## Abstract algebra questions

Forum for the GRE subject test in mathematics.
actuaryanalyst
Posts: 2
Joined: Sat May 23, 2009 6:19 pm

### Abstract algebra questions

guys,why automorphisms of rational numbers are trivial ?

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Re: Abstract algebra questions

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

zombie
Posts: 27
Joined: Thu Nov 20, 2008 2:30 am

### Re: Abstract algebra questions

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
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Re: Abstract algebra questions

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

actuaryanalyst
Posts: 2
Joined: Sat May 23, 2009 6:19 pm

### Re: Abstract algebra questions

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
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Re: Abstract algebra questions

How come "1" must be mapped to "1"?

mk
Posts: 13
Joined: Fri May 22, 2009 1:14 am

### Re: Abstract algebra questions

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

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Re: Abstract algebra questions

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

mk
Posts: 13
Joined: Fri May 22, 2009 1:14 am

### Re: Abstract algebra questions

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
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Re: Abstract algebra questions

Thanks for clarification.

mstrfrdmx
Posts: 12
Joined: Sat Jan 07, 2012 3:35 am

### Re: Abstract algebra questions

This is old but never really received satisfactory treatment.

lime wrote: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.