# Theorem 23.18
Theorem 23.18: Suppose $E$ is a field extension of $F$, then the following are equivalent: \\
a) $E$ is a finite, normal, separable field extension of $F$.
b) $E$ is a splitting field over $F$ of a separable polynomial.
c) $E_G = F$ for some finite group $G$ of automorphisms over $E$. \\ \\
Proof: \\
For a $\Rightarrow$ b. \\
$E$ is separable $\Rightarrow$ if $e \in E, \, \exists \, p_e(x) \in F[x]$ such that $p_e$ is separable and $p_e(e) = 0$. $E$ is normal $\Rightarrow$ if $p(\alpha_1)=0$ with $p$ monic and $\alpha \in E$ then $p(x) = \prod_{i=1}^n (x-\alpha_i)$, for some set of $\{\alpha_i\}_{i=1,..,n} \subset E$. Since $E$ is a finite extension and separable, then $\exists \alpha \in E$ such that $F(\alpha)=F(\alpha_1,...,\alpha_n) = E$ by Theorem 23.12. So then there is an irreducible $p \in F[x]$ such that $p(\alpha)= p(\alpha_i) = 0$ and p splits. $p$ splits and defines this splitting extension, so $E$ is the splitting field over $F$ wrt $p$. \\ \\
b $\Rightarrow$ c \\
$E$ is a splitting field over $F$, so $G(E/F)$ is finite and $E_{G(E/F)} = F$, by Proposition 23.16. Thus such a finite $G$ exists and c is implied. \\ \\
c $\Rightarrow$ a \\
$|G| < \infty$, by Lemma 23.17 $[E:F] \leq |G| < \infty$ so $E$ is a finite extension of $F$. Recall: \\
\[G(E/F) = \{ \sigma \in \text{Aut}(E) \mid \sigma(\alpha) = \alpha \, \, \, \forall \alpha \in F \}\]
and
\[ E_G = \{ \alpha \in E \mid \sigma(\alpha) = \alpha \, \, \, \forall \sigma \in G \} \] \\
So if $E_G = F$ and $\sigma \in G$, then $\forall \alpha \in F$, $\sigma(\alpha) = \alpha$ $\Rightarrow \sigma \in G(E/F)$. So then $G \subseteq G(E/F)$.
\\ \\ Take $\alpha_1 \in E-F$ with minimal polynomial $p(x) \in F[x]$ (we know this exists by theorem 21.15). Every other root of $p(x)$ is conjugate to $\alpha_1$ and none are in $F$ (by it being minimal), to show that $E$ is normal, I must show that all conjugates to $\alpha_1$ are in $E$.
\\ \\
OUTLINE TIME:
I wish to show that the minimal polynomial of $\alpha_1$ has $n$ distinct roots which I can obtain from using the automorphisms in $G$. We know the elements of $G$ permute roots, if I can find all of them using these permutations then I have shown that if $\alpha_1 \in E$ then all roots of $p$ are in $E$, showing normality. Showing that these are necessarily distinct shows then that $E$ is also separable, completing the proof. \\ \\
\section{Properties of $GF(3^3)$}
\subsection{Elements and Additive Inverses}
The monic polynomial I will use is: $p(x) = x^3 + 2x+1$. The elements of $GF(3^3)$ are degree 2 polynomials in $\mathbb{Z}_3[x]$. So the elements and their additive inverses are as follows (shortening the list by noting that we know $-(-x) = x$:
