# Vague definitions of ramified, split and inert in a quadratic field

Our lecturer defined the following:

Let $K=\mathbb Q(\sqrt d)$ be a quadratic field and $p$ a prime number, then

(1) $\ p$ is ramified in $K$ if $\mathcal O_K⁄(p)\cong \mathbb F_p [x]⁄(x^2)$
(2) $\ p$ is split in $K\ \ \ \ \ \$ if $\mathcal O_K⁄(p)\cong \mathbb F_p[x]^2$
(3) $\ p$ is inert in $K\ \ \ \ \ \$ if $\mathcal O_K⁄(p)\cong \mathbb F_{p^2}$

And he stated:

Let $K$ and $p$ be as above. Then:

$p$ is ramified $\iff p$ divides the discriminant disc$(K)$ of $K$
$p$ split $\iff p\nmid \text{disc(K)}$ and d is a QR mod p
$p$ is inert $\iff p\nmid \text{disc(K)}$ and d is not a QR mod p

To be honest, I have a little bit of trouble wrapping my head around these definitions. Also, I don't seem to be able to find any useful sources online, which provide examples or (not to mention) use similar definitions. All I can find are things like "totally split", "remains prime in", etc. and a lot of different notation.

Hence my question, are these definitions actually common and is there any way of explaining them in a more approachable manner?

edit:
Just in case somebody is interested, the following texts, which I found after further research, appeared quite useful to me:

• Yes, these are very important definitions in algebraic number theory. What is vague about them? Apr 29, 2013 at 20:25
• Let $K=\mathbb Q(\sqrt{-3})$, then $\mathcal O_K$ are the Eisenstein integers. Now let $p=2,\ 3,\ 5,$ or $7$. Then I can use the 2nd bunch of statments to determine whether these are ramified/split/inert, but what does this tell me exactly and how is this correlated to the actual definition? Apr 29, 2013 at 20:35
• Shouldn't the $\mathbb{F}_p[x]^2$ be $\mathbb{F}_p^2$ instead? Oct 17, 2016 at 14:42

First let me say your lecturer is only right assuming that $p$ is an odd prime number.

IMHO this business of bringing in the discriminant is really confusing, at least for a quadratic field. Let me now give you the usual definition (as found in Neukirch, Marcus) of these terms and then proceed to show you how they are equivalent to your definition above.

Let $K$ be an algebraic number field. Choose a prime $p \in \Bbb{Z}$, we know that $\mathcal{O}_K$ is a Dedekind domain and hence $(p)$ factors into primes $\mathfrak{p}_1^{e_1}\ldots \mathfrak{p}_n^{e_n}$ in $\mathcal{O}_K$. We say that

1. $p$ is ramified in $\mathcal{O}_K$ if there is at least one $j$ for which $e_j > 1$.

2. We say $p$ is totally split if the $e_i = 1$ for all $1 \leq i \leq n$ and $n = [K : \Bbb{Q}]$

3. We say $p$ is inert if it does not split at all in $\mathcal{O}_K$.

Thus in the case that $K$ is a quadratic field, we see a prime $p$ is ramified if $(p)$ factors into $\mathfrak{p}^2$, totally split if it factors as $\mathfrak{p}_1\mathfrak{p}_2$ and inert if it just remains the same.

Now how does this business about the discriminant come in? The discriminant of $\mathcal{O}_K$ when $K = \Bbb{Q}(\sqrt{n})$ is $n$ if $n \equiv 1 \mod{4}$, otherwise it is $4n$. Now the main reason why I say bringing in the discriminant is misleading is because the key point that $p|n \implies$

$$(p) = (p, \sqrt{n})^2$$

in $\mathcal{O}_K$ gets piled under a heap of abstraction! How do you see the equality above? Well expand the product of the two ideals, see what you get!

What about the case when $p \nmid n$? Check for yourself that $p$ splits as

$$(p) = (p, m + \sqrt{n})(p,m - \sqrt{n})$$

where $n \equiv m^2 \mod p$. What about the last case when $p \nmid n$ and $n$ is not a quadratic residue mod $p$? To see that $p$ is inert I suggest that you compute the quotient $\mathcal{O}_K/(p)$ explicitly and see that it is a field (thus integral domain).

• Thanks for this answer. just a few follow-up questions: 1. When you choose a prime $p\in\mathbb Q$, do you mean $p\in\mathbb Z$? 2. When you factor $(p)$ into $\mathfrak p_1^{e_1}\ldots\mathfrak p_n^{e_n}$ then $n$ is not necessarily the degree of the extension? Only if $(p)$ splits? And I think you made a little error: "it factors as $\mathfrak p_1$" should read $\mathfrak p_1\mathfrak p_2$? 3. Is it correct that if for $p\in\mathbb Z$ the ideal $(p)$ is inert, then $p$ remains a prime in $\mathcal O_K$? Apr 30, 2013 at 12:05
• @Thomas I have made some edits. First $n$ is not necessarily the degree of the extension; we only know that it is at most the degree. Finally the answer to your last question is yes.
– user38268
May 4, 2013 at 15:12
• Do you assume that $n$ is squarefree in the calculation of the discriminant? Dec 6, 2015 at 12:30
• The definition 1. should read "$p$ is ramified in $\mathcal{O}_{K}$ if there is at least one $j$ for which $e_{j} > 1$." Feb 23, 2016 at 14:46