I'm having troubles with the following problem:

Let $p$ be a prime number in $\mathbb{Z}$, and $\alpha\in\mathbb{Z}\left[\sqrt{p}\right]$ which is not a unit. Prove that $\alpha$ have a factorization in irreducible elements of $\mathbb{Z}\left[\sqrt{p}\right]$.

At the beginning I though that that ring was a Euclidean Domain, but that fails for $\mathbb{Z}\left[\sqrt{5}\right]$. So, I don't know where to start now.

Thanks in advance.


Hint: Consider the norm $N(a+b\sqrt{p})=|a^2-b^2p|$. Prove that if $\delta$ divides $\alpha$, then $N(\delta) \le N(\alpha)$. Then use induction on $N(\alpha)$.

Or, more sophisticatedly, argue that $\mathbb{Z}\left[\sqrt{p}\right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.

  • 2
    $\begingroup$ @MattSamuel That's a rather serious gap. What algebra textbook(s) did you use? $\endgroup$ – Bill Dubuque Dec 10 '18 at 21:43
  • $\begingroup$ @Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include. $\endgroup$ – Bill Dubuque Dec 10 '18 at 23:56

Suppose $\alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that $$\alpha=a_1a_2$$ If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so $$\alpha = a_3a_4a_2$$ In general there is an increasing chain of ideals $$(a_1)\subseteq (a_3)\subseteq \cdots$$ Since $\mathbb{Z}[\sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing, $$\alpha = a_1a_2$$ with $a_1$ irreducible. Thus you can iterate this to write $$\alpha = a_1\cdots a_n$$ with $a_1,\ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals $$(a_{n,1})\subseteq (a_{n+1,2})\subseteq \cdots$$ This must also stabilize, until you end up with a factorization into irreducible elements.

  • 1
    $\begingroup$ At this level, one should justify the Noetherian claim. $\endgroup$ – Bill Dubuque Dec 10 '18 at 21:45

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