# How else can I tell I can do this with $5$ but not $2$ or $3$ in $\textbf{Z}[\sqrt{30}]$?

In $\textbf{Z}[\sqrt{30}]$, the number $5$ splits, since, for example, $N(5 + \sqrt{30}) = -5$. But the ideal $\langle 5 \rangle$ is a ramifying ideal, since it is equal to $\langle 5, \sqrt{30} \rangle^2$. It should have been immediately obvious to me (but it wasn't) that $(5 + \sqrt{30}) \in \langle 5, \sqrt{30} \rangle$.

Given the factorization of $30$, it is clear that $\langle 2 \rangle$ and $\langle 3 \rangle$ are also ramifying ideals. By examining the least significant digits of the squares, it is also clear that, as numbers, $2$ and $3$ are inert and therefore irreducible.

But by what other way can I quickly verify that $2$ and $3$ are inert but generate ramifying ideals? And can that other way be readily carried over to a domain like $\textbf{Z}[\sqrt{42}]$, where the least significant digits of the squares might not necessarily help in discerning primes that split but generate ramifying ideals from those that are inert but also generate ramifying ideals?

EDIT: I had a stray exponent $2$ at first. Sorry for the confusion.

• I’m not sure I understand your question. The discriminant is $120$, and every prime that divides that is ramified. I also don’t understand your use of the word “inert” here. Isn’t that word usually restricted to unramified primes? And finally, I do not believe that $5+\sqrt30\in(5,\sqrt30\,)^2$. May 7, 2016 at 5:47
• @Lubin I think he meant $\langle 5, \sqrt{30} \rangle$, not $(5, \sqrt 30)^2$. Also, maybe he should've bolded "the number" and "as numbers", or used the word "elements" instead. May 7, 2016 at 16:46
• The ideal generated by each prime that divides $120$ is ramified, but the prime number itself is either inert (like $2$ or $3$ here) or it splits (like $5$ does). But I doubt there's any other way to discover this fact besides trying to solve $x^2 - 30y^2 = \pm p$. It's a very different situation from when the discriminant is $4p$ or $p$ itself, like in $\mathbb{Z}[\sqrt{79}]$, where we see that $79 = (\sqrt{79})^2$. May 7, 2016 at 21:34

First you should mind that a prime in a quadratic ring of integers either splits, ramifies or is inert. In your example above we have

$$(5+\sqrt{30})\cdot(-11+2\sqrt{30})=5-\sqrt{30}$$

and obviously

$$(5-\sqrt{30})\cdot(-11-2\sqrt{30})=5+\sqrt{30}$$

so that the ideals $\langle 5+\sqrt{30} \rangle$ and $\langle 5-\sqrt{30} \rangle$ are equal and the number $11+2\sqrt{30}$ is a unit. Hence

$$\langle 5+\sqrt{30} \rangle^2 = \langle 5 \rangle.$$

Further more

$$(5+\sqrt{30})\cdot(6-\sqrt{30})=\sqrt{30}$$

and

$$(5+\sqrt{30})\cdot(-5+\sqrt{30})=5$$

and therefore

$$\langle 5+\sqrt{30} \rangle = \langle 5-\sqrt{30} \rangle = \langle 5,\sqrt{30} \rangle.$$

All in all the prime 5 ramifies to the principal ideal $\langle 5+\sqrt{30} \rangle^2$ and it does not split.

By Dedekind's theorem a prime ramifies if the prime divides the discriminant of the ring of integers. The only question that can be asked is whether the ideal is principal or whether it has two generators. If the prime $2$ factorizes to a principal ideal $\langle a+b\sqrt{30} \rangle^2 = \langle 2 \rangle$ the generator $a+b\sqrt{30}$ of this ideal then divides the number $2$. But then the number $a-b\sqrt{30}$ also divides the number $2$ and we must solve the equation

$$(a+b\sqrt{30})\cdot(a-b\sqrt{30})=a^2-30b^2=\pm 2$$

in order to determine if the ideal is a principal one.

The ideal $\langle q \rangle$ of an inert prime $q$ is prime!

Şaban Alaca and Kenneth Williams have written a very good introductory book Introductory Algebraic Number Theory with many examples. They particularly focus on algebraic integers, not only ideals. If you are in need of a book that is as abstract as interesting and that only deals with quadratic and cyclotomic integers you should take a look into Fermat's Last Theorem of Edwards.