[rkowalick]

I am a little confused by your example, but the definition

  σf(x_1,...,x_n) = f(σx_1,...,σx_n)

is certainly consistent and I'm not sure what you were hoping to show with your example functions.

I'm really not sure what σ(2) is supposed to represent? σ acts on rational functions like f and g, so

  σf(5, 3) = f(3, 5) = -2

and

  σg(5,3) = g(3, 5) = 2

Of course σf and f might not be equal, but I don't see how that is a contradiction? Happy to try and clear things up.

[monfrere]

> σ acts on rational functions like f and g

This is what I was missing. But the paper also says σ is an automorphism of Q(x_1, ... x_n). Which is weird, since I thought Q(x_1, ... x_n) was a subfield of the reals (not a field of rational functions). So I still don't get what's going on.

Sigh, I think I've forgotten more since school than I thought.

[ky3]

You're doing fine. Do cheer up.

No field of rational functions here, that's waaaay off given what Stillwell intends to do.

Also, subfields of reals are a bit restrictive, don't you think?

[monfrere]

Oh, right, subfield of the complex numbers. Now sigma is supposed to be an automorphism on that subfield. Which means sigma does take scalar values as arguments, contrary to what stablemap said...

[ky3]

> Oh, right, subfield of the complex numbers

Right-o!

> Now sigma is supposed to be an automorphism on that subfield. Which means sigma does take scalar values as arguments, contrary to what stablemap said...

There's that ol' abuse of notation going on here. The first sigma is just a permutation on the roots: {x_i} -> {x_i}. In particular, this skinny sigma is not defined on scalars.

It embiggens into a second sigma that's your automorphism: Q({x_i}) -> Q({x_i}). This fat sigma now maps scalars.

Now identify the first and second sigmas, and the abuse is complete.