Proof of Homomorphism Property: Field Condition Implies 1-1 or Zero Mapping

[Bachelier]

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 $\in$ img (g)

now g(a)-g(b) = g(a-b) = 0

i) if a $\neq$ b, then g the zero mapping
ii) if a = b, then g is one to one.

what do you think?

thanks

 

[micromass]

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 $\in$ img (g)

now g(a)-g(b) = g(a-b) = 0

i) if a $\neq$ b, then g the zero mapping
ii) if a = b, then g is one to one.

what do you think?

thanks

I would be very suspicious with your proof. Why?? You didn't use anywhere that R is a field!

Where you went wrong is in (i) and (ii). I just don't know what you did there.

 

[Deveno]

you can't say that because g(a) = g(b), but that a ≠ b, implies g is the 0-map.

for example: g:Z--->Z/(n) given by g(k) = k (mod n) has the property that

g(n) = g(2n), but g is NOT the 0-morphism.

all you have shown is that g is injective, or it's not. which isn't saying much, and requires no proof in the first place.

you should be thinking about the ideal ker(g).

 

[MarcoD]

From the mathematical properties of the assumptions you should prove the properties of the theorem or conclusion. That also implies that with little knowledge of the specific mathematical domain, your proof should have been readable to me. Which it isn't.

Write down the assumptions and write down the goal and make the road from assumptions to conclusion as clear as possible.

 

[lavinia]

Assume g(a) = g(b)

So a and b are only two points of the field. If g(a) = g(b) then the only way this implies that g is zero is if the field has only 2 elements in it.

since the image (g) is a subring, it is closed under subtraction, then g(a) - g(b)= 0 $\in$ img (g)

if g(a) = g(b) then g(a) - g(b)= 0. It has nothing to do with subrings.

now g(a)-g(b) = g(a-b) = 0

i) if a $\neq$ b, then g the zero mapping
ii) if a = b, then g is one to one.

Just because g(a-b) = 0 doesn't mean that g can not be non zero on some other field element.

 

[Bachelier]

Thank you folks.

How about this.

Assume that g is not 1-1 and not the zero mapping.
Then ∃ a ∈ R s.t. g(a) = 0 and a ≠ 0

Since R is a field, a.a$^{-1}$ = 1

So now: g (1) = g(a.a$^{-1}$) = g(a). g(a$^{-1}$) = 0 (since g is an homomorphism)

Then ∀ b ∈ R, g(b) = g(1.b) = g(1).g(b) = 0

Hence g is the zero mapping. $\otimes$

Hence g is the zero mapping.

Does this prove it is 1-1 as well?

 

[Deveno]

Thank you folks.

How about this.

Assume that [STRIKE]g is not 1-1 and not the zero mapping

Then[/STRIKE] (you don't need this part) ∃ a ∈ R s.t. g(a) = 0 and a ≠ 0

Since R is a field, a.a$^{-1}$ = 1

So now: g (1) = g(a.a$^{-1}$) = g(a). g(a$^{-1}$) = 0 (since g is an homomorphism)

Then ∀ b ∈ R, g(b) = g(1.b) = g(1).g(b) = 0

Hence g is the zero mapping. $\otimes$

i believe it does, but you should add:

otherwise, there is no such a, so ker(g) = {0}, and g is 1-1.

 

[Bachelier]

i believe it does, but you should add:

otherwise, there is no such a, so ker(g) = {0}, and g is 1-1.

Thank you.

 

[lavinia]

Thank you folks.

How about this.

Assume that g is not 1-1 and not the zero mapping.
Then ∃ a ∈ R s.t. g(a) = 0 and a ≠ 0

Since R is a field, a.a$^{-1}$ = 1

So now: g (1) = g(a.a$^{-1}$) = g(a). g(a$^{-1}$) = 0 (since g is an homomorphism)

Then ∀ b ∈ R, g(b) = g(1.b) = g(1).g(b) = 0

Hence g is the zero mapping. $\otimes$

Hence g is the zero mapping.

Does this prove it is 1-1 as well?

more simply, 0 = g(b) = b.g(1) so g(1) = 0 since we can divide by b.