# 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.

$\mathrm{GF}(2^2)$ 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}

There are a number of ways to represent elements of the field; we'll start by representing them as polynomials with degree at most 1, and with integer coefficients modulo 2. There are four such polynomials: {0, 1, x, x + 1}.

Here are the addition and multiplication tables:

+   0   1   x   x + 1
0   0   1   x   x + 1
1   1   0   x + 1   x
x   x   x + 1   0   1
x + 1   x + 1   x   1   0
⋅   0   1   x   x + 1
0   0   0   0   0
1   0   1   x   x + 1
x   0   x   x2 mod m    x2 + x mod m
x + 1   0   x + 1   x2 + x mod m    x2 + 1 mod m


Hold on. What's that funny-looking m?

It's a "reduction polynomial" which brings the product back down to degree 1 or less. It has to be a polynomial of degree 2. There are four such polynomials: let's try each and see what we get.

x2
0   x
x   1
x2 + 1
1   x + 1
x + 1   0
x2 + x
x   0
0   x + 1
x2 + x + 1
x + 1   1
1   x


Note that the first three polynomials all factor into products of lower-degree polynomials: x2 = x(x), x2 + 1 = (x + 1)(x + 1), x2 + x = x(x + 1). Only x2 + x + 1 is prime; and this prime reduction polynomial generates a complete multiplication table with no 0s. This is a necessary condition to be a field. Our final tables are:

+   0   1   x   x + 1
0   0   1   x   x + 1
1   1   0   x + 1   x
x   x   x + 1   0   1
x + 1   x + 1   x   1   0
⋅   0   1   x   x + 1
0   0   0   0   0
1   0   1   x   x + 1
x   0   x   x + 1   1
x + 1   0   x + 1   1   x


We can also write our elements in binary form: 0 => 00, 1 => 01, x => 10, and x + 1 => 11. In this notation our tables become:

+   00  01  10  11
00  00  01  10  11
01  01  00  11  10
10  10  11  00  01
11  11  10  01  00
⋅   00  01  10  11
00  00  00  00  00
01  00  01  10  11
10  00  10  11  01
11  00  11  01  10

• Downvoted because it is a very late answer to a question which already had a good answer and still needs some formatting to become, well... readable, sorry – VincBreaker Apr 27 '18 at 20:36
• @VincBreaker Your comment was definitely on-point, but you might as well have edited the answer while you were at it… ;) Anyway, I just formatted those tables to make things a bit more readable. – e-sushi Apr 28 '18 at 12:12

01=1; 10= 2; 11=3 ; 01* 10= 1*2=2= 10 Just convert the two bits into decimal and apply multiplication. The result of multiplication should be converted into binary again and hence , inserted within the table.