Galois theory: Finite fields

[@TheArangol]

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

[@707107]

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.

[@sachalucienmoserferreira2233]

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

[@romywilliamson4981]

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.

[@joeseppe1398]

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

[@romankrylov605]

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

[@nostalgia_1439]

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

[@PawelKasprzak864]

Professor Borcherds, thanks for possting these videos.

[@JPK314]

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?

[@tiop52]

Yes he means p not 0.

[@Aeropig_]

He means that the number p equals 0 in the ring (i.e the ring has characteristic dividing p), so it does hold (by the binomial theorem). I think you are confusing the element p in the ring with the characteristic of the ring, you're right that it does not hold if char F = 0.

[@tiop52]

@@Aeropig_ Ah yes this makes sense.

[@JPK314]

@@Aeropig_ ah this makes sense, thanks. Yeah it's hard to talk about p both as an integer and as its image in the map of Z into the field without some ambiguity

[@husseinshimal7567]

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.

[@余淼-e8b]

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

[@pronaybiswas7524]

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

[@drewduncan5774]

The better you get at math, the worse you get at arithmetic

[@loveihcnurnat664]

14:57 I guess it should be a square bracket, not ()

[@jacobfertleman1980]

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

[@andrewxiwu]

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

[@f5673-t1h]

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.

[@erichahn3336]

You mean obvious kicks in

[@JPK314]

@@erichahn3336 ...no, I'm pretty sure he means induction. I will say though that the proof provided in the OP uses (x+y)^p mod p = x^p+y^p mod p, at which point you may as well just say (x+y)^q mod q= x^q+y^q mod q and, because a mod bc = a mod b (this is true for any a,b,c in N) we have our result without any induction by letting b=p,c=p^(n-1)

[@brendawilliams8062]

Thankyou

[@mark_tilltill6664]

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: kzbin.info/www/bejne/rnbccnp_o9ecfqc

[@queenpost]

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

[@migarsormrapophis2755]

yeeeeee baby

[@OhadAsor]

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

[@bryenico]

That's not what canonical means

[@Vidrinskas]

..because his name is always last on joint papers hahahahahah