## 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"?
We are speaking about Q under addition, aren't we?

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

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 are linear transformations of the form However, since we're talking about ring automorphisms, we have and hence the result. More interestingly, is trivial.

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