Irreducible in $\mathbb{Z}[\sqrt{-5}]$

Question

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.

Answer 1

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 $N(A)=3$.

Answer 2

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.

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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.
http://www.ams.org/proc/1996-124-12/S0002-9939-96-03387-4
http://www.math.ndsu.nodak.edu/faculty/coykenda/paper3.pdf

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