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

Question

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:

• An Introduction to Number Theory, by G. Everest:
  http://www.amazon.co.uk/dp/1852339179
  Specifically Chapter 4.4 The Ideal Class Group and Section 4.4.1 Prime Ideals

• The lecture notes of Robin Chapman:
  http://empslocal.ex.ac.uk/people/staff/rjchapma/notes/ant.pdf (p. 16-18) http://empslocal.ex.ac.uk/people/staff/rjchapma/notes/ant2.pdf (p. 38-39 & Chapter 5)

Answer 1

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).