorisation when using non-commutative coefficients since the order matters. It's clear now, thanks to you. The last thing I need is if there is a property on the numbers or the multiplicity of roots.
The whole problem appeared when I was studying some algebra book. The author starts by showing that the multiplicative group of a finite field is always cyclic, but he's using a commutative property to do so. Atm he has not demonstrate the Wedderburn theorem so we must assume the fiel is not commutative, hence faggotized his demonstration. The main point was : X^d-1 has d or less roots in a field. So in non-commutative business this is wrong and I was fucked up.
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4tran2009-12-03 8:21
I'm pretty sure left/right inverses are the same in associative algebras.
Lemma: identity element is unique (if there exists a left and right identity)
Suppose there are left/right identities: L,R, respectively.
Then L = LR = R by the definition of the corresponding identities. Thus, L = R, and there is only 1 identity.
Suppose there are left/right inverses of an element g: l,r, respectively. Then
rg = Identity(rg) = (lg)(rg) = l(gr)g = lg = Identity
Thus, right inverses are also left inverses.
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Anonymous2009-12-03 11:03
>>9
i have no idea about number of roots of a polynomial over non-commutative ring, maybe one of these books might have something.
wikipedia says there are left/right euclidean algorithms over these rings. but you should remember that polynomials dont always factor completely into their roots, just irreducible factors (i'm guessing left/right irreducible in non-commutative case), and the factorization might not be unique.