Question

a) Find all the units of Z[ − 7]. b) Check whether or not < 8x + 6x − 9x + 24 > [x] 3 2 Q is a field. c) Construct a field with 125 element

Answer 1

(a) I know some vague facts about this, but nothing too concrete. I know that if $(a+b\sqrt{7})$ is invertible, there exists some $(c+d\sqrt{7})$ such that $(ac+7bd)+(ad+bc)\sqrt{7}=1$ , so $ac+7bd=1$ and $ad+bc=0$ . I'm also aware that there's a natural norm on $Z[\sqrt{7}]$ , namely $N(a+b\sqrt{7})=a^2−7b^2$ . As the norm is multiplicative, we know that for $x∈Z[\sqrt{7}]\times N(x)=±1$ .

(b).With p = 3, we have $a_0 = a_1 = a_2 = 0$ (mod p) since 24 = −9 = 6 = 0

(mod 3),$a_n = a_3 \neq 0 (mod 3)$ since $a_3 = 8 = 2 (mod 3)$ , and $a_0 \neq 0 (\text{ mod }p^2)\text{ since } a_0 = 24 = 6 (mod 9).$ So, by the Eisenstein Criterion, $8x^3 + 6x^2 − 9x + 24$ is irreducible over Q.

$< 8x^3 + 6x^2 − 9x + 24 > Q[x]$ is not a field,.

(c)

If a cubic polynomial of $F_5[x]$ is reducible, then it splits into a linear factor and a quadratic factor or into the product of three linear factors. Linear factors are very easy to test for, as (x−a) is a factor of f if and only if f(a)=0.

So you might choose a random degree 3 polynomial and test for the five possible roots.

For instance, I'm tempted to try $f(x)=x^3+2x+1$ . Then

$f(0)=1,\\f(1)=4,\\f(2)=8+4+1≡3,\\f(3)=27+6+1≡4,\\f(4)=64+8+1≡3$ .

As none of these are zero, f(x) is an irreducible cubic.