How can I prove that $2+\sqrt{-5}$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$? I tried to show by $2+\sqrt{-5}=(a+b\sqrt{-5})(c+d\sqrt{-5})$ but I could not get a contradiction.


2 Answers 2


If $2+\sqrt{-5} = AB$, then $9=N(2+\sqrt{-5})=N(A)N(B)$ where $$N(u+v\sqrt{-5})=(u+v\sqrt{-5})(u-v\sqrt{-5})=u^2+5v^2$$ is the norm. If $N(A)=1$ or $N(B)=1$, then $A$ is a unit or $B$ is a unit, respectively. So if $2+\sqrt{-5}$ is reducible, there must be $A,B$ with $N(A)=N(B)=3$. Show that there can't be any $A$ such that $N(A)=3$.

  • $\begingroup$ how do you show correctly that there can't be any $A$ such that $N(A) = 3$? $\endgroup$
    – user100463
    Apr 26, 2017 at 20:20
  • 3
    $\begingroup$ Show there are no integer solutions to $u^2+5v^2=3$. @ALannister Clearly, $v$ must be zero, or $u^2+5v^2\geq 5$. So now you need $u^2=3$. $\endgroup$ Apr 26, 2017 at 21:22
  • $\begingroup$ logically, I thought that's how it should be done, but since I'd never actually seen anyone do it before, I wasn't sure. Thank you! $\endgroup$
    – user100463
    Apr 26, 2017 at 22:27
  • $\begingroup$ So by this logic, all elements of $\mathbb{Z}[\sqrt{-5}]$ of norm $9$ are irreducible? $\endgroup$
    – Vadim
    May 20, 2021 at 11:57
  • 1
    $\begingroup$ Yes, but there really aren't that many elements of norm $9.$ $\pm 2\pm\sqrt{-5}$ and $\pm 3.$ @Vadim $\endgroup$ Oct 3, 2022 at 17:23

While you already have a very big hint, I thought it would be worthwhile to expand a bit on the conceptual motivation. Generally, when studying factorization theory in rings of algebraic integers one can deduce a great deal from studying the factorizations of their corresponding norms - which form a "simpler" multiplicative submonoid of the integers (cf. the method of simpler multiples). This is true simply because the norm map is multiplicative $\,N(xy) = N(x)N(y)\,$ so it preserves many properties related to factorization. For example, in many favorable contexts (e.g. Galois) a number ring enjoys unique factorization iff its monoid of norms does. For references (Bumby and Dade, Lettl, Coykendall) see my sci.math post on 19 Dec 2007 partially excerpted below (see the post for AMS reviews of related interesting papers). See also here for analogous methods for factoring polynomials via their "simpler" evaluations.

Let $d\,$ be a squarefree positive integer, with $\,d > 1,\,$ and let $R\,$ be the ring of integers of the real quadratic number field $\,\Bbb Q(\sqrt d).\,$ Let $\,z \in R\,$ be such that $\,N(z)\,$ is composite.

Question $(1)\!:\ $ Must $z$ be reducible in $R\,?$

No, e.g. inert primes $p\,$ have $\, N(p) = p^2.$

A number ring is a PID iff its atoms have prime power norms, i.e. exactly one prime $\,p\,$ occurs in norms of an irreducible element $\,N(q) = p^n,\, p,q\,$ atoms in resp. rings. The proof is easy.

Question $(2)\!:\ $ If there exist nonunits $\,x,y \in R\,$ such that $\,N(x) N(y) = N(z),\,$ must $\,z\,$ be reducible in $\,R\,?$

No, e.g. $\,x = y = 3,\, z = 5 + 2 \sqrt{-14}.$

Bumby and Dade [2] classified those quadratic number fields $K$ where reducibility depends only on the norm. Denoting the class group by $H,$ they proved that $K$ satisfies this property iff

(a) $\ H\,$ has exponent $2,\,$ or
(b) $\ H\,$ is odd, $\,$ or
(c) $\ K\,$ is real with positive fundamental unit and the $2$-Sylow subgroup of the narrow class group is cyclic.

Note the above example $\,\Bbb Q(\sqrt{-14})$ has class group $\,K = {\rm C}(4)\,$ with exponent $4$.

See Coykendall's paper [1] for a recent discussion of related results and generalizations.

[1] Jim Coykendall. Properties of the normset relating to the class group.

[2] 35 #4186 10.65 (12.00)
Bumby, R. T. Irreducible integers in Galois extensions.
Pacific J. Math. 22 1967 221--229.


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