Results

We now want to consider the subgroups A_j contained in our family $\mathcal{F}$. Define $A_0=\{(a,0,0)\}$ and $A_\infty=\{(0,0,a)\}$and $A_t=\{(a,f_t(a)+\displaystyle\frac{t}{2}a^2,ta):\;a\in \mathbb{F}\}$, where f_t(a) is an additive function. It can be shown that $A_0A_\infty\bigcap A_t=(0,0,0)$,satisfying P₃, if and only if $f_t(a)\ne\displaystyle\frac{ta^2}{2}$, where f_t(a) is an additive function. So it needs to be shown what types of function are additive. The next theorem shows that there is only one type of additive function.

Theorem: Let $\mathbb{F}$ be the Galois field, GF(q)where q=p^e. Then if $f:\mathbb{F}\to\mathbb{F}$ is an additive function, then

f(b)=b+a₁b^p+a₂b^p2+...+a_e-1b^pe-1

for any $a,b\in \mathbb{F}$.

proof:

Let $\mathcal{F}=\{f:\mathbb{F}\to\mathbb{F} :f(a+b)=f(a)+f(b)\mbox{ and } a,b\in \mathbb{F}\}$.

Let $\sigma:a\to a^p$. Then $\sigma\in G=Aut(\mathbb{F} )$ and $\mid G-\{0\}\mid=e$.

For $a_0,a_1,...,a_{e-1}\in \mathbb{F}$, define f to be the additive function such that

f(b)=a₀b+a₁b^p+...+a_e-1b^pe-1

Because f is additive, if f(b)=0 for all $b\in\mathbb{F}$, then a₀=a₁=...=a_e-1=0. Thus, $\{a_0b,...,a_{e-1}b^{e-1}\}$ are linearly independent and form a basis. Therefore we can construct q^e distinct additive functions of this type. It is left to show that these are all the possible additive functions.

Think of $\mathbb{F}$ as a vector space of dimension e over $\ensuremath{\mathbb{Z} } _{p}$and let $\mathcal{F}$ be the additive group of all linear operators on $\mathbb{F}$. What is the dimension of $\mathcal{L}\left(\mathbb{F} /\ensuremath{\mathbb{Z} } _{p}\right)$?

The dimension of all operators $T:\mathbb{F}\to\mathbb{F}$ is the dimension of the set of all $n\times n$ matrices. Here n=e and so the dimension=e². Therefore, we have pe²=p(e)^e=q^e and the dimension of $\mathcal{L}\left(\mathbb{F} /\ensuremath{\mathbb{Z} } _{p}\right)=q^e$. But this was the number of additive functions of the type

f(b)=a₀b+a₁b^p+...+a_e-1b^pe-1

Therefore this are the only possible additive functions.

Q.E.D.

Corollary: Let $\mathbb{F} =GF(9)$ be the galois field of order 9=3² and let $\mathcal{F}$ be the family of all additive functions over $\mathbb{F}$. Then

$$\mathcal{F}=\{f:\mathbb{F}\to \mathbb{F} :f_1(b)=b+a_tb^3,a,b\in\mathbb{F}\}\;\;\;\;(*)$$

Result follows from above theorem.


We can use this result to discover the possible additive functions satisfying equation (*).

Consider the galois field $GF(3)\cong\ensuremath{\mathbb{Z} } /3\ensuremath{\mathbb{Z} }$. Extend this field to the factor ring over the ideal generated by the irreducible polynomial x²+1. The result is

$$\mathbb{F} =\ensuremath{\mathbb{Z} } _3[x]/<x^2+1>=ax+b +<x^2+1>\mbox{ for }a,b\in\ensuremath{\mathbb{Z} } _3$$

Because all multiples of x²+1 are absorbed into the ideal we can think of x²+1 as being congruent to 0. Therefore x²=-1. Now we can create the multiplicative group from the primitive element $\alpha=x+1$.

$$\begin{eqnarray*}\alpha&=&x+1\\ \alpha^2&=&(x+1)^2=x^2+2x+1=2x=-x\\ \alpha^3&=... ...=x(x+1)=x^2+x=x-1\\ \alpha^8&=&(x+1)^8=(x-1)(x+1)=x^2-1=-2=1\\ \end{eqnarray*}$$

The group is $G=\bigg\{(x+1),-x,(1-x),-1,(-x-1),x,(x-1),1\bigg\}$.

