Відео

I use microsoft whiteboard and record with OBS studio. OBS is free and records via window capture. Whiteboard is not great but gets the job done haha

Thanks bro 👍 You are doing great job

Thanks very much!

Thanks :)

Good idea, let me put some of the Galois theory vids together. Thanks for the support

Thanks very much! I think one reason is because the topics are niche. Only so many people who want to learn number theory/galois theory/etc. Hoping to get more in the future though

I can try, what is your question?

Thanks! Not sure if you found it already but there's a video about that on my channel.

In this case it's the most straightforward way. If you're talking about computational complexity, then I believe some algorithms still factor the polynomial to compute the group, so I don't know if there's a way around it in general.

I would like to make a video eventually that covers that in more detail. Thanks for the suggestion :)

Still finding your videos helpful and clearly presented! Looking forward to more. Thank you!

Hey! I really appreciate it :) UA-cam should have a private messaging button still? You could use that if you want to chat more/have any questions.

Good question. If you try to use p=2 the polynomial becomes x^3, which is reducible. If you reduce mod 3 you get x^3+x+2, which is reducible because 2 is a root. I ended up with 5 because that's the next smallest choice that gives you something irreducible.

Thanks for the support! I have a few Galois theory videos upcoming.

Yes, there are some details missing from that part of the video... hopefully it is still helpful :O

For sure! Glad it was helpful

I'm glad the video was helpful :)

I wasn't able to find a copy of that book easily accessible. I'm guessing the author just has a different definition of cubic resolvent. The one I'm using should be fine for computing Galois groups though.

Thanks! That result is always true, see en.wikipedia.org/wiki/Degree_of_a_field_extension#The_multiplicativity_formula_for_degrees

Thanks! They are an interesting and still active topic. The first example you gave should follow from a similar argument. In particular, the ideals <2> and <3> are still maximal ideals in that ring (you can check this using Legendre symbols for any prime p). The second example is harder because <2> isn't prime in that ring.

@@coconutmath4928 you mean like e.g. 2 divides (1+3√-13)(1-3√-13) but none of these numbers, which are both irreducible. And 2 is also irreducible.

Thank you!

I appreciate it! I use microsoft whiteboard (which can be slow but is generally reliable).

Yes that's true... transitivity only guarantees a p-cycle when p is prime.

Can you clarify this? We did find the automorphisms of G explicitly.

@@coconutmath4928 i mean why we use the galois group for example to solve a cubic equation when we can solve it easily using cardon's method which involve the real and imaginary solutions

The point of Galois theory is not to find the solutions of polynomials, but rather to find intermediate fields in a (finite) Galois extension. For example, in this case we know there is only one intermediate field ℚ ⊂ M ⊂ K, since the Galois group ℤ/4ℤ only has one proper nontrivial subgroup. I feel like a lot of expository material focuses a lot on the insolubility of the quintic. While this is an elegant application of the theory, thinking that is all kind of misses the point.

Of course! For anyone who might be wondering, for any z such that 0 < k{z} <= 1 we have kz = k{z}+k*floor(z). Therefore kz can be written as an integer plus a positive number less than or equal to 1, which forces that number (k{z}) to equal {kz}.

Thanks! Yes, to do this you need to differentiate the eta function (which is like an alternating zeta function) and then use the Wallis product to evaluate the resulting sum. en.wikipedia.org/wiki/Wallis_product#Derivative_of_the_Riemann_zeta_function_at_zero

This is a subgroup of S5, but it isn't generated by a 2 cycle and a 5 cycle. See here groupprops.subwiki.org/wiki/Subgroup_structure_of_symmetric_group:S5. The issue is that there a lot of subgroups of D5 that are isomorphic to Z2 and not all of them contain a 2-cycle, so knowing we have a transposition limits our options by a lot more.

@@coconutmath4928 aaah indeed, it's a double transposition that generated D5. Thank you for pointing that out.

Of course! Subgroups of S_n are tricky.