Galois theory: Finite fields

This lecture is part of an online graduate course on Galois theory.

We use the theory of splitting fields to classify finite fields: there is one of each prime power order (up to isomorphism).

We give a few examples of small order, and point out that there seems to be no good choice for a standard finite field of given order: this depends on the choice of an irreducible polynomial.

Finally we show how to count the number of irreducible polynoials of given degree in a finite field (using the field of order 64 as an example).

The lecture mentioned that there does not seem to be a clear choice for the "best" of the three irreducible polynomials of degree 4 over F2. Here is a poll where viewers can vote on which they prefer and see how others have voted:

[TheArangol]

Professor Borcherds, these videos are fantastic. Thank you very much for taking the time to post them.

Dear Professor Borcherds, your name reminded me of the days when I took Modular Forms class by Professor Duke. In the class, the j function and its product form were discussed. Thank you very much for your effort in putting your videos. This will be a good place that I can go back to review Commutative Algebra and Algebraic Geometry.

[romywilliamson]

I have my Galois theory exam in 2 weeks, I'm going to be imagining the mathematicians on Alpha Centauri whenever a finite field comes up :D
seriously, thanks for demystifying this topic for me.

[romankrylov]

19:36 Thumbs up for the Galactic Standards Organization proposal.

[sachalucienmoserferreira]

Thanks for share your knowledgment with us! Its a pleasure watching your video course

[downinthehole]

The ISO framing device was hilarious (Aaron Aardvark & Zach Zigmund), thank you for that!

[joeseppe]

thanks for the video
as it solves a long lasting question in simple and elegant way...

[pronaybiswas]

I laughed so hard when he said "can't do Arithmetic". T H E P A I N I S R E A L

[PawelKasprzak]

Professor Borcherds, thanks for possting these videos.

[nostalgia_]

13:30 how does one remember these? - thanks for the video by the way.

[f-th]

For closure of addition: (x+y)^q = (x+y)^p^n = ((x+y)^p)^p^(n-1) = (x+y)^p^(n-1), induction kicks in.

[andrewxiwu]

ISO feels a lot of pressure when it shows up in a Galois theory course

[余淼-eb]

Dear Prof. Borcherds, would you mind recommending some books for Galois theory and abstract algebra? Thanks so much.

[queenpost]

Now, for me it seems the course is over; the following Videos are private. This is very sad, because i like it.

[husseinshimal]

I do not understand the addition closure part. Where to look ?

At 26:20, we hear "There are the 2 elements in the field of order 2, and these satisfy an irreducible polynomial of degree 1." Given that both of these elements are distinct, how can a polynomial of degree 1 have both of these distinct elements as roots? Should the polynomial they satisfy not be x * (x - 1), which has degree 2, not 1? These elements are indeed the possible quotients of modding out F_2[X] by any of the two, x or x - 1, polynomials of degree 1 in F_2[X], but they do not both satisfy either x or x - 1. In particular one of these elements is 0, the other is 1. What am I missing? Is this phrase instead talking about the *minimal* polynomial of these elements, and saying that for every element in this copy of F_2, its *minimal* polynomial has degree 1? This much is true, but of course the minimal polynomials of each of these two elements do not agree, for 0 it is x, for 1 it is x - 1. If that's the case then perhaps we can phrase it as "These elements satisfy irreducible polynomials of degree 1", not "These elements satisfy _an_ irreducible polynomial of degree 1", if these polynomials they satisfy vary with the element.

[jacobfertleman]

Wait but what about q= 4
2 of the roots added equal -1 which isn't a cube root of unity so x^4-x

[OhadAsor]

please make a video about GF[2^2^n] under the setting of so-called "nim multiplication" :)

[JPK]

7:24 you clearly state "if p=0" so I don't know how it could be a typo/misspeak, but I don't see how any of this holds if p=0, and, of course, it *does* hold if p≠0, right?

[mark_tilltill]

For 3^2 irreducible x^2+1
I also puzzled over square_root function. The main diagonal of the multiplication table contains
1, 2, 6, 3
Square_root(6) = 8 and 4 these are conjugate analogs
I think that the quadratic formula can be made to work on this also.
Interesting question:
What is square_root(4).
Is it analog to imaginary numbers?
Doesn't the fundamental theorem of algebra hold here?
Here are the tables that sent me down this fun rabbit hole: