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Gallian's 'Contemporary Abstract Algebra', Chapter 13 Problem 44.

Question has been asked here already, but the only answer involves factor rings, which haven't been covered at this point in the book yet.

If $k^2$ is a square mod p, we want to show $Z_p[\sqrt{k^2}] = Z_p[k] $ is not a field. Since a field is equivalent to an integral domain, we can look for nonzero elements $a + bk $ and $c + dk $ such that their product $ac + (ad+bc)k + bdk^2 = 0$ in $Z_p$. I am stuck at this point.

If k is not a square mod p, we want to show that $a + b\sqrt{k} $ has an inverse in $Z_p[\sqrt k]$. This would be true if we could rationalize $\frac{1}{a + b\sqrt{k}} $, and for that we need $(a + b\sqrt{k})(a - b\sqrt{k}) = a^2 - b^2k = 1 $ mod $ p$, or $p | a^2-b^2k-1$ for any a,b in $Z_p$. I am stuck at this point.

Daniel
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  • If you haven't covered quotients of poynomial rings by an ideal, it is the best opportunity to do it now (it is very helpful, and anyway will come up soon). The answer by lhf from the duplicate is perfect, I think. You only need the definition of $\Bbb F_p[x]/(x^2-k)$, which is your $\Bbb F_p[\sqrt{k}]$. – Dietrich Burde Feb 17 '24 at 19:30
  • This is more of a problem of notation. What is $\mathbb{Z}_p[k]$? What's the element in the parenthesis and how does it relate to $k$? What would $\mathbb{Z}_p[\sqrt{k}]$? Can the element in parenthesis be treated as an element of the ring itself? What would $\mathbb{Z}_p[x]/(x^2-k^2)$ be instead? – Kolja Feb 17 '24 at 19:33
  • See the Lemma in the linked dupe. – Bill Dubuque Feb 17 '24 at 19:41

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