V1b: Mathematical prerequisites (Kyber and Dilithium short course)

[@cryptography101-alfred]

Errata: Slide 29: a*b^T should be a^T*b.

[@shravanmishra9992]

Very clear and nice ❤

[@cryptography101-alfred]

Thanks!

[@everaldoantoniomoreiraalve1023]

Excellent! Thank you very much for sharing your knowledge. I have a question about slide 31. Wouldn't mods q be equivalent to mod+- q ?

[@cryptography101-alfred]

mods and mod+- are the same thing. The NIST standards and research papers use the mod+- q notation. I chose to use the mods q notation because it's easier to say and write, and has been used before ("mods" stands for "symmetric mod").

[@Usinuid]

Hello Alfred, could you please help me understand the f(x) - g(x) result at 3:16 (why 4x?). Thank you :)

[@cryptography101-alfred]

At 3:11: The difference of the "x" terms in f(x) and g(x) is 0-3x = -3x. Since -3 is congruent to 4 modulo 7, we can write -3x as 4x.

[@Usinuid]

@@cryptography101-alfred Thank you, that helped me. :) On the following line of the same slide, how do you obtain x, I reckon you obtain 2 by (5*6)%7 but then how can we obtain the lonely x, shouldn't I be doing (0*3x)%7?

[@cryptography101-alfred]

@@Usinuid When multplying f(x) and g(x), you get an x term by multiplying the 5 term in f(x) and the 3x term in g(x). The result is 15x, which is the same as x (modulo 7).