- #1
Bachelier
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The question is:
Let g: R --> T be a homomorphism. If R is a Field, show that g is either 1-1 or the zero mapping
So I used a direct proof with cases.
Assume g(a) = g(b)
since the image (g) is a subring, it is closed under subtraction, then g(a) - g(b)= 0 [itex]\in[/itex] img (g)
now g(a)-g(b) = g(a-b) = 0
i) if a [itex]\neq[/itex] b, then g the zero mapping
ii) if a = b, then g is one to one.
what do you think?
thanks
Let g: R --> T be a homomorphism. If R is a Field, show that g is either 1-1 or the zero mapping
So I used a direct proof with cases.
Assume g(a) = g(b)
since the image (g) is a subring, it is closed under subtraction, then g(a) - g(b)= 0 [itex]\in[/itex] img (g)
now g(a)-g(b) = g(a-b) = 0
i) if a [itex]\neq[/itex] b, then g the zero mapping
ii) if a = b, then g is one to one.
what do you think?
thanks