We now want to see what each subgroup in the four-gonal family looks like, in particular what each A-t looks like. Once we have a possible A_t, we can check that $A_0A_{\infty}\bigcap A_t=\{0,0,0\}$.

Remember that the subgroup A_t in the fourgonal family looks like

$$\left(a,f_t(a)+\displaystyle\frac{1}{2}a^2,ta\right)\mbox{ for }a\in \mathbb{F}$$

In order to satisfy equation (*), we need that $f_t(b)\ne \displaystyle\frac{b^2}{2}$. Suppose that f₁(a)=a+(x+1)a³. Then

$$\begin{eqnarray*}f_1(x+1)&=&(x+1)+(x+1)(x+1)^3\\ &=&(x+1)+(x+1)^4\\ &=&x\\ \\ [1mm] f_1(-x)&=&(-x)+(x+1)(-x)^3\\ &=&-1\\ \end{eqnarray*}$$

$$\begin{eqnarray*}f_1(-x+1)&=&(-x+1)+(x+1)(-x+1)^3\\ &=&x+1\\ \\ [1mm] f_1(-x-1... ...&.&\\ f_1(1)&=&1+(x+1)(1)^3\\ &=&x-1\\ \\ [1mm] f_1(0)&=&0\\ \end{eqnarray*}$$

But (-x-1)²=-x and $$\frac{1}{2}\equiv2^{-1}\equiv(-1)^{-1}=1$$. So immediately we might have noticed that $f_1(x+1)=\displaystyle\frac{1}{2}(x+1)^2$. Even without this observation when constructing the group A₁ the first element we find is (x+1,0,x+1) which is an element of $A_0A_\infty$, and so $A_0A_\infty\bigcap A_1$ nontrivially. Therefore, the additive function

f₁(b)=b+(x+1)b³

does not satisfy equation (*).

Now look at an additive function that satisfies (*). Let f₁(a)=a+(x-1)a³. After some calculations we arrive at the following.

$$\begin{eqnarray*}f_1(x+1)&=&1\\ f_1(-x)&=&x-1\\ f_1(-x+1)&=&-x-1\\ f_1(-1)&=&... ...\ f_1(x)&=&1-x\\ f_1(x-1)&=&x+1\\ f_1(1)&=&x\\ f_1(0)&=&0\\ \end{eqnarray*}$$

Using this information we can now construct a subgroup A₁using this additive function. Remember that because $$\frac{1}{2}\equiv-1$$ we get the following

$$A_1=\{(a,f_1(a)-a^2,a)\}$$

Calculations involving the nine possibilities for f₁(a) tell us that there are only three choices satisfying equation (*),

$$\begin{eqnarray*}f_1(a)&=&a-xa^3\\ f_1(a)&=&a+xa^3\\ f_1(a)&=&a+(x-1)a^3\\ \end{eqnarray*}$$

with corresponding subgroups

$$\begin{eqnarray*}A_1&=&\bigg\{(a,a-xa^3-a^2,a): a\in\mathbb{F}\bigg\}\\ A_1&=&\... ...\}\\ A_1&=&\bigg\{(a,a+(x-1)a^3-a^2,a):a\in\mathbb{F}\bigg\}\\ \end{eqnarray*}$$

The next step to show that the only GQ over GF(9) is the classical example is to find all possible additive functions f₂(a) that satisfy properties P₁,...,P₅. From the work Laurel has done with her mathematica program, we expect to eliminate all possible additive functions after checking the possibilities for the additive function f₂(a). Hopefully, after proving the conjecture through elimination we will find the intuition to give a mathematical proof which is not dependent on these calculations. That is the goal of this project.