Finding how multiplication and addition behave on $\mathbb{F}_4$ without any result

Question

I'm currently tutor in an undergraduate course and the students are asked to find how addition and multiplication behave on the field with 4 elements: $\{0,1,x,y\}$. In the solution the teacher gave me, he uses the fact that $(\mathbb{F}_4^*, \cdot )$ is cyclic with order 3 to find that $x$ and $y$ are both generators and that $(\mathbb{F}_4,+)$ is a group of order 4 to deduce that $0=1+1+1+1=(1+1)(1+1)$ and hence, $(1+1)=0$ since a field is integral.

However, I'm not satisfied with this reasoning because this question is asked to student in first year and they juste have been introduced to fields (and this was part of an analysis course so they won't see any result soon). So I tried to find how it behaves by myself, using as few results as possible.

Multiplication is the easy part, $0\cdot a$ and $1 \cdot a$ is trivial for all $a\in \mathbb{F}_4$. Then, $x\cdot y\not = 0$ because it is invertible, and it can't be equal to $x$ or $y$ because that would imply the other element to be equal to $1$. Hence, $x\cdot y=1$. We are left to find what is $x^2$ (and similarly, $y^2$). $x^2$ cannot be $0$, nor $x$. If $x^2=1$, then $y=(x^2)y=x(xy)=x\cdot 1=x$. Contradiction. Hence $x^2=y$ and by the same reasoning, we have $y^2=x$.

It was easy with the multiplication, but I can't figure out any non-trivial result concerning addition. My intuition tells me that distributivity should be used here but I'm feeling like it's just converting a problem into another one because we don't know anything about addition.

Does anyone here have an idea of how addition can be "found" without using any advanced result (or as few results as possible)?

Thanks in advance.

Answer 1

The only "new" fact you could use here, is that the characteristic of a field with $4$ elements must be $p=2$. This is not difficult, but you need to introduce the characteristic of a field. Then addition is clear, with $\{0,1,a,b\}$: $\;a+b=0\implies a=-b=b\;$ , since the field has characteristic $2$. It also can't be $\;a+b=a\;,\;\;a+b=b\;$, else $\;b=0\;$ or $\;a=0\;$. Thus it must be $$\;a+b=1\implies b=1-a=1+a\;$$

Generalize the above and get the addition table.

Answer 2

Going off what you were given, if $(\mathbb{F}_4^*, \cdot )$ is cyclic with order $3$ then you know the cyclic subgroup generated by $x$ is: $$\langle x\rangle=\{x^1,x^2=y,x^3=1\}$$ since cycling through the powers of $x$ cycles through all the nonzero elements of $\mathbb{F}_4$, and these are $x^1=x$, $x^3=1$ (since $(\mathbb{F}_4^*, \cdot )$ has order $3$) which leaves $x^2$ which must equal $y$ since we are also told $(\mathbb{F}_4^*, \cdot )$ is cyclic. By the same reason $\langle y\rangle=\{y^1,y^2=x,y^3=1\}$, and we have both $x$ and $y$ generators of the cyclic group $(\mathbb{F}_4^*, \cdot )\cong C_3$.

Now you have $(\mathbb{F}_4,+)$ is an additive group of order $4$. Therefore the order of the subgroup formed by $1$ is either $2$ or $4$ (since by Lagrange's theorem the order of a subgroup has to divide the order of the group). So look at $(1+1)(1+1)=1+1+1+1=0$. If the order is $4$, then $(1+1)\neq0$, but then this contradicts $(1+1)(1+1)=0$, since we are in a finite field (which is obviously an integral domain (ID), and so admits no zero divisors: in an ID if $ab=0$ then either $a=0$ or $b=0$). Hence $1+1=0$, and $1$ has order $2$.

The reason behind this is that $\mathbb{F}_4=\mathbb{F}_{2^2}$ is a finite extension of degree $2$ of the prime subfield $\mathbb{F}_2$ with two elements (the prime subfield of a field $F$ is the subfield of $F$ generated by the multiplicative identity $1_F$ of $F$, and is isomorphic to $\mathbb{Q}$ (if $\operatorname{char}(F)=0$) or $\mathbb{F}_p$ (if $\operatorname{char}(F)=p$)). So when we say $1$ has order $2$ in $\mathbb{F}_4$ it is because $\mathbb{F}_4$ has characteristic $2$.

Now since $x$ is not $0$ or $1$, then $x+1\neq x$, and also $x+1\neq0$, for then $x=-1$, but $-1=1$ and we gain a contradiction. Hence $x+1=y$ and our addition and multiplication tables are:

$$ \begin{array}{c|cccc} + & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 1 & x & x+1 \\ 1 & 1 & 0 & x+1 & x \\ x & x & x+1 & 0 & 1\\ x+1 & x+1 & x & 1 & 0 \end{array} \begin{array}{c|cccc} \cdot & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & x & x+1 \\ x & 0 & x & x+1 & 1\\ x+1 & 0 & x+1 & 1 & x \\ \end{array} $$