# behavior of a rational prime in quadratic extension (definition)

Let $\mathbb Q \subset K=\mathbb Q (\sqrt{-n}) \subset L$, where $K/ \mathbb Q$ is a finite extension (i.e. $K$ is a number field) and $L/K$ is a maximal uramified abelian extension. If $p \mathbb Z$ is a prime rational integer then :

Q $1$: What does it mean $p$ ramifies, splits, split completely in $K$ ?

Q $2$: What does it mean $p$ ramifies, splits, split completely in $L$

Basically, the question I have is: when we refer to $p$ we mean the ideal $(p):= p \mathbb Z$ ?

What I understand is that:

A $1$. We consider the ideal $p \mathfrak O_K$ of the ring of integers $\mathfrak O_K$ , and the behavior of $p$ refers to the behavior of this ideal.

A $2$.Similar we consider the ideal $p \mathfrak O_L$ of $\mathfrak O_L$.

Q 3: Why we don't write $p \mathbb Z \mathfrak O_K$ ( and similar for $L$ )and we write just $p \mathfrak O_K$ ?

I suppose that the fact that $L$ is maximal unramified abelian extension, doesn't play any role. I just put it like this, because it is how I find it in the context.

I would really appreciate if someone can verify that what I understanding is correct or not.

One reason that one does not write $p \mathbb{Z} \mathfrak{O}_K$ is that it is somewhat redundant, since $\mathbb{Z} \mathfrak{O}_K = \mathfrak{O}_K$, so $p\mathbb{Z} \mathfrak{O}_K = p\mathfrak{O}_K$ anyway.

However, you have a point that it can be more systematic to talk consistently about ideals and not to indentifiy elements with the principal elements they generate.

• Thank you for your reply? So, what understand in Q.1 and Q.2 is correct right ? One last thing abour Q.2 : why we don't write $p \mathfrak O_K \mathfrak O_L$ for the ideal in $\mathfrak O_L$ ? It is the same thing that you wrote for $\mathbb Z$ and $\mathfrak O_K$ right ? – passenger May 6 '15 at 13:37
• Yes you understand correctly. Yes, the reason is the same. More generally for any domains $R \subset S$ you have $RS = S$ so that $aRS = aS$. That is whether you consider the principlal ideal generated by $a$ in $S$ or the idea generated in $S$ by the principla ideal generated by $a$ in $R$ it is the same thing. Yet, some results you might have seen will fit better with the situation when you indeed think about the principla ideal generated by $p$ rather than the elemenet $p$. – quid May 6 '15 at 13:40
• Thank you very much for your time ! – passenger May 6 '15 at 13:45

When referring to an element of a ring, such as $p\in \mathbb{Z}$, one simply writes the name of the element: $p$. When referring to the ideal generated by $p$, one can write either $(p)$ or $p\mathbb{Z}$, as you point out. Since the integers are a PID, it is fairly common to see the two notations conflated, although you must always be clear in your mind as to what you mean.

If $R$ is a ring and $x\in R$, then the principal ideal generated by $x$ is defined as $(x) = xR$; that is, $\{xr\,\mid\, r\in R\}$. In your case, where $R = \mathcal{O}_K$, we also have $p\in R$ since $\mathbb{Z}\subseteq \mathcal{O}_K$, so that $p\mathcal{O}_K$ is the principal ideal of $\mathcal{O}_K$ generated by $p$. Simply writing $(p)$ for this ideal is potentially (probably) confusing, so $p\mathcal{O}_K$ is much clearer.

And by the way, $p\mathbb{Z}\mathcal{O}_K = p\mathcal{O}_K$ (write out what each of these means in terms of element products if you don't see this), so what you write is correct but not normally done.