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Topic: A field's underlying groups (Read 637 times) 

Pietro K.C.
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A field's underlying groups
« on: Feb 3^{rd}, 2003, 6:56pm » 
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Something I read off an algebra problem set from a class named MATH 129, taught god knows where. Not very difficult. If (K, +, *) is a field, show that the two groups (K, +) and (K{0}, *) are NOT isomorphic. Note: A field is an ordered triple (K, +, *), where K is a nonempty set and +,* are binary operations on K, that is, functions from the cartesian product K X K into K itself. Additionally, the operations must satisfy what are called the field axioms: http://mathworld.wolfram.com/FieldAxioms.html In the problem statement, "0" refers to the identity element with respect to +.


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Eigenray
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Re: A field's underlying groups
« Reply #1 on: Apr 25^{th}, 2003, 1:23am » 
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Claim: 0 != 1. If 0 = 1, then for any x in K, x = 1*x = 0*x = (0+0)*x = 0*x + 0*x = 1*x + 1*x = x + x x + x = x + x + x 0 = x So all elements are equal; then {} and {0} would be isomorphic, which is just plain nutty. Claim: For any x, 0*x = 0 0*x + 0*x = (0+0)*x = 0*x So 0*x = 0. Claim: (1)*x = x (1)*x + x = (1 + 1)*x = 0*x = 0 Claim: xy = 0 iff x=0 or y=0. If x=0 or y=0, then xy = 0. If xy=0, and x != 0 and y != 0, then they both have inverses, and then 0 = 0*(y^1*x^1) = xy*y^1*x^1 = x*x^1 = 1, which is false. Okay, the actual proof: x^2 = 1 iff x^2  1 = 0 x^2  1 = x^2 + x + x 1 = (x+1)*(x+ 1) So x^2 = 1 has exactly two solutions, 1 and 1, unless 1+1=0, in which case there's one solution. Suppose y + y = 0. If y != 0, it has an inverse, so 0 = y^1*0 = y^1*(y + y) = 1 + 1 Then for any x in K, 0 = x*0 = x*(1+1) = x + x, so every element of K satisfies x+x=0. Since there is a bijection between K\{0} and K, K must be infinite. Therefore the equation x+x=0 has either 1 solution if 1+1 !=0, or infinitely many solutions otherwise. Suppose we had an isomorphism, f: K* > K+. i.e., f : K\{0} > K is a bijection, and f(xy)=f(x)+f(y) for all x,y in K. f(1) = f(1*1) = f(1)+f(1), so f(1) = 0. x*x = 1 <> f(x*x) = f(x)+f(x) = f(1) = 0. Thus there is a bijection between solutions of x*x=1 and y+y=0, which contradicts the above.


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