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Consider the integer ring of $\mathbb{Q}[\sqrt{5}]$, i.e. $\mathbb{Z}[(\sqrt{5}-1)/2]$. Then if $N(x)$ denotes the field norm of $x\in\mathbb{Z}[(\sqrt{5}-1)/2]$, then $N(x) = p$ for a rational prime $p$ implies that $x$ is a prime in the integer ring.

My question is when the converse is also true? I.e. when will the implication $$x\in\mathbb{Z}[(\sqrt{5}-1)/2] \textit{ is a prime } \quad\Rightarrow\quad N(x)\in\mathbb{Z}\textit{ is a rational prime }$$ hold? What would be an example that illustrates that this not generally true?

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  • $\begingroup$ In order too give a fitting answer; what is your definition of $N(x)$? $\endgroup$ – Servaes Feb 11 at 0:20
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    $\begingroup$ Sure, that's a fine definition. There are multiple (equivalent) ways to define it, but at an introductory level it is not evident that these different ways are in fact equivalent, which is why I asked. I'll write up an answer. $\endgroup$ – Servaes Feb 11 at 0:24
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    $\begingroup$ I misjudged; this is too much to do by elementary methods. I can tell you that $7\in\Bbb{Z}[\tfrac{\sqrt{5}-1}{2}]$ is a prime with $N(7)=7^2$ not a rational prime. More generally, any rational prime $p\equiv2,3\pmod{5}$ is also prime in $\Bbb{Z}[\tfrac{\sqrt{5}-1}{2}]$ and has $N(p)=p^2$. $\endgroup$ – Servaes Feb 11 at 1:18
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    $\begingroup$ @JyrkiLahtonen Or $p=5$. $\endgroup$ – Servaes Feb 11 at 6:49
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    $\begingroup$ No harm done @Servaes. I was just a bit surprised to see you state the complementary case (containing exactly the same information). But I forgot that not all primes are coprime to five! Also, I am convinced you did not need my comment at all. $\endgroup$ – Jyrki Lahtonen Feb 11 at 6:58
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Let $\alpha:=\tfrac{\sqrt{5}-1}{2}$ and let $R:=\Bbb{Z}[\alpha]$. For every $x\in R$ there exist unique $u,v\in\Bbb{Z}$ such that $x=u+v\alpha$. The minimal polynomial of $\alpha$ over $\Bbb{Z}$ is $X^2+X-1$, so $R$ has a ring automorphism $$\overline{\hphantom{\,}\cdot\hphantom{\,}}:\ R\ \longrightarrow\ R:\ u+v\alpha\ \longmapsto\ u-v(1+\alpha),$$ where $u,v\in\Bbb{Z}$. Denote this by $\overline{u+v\alpha}=u-v(1+\alpha)$, just like complex conjugation. Then for any $x\in R$ $$N(x)=x\overline{x}=(u+v\alpha)(u-v(1+\alpha))=u^2-uv-v^2.$$

Lemma: If $x\in R$ is prime then $N(x)=\pm p^k$ for some rational prime $p\in\Bbb{Z}$ and $k\in\{1,2\}$.

Let $x\in R$ be prime. Then $N(x)$ is a nonzero rational integer, and if $N(x)=ab$ for coprime integers $a,b\in\Bbb{Z}$ then $x\mid ab$ and hence either $x\mid a$ or $x\mid b$. Without loss of generality $x\mid a$, so $a=xy$ for some $y\in R$, hence $$\overline{x}\,\overline{y}=\overline{xy}=\overline{a}=a,$$ which shows that also $\overline{x}\mid a$. It follows that $x\overline{x}\mid a^2$. Because $x\overline{x}=N(x)=ab$ and $a$ and $b$ are coprime, it follows that $b=\pm1$. This shows that $N(x)=\pm a$ is a prime power, up to sign, say $a=p^k$. Then $$x\mid p^k\qquad\text{ and so }\qquad x\mid p,$$ because $x$ is prime. As before it follows that $x\overline{x}\mid p^2$, and so $N(x)\mid p^2$ so $k\leq2$. Note that $k=0$ is impossible as then $x\overline{x}=N(x)=p^0=1$, meaning that $x$ is a unit. But $x$ is a prime, hence not a unit, a contradiction. Hence either $N(x)=\pm p$ or $N(x)=\pm p^2$.$\hspace{30pt}\square$

Proposition: The norm of a unit in $R$ is a rational unit.

If $u\in R^{\times}$ is a unit then $uv=1$ for some $v\in R^{\times}$, and hence $$\overline{u}\,\overline{v}=\overline{uv}=\overline{1}=1,$$ so $\overline{u}$ is also a unit. It follows that the rational integers $N(u),N(v)\in\Bbb{Z}$ satisfy $$N(u)N(v)=u\overline{u}v\overline{v}=uv\overline{uv}=1,$$ and so $N(u)=N(v)=\pm1$.$\hspace{30pt}\square$

Result: For a prime $x\in R$ the norm $N(x)$ is a rational prime if and only if $x$ is not of the form $x=up$, with $u\in R^{\times}$ a unit and $p\in\Bbb{Z}$ a rational prime.

