# When will a prime element of $\Bbb{Z}[(\sqrt{5}-1)/2]$ have field norm equal to a rational prime?

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?

• In order too give a fitting answer; what is your definition of $N(x)$? – Servaes Feb 11 at 0:20
• 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. – Servaes Feb 11 at 0:24
• 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 $N(p)=p^2$. – Servaes Feb 11 at 1:18
• @JyrkiLahtonen Or $p=5$. – Servaes Feb 11 at 6:49
• 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. – Jyrki Lahtonen Feb 11 at 6:58

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 $$k\in\Bbb{Z}$$.

• 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? – gen Feb 11 at 10:20
• 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 $p\equiv\pm2\pmod{5}$. – Servaes Feb 11 at 10:27
• Is this just due to the fact that $p$ is inert in R if and only if $p \equiv \pm 2\ (\operatorname{mod}\ 5)$? – gen Feb 11 at 10:50
• Yes it is; this means precisely that there is no $x\in R$ with $N(x)=p$. – Servaes Feb 11 at 10:52
• OK, I think I should be able to sweat that result myself. Thanks again. – gen Feb 11 at 10:53