-1
$\begingroup$

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

  • 2
    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
  • 4
    Is $\mathbb Z_5$ the ring with five elements or the $5$-adic integers?2012-03-09
  • 0
    @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
  • 0
    @JyrkiLahtonen: No, I don't really know that stuff... :-) (I am familiar with the terms, but I didn't know that!)2012-03-09
  • 0
    @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
  • 0
    @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
  • 0
    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

2 Answers 2

4

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 $F_{25}$.

  • 0
    Thank you for the thoughts. I will work on it and come back with any concerns!2012-03-10
3

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.

  • 1
    Actually, that should be $(a^2+a+2b^2+1) +(2ab+b)\sqrt{2}$.2012-03-09
  • 0
    @Thomas: Thank you; quite so.2012-03-09
  • 0
    If you meant the $5$-adic integers, note that a root in $\mathbb{Z}/5\mathbb{Z}[\sqrt{2}]$ will lift to a root in $\mathbb{Z}_5[\sqrt{2}]$ in the "obvious" way.2012-03-09
  • 0
    May be I should delete my first comment under the OP, because, when read carefully, it gives away the solution, and is getting in the way of your pedagogical effort? You already had my +1 anyway :-)2012-03-09
  • 0
    @JyrkiLahtonen: Actually, your comment is fine, as is your answer.2012-03-10
  • 0
    Thank you as well for the thoughts. I will work on it and come back with any concerns!2012-03-10