If there exists a unit $u\in R^{\times}$ such that $x=up$ for some rational prime $p\in\Bbb{Z}$, then $$N(x)=x\overline{x}=u\overline{u}p\overline{q}=N(u)p^2=\pm p^2,$$ where the last identity holds by the proposition above. This shows that $N(x)$ is not a rational prime.

Coversely, if $x\in R$ is prime such that $N(x)$ is not a rational prime then by the lemma above $N(x)=\pm p^2$ for a rational prime $p\in\Bbb{Z}$. Then $x\mid p^2$ and hence $x\mid p$ because $x$ is prime, say $p=xy$. Then as before $p=\overline{xy}$ and so $$p^2=xy\overline{xy}=x\overline{x}y\overline{y}=N(x)N(y)=\pm p^2N(y),$$ which shows that $y\overline{y}=N(y)=\pm1$. This means $y$ is a unit and hence so is $u:=y^{-1}$, and we have $x=up$.$\hspace{30pt}\square$


Fact 1: The units of $R$ are precisely the element of the form $\pm\alpha^k$ with $k\in\Bbb{Z}$, where $\alpha^{-1}=1+\alpha$. I do not know of a proof that there are no other units that doesn't rely on Dirichlets unit theorem.

Fact 2: A rational prime $p\in\Bbb{Z}$ is of the form $p=N(x)$ for some $x\in R$ if and only if $p\equiv\pm1\pmod{5}$ or $p=5$. I do not know of a proof that does not rely on the Kummer-Dedekind theorem and quadratic reciprocity.


EDIT: Using some more machinery, we get some more results.

If $x$ is prime then by the lemma above $N(x)=\pm p^k$ with $k\in\{1,2\}$. If $x=u+v\alpha$ with $u,v\in\Bbb{Z}$ this means that $$u^2-uv-v^2\equiv0\pmod{p}.$$ If $u,v\equiv0\pmod{p}$ then $x=yp$ for some $y\in R$, and $N(x)=N(y)p^2$, so $y$ is a unit.

Otherwise $u,v\not\equiv0\pmod{p}$ and hence $uv^{-1}\in\Bbb{F}_p$ is a root of $$X^2-X-1\in\Bbb{F}_p[X].$$ In particular this polynomial splits, so its discriminant $\Delta=5$ is a square in $\Bbb{F}_p$. By the law of quadratic reciprocity, for $p\neq2,5$ this is equivalent to $p$ being a square mod $5$, i.e. $p\equiv\pm1\pmod{5}$.

As for the primes $p=2,5$; it is easily checked that $N(2-\alpha)=5$, and that $N(u+v\alpha)=\pm2$ has no solutions as $$u^2-uv-v^2\equiv0\pmod{2},$$ implies that both $u$ and $v$ are even, hence $N(u+v\alpha)\equiv0\pmod{4}$. This shows that for any $x\in R$, the norm $N(x)$ is a rational prime if and only if $N(x)$ a rational prime that is a quadratic residue modulo $5$. This also implies that if $N(x)=\pm p^2$ for a prime $x\in R$, then $p\equiv\pm2\pmod{5}$ and $x=up$ for some unit $u\in\Bbb{R}^{\times}$. By Dirichlets unit theorem every unit is of the form $u=\pm\alpha^k$ for some $k\in\Bbb{Z}$.

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  • $\begingroup$ This is a nicely explained answer, I also appreciate your effort to set the level of depth at that of the OP. I have just two questions: i) where do you use Fact 2? ii) the fact that elements $x = up: \ u\in R^{\times} \textit{ unit},\ p\textit{ rational prime}$ behave differently only matters if such elements can be primes in $R$. Is it true that all such elements are prime in $R$, or is it just a fraction of them that happen to be prime? $\endgroup$ – gen Feb 11 at 10:20
  • $\begingroup$ I do not use the two facts anywhere; I just state them to give a broader picture than elementary methods allow me to construct. And I will try to include a proof that $up$ with $u\in\Bbb{R}^{\times}$ and $p$ a rational prime is a prime in $R$ if and only if $p\equiv\pm2\pmod{5}$. $\endgroup$ – Servaes Feb 11 at 10:27
  • $\begingroup$ Is this just due to the fact that $p$ is inert in R if and only if $p \equiv \pm 2\ (\operatorname{mod}\ 5)$? $\endgroup$ – gen Feb 11 at 10:50
  • $\begingroup$ Yes it is; this means precisely that there is no $x\in R$ with $N(x)=p$. $\endgroup$ – Servaes Feb 11 at 10:52
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    $\begingroup$ OK, I think I should be able to sweat that result myself. Thanks again. $\endgroup$ – gen Feb 11 at 10:53

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