20241127_DiskMat

other exercises: 20241128_DiskMat_Serie10.pdf
further references:

https://math.stackexchange.com/a/1051037

5.10 a)
To show that $F = Z_{3} [x]_{x^{3} + 2 x^{2} + 1}$ is a field, we have to show that $m (x) = x^{3} + 2 x^{2} + 1$ is irreducible (Theorem 5.37).

\begin{aligned} m (0) & = 0 + 0 + 1 = 1 \\ m (1) & = 1 + 2 + 1 \equiv_{3} 1 \\ m (2) & = 8 + 8 + 1 \equiv_{3} 2 \end{aligned}

Thus, $m (x)$ does not have any roots. Since $m (x)$ is a polynomial of degree 2 and has no root, $m (x)$ is irreducible (Corollary 5.30).
Thus, $F$ is a field (Theorem 5.37). $◻$

5.10 b)
We will find a generator of $F^{*}$ , where

\begin{aligned} F^{*} & = F ∖ {0}, & (Def. 5.26) \\ F & = Z_{3} [x]_{x^{3} + 2 x^{2} + 1}, \\ m (x) & = x^{3} + 2 x^{2} + 1. \end{aligned}

Since $m (x)$ is a polynomial of degree $3$ over $Z_{3}$ , and $Z_{3}$ has $3$ elements,

\begin{aligned} | F | & = 3^{3} = 27 & (Lemma 5.34) \\ | F^{*} | & = | F | - 1 = 26 \end{aligned}

Thus, an element $d \in F^{*}$ needs to have

o r d (d) \in {1, 2, 13, 26} (Corollary 5.9)

We assume $x$ is a generator of $F^{*}$ :

\begin{aligned} x^{1} & = x & ⟹ o r d (x) \neq 1 \\ x^{2} & = x^{2} & ⟹ o r d (x) \neq 2 \\ x^{4} & \equiv_{m (x)} x^{4} + 2 x (x^{3} + 2 x^{2} + 1) \equiv_{3} x^{3} + 2 x \\ \equiv_{m (x)} x^{3} + 2 x + 2 (x^{3} + 2 x^{2} + 1) \equiv_{3} x^{2} + 2 x + 2 \\ x^{8} & = x^{4} x^{4} = (x^{2} + 2 x + 2) (x^{2} + 2 x + 2) = x^{4} + 4 x^{3} + 8 x^{2} + 8 x + 4 \\ \equiv_{3} x^{4} + x^{3} + 2 x^{2} + 2 x + 1 \equiv_{m (x)} x^{4} + x^{3} + 2 x^{2} + 2 x + 1 + 2 x (x^{3} + 2 x^{2} + 1) \\ \equiv_{3} 2 x^{3} + x^{2} + 2 x + 1 \equiv_{m (x)} 2 x^{3} + x^{2} + 2 x + 1 + (x^{3} + 2 x^{2} + 1) = 2 x + 2 \\ x^{12} & = x^{4} x^{8} = (x^{2} + 2 x + 2) (2 x + 2) = x^{3} + 6 x^{2} + 8 x + 4 \equiv_{3} x^{3} + 2 x + 1 \\ \equiv_{m (x)} x^{3} + 2 x + 1 + 2 (x^{3} + 2 x^{2} + 1) \equiv_{3} x^{2} + 2 x + 1 \\ x^{13} & = x x^{12} = x^{3} + 2 x^{2} + x \equiv_{m (x)} x^{3} + 2 x^{2} + x - (x^{3} + 2 x^{2} + 1) \equiv_{3} x + 2 & ⟹ o r d (x) \neq 13 \end{aligned}

Since $o r d (x) \notin {1, 2, 13}$ , $o r d (x) = 26$ .
Since $⟨ x ⟩$ generates a finite field and has the same cardinality as $F^{*}$ , $⟨ x ⟩$ is isomorphic to $F^{*}$ (Theorem 5.39).
Thus, $⟨ x ⟩$ is a generator of $F^{*}$ .

5.10 c)
Let $a (y) = y^{2} + 2 y (x^{2} + x) + (x^{2} + 2)$ in $F [y]$ ( $F = Z_{3} [x]_{x^{3} + 2 x^{2} + 1} [y]$ ).

\begin{aligned} | F | & = 3^{3} = 27 & (Lemma 5.34) \end{aligned}

The roots of $a (y)$ are elements $r \in F$ , for which $a (r) = 0$ (Def. 5.33).
Since $a (y)$ is a polynomial of degree $2$ , $a (y)$ has at most $2$ roots (Theorem 5.31).

