# How Multiplication Table is generated for GF(2^2) field

I was unable to solve the multiplication table given in the book for $\mathrm{GF}(2^2)$.However, I have managed to solve the addition table.

Acoording to the Book multiplication is the AND operation, but when I applied this I did not get the answer given in the book.

This is the Table given in the book:

I couldn't replicate the same answers for 10x10 or 11x11.

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$\mathrm{GF}(2^2)$ is the finite field of 4 elements, and has minimal polynomial $x^2+x+1$. Throughout this question I will use $ab$ to denote $ax+b$ (ie $10=1*x+0$) - this is standard notation when considering finite fields over $\mathbb F_2$ since it aligns with how we consider bits in bytes.
$ab+cd := ab\oplus cd$
For multiplication, we do standard polynomial multiplication, but then must reduce by the minimal polynomial. That is, we use the identity that $x^2=x+1$. So:
\begin{aligned} ab\cdot cd &= (a*x+b)(c*x+d) \\ &= (a * c)x^2+(a* d + b * c)x + b*d \\ &= (a* d + b * c + a * c) x + [b*d + a * c] \end{aligned} Now, as discussed above + is xor (denoted $\oplus$), and the hint given by your book is that multiplication of the coefficients is the same as an AND operation (which I'll denote with $\&$):
\begin{aligned} ab\cdot cd &= (a\&d \oplus b\&c \oplus a\&c) x \oplus [b\&d \oplus a\&c] \\ &= [a\&d \oplus b\&c \oplus a\&c][b\&d \oplus a\&c] \end{aligned}