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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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    If $\mathbb{Z}_5$ is the field of 5 elements, then $2=-3$. Hence also $\sqrt{2}=\sqrt{-3}$. But your polynomial has $(-1\pm\sqrt{-3})/2$ as zeros, so something is wrong?2012-03-09
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    Is $\mathbb Z_5$ the ring with five elements or the $5$-adic integers?2012-03-09
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    @AsafKaragila, undoubtedly you know it, but if $\mathbb{Z}_5$ stands for the ring of 5-adic integers, then $\mathbb{Z}_5[\sqrt2]$ is its integral closure in the unique unramified quadratic extension of $\mathbb{Q}_5$. And that ring also has all the cubic roots of unity. So thinking in terms of 5-adics does not change the conclusion that the printed polynomial is still reducible. May be the polynomial was meant to be cubic, and there is a typo?2012-03-09
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    @JyrkiLahtonen: No, I don't really know that stuff... :-) (I am familiar with the terms, but I didn't know that!)2012-03-09
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    @TiredSophomore You mean you are viewing $x^2 + x + 1$ as an element of the ring $\bigg(\mathbb{Z}/5\mathbb{Z}[\sqrt{2}]\bigg)[x]$?2012-03-09
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    @BenjaminLim: Yes; it is common to talk about a polynomial $f(x)$ that has coefficients in a ring containing $R$ as being "irreducible over $R$" (or sometimes "in") to mean, irreducible as an element of $R[x]$.2012-03-10
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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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