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. In lambda notation:* sigma
by topynate 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. In lambda notation:*
sigma f = \a b -> f (sigma a) (sigma b) = f b a
That means sigma f = \a b -> b - a. On the other hand, g := \a b -> 2. So sigma g = g.
Looking again at your equality:
> sigma f(5, 3) = sigma (5 - 3) = sigma 2 = f(3, 5) = -2
We should read sigma f(5, 3) as (sigma f)(5,3) = (3 - 5) = -2. Note: sigma f is not equal to sigma (5 - 3), because f is not the same rational function as "5 - 3"!
* This is an abuse of notation, because a and b are bound variables inside the lambda. In this case, being more precise would probably be less clear.
- 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.