I need help proving that $x^2+x+1$ is irreducible in $\mathbb{Z}_{5}[\sqrt{2}](x)$. Anyone be willing to at least help me get a good start?
--edit: typo, added the (x) for $\mathbb{Z}_{5}[\sqrt{2}](x)$
I need help proving that $x^2+x+1$ is irreducible in $\mathbb{Z}_{5}[\sqrt{2}](x)$. Anyone be willing to at least help me get a good start?
--edit: typo, added the (x) for $\mathbb{Z}_{5}[\sqrt{2}](x)$
There is only one quadratic extension of the field $\mathbb{Z}_5$, namely $F_{25}$, so $\mathbb{Z}_5[\sqrt2]\simeq F_{25}$ and all quadratic polynomials with coefficients in $\mathbb{Z}_5$ become reducible over $F_{25}$.
Since it is a quadratic polynomial, it is irreducible if and only if it has no roots.
Let $a+b\sqrt{2}\in\mathbb{Z}_5[\sqrt{2}]$, with $a,b\in\mathbb{Z}_5$. Plugging in and evaluating, you obtain $(a^2+a+2b^2+1) + (2ab+b)\sqrt{2}.$
Determine whether that can be zero or not.