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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.

Bill Dubuque
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user20353
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3 Answers3

19

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 $N(A)=3$.

Thomas Andrews
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  • how do you show correctly that there can't be any $A$ such that $N(A) = 3$? –  Apr 26 '17 at 20:20
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    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 $u^2=3$. – Thomas Andrews Apr 26 '17 at 21:22
  • 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! –  Apr 26 '17 at 22:27
  • So by this logic, all elements of $\mathbb{Z}[\sqrt{-5}]$ of norm $9$ are irreducible? – Vadim May 20 '21 at 11:57
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    Yes, but there really aren't that many elements of norm $9.$ $\pm 2\pm\sqrt{-5}$ and $\pm 3.$ @Vadim – Thomas Andrews Oct 03 '22 at 17:23
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While you already have a big hint, it is instructive to expand a bit on the conceptual motivation. Generally, when studying factorization theory in rings of algebraic integers we can usually 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 that 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 (i.e. irreducibles) have prime power norms, i.e. exactly one prime $\,p\,$ occurs in norms of atoms: $\,N(q) = p^n,\, p,q\,$ 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], [3] classified those quadratic number fields $K$ where reducibility depends only on the norm, i.e. when $\,\alpha\,$ irreducible and $\,N(\beta) = N(\alpha)\,\Rightarrow\, \beta$ irreducible. Denoting the class group by $H,$ they proved that $K$ satisfies this property iff

  • $\ H\:\!$ has exponent $2,\,$ or
  • $\:\! H\:\!$ is odd, $ $ or
  • $\:\! K\:\!$ is real with positive fundamental unit and cyclic $2$-Sylow subgroup of narrow class group.

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] Bumby, R. T., Dade, E. C.
Remark on a problem of Niven and Zuckerman
Pacific J. Math. 22 1967 15--18

[3] 35 #4186 10.65 (12.00)
Bumby, R. T. Irreducible integers in Galois extensions.
Pacific J. Math. 22 1967 221--229.

Bill Dubuque
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I wanted to note that it is possible to show the irreducibility of $2+\sqrt{-5}$ without using the norm.

We can solve $c,d$ in terms of $a,b$ in OP's equation: $$c=\frac{2a+5b}{a^2+5b^2}\tag1$$ $$d=\frac{a-2b}{a^2+5b^2}\tag2$$ Without loss of genarilty, let $b\in\Bbb Z^+.$ Then, $$1\leq c-2d=\frac{9b}{a^2+5b^2}\leq\frac9{5b}$$ and hence $b=1.$ Can you continue from here?

Bob Dobbs
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  • But you do effectively use a $,\color{#c00}{{\rm norm}\ n},$ when $\rm\color{#c00}{rationalizing}$ the denominator when solving for $,c,d,,$ i.e. $$,c + d\sqrt{-5,} = ,\frac{2+\sqrt{-5}}{a+b\sqrt{-5}}, = ,\frac{(2+\sqrt{-5})(a-b\sqrt{-5})}{\underbrace{\color{#c00}{a^2+5b^2}}_{\large\color{#c00} n}}\ \Rightarrow\ \begin{align}c = (2a!+!b)/n\ d = (a!-!2b)/n\end{align}\qquad\qquad$$ – Bill Dubuque May 06 '25 at 06:26
  • i.e. using $,\color{#0a0}{\alpha\mid\beta \iff N(\alpha)!=!\alpha\bar\alpha\mid \beta\bar\alpha},,$ which is the standard norm-based method of simpler multiples to reduce division by an algebraic integer to division by a rational integer. But it ends up being simpler to use instead:
    $\ \color{#0a0}{\alpha\mid\beta\ \Rightarrow\ N(\alpha)\mid N(\beta)},,$ i.e. $,\color{#0a0}{\alpha\bar\alpha\mid \beta\bar\beta},,$ here $,a^2+5b^2\mid 9,,$ which exploits norm multiplicativity to the hilt (for both $\alpha$ & $\beta$). $\ \ $     – Bill Dubuque May 06 '25 at 06:27
  • My answer elaborates on the effectiveness of such "divisibility simplification" maps (which formally come to the fore when one studies divisor theory). $\ \ $ – Bill Dubuque May 06 '25 at 06:39
  • I assumed that the norm "$n$" is not historically defined yet. @BillDubuque – Bob Dobbs May 06 '25 at 08:01
  • To properly assess your "without" claim we need to see the omitted steps which show how you compute $,c,$ and $,d,,$ esp. how you obtained $,a^2+5b^2.,$ Please add those to your answer. $\ \ $ – Bill Dubuque May 06 '25 at 08:12
  • I think you did something like that in the comments. Thanks for your great assistance. @BillDubuque – Bob Dobbs May 06 '25 at 08:23
  • Ok, if that's how you did it then you do essentially use the concept of the norm. The only "without" is that you do so only for $\alpha$ (not also for $\beta),,$ which makes the argument more complicated, as I explained above. $\ \ $ – Bill Dubuque May 06 '25 at 08:31