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Abstract Algebra, Division Ring question |
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| Oct31-10, 12:15 PM | #1 |
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Abstract Algebra, Division Ring question
1. The problem statement, all variables and given/known data
Let R = { [ a + b*sqrt(m) c + d*sqrt(m) ] } [ n(c - d*sqrt(m)) a - b*sqrt(m) ] (Sorry if the matrix is unclear... I can't get it space nicely. r11 = a + b*sqrt(m) r12 = c + d*sqrt(m) r21 = n(c -d*sqrt(m)) and r22 = a-b*sqrt(m) ) where a, b, c, d are elements of the rationals, and m, n elements of the integers, m and n not zero. Show R is a division Ring if and only if the equation x^2 - m*y - n*z^2 +m*n*t = 0 has no other rationals solution except x = y = z = t = 0 ( The trivial solution.) 2. Relevant equations Going to use the determinant in the forward direction of the proof. Possibly need linear independence. A division Ring is simply a field without commutativity.( I know a lot of people call it a skew-field) 3. The attempt at a solution (====>) So i working on the forward direction and the only way R is a division ring if that matrix is invertible. So the determinant of that matrix would be a^2 - m*b^2 - (n*c^2 -n*d^2*m). So we have a^2 -m*b^2 -n*c^2 +n*m*d^2 = 0. So i set it equal to 0. (that equation looks pretty similar to the one I need.) So i guess i need to show that a = b = c = d = 0 is the only rational solution. It seems i need to show a,b,c,d are linearly independent maybe but I am stuck but I feel its so close to being finished. (<===) So assume the equation x^2 - m*y - n*z^2 +m*n*t = 0 has only the trivial solution. So we have a linear combination set equal to zero and the only solution is the trivial solution right? So i guess then x, y, z, t are linearly independent? I am not sure where this is going to get me to that R is a division Ring I am really completely stuck on this one. |
| Oct31-10, 02:52 PM | #2 |
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I have no idea why no one is willing to help me with this problem I am extremely close.
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