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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.

H. Potter
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  • A possibility might be to use subtraction(additive inverse) as well. The axioms imply that there must be an element $a$ such that $1+a=0$, or $a=-1$. If you first show (use distributivity) that $(-1)^2=1$ you get that $a^2=1$. But the multiplication table you already produced then shows that the only alternative is $a=1$. The rest is easy. – Jyrki Lahtonen Sep 22 '16 at 06:11

2 Answers2

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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.

Dietrich Burde
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  • How do you show the characteristic is not 3 or 4? – Mees de Vries Sep 21 '16 at 19:16
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    Suppose the characteristic is $4$. Then, $1 + 1 \neq 0$, but $$(1 + 1)(1 + 1) = 1 + 1 + 1 + 1 = 0,$$ so $1 + 1$ is a zero divisor, a contradiction, and hence the characteristic is $2$ (do you see why it cannot be $3$?) – Dietrich Burde Sep 21 '16 at 19:17
  • Oh, right. But what about 3? That seems more difficult. – Mees de Vries Sep 21 '16 at 19:19
  • No, that's even easier. The order of $1$ in the group $(\mathbb{F}_4,+)$ divides $4$, so it is either $2$ or $4$, but not $3$. – Dietrich Burde Sep 21 '16 at 19:27
  • That's the obvious answer. But that uses (basic) group theory, which we were trying to avoid. You can prove it's not three by first proving that the operation $y \mapsto x + y$ is an bijection for each $x$, and then show it's the identity if and only if $x = 0$, and then go from there... but that's a lot of work. – Mees de Vries Sep 21 '16 at 19:34
  • The order of an element is much easier than the concept of a finite field itself, so I am not sure, that we have to avoid it. Studying finite fields without groups does not make sense to me. – Dietrich Burde Sep 21 '16 at 19:38
  • From the context in the question I assumed that this question was for first years, who will only work in $\mathbb R$ or $\mathbb C$ for another couple of months, and this is just an exploratory exercise to see that fields don't have to look like the "numbers" you are used to. I feel like the question is explicitly trying to avoid any background group theory. But perhaps that is more for the original question asker to decide. – Mees de Vries Sep 21 '16 at 19:54
  • @MeesdeVries That's exactly the conclusion I came up with after thinking for a while about the purpose of this question. After seeing all these suggestions, I think that introducing the characteristic of a field or that the order of an element divides the cardinal of the group must be the best ideas. Or maybe use Jirky's suggestion in the comments of the question. Thanks for all you answers! – H. Potter Sep 22 '16 at 06:41
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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:

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

Daniel Buck
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