HS2024_DiskMat

1)d)
suppose $a \neq b$ for $a, b \in X$

\begin{aligned} \dot{⟹} (f \circ g \circ f) (a) \neq (f \circ g \circ f) (b) & (f \circ g \circ f is injective) \\ \dot{⟹} f (g (f (a))) \neq f (g (f (b))) & (def. \circ) \\ \dot{⟹} f (a) \neq f (b) & (since g (f (a)) = a) \\ \dot{⟹} f is injective \end{aligned}

unknown)e)

To prove that $Z_{3} [x]_{x^{4} + x + 2}$ is a filed, we need to show that $m (x) = x^{4} + x + 2$ is irreducible.

We need to show that $m (x)$ can be factored into polynomials of degree 1 and 3, or two polynomials of degree 2.

Degree 1 and 3: Since $0, 1, 2$ are not roots of $m (x)$ , $m (x)$ has no linear factors. -> not possible

Degree 2: Suppose $m (x) = (x^{2} + a x + b) (x^{2} + c x + d)$

Expanding:

x^{4} + (a + c) x^{3} + (a c + b + d) x^{2} + (a d + b c) x + b d \overset{!}{=} x^{4} + x + 2

This gives a system of equations:

\begin{aligned} a + c & = 0 & I \\ a c + b + d & = 0 & I I \\ a d + b c & = 1 & I I I \\ b d & = 2 & I V \end{aligned}

\begin{aligned} - a^{2} + b + d & = 0 & V \\ a (d - b) & = 1 & V I \end{aligned}

Die einzigen mögliche Werte für $b, d$ sind $1, 2$ oder umgekehrt

case $b = 1, d = 2$ :
in $V, V I$ :

\begin{aligned} - a^{2} + 3 = 0 & ⟹ a = 0 \\ a \cdot 1 = 1 & ⟹ a = 1 & (contradiction) \end{aligned}

case $b = 2, d = 1$ :
in $V, V I$ :

\begin{aligned} - a^{2} + 3 = 0 & ⟹ a = 0 \\ a \cdot (- 1) = 0 & ⟹ a = - 1 & (contradiction) \end{aligned}

Since there are no valid $a, b, c, d$ satisfiying all equations, $m (x)$ cannot be factored into two quadratic polynomials.

$⟹$ $m (x)$ is irreducible $⟹$ $Z_{3} \dots$ is a field. $◻$