Show that the elements 2,3, and $1 \pm \sqrt{-5}$ are irreducible elements of $\mathbb{Z}[\sqrt{-5}]$. I have never seen this notation before. From another post I am interpreting this to mean the following:

$\mathbb{Z}[\sqrt{-5}] = \{a_{0}+a_{1}\sqrt{-5}+ \dots + a_{n}(\sqrt{-5})^{n} \colon a_{i} \in \mathbb{Z} \}$.

Am I correct or is it something else? I do not need the proof, just verification of notation.

  • $\begingroup$ Note that $\sqrt{-5}^2 = -5$, so you don't need all those terms. Also, the power should not be under the root - that will cause confusion on the future. $\endgroup$ – Thomas Andrews Feb 1 '16 at 16:37

You are correct, but it it suffices to see that $\mathbb Z [\sqrt{-5}] = \{a_0 + a_1\sqrt{-5}: a_i \in \mathbb Z \}$

  • 1
    $\begingroup$ Why are the other powers of $\sqrt{-5}$ not needed? $\endgroup$ – Jack Feb 1 '16 at 16:38
  • 3
    $\begingroup$ Note that $(\sqrt{-5})^{2k} \in \mathbb Z$ and $(\sqrt{-5})^{2k+1} \in \sqrt{-5} \mathbb Z$ for every $k$ $\endgroup$ – flawr Feb 1 '16 at 16:41

It's somehow correct, but it is misleading. The correct way to understand this notation is $$\mathbb Z[\sqrt{-5}]=\{a+b\mathrm i\sqrt5\,\mid a,b\in\mathbb Z\}$$

  • 1
    $\begingroup$ There's no reason to write $\sqrt{-5}$ as $i\sqrt{5}$... $\endgroup$ – paul garrett Feb 1 '16 at 16:47
  • $\begingroup$ @paulgarrett. The OP meant $\mathbb Z[\alpha]$ such that $\alpha^2=-5$. The notation $\sqrt{-5}$ is obsolete and misleading since $\sqrt{-5}\sqrt{-5}\neq\sqrt{(-5)(-5)}$. $\endgroup$ – Tom-Tom Feb 1 '16 at 19:50
  • $\begingroup$ No, it is certainly not obsolete at all. (The possibility of mis-using products of square roots is irrelevant, and bringing $\sqrt{-1}$ into the picture doesn't help anything.) $\endgroup$ – paul garrett Feb 1 '16 at 20:20
  • $\begingroup$ "bringing $\sqrt{-1}$ into the picture doesn't help anything". I can only agree with that, since this notation makes no more sense than $\log(-1)/37\pi$ or $0/0$. $\endgroup$ – Tom-Tom Feb 1 '16 at 21:03
  • $\begingroup$ I guess you and I are in different worlds of mathematical usage. I agree that square roots of negative numbers oughtn't be introduced to early... but in introductory abstract algebra or introductory algebraic number theory, writing $\sqrt{-5}$ or $\sqrt{-1}$ is certainly conventional usage. Not for grade-schoolers, no. Also, $\log(-1)$ is $\pi i$ ambiguous by multiples of $2\pi i$. Care is required, but it has sense. I think at the level of the question, $\mathbb Z[\sqrt{-5}]$ is apt, and is at least completely standard. $\endgroup$ – paul garrett Feb 1 '16 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.