3 ms·
> Your sigma is a bijection on rational functions of a, b, not on the evaluations of those functions at particular values of a and b. I see, this is the key po
by monfrere 9y ago
> Your sigma is a bijection on rational functions of a, b, not on the evaluations of those functions at particular values of a and b.
I see, this is the key point I was missing. But now I am confused as to whether Q(x_1, x_2) is supposed to be a subfield of the real numbers or a field of rational functions of x_1, x_2.
- bollu 9y agoQ(x_1, x_2) is the smallest field containing Q, x_1, x_2. So, it's a subfield of the reals, assuming x_1, x_2 \in R
- JadeNB 9y ago> But now I am confused as to whether Q(x_1, x_2) is supposed to be a subfield of the real numbers or a field of rational functions of x_1, x_2. The latter. However, wherever you get `x_1` and `x_2`, if `\{x_1, x_2\}` is algebraically independent over `\mathbb Q`, then `\mathbb Q(x_1, x_2)` is isomorphic to a field of rational functions. This allows you to realise the same ground field inside many different larger fields.