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   A field's underlying groups
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   Author  Topic: A field's underlying groups  (Read 637 times)
Pietro K.C.
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A field's underlying groups  
« on: Feb 3rd, 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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Re: A field's underlying groups  
« Reply #1 on: Apr 25th, 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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