# Addition and Multiplication in $F_4$

Could anyone explain the example below? Why is $F_4 =$ {$0,1,x,x+1$}? (I was learning that it should be $F_4 =$ {$0,1,2,3$}). And how do we get the two tables?

• @RobArthan Indeed 2 has no multiplicative inverse, not thinking things through... – Doug M May 11 '18 at 19:41

$\Bbb{F}_4$ is the finite field of order $4$. It is not the same as $\Bbb{Z}_4$, the integers modulo 4. In fact, $\Bbb{Z}_4$ is not a field. $\Bbb{F}_4$ is the splitting field over $\Bbb{F}_2 = \Bbb{Z}_2$ of the polynomial $X^4 - X$. You get the addition table by observing that $\Bbb{F}_4$ is a 2-dimensional vector space over $\Bbb{F}_2$ with basis $1$ and $x$ where $x$ is either of the roots of $X^4 - X = X(X - 1)(X^2 + X + 1)$ that is not in $\Bbb{F}_2$. You get the multiplication table by using $x^2 + x + 1= 0$ to simplify the expressions for the products of $x$ and $x + 1$.
• I'm self-studying this topic, and was a bit lost when the author of the note above jumped straight from prime fields to non-prime fields. It is my observation that if we classify the finite fields $\mathbb{F_n}$ by their cardinalities $n$, then if $n$ is a prime number $p$, $\mathbb{F_n}$ coincides with the prime field $\mathbb{F_p}$. Otherwise, if $n$ is not prime, we must find $n$ polynomials in place of natural numbers $\{0,1,2,...,(n-1)\}$ to construct the field. Is this observation correct? – ensbana May 12 '18 at 9:25
• There is a finite field with $n$ elements iff $n$ is a prime power. The field with $p^k$ elements is the splitting field of $x^{p^k} - x$ and its elements comprise precisely the $p^k$ distinct roots of that polynomial. See en.wikipedia.org/wiki/Finite_field for more info. – Rob Arthan May 12 '18 at 11:30
The addition and multiplication tables are essentially forced on us. Suppose we have a field with four elements. Under addition it is an abelian group. We must have the elements zero and one. The first question is what is the additive order of one? It must be a divisor of four, so it is either four or two. In the first case, we have that $\;\{0,1,2:=1+1,3:=2+1\}\;$ are the four elements. However, since we have that $\;2\times 2=4=0,\;$ zero divisors exist and thus it can't be a field.
The only other possibility is that $\;1+1=0\;$ which implies that there are two other elements. Call one of them $\;x\;$ and the other is $\;x+1\;$ since it can't equal any of the other three elements. In the same way, $\;x+x=0\;$ since it can't be any of the other three elements. This is enough to uniquely determine the sixteen entries in the addition table.
For multiplication, the characteristic properties of zero and one determine twelve of the entries in the multiplication table. We get $\;x\times x = 1+x\;$ since it can't equal any of the other three elements. We also get $\;x \times(1+x) = 1\;$ since the multiplicative inverse of $\;x\;$ can't equal any of the other three elements. Finally, $\;(1+x)\times(1+x)=x\;$ and we are done.