\begin{equation*}
\begin{split}
-(0) & = 0 \\
-(1) & = 2 \\
-(\alpha) & = 2\alpha \\
-(\alpha+1) & = 2\alpha+2 \\
-(2\alpha+1) & = \alpha+2 \\
-(\alpha^2) & = 2\alpha^2 \\
-(\alpha^2+1) & = 2\alpha^2 + 2 \\
-(\alpha^2 + 2) & = 2\alpha^2 + 1 \\
-(\alpha^2 + \alpha) & = 2\alpha^2 + 2\alpha \\
-(2\alpha^2 + \alpha) & = \alpha^2 + 2\alpha \\
-(\alpha^2 + \alpha + 1) & = 2\alpha^2 + 2\alpha + 2 \\
-(\alpha^2 + \alpha + 2) & = 2\alpha^2 + 2\alpha + 1 \\
-(\alpha^2 +2\alpha + 1) & = 2\alpha^2 + \alpha + 2 \\
-(2\alpha^2 + \alpha + 1) & = \alpha^2 + 2\alpha + 2 \\
\end{split}
\end{equation*}
\subsection{Powers of $\alpha$}
For all multiplication I used the following code to multiply:
\begin{lstlisting}[language=Python]
import numpy as np
def multpolys(a,b):
c = np.zeros(5)
for i in range(3):
for j in range(3):
c[i+j] += a[i]*b[j]
return np.mod(c,3)

def modp(a,p):
for i in range(2):
a -= np.mod(a[4-i]*np.roll(p,1-i),3)
a = np.mod(a,3)
return a[0:3]

def gf27mult(a,b,p):
return modp(multpolys(a,b),p)
\end{lstlisting} and then the following to find powers of $\alpha$

\begin{lstlisting}[language=Python]
a = np.array([0,1,0])
b = a
count = 1
while not (b[0] == 0 and b[1] == 1 and b[2] == 0) or count == 1 :
print("\\alpha^{" + str(count) + "} & = " + stringpoly(b) + "\\\\")
b = gf27mult(b,a,p)
count += 1
\end{lstlisting}
\begin{equation*}
\begin{split}
\alpha^{1} & = \alpha \\
\alpha^{2} & = \alpha^2 \\
\alpha^{3} & = \alpha + 2\\
\alpha^{4} & = \alpha^2 + 2\alpha \\
\alpha^{5} & = 2\alpha^2+ \alpha + 2\\
\alpha^{6} & = \alpha^2 + \alpha + 1\\
\alpha^{7} & = \alpha^2 + 2\alpha + 2\\
\alpha^{8} & = 2\alpha^2+ 2\\
\alpha^{9} & = \alpha + 1\\
\alpha^{10} & = \alpha^2 + \alpha \\
\alpha^{11} & = \alpha^2 + \alpha + 2\\
\alpha^{12} & = \alpha^2 + 2\\
\alpha^{13} & = 2\\
\alpha^{14} & = 2\alpha \\
\alpha^{15} & = 2\alpha^2\\
\alpha^{16} & = 2\alpha + 1\\
\alpha^{17} & = 2\alpha^2+ \alpha \\
\alpha^{18} & = \alpha^2 + 2\alpha + 1\\
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
\alpha^{19} & = 2\alpha^2+ 2\alpha + 2\\
\alpha^{20} & = 2\alpha^2+ \alpha + 1\\
\alpha^{21} & = \alpha^2 + 1\\
\alpha^{22} & = 2\alpha + 2\\
\alpha^{23} & = 2\alpha^2+ 2\alpha \\
\alpha^{24} & = 2\alpha^2+ 2\alpha + 1\\
\alpha^{25} & = 2\alpha^2+ 1\\
\alpha^{26} & = 1\\
\end{split}
\end{equation*}
\subsection{Multiplicative Inverses}
I found these with the following code (however, an easier version would be to just find the $m$ such that $\beta = \alpha^m,$ and $26-m = n$, then $\beta^{-1} = \alpha^n$. I made this code before doing the powers though, so this is the code.)
