# If $\vert d\vert\geq 3$, is $-1+\sqrt{d}$ an irreducible element of $\mathbb{Z}[\sqrt{d}]$?

Let $$d\in\mathbb{Z}$$ be an integer which is not a square (it does not have to be squarefree, though).

Question. Assume that $$\vert d\vert\geq 3$$ to avoid special cases. Is is true that $$\pi=-1+\sqrt{d}$$ is irreducible in $$R=\mathbb{Z}[\sqrt{d}]$$ ?

Remark. I can show this it is true in the following cases:

• $$d<0$$

• $$d=1\pm p$$, where $$p$$ is prime.

Apart from these cases, I have no idea whether $$\pi$$ might be irreducible in full generality or not when $$d>0$$. (I tried to use a CAS to produce examples and counterexamples, but I am really bad at programming so I didn't get very far.)

Addendum. I tried to determine first the values of $$d$$ for which $$\pi$$ is a prime element. This happens to be the case exactly when $$d=1\pm p$$, $$p$$ prime, so this does not give any new insight.

Update (February 21,2024). In his answer, Keith Conrad gives examples of integers $$d\not\equiv 1 \ [4]$$ for which $$-1+\sqrt{d}$$ is not irreducible.

When $$d\equiv 1 \ [4]$$, there are also such examples. For example, $$(-1+\sqrt{41})=(7+\sqrt{41})(-6+\sqrt{41})$$ is a non trivial factorization (the first factor has norm $$8$$ and the second one has norm $$-5$$).

Another example is $$-1+\sqrt{57}=(7+\sqrt{57})(8-\sqrt{57})$$.

• I don't know about the general question, but I tried the first example you can't prove ($d=5$) and found $-1+\sqrt{5} = (7-3\sqrt{5})(2+\sqrt{5})$. Literally just by guessing - sorry that I don't have a more informative method! Feb 12 at 12:59
• @preferred_anon the number $2+\sqrt{5}$ is a unit, with inverse $\sqrt{5}-2$, so that factorization does not show $-1+\sqrt{5}$ is reducible. In fact, $-1+\sqrt{5}$ is irreducible in $\mathbf Z[\sqrt{5}]$.
– KCd
Feb 12 at 13:01

Suppose $$\mathbf Z[\sqrt{d}]$$ has unique factorization, so it is the ring of integers in $$\mathbf Q(\sqrt{d})$$ (because UFDs are integrally closed), which implies $$d$$ is squarefree.

In a UFD, prime and irreducible elements are the same thing. An element $$\alpha$$ is prime exactly when the ideal $$(\alpha)$$ is a prime ideal, which makes $$(\alpha)$$ a maximal ideal: the residue ring $$\mathbf Z[\sqrt{d}]/(\alpha)$$ is finite and a finite integral domain is a field. The size of $$\mathbf Z[\sqrt{d}]/(\alpha)$$ is $$|{\rm N}(\alpha)|$$ and a finite field has prime-power order, so when $$\mathbf Z[\sqrt{d}]$$ has unique factorization, a necessary condition that an element $$\alpha$$ in this ring be irreducible is that the absolute value of its norm is a prime power.

Since $$|{\rm N}(-1+\sqrt{d})| = |1 - d| = |d-1|$$, when $$\mathbf Z[\sqrt{d}]$$ has unique factorization and $$|d-1|$$ is not a prime power, $$-1+\sqrt{d}$$ is not irreducible.

Example. When $$d = 7, 11, 19, 22, 23, 31, 43, 46$$, and $$47$$, $$\mathbf Z[\sqrt{d}]$$ is the ring of integers in $$\mathbf Q(\sqrt{d})$$ and has class number $$1$$, so it is a PID and thus has unique factorization. In these cases, $$|d-1| = d-1$$ is not a prime power.

In the first case, $$d = 7$$, an explicit factorization of $$-1+\sqrt{7}$$ into two non-units is $$(\sqrt{7}+3)(2\sqrt{7}-5)$$, where the factors are not units since they have norm $$2$$ and $$-3$$, respectively.

In the second case, $$-1+\sqrt{11} = (3+\sqrt{11})(7-2\sqrt{11})$$, where the factors have norm $$-2$$ and $$5$$.

• Thank you. Do you know what happens when $d\equiv 1 \ [4]$ ? (so $\mathbb{Z}[\sqrt{d}]$ is not a UFD anymore) Feb 12 at 14:21
• After reading over my post, have you tried looking at $d \equiv 1 \bmod 4$ yourself?
– KCd
Feb 12 at 14:57
• Well, if $d\equiv 1 [4]$, $\mathbb{Z}[\sqrt{d}]$ is not a UFD, and irreducible and prime elements are not the same. I know when $-1+\sqrt{d}$ is prime, but that's it...If there is something i can use in your answer in this case, i missed it , i'm sorry. Maybe I could work out the potential irreducible factors of $-1+\sqrt{d}$ in $\mathbb{Z}[\frac{-1+\sqrt{d}}{2}]$ and see if any of them lie in $\mathbb{Z}[\sqrt{d}]$. Is that what you have in mind ? Feb 12 at 15:13
• For what is worth, i think i can show that $2$ is always irreducible in $\mathbb{Z}[\frac{-1+\sqrt{d}}{2}]$, and that $\frac{-1+\sqrt{d}}{2}$ is prime if and only if $d=1+4p$, where $p$ is prime. In particular, this should be enough to show that if $d=1+4p$, then $-1+\sqrt{d}$ is irreducible in $\mathbb{Z}[\sqrt{d}]$ as soon as$\mathbb{Z}[\frac{-1+\sqrt{d}}{2}]$ is a UFD . However, i cannot find any counterexample yet if $\frac{d-1}{4}$ is composite, as solving a generalized Pell equation is not a thing I am comfortable with. Feb 12 at 15:44