Prime ideal and ring of integers I am reading an introduction into algebraic number theory and struggling on the following sentence:

Let $M/F$ be an extension of number fields, and $\mathfrak p$ an prime ideal of $\mathcal{O}_F$. Then $\mathcal{O}_M \mathfrak p$ is an ideal in $\mathcal{O}_M$ which is no longer a prime ideal in general.

My question is: what does the notation $\mathcal{O}_M\mathfrak p$ mean? I think it means the product of the two ideals, but then why is $\mathfrak p$ also an ideal of $\mathcal{O}_M$? Why is $\mathcal{O}_M\mathfrak p$ even an ideal of $\mathcal{O}_M$? 
Any help is appreciated.
 A: Question:"Why is $O_Mp$ even an ideal of $O_M$?"
Answer: If $\phi: A\rightarrow B$ is a map of commutative unital rings and $I$ is an ideal in $A$, you define $IB$ (or $BI$) to be the following ideal:
I1. $IB:=\{\sum_i b_ix_i$ with $b_i \in B, x_i:=\phi(y_i)$ with $y_i\in A\}.$
You may check that $IB\subseteq B$ is an abelian group with the property that  for
all $x\in IB, b\in B$ it follows $bx\in IB$. Hence $IB$ is an ideal in $B$.
As mentioned above: The standard example of a ring extension where the extension of a prime ideal is no longer a prime ideal is the extension
$\phi: A:=\mathbb{Z} \subseteq \mathbb{Z}[i]:=B$
let $I:=(2) \subseteq A$. The ideal $I$ is prime in $A$ but in $B$ there is a factorization $2=(1-i)(1+i)$, hence the ideal $IB$ equals $(1-i)(1+i)B$ which is no longer prime. If $p\neq 2$ is a prime with $p\equiv 1 (mod 4)$ there is an equality
$p=a^2+b^2=(a-ib)(a+ib)$ in $B$ for two integers $a,b\geq 1$
Hence the prime ideal $(p)B$ is no longer prime in $B$. This is Theorem 1.1.1 in Neukirch's book "Algebraic number theory". In Theorem 1.4 in the same book the prime elements in $B$ are classified, and a prime number $p \in A$
remains prime in $B$ iff $p\equiv 3 (mod 4)$. If $p \equiv 1(mod 4)$ with $p=a^2+b^2$ it follows the elements $a+ib, a-ib$ are prime elements in $B$,
hence
I1. $(p)B=(a+ib)(a-ib)B$ gives rise to the decomposition
I2. $(p)B=\mathfrak{p}_1\mathfrak{p}_2$
with $\mathfrak{p}_1=(a+ib)$ and $\mathfrak{p}_2=(a-ib)$. Here $\mathfrak{p}_i$ are prime ideals in $B$.
You get a canonical map of rings
I3. $A/(p) \rightarrow B/(p)B \cong B/\mathfrak{p}_1\mathfrak{p}_2 \cong B/\mathfrak{p}_1 \oplus B/\mathfrak{p}_2$
and since $\mathfrak{p}_i$ are non-zero prime ideals it follows they are maximal ideals. Hence the ring $B/(p)B$ is a direct sum of the fields $B/\mathfrak{p}_i$. Since $A/(p):=\mathbb{F}_p$ is the finite field with $p$
elements it follows there is an isomorphism
I4. $B/\mathfrak{p}_i \cong \mathbb{F}_{p^{r_i}}$
for some integer  $r_i \geq 2$. You may calculate the integer $r_i$ explicitly.
