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Kimi
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Topology

Postby Kimi » Tue, 28 Jun 2005 10:20 pm

To show you the math behind IT which doesn't seem to be related huh?

Let R be a ring with identity which is not necessarily commutative. In view of Theorem A, it is natural to conjecture that R is a division ring <=> it has no non-trivial ideals. Try to prove this conjecture by the method used in the proof of Theorem A. At what precise point does this attempted proof break down? How much of the conjecture can you prove?

Theorem A: If R is a commutative ring with identity, then R is a field <=> it has no non-trvial ideals.
We first assume that R is a field, and we show that is has no non-trivial ideals. It suffices to show that if I is a non-zero ideal in R then I=R. Since I is non-zero, it must contain some element a != 0. R is a field, so a has an inverse a and I (being an ideal) contains 1=inverse a multiplied by a. Since I contains 1, it also contains x=x1 for every x in R, and there I=R.
We now assume that R has no non-trivial ideals, and we prove that R is a field by showing that if x is a non-zero element in R, ten x has an inverse. The set I={yx: yE(element)R} of all multiplies of x by elements of R is easily seen to be an ideal. Since I contains x=1x, it is a non-zero ideal, and it consequently equals R. We conclude from this that I contains 1, and therefore that there is an element y in R such that yx=1. This shows that x has an inverse, so R is a field.

Anybody can prove it maybe? I even a PhD friend I know couldn't get this one right :(

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Postby Guest » Tue, 28 Jun 2005 10:35 pm

brother, it's time for you go to TT... yeah, be careful

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Kimi
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Postby Kimi » Tue, 28 Jun 2005 10:36 pm

TT??


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