What does "compute Aut(G)" mean?

[Hill]

Homework Statement

Let G = {a+b√5 : a,b ∈ Q} ⊂ R.
Prove that G+ = G is a subgroup of R under addition, and compute Aut(G+).

Relevant Equations

Aut(G+)

I know that Aut($G^+$) is the group of automorphisms on $G^+$. But what does it mean to "compute" it?

 

[Hill]

I think that it means

$Aut(G^+)=Q^* \times Q^*$

where

$Q^*=Q\setminus \{ 0 \}$.

Is it correct?

 

[sbrothy]

You wouldn't be from Arizona would you?

https://math.arizona.edu/~cais/594Page/soln/solnexam1.pdf

Nah, I'm on very thin ice here... Sorry.

 

[Hill]

You wouldn't be from Arizona would you?

https://math.arizona.edu/~cais/594Page/soln/solnexam1.pdf

Nah, I'm on very thin ice here... Sorry.

No, I'm not from Arizona.

Also the link doesn't work for me. I get this:

 

[haruspex]

I think it is a matter of spotting that G+ is isomorphic to a much more familiar object and applying a standard result. The word "nonsingular" is a hint.

 

[Hill]

I think that any automorphism on $G^+$ has a form $a+b\sqrt5 \to qa+rb\sqrt5$, where $q,r \in \mathbb{Q} \setminus \{ 0 \}$. Then
$Aut(G^+)=\mathbb{Q} \setminus \{ 0 \} \times \mathbb{Q} \setminus \{ 0 \}$.

 

[haruspex]

I think that any automorphism on $G^+$ has a form $a+b\sqrt5 \to qa+rb\sqrt5$, where $q,r \in \mathbb{Q} \setminus \{ 0 \}$.

Why exclude 0?
Edit: And what about $a+b\sqrt5 \to qb+ra\sqrt5$?

 

[Hill]

Why exclude 0?

Because say $a+b\sqrt5 \to rb\sqrt5$ is not bijective.

 

[haruspex]

Because say $a+b\sqrt5 \to rb\sqrt5$ is not injective.

I think you mean it is not surjective. Or to put it another way, that transformation collapses the plane (x, y) into the line (0, y). What other lines might a linear transformation collapse the plane into?

Also, please see my edit to post #7.

 

[Hill]

I think you mean it is not surjective. Or to put it another way, that transformation collapses the plane (x, y) into the line (0, y). What other lines might a linear transformation collapse the plane into?

Also, please see my edit to post #7.

Besides $(x, 0)$, I don't see any.

Edit: And what about $a+b\sqrt5 \to qb+ra\sqrt5$?

It makes the answer rather $Aut(G^+)=\mathbb{Q} \setminus \{ 0 \} \times \mathbb{Q} \setminus \{ 0 \}\times \mathbb{Z}_2$.

 

[haruspex]

Besides $(x, 0)$, I don't see any.

What do you know about linear transformations and collapse of dimensions?

 

[Hill]

What do you know about linear transformations and collapse of dimensions?

I think I know what it is. The group of invertible linear transformations, $GL(2, \mathbb{Q})$.

 

[haruspex]

I think I know what it is. The group of invertible linear transformations, $GL(2, \mathbb{Q})$.

I agree.