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I'm currently working through some extra problems while studying for an algebra exam. Right now, I am having trouble showing that $4+\sqrt{10}$ is not prime in $\mathbb{Z}[\sqrt{10}]$. I've been trying to find elements $a,b \in \mathbb{Z}[\sqrt{10}]$ such that $ab \in (4 + \sqrt{10})$ but neither $a$ nor $b$ are in the ideal. This method worked well for showing that 2 and 3 are not prime in this ring, but I'm not having much success for this example. I know that I could also show that $\mathbb{Z}[\sqrt{10}] / (4 + \sqrt{10})$ is not an integral domain, but I don't think that simplifies anything. Is there another approach that I'm missing?

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  • $\begingroup$ The norm of a prime element must be the power of a prime number, while $N(4+\sqrt{10})=6$ is not. $\endgroup$ – Crostul Dec 13 '17 at 20:52
  • $\begingroup$ @Crostul $N(4+\sqrt{10})=6$. $\endgroup$ – Lord Shark the Unknown Dec 13 '17 at 20:53
  • $\begingroup$ How about using variabels like $ (a + b \sqrt 10 )(C + d \sqrt 10) = 4 + \sqrt 10 $ ? $\endgroup$ – mick Dec 13 '17 at 20:54
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    $\begingroup$ @mick I should mention that I've already shown that $4 + \sqrt{10}$ is irreducible, so one of those factors would be a unit meaning that the other produces the same ideal as $4 + \sqrt{10}$. I'd need to look for a product that gives me a multiple of $4 + \sqrt{10}$ - which was giving me too many variables to work with! $\endgroup$ – emma Dec 13 '17 at 20:58
  • $\begingroup$ Why must the norm of a prime be a prime power ? Assume it is 6 and 2,3 do not occur as norms ?? $\endgroup$ – mick Dec 13 '17 at 20:58
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You mentioned $2$ and $3$ just after saying you needed $a$, $b$ with $a\notin I$, $b\notin I$ and $ab\in I$.

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  • $\begingroup$ That I did. Now it makes sense why these numbers all belonged to the same problem.... thanks! $\endgroup$ – emma Dec 13 '17 at 20:56
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I like your second approach (showing that $\mathbb{Z}[\sqrt{10}] / (4 + \sqrt{10})$ is not a domain), since it also tells you something about the factorization of the ideal $(4 + \sqrt{10})$. By the Third Isomorphism Theorem and the Chinese Remainder Theorem, we have \begin{align*} \frac{\mathbb{Z}[\sqrt{10}]}{(4 + \sqrt{10})} &\cong \frac{\mathbb{Z}[x]/(x^2 -10)}{(4 + x, x^2 - 10)/(x^2 - 10)} \cong \frac{\mathbb{Z}[x]}{(x + 4, x^2 - 10)}\\ &\cong \frac{\mathbb{Z}[-4]}{((-4)^2 - 10)} = \frac{\mathbb{Z}}{(6)} \cong \frac{\mathbb{Z}}{(2)} \times \frac{\mathbb{Z}}{(3)} \, . \end{align*} Since $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ is not a domain, then $(4 + \sqrt{10})$ is not prime. In fact, one can show that $$ (4 + \sqrt{10}) = (2, 4 + \sqrt{10})(3, 4 + \sqrt{10}) \, . $$

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$$(1+\sqrt{10})(2+\sqrt{10})\in (4+\sqrt{10})$$

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