Galois fields are used in crypto a bit, and I recommend this lecture to understand it in the context of crypto https://www.youtube.com/watch?v=x1v2tX4_dkQ PS: He has an entire series of lectures on his channel. Highly recommend.
Thanks ;)
Galois Theory for Beginners (2010) [pdf]
31–36 of 36 posts
Re: Galois Theory for Beginners (2010) [pdf]
#32Earlier quoted context omitted.
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…
> 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.
Q(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
Re: Galois Theory for Beginners (2010) [pdf]
#33Fun fact: Evariste Galois made major contributions to math in his teens, before dying in a duel at age 20. https://en.wikipedia.org/wiki/%C3%89variste_Galois
That's fantastic. Not him dying, but you're not likely to hear a story like that these days.
Re: Galois Theory for Beginners (2010) [pdf]
#34Earlier quoted context omitted.
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…
> 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.
> 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.
Re: Galois Theory for Beginners (2010) [pdf]
#35I have never studied group theory, so this is way beyond what I'm ready for. But I scanned over a few sections and read to the point that I got lost.
The part that surprised me is that it seems to focus on rational numbers. I always assumed that Galois Groups were focused on more abstract concepts of sets. Is Galois Theory mostly about rational number (or even real numbers), or is the author just using the rationals to keep the paper focused on beginners?
Re: Galois Theory for Beginners (2010) [pdf]
#36Galois theory without functor... Without fundamental theorem of algebra. Are you kidding?
> Galois theory without functor...
The category-theoretic accretion to Galois theory is a much later addition; Galois certainly didn't think in those terms, and I think that it is not obligatory for an expository (or even a mathematical!) treatment of the subject to do so.