A group of sheaves of grain stood on end in a field!
The description makes this challenge sound complicated, with abstract algebra and algebraic geometry but the challenge is actually quite simple
The cipher is basically doing \(c=\frac{k_1}m+k_2\pmod b\) where these are polynomials instead of integers, however most normal operations works for polynomials.
We are also given a lot of m,c pairs, though only 2 was needed
Furthermore everything is done in \(\mb F_2\), which simplifies a lot, addition and subtraction are also the same thing which is quite convenient
Using 2 pairs of m,c, we get(under mod \(b\)):
\begin{align*}
k_2 &= \frac{c_1m_1+c_2m_2}{m_1+m_2}\\\
k_1 &= m_1(c_1+k_2)
\end{align*}
Now using these keys, we can easily retrive the flag
Flag:
CCTF{GF2_F1nI73_Crc13_f1elds}