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Let $D\in \mathbb{Z}$ where $D$ is not a perfect square. Prove that if $\alpha\in \mathbb{Z}[\sqrt{D}]$, and $\alpha= a+b\sqrt{D}$ with: $|a^2-Db^2| = p$, a rational prime, then $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{D}]$. Is the converse true?

So, for the if statement I figure I should assume $\alpha$ isn't irreducible, say $\alpha=\beta\gamma$ where they are not units, then $N(\beta)N(\gamma)=p$, but then I feel it is incorrect to assume a norm exists, because this would make it a Euclidean domain? is that the purpose of the absolute value?

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  • $\begingroup$ If you assume that a norm exists when it doesn't, it could lead to a contradiction that proves your point. $\endgroup$
    – user153918
    Commented Jun 2, 2015 at 17:16

3 Answers 3

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If $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{D}]$, is then $|\mathsf{N}(\alpha)| = p$ certain to be a prime number in $\mathbb{Z}$? (I'm stating the converse to make sure we're on the same page; if I've misunderstood what the converse is here then we'll never hear the end of it).

No. Consider $D = -10$. It turns out that $\sqrt{-10}$ is irreducible, yet it has a norm of 10. The formula for norm in this domain works out to be $$\mathsf{N}(a + b\sqrt{-10}) = a^2 - (-10)b^2 = a^2 + 10b^2.$$ That's all a norm is: a function that enables you to compare numbers from other domains within the familiar framework of $\mathbb{Z}^+ \bigcup \{0\}$. The Euclidean algorithm is another familiar thing from $\mathbb{Z}$, but, unlike the norm, it can't always be carried over.

The norm can never be a negative number here, so there's no need to specify absolute value (quite a different story if $D$ is positive). If $\sqrt{-10}$ is reducible, then we can solve $\mathsf{N}(\beta) = 2$ and $\mathsf{N}(\gamma) = 5$. Except we can't. The possible norms in this domain are 0, 1, 4, 9, 10, 11, 14, 16, 19, 25, ... (see Sloane's http://oeis.org/A020673). Either $\beta$ or $\gamma$ is a unit.

Two prior answerers already mentioned the norms of rational primes. I mention it again for the sake of completeness.

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  • $\begingroup$ Thank you, I will take a closer look at quadratic forms to complete my knowledge. $\endgroup$ Commented Jun 3, 2015 at 9:07
  • $\begingroup$ It's a very deep topic. A math professor actually said to me once that there's still a lot we don't know about quadratic forms. And he knows a lot more about these than I do. $\endgroup$ Commented Jun 3, 2015 at 12:33
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Given a number field $K$, the field norm $\mathsf{N}_{K/\mathbb{Q}}$ is defined regardless of whether the ring of integers $\mathcal{O}_K$ is a Euclidean domain. You can prove that an element $\alpha\in\mathcal{O}_K$ is a unit iff $\mathsf{N}_{K/\mathbb{Q}}(\alpha)=\pm1$.

Let $K=\mathbb{Q}(\sqrt{-5})$, so $\mathcal{O}_K=\mathbb{Z}[\sqrt{-5}]$. The element $2$ is irreducible in $\mathcal{O}_K$, but $$\mathsf{N}_{K/\mathbb{Q}}(2)=2^2-(-5)0^2=4$$ is not a prime. You can see that $2$ is irreducible because, supposing for the sake of contradiction that $2=\beta\gamma$ for two non-units, we have $$4=\mathsf{N}_{K/\mathbb{Q}}(2)=\mathsf{N}_{K/\mathbb{Q}}(\beta)\cdot \mathsf{N}_{K/\mathbb{Q}}(\gamma)=\pm2\cdot\pm 2$$ but there are no elements of $\mathcal{O}_K$ that are of norm $2$ or $-2$ because $$\mathsf{N}_{K/\mathbb{Q}}(a+b\sqrt{-5})=a^2-(-5)b^2=a^2+5b^2$$ always positive, it is $\geq 5$ if $b\neq 0$, and can only be a square number if $b=0$.

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  • $\begingroup$ Thank you for clarifying the norm for me. Also, thank you for the norm of 2 counterexample...I went that direction right away as well, but it's nice to see some handy work, as I am bit rusty right now. $\endgroup$ Commented Jun 2, 2015 at 7:55
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The converse can be true only when the norm is the square of a rational prime. If $D$ is negative, then $\alpha$ is purely real.

The norm function is available regardless of whether the domain is Euclidean or not. The purpose of the absolute value is so that the function does not return a negative number as it's result. For example, $N(1 - \sqrt{3}) = -2$, but with the absolute value, $|N(1 - \sqrt{3})| = 2$, so that it meets one of the requirements to be a Euclidean function (its a mapping to the positive integers and $0$).

Whether a particular positive integer can be a norm in a particular domain is a different issue. For example, in $\textbf{Z}[\sqrt{3}]$, no number has a norm of $5$ (and $5$ itself has a norm of $25$.

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  • $\begingroup$ Thank you, yes I have been studying Pell's equation and diophantine equations to aid my research...glad to see it is useful, and I suspect it will be useful in determining factorizations. On another note, are you proposing that for positive $D$, if we do have a UFD, our Euclidean function will always be the norm with absolute value? $\endgroup$ Commented Jun 3, 2015 at 9:00
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    $\begingroup$ I don't know if that's what Albert is proposing, but it's very close to correct. Definition 2.11 in Huishi Li's An Introduction to Commutative Algebra says a Euclidean function is $R^\times\to\mathbb{N}$, which presumably includes 0, but which certainly excludes negatives. As Albert showed, the norm function by itself can give negative values if $D$ is positive. There are some real UFDs for which the norm function requires yet another adjustment to be a Euclidean, and still others where the Euclidean function is known by only very few or maybe none. See oeis.org/wiki/Euclidean_domains $\endgroup$
    – Mr. Brooks
    Commented Jun 4, 2015 at 21:24
  • $\begingroup$ @Mr. Brooks , thanks for your thoughts:) $\endgroup$ Commented Jun 25, 2015 at 5:50
  • $\begingroup$ You're welcome. $\endgroup$
    – Mr. Brooks
    Commented Jun 29, 2015 at 20:55

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