The discriminant of the ring $\mathbb{Z}[\sqrt{10}]$ is 40. Since 2 divides 40, 2 is ramified in this ring. I know that the ideal $(2)$ factors as $(2,\sqrt{10})^2$, but I cannot find a factorization of 2 in this ring. The equation $x^2-10y^2=\pm 2$ has no solutions. How would I find a factorization of 2 in this ring (or is there none)? I would like a method that can be generalized to other quadratic integer rings.


1 Answer 1


I assume that you mean a factorization of $2$ into irreducible elements in this ring, then note that the fact that $x^2-10y^2= \pm 2$ has no solutions, i.e. there are no elements of norm $2$ in $\Bbb Z[\sqrt{10}]$ implies that $2$ is an irreducible element in $\Bbb Z[\sqrt{10}]$, since if we ahve $2=ab$, then taking norms gives $4=N(2)=N(a)N(b)$, but since there are no elements of norm $\pm 2$, we get that $N(a)=\pm 1$ or $N(b) = \pm 1$, so either $a$ or $b$ is a unit.

  • $\begingroup$ So the ideal $(2)$ can factor, while the element $2$ does not? $\endgroup$
    – weux082690
    Jun 2, 2018 at 1:34
  • $\begingroup$ @weux082690 yes $\endgroup$ Jun 2, 2018 at 1:35
  • $\begingroup$ Is there a way to tell if the element factors? $\endgroup$
    – weux082690
    Jun 2, 2018 at 1:37
  • $\begingroup$ @weux082690 arguing with the norm as in this example is often helpful. You can also try to factor the ideal generated by that element first, if that factors as a product of principal ideals, this will also give you a factorization of the element. $\endgroup$ Jun 2, 2018 at 1:46
  • $\begingroup$ Also note that if the class number is greater than one, there will be element with non-unique factorizations. $\endgroup$ Jun 2, 2018 at 1:46

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