Find how $p=2,3,5,7$ splits in $\mathbb{Q}(\sqrt{-5})$ (i.e. find those $e_i,f_i$ for $1 \leq i \leq r$).

Can somebody please explain how this is done? My attempt is the following:

Let K = $\mathbb{Q}(\sqrt{-5})$ be a number field and let $f(x) = x^2+5$ be the minimal polynomial. Then the discriminant is disc$(f)= 0^2-4(5)=-20$. Since the discriminant is not square-free, then we consider the four cases:

  1. when $p=2$, notice that $2|(-20)$ so 2 is ramified in $K$.
  2. when $p=3$, $3 \nmid -20$, so 3 is unramified in $K$.
  3. when $p=5$, $5 | -20$, so 5 is ramified in $K$.
  4. when $p=7$, $7 \nmid -20$, so 7 is unramified in $K$.

My question is this, how does knowing whether a prime is ramified help in determining if it splits in a given number field? I apologize if this question seems very elementary, but I am not understanding what it means to split, ramify, or be inert. I am taking a course in algebraic number theory, but we are only presented with theorems and, "facts" about number fields without any proofs whatsoever. I try to fill in the gaps, but this is difficult when the books keep mentioning that these are "obvious" results.

Thank you in advance.


2 Answers 2


If a prime does not ramify, then it splits or is inert. We're dealing with three mutually exclusive possibilities here. You're not alone in being confused about the meaning of splitting, ramifying and inertia. It might help to visualize some kind of physical movement representation of what happens when you move an integer with no imaginary part from $\mathbb{Z}$ to an imaginary ring like $\mathbb{Z}[\sqrt{-5}]$.

So, 3 is irreducible in $\mathbb{Z}$. It's also irreducible in $\mathbb{Z}[\sqrt{-5}]$. We could say that in a way, it stays the same, it is inertial.

Well, maybe inertial is straightforward enough, but ramifying and splitting are confusing because they mean different things but they sound like they should mean the same thing.

We see that 5 is irreducible in $\mathbb{Z}$ but in $\mathbb{Z}[\sqrt{-5}]$ we have $(-1)(\sqrt{-5})^2 = 5$. Since 5 divides the discriminant, it can't be inertial. It is said to ramify, but I can't think of any neat mnemonic to help you remember this.

I skip over 7 because like 3 it is inertial. In fact, let's jump all the way over to 29. Clearly 29 is coprime to the discriminant, so it can't ramify, but $-20$ is a quadratic residue modulo 29, suggesting 29 can't be inert either. Indeed $(3 - 2\sqrt{-5})(3 + 2\sqrt{-5}) = 29$. Hence it is said to split.

Chapter 9 in Niven and Zuckerman's Introduction to the Theory of Numbers uses some deprecated notations but if you get past that you'll probably find the explanations and proofs are quite clear.

  • $\begingroup$ Thank you very much for the brief explanation; good to know that I am not the only one who finds these concepts difficult to understand. $\endgroup$
    – Jamil_V
    Oct 27, 2014 at 2:54

There are many references on the web for this topic; one good one is here.

In summary, if $K = \mathbb{Q}(\sqrt{d})$ for $d$ squarefree and $p$ is an odd prime, then

  • $p$ is ramified if $p\mid d$ (as you point out)
  • $p$ splits if $\left(\frac{d}{p}\right) = 1$
  • $p$ is inert if $\left(\frac{d}{p}\right) = -1$

where $\left(\frac{d}{p}\right)$ is the Legendre symbol.

For $p=2$, the corresponding statements are:

  • $2$ is ramified if $d\equiv 2, 3\pmod{4}$
  • $2$ is split if $d\equiv 1\pmod{8}$
  • $2$ is inert if $d\equiv 5\pmod{8}$
  • 1
    $\begingroup$ So the $d$ in your first bullet point is the discriminant of the minimal polynomial? Or is it just some arbitrary integer with the condition that it needs to be square-free? $\endgroup$
    – Jamil_V
    Oct 25, 2014 at 23:04
  • 1
    $\begingroup$ $d$ is the $d$ in $K = \mathbb{Q}(\sqrt{d})$. $\endgroup$
    – rogerl
    Oct 26, 2014 at 0:25

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