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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 $\mathbb{Z}_{5}[\sqrt{2}](x)$

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    Oh, yes! Typo. I mis-read one of the replies. It's supposed to be $\mathbb{Z}_5 [\sqrt{2}](x)$ I will make the change above.2012-03-12

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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 $F_{25}$.

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    Thank you for the thoughts. I will work on it and come back with any concerns!2012-03-10
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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.

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    Thank you as well for the thoughts. I will work on it and come back with any concerns!2012-03-10