$x^{2} + 1$ is irreducible in $\mathbb{Z}_{3} [x]$, and so K = $\mathbb{Z}_{3} [x] / \langle x^{2}+ 1\rangle$ is a field with 9 elements. Let $a$ in K be a root of $f(x)$. Find irreducible polynomials in $\mathbb{Z}_{3} [x]$ having $a + 1$ and $a - 1$ as roots respectively.
No clue... is it simply plugging $a + 1$ into $f(a) = a^{2}+ 1$, so that $g(a) = (a + 1)^{2}+ 1$, hence $g(x) = x^{2}+ 2x + 2$, which is irreducible in $\mathbb{Z}_{3} [x]$?
Thanks!!
Well, your field $K$ is a quadratic extension of $\mathbb{F}_3$, so we should try squaring $a+1$ to find some relation. $(a+1)^2=a^2+2a+1 = 2a$ (since $a^2 +1 = 0$), so we have the relation $(a+1)^2 - 2a = 0$. But we want this to be a polynomial in $a+1$, so we can rewrite this as $$(a+1)^2 -2a -2 +2=0$$ or $$(a+1)^2 -2(a+1)+2,$$ so $a+1$ satisfies the polynomial $x^2-2x+2$.
I think your error is that $a+1$ is not not necessarily a root of $f$, which is what you seem to be trying.