Multiplication operation on field with four elements whose underlying set is $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$?

Question

Exercise from Aluffi, Alg: Chap. $0$

I know $\mathbb F_4 \cong (\mathbb Z/2\mathbb Z)[x]/(x^2+x+1)$. So, we must have $0 \leftrightarrow (0,0)$, $1 \leftrightarrow (1,1)$ and either $$x \leftrightarrow (0,1), \ \text{ and }, \ 1+x \leftrightarrow (1,0), \text{ or }$$ $$x \leftrightarrow (1,0), \ \text{ and }, \ 1+x \leftrightarrow (0,1).$$

Since $x^2=1+x$ and $(1+x)^2=x$ and $x(1+x)=1$, I know what the answer to the exercise is. However, what is multiplication on $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$ in general?

Componentwise multiplication does not work.

$(a,b)(c,d)=(ac+bd, ad+bc)$ also does not work.

How should multiplication be defined?

Answer 1

Well say $x=(1,0)$ for instance. Then $x^2=x+1$ tells you that $(1,0)(1,0) = (1,0)+(1,1) =(0,1)$. Similarly, $x(x+1) = x^2+x = 1$ tells you $(1,0)(0,1) = (1,1)$. Finally $(0,1)(0,1) = (x+1)^2 = x^2+1 = x$, therefore $(0,1)^2 = (1,0)$ (note : I have repeatedly used the fact that we were in characteristic $2$)

Therefore you have the following table :

$(0,0)$ times anything is $(0,0)$.

$(1,1)$ times anything is that thing

$(1,0)(1,0) = (0,1)$; $(1,0)(0,1) = (1,1)$

$(0,1)(0,1) = (1,0)$

If you want something more like a formula than a table, you have $(a,b)(c,d) = (a(1,0)+b(0,1))(c(1,0)+d(0,1))= ac(0,1) + bd(1,0) + (ad +cb)(1,1) = (bd,ac)+ (ad+cb, ad+cb) = (bd+ad+cb, ac + ad + cb)$,

all in all addition may be defined as $$(a,b)(c,d) = (bd+ad+cb, ac+ad+cb)$$

Answer 2

Simply fill in the following table:

\begin{array}{c|c|c|c|c} \times & (0,0) & (0,1) & (1,0) & (1,1) \\ \hline (0,0) & & & & \\ \hline (0,1) & & & & \\ \hline (1,0) & & & & \\ \hline (1,1) & & & & \end{array}

We know that the additive identity ($(0,0)$) times anything must be $0$. Then we know that $1$ times anything must be itself. In the additive group $(\mathbb{Z}/2\mathbb{Z})^2$, there is nothing that distinguishes $(0,1), (1,0)$ or $(1,1)$ from each other so you can pick any of them to be the multiplicative identity. That gives you 4 spots to fill. Remember that multiplication must be commutative and elements must have inverses. Check associativity last.