Abstract algebra questions

[actuaryanalyst]

guys,why automorphisms of rational numbers are trivial ?

[lime]

Who said that? x -> 2x is automorphism, isn't it? But it is not trivial.

[zombie]

I think it depends on whether we interpret Q as an additive group or a field.

f(x+y) = f(x) + f(y) for x,y ε Q iff for ax + b, b = 0; also, if f is just an additive group isomorphism from Q onto Q, then a would trivially be 1. Isn't this a practice test question? I recall originally thinking Q had ∞ automorphisms then rethinking that it must only be 1.

[lime]

x -> 2x is also "onto" and "one-to-one", isn't it?

[actuaryanalyst]

f(x) --> 2x cant be an automorphism since 1 has to go to 1 and zero has to go to zero.
for integers its easy to prove it as utilizing the axioms for homo plus the fact that 1 goes to 1 we get :
f (n) = f (1 + 1 + .... + 1) // add 1 n times // = f(1) + f(1) + ... f(1) = 1 + 1 + ... + 1 = n - done
but for Q ...

[lime]

How come "1" must be mapped to "1"?
We are speaking about Q under addition, aren't we?

[mk]

He must mean Q as a ring with + and *.

[lime]

Speaking about morphism we basically imply only one operation being preserved, don't we?

[mk]

Yes, but it's a theorem that if a ring homom is onto then it preserves the "1" element, so any automorphism must send 1 to 1.

Edit: Actually ring homoms are functions s.t. f(a+b) = f(a) + f(b) and f(ab) = f(a)f(b), and from this you can deduce that onto homoms preserve the "1" element.

[lime]

Thanks for clarification.

[mstrfrdmx]

This is old but never really received satisfactory treatment.

x -> 2x is also "onto" and "one-to-one", isn't it?

Yes, but it doesn't preserve the multiplicative structure of $$\mathbb{Q}.$$

f(x+y) = f(x) + f(y) for x,y ε Q iff for ax + b, b = 0; also, if f is just an additive group isomorphism from Q onto Q, then a would trivially be 1. Isn't this a practice test question? I recall originally thinking Q had ∞ automorphisms then rethinking that it must only be 1.

All automorphisms of $$\langle \mathbb{Q} , + \rangle$$ are linear transformations of the form $$f(x)=f(1)x$$ However, since we're talking about ring automorphisms, we have $$f(1)=1$$ and hence the result. More interestingly, $$\textrm{Aut} (\mathbb{R} / \mathbb{Q})$$ is trivial.