Because $m (x) = x^{3} + 2 x^{2} + 1$ is a polynomial with degree $3$ , elements of $F [y]$ are polynomials with degree $\leq 2$ (Def. 5.34) with coefficients in $Z_{3}$ .
Thus, each element in $F [y]$ can be written as

a x^{2} + b x + c

where $a, b, c \in Z_{3}$ .
By substituting each of the 27 elements of $F [y]$ for $y$ in $a (y) = 0$ , we can find the roots of $a (y)$ . Since it is quite time-consuming to do the calculations by hand, I have written a python script for this purpose [[diskmat_serie10_5c.py]]:

from sympy import symbols, div, Poly, GF

# Define the variables
x,y = symbols('x y')

# define all (27) elements
elements = [0,1,2,x,2*x,x**2,2*x**2, x**2+1, x**2+2, 2*x**2+1, 2*x**2+2,
    x**2+x, 2*x**2+x, x**2+2*x+1, x**2+2*x+2, x**2+x+1, x**2+x+2,
    2*x**2+2*x+1, 2*x**2+2*x+2, 2*x**2+x+1, 2*x**2+x+2, 2*x+1, 2*x+2,
    x+1, x+2, x**2+2*x, 2*x**2+2*x]
denominator = Poly(x**3+2*x**2+1,x)

# compute all elements
for idx, y in enumerate(elements):
    numerator = Poly(y**2+y*2*(x**2+x)+(x**2+2), x)

    # Perform division in Z_3
    quotient, remainder = div(numerator, denominator, domain=GF(3))

    # the remainder of the division is the same as modulo polynomial
    print(f"{str(idx).zfill(2)}: {y}".ljust(20), remainder)

The results of the inserted elements:

00: 0                Poly(x**2 - 1, x, modulus=3)
01: 1                Poly(-x, x, modulus=3)
02: 2                Poly(-x**2 + x, x, modulus=3)
03: x                Poly(0, x, modulus=3)
04: 2*x              Poly(x**2 + 1, x, modulus=3)
05: x**2             Poly(0, x, modulus=3)
06: 2*x**2           Poly(x**2 + x - 1, x, modulus=3)
07: x**2 + 1         Poly(x**2 - x + 1, x, modulus=3)
08: x**2 + 2         Poly(-x**2 + x + 1, x, modulus=3)
09: 2*x**2 + 1       Poly(x**2, x, modulus=3)
10: 2*x**2 + 2       Poly(x**2 - x, x, modulus=3)
11: x**2 + x         Poly(x**2 - 1, x, modulus=3)
12: 2*x**2 + x       Poly(x**2 + x - 1, x, modulus=3)
13: x**2 + 2*x + 1   Poly(-x**2 - 1, x, modulus=3)
14: x**2 + 2*x + 2   Poly(-1, x, modulus=3)
15: x**2 + x + 1     Poly(-x**2 + x, x, modulus=3)
16: x**2 + x + 2     Poly(-x, x, modulus=3)
17: 2*x**2 + 2*x + 1 Poly(x, x, modulus=3)
18: 2*x**2 + 2*x + 2 Poly(x, x, modulus=3)
19: 2*x**2 + x + 1   Poly(x**2 - x, x, modulus=3)
20: 2*x**2 + x + 2   Poly(x**2, x, modulus=3)
21: 2*x + 1          Poly(-1, x, modulus=3)
22: 2*x + 2          Poly(-x**2 - 1, x, modulus=3)
23: x + 1            Poly(-x**2 + x + 1, x, modulus=3)
24: x + 2            Poly(x**2 - x + 1, x, modulus=3)
25: x**2 + 2*x       Poly(x**2 + 1, x, modulus=3)
26: 2*x**2 + 2*x     Poly(x - 1, x, modulus=3)

As we can see, $a (x) = 0$ and $a (x^{2}) = 0$ . To verify the roots:

for $y = x$ :

\begin{aligned} a (x) & = x^{2} + 2 x (x^{2} + x) + (x^{2} + 2) \equiv_{3} 2 x^{3} + x^{2} + 2 \equiv_{m (x)} 2 x^{3} + x^{2} + 2 + (x^{3} + 2 x^{2} + 1) \equiv_{3} 0 \end{aligned}

for $y = x^{2}$ :

\begin{aligned} a (x^{2}) & = (x^{2})^{2} + 2 x^{2} (x^{2} + x) + (x^{2} + 2) \equiv_{3} 2 x^{3} + x^{2} + 2 \equiv_{m (x)} 2 x^{3} + x^{2} + 2 + (x^{3} + 2 x^{2} + 1) \equiv_{3} 0 \end{aligned}

Since $a (x) = 0$ and $a (x^{2}) = 0$ , $x, x^{2}$ are roots of $a (y)$ (Def 5.33). Due to (Theorem 5.31) $x, x^{2}$ are all roots of $a (y)$ .