Please show that x^3-2 is irreducible in Z/19Z [x]/<x^2-2>. There HAS to be a better way than plugging and chugging 361 polynomials.
Thanks!
Julian
It is sufficient to show $\displaystyle t^3 - 2$ has no roots in $\displaystyle \mathbb{Z}_{19}[x]/(x^2-2)$. Let $\displaystyle \alpha = x + (x^2-2)$. Then we see that $\displaystyle \alpha^2 = 2$.
Now let $\displaystyle a+b\alpha$ be an element in this field with $\displaystyle a,b\in \mathbb{Z}_{19}$. Now show that $\displaystyle (a+b\alpha)^3 - 2 =0$ is impossible by expanding out this the expression and using the fact $\displaystyle \alpha^2 = 2$, and comparing coefficient.