\begin{lstlisting}[language = Python]
p = np.array([1,2,0,1,0])
mult = np.zeros((27,27,3))
for i1 in range(3):
for j1 in range(3):
for k1 in range(3):
a = np.array([i1,j1,k1])
for i2 in range(3):
for j2 in range(3):
for k2 in range(3):
mult[i1+3*j1+9*k1][i2+3*j2+9*k2]
= modp(multpolys(a,np.array([i2,j2,k2])),p)
mult2 = np.zeros((27,27))
w = np.array([1,3,9])
for i in range(27):
for k in range(27):
mult2[i][k] = sum(mult[i][k][:]*w)
inv = np.argwhere(mult2 == 1)
def invertnum(n):
x = np.zeros(3)
x[0] = n\%3
x[1] = ((n-x[0])/3) \% 3
x[2] = ((n-3*x[1]-x[0])/9)
return x
for i in range(26):
print("("+stringpoly(invertnum(inv[i][0]))+")^-1
= " + stringpoly(invertnum(inv[i][1])))
\end{lstlisting}

\begin{equation*}
\begin{split}
(1)^{-1} & = 1 \\
(2)^{-1} & = 2 \\
(\alpha )^{-1} & = 2\alpha^2 + 1 \\
(\alpha + 1)^{-1} & = 2\alpha^2 + \alpha \\
(\alpha + 2)^{-1} & = 2\alpha^2 + 2\alpha \\
(2\alpha )^{-1} & = \alpha^2 + 2 \\
(2\alpha + 1)^{-1} & = \alpha^2 + \alpha \\
(2\alpha + 2)^{-1} & = \alpha^2 + 2\alpha \\
(\alpha^2 )^{-1} & = 2\alpha^2 + 2\alpha + 1\\
(\alpha^2 + 1)^{-1} & = 2\alpha^2 + \alpha + 2\\
(\alpha^2 + 2)^{-1} & = 2\alpha \\
(\alpha^2 + \alpha )^{-1} & = 2\alpha + 1\\
(\alpha^2 + \alpha + 1)^{-1} & = 2\alpha^2 + \alpha + 1\\
(\alpha^2 + \alpha + 2)^{-1} & = 2\alpha^2 \\
(\alpha^2 + 2\alpha )^{-1} & = 2\alpha + 2\\
(\alpha^2 + 2\alpha + 1)^{-1} & = 2\alpha^2 + 2\\
(\alpha^2 + 2\alpha + 2)^{-1} & = 2\alpha^2 + 2\alpha + 2\\
(2\alpha^2 )^{-1} & = \alpha^2 + \alpha + 2\\
(2\alpha^2 + 1)^{-1} & = \alpha \\
(2\alpha^2 + 2)^{-1} & = \alpha^2 + 2\alpha + 1\\
(2\alpha^2 + \alpha )^{-1} & = \alpha + 1\\
(2\alpha^2 + \alpha + 1)^{-1} & = \alpha^2 + \alpha + 1\\
(2\alpha^2 + \alpha + 2)^{-1} & = \alpha^2 + 1\\
(2\alpha^2 + 2\alpha )^{-1} & = \alpha + 2\\
(2\alpha^2 + 2\alpha + 1)^{-1} & = \alpha^2 \\
(2\alpha^2 + 2\alpha + 2)^{-1} & = \alpha^2 + 2\alpha + 2\\
\end{split}
\end{equation*}
\subsection{Characteristic}
The characteristic of the field is 3. Since $ 1 + 1 + 1= 3(1) = 0 $.
\subsection{Finding all subrings}
All subrings are closed under multiplication and subtraction. So, recalling that $\beta = \alpha^n$ for some $n$ for all $\beta \in GF(3^3)-\{0\}$ then if $R \subseteq GF(3^3)$ then $\beta^m = \alpha^{nm} \in R, \forall m \in \mathbb{Z}$. Notice then that if $\beta = \alpha^n \in R$ then for it to be closed and not $\{0\}$ or $GF(3^3)$ it is necessary for $gcd(n,26) > 1$ (else $\beta$ will generate $\alpha$ which will generate all elements of $GF(3^3)-\{0\}$ via multiplication. We then must then check that it is closed also under subtraction. 23 = 2*13, so it is sufficient to check only $\alpha^2, \alpha^{13}$. $-\alpha^2 = 2\alpha^2 = (\alpha)^{15}$. Then, since $gcd(15,26) = 1$, $\alpha^{15}$ will generate all elements of $GF(3^3)-\{0\}$. Checking $\alpha^{13} = 2$, $-\alpha^{13} = 1 = \alpha^{26}$. This with $0$ gives $\mathbb{Z}_{3}$ which is also a subfield of $GF(3^3)$. So the subrings are $\{\{0\}, \mathbb{Z}_{3}, GF(3^3)\}$, which all happen to be fields.
\subsection{All other Irreducibles} Since we only have to find a cubic irreducible polynomial, then we only have to find all cubics in $\mathbb{Z}_3[x]$ with no roots in $\mathbb{Z}_3$. These are:
\begin{equation*}
\begin{split}
&x^3 +2x + 1\\
&x^3 +2x + 2\\
&x^3 +x^2 + 2\\
&x^3 +x^2 + x + 2\\
&x^3 +x^2 + 2x + 1\\
&x^3 +2x^2 + 1\\
&x^3 +2x^2 + x + 1\\
&x^3 +2x^2 + 2x + 2\\
\end{split}
\end{equation*}