Question 9 in Marcus book.

Let $K$ and $L$ be the number field such that $K\subset L$ and let $R,S$ be their algebraic integers, respectively.

a) Let $I$ and $J$ be ideals in $R$, and suppose $IS|JS$. Showw that $I|J$.

(Suggestion(from the book): Factor $I$ and $J$ in primes in $R$)

b) Show that for each ideal $I$ in $R$, we $I=IS\cap R$.

(Set $J=IS\cap R$ and use a)

c)characterise those ideals $I$ of $S$ such that $I=(I\cap R)S$.

Now, I need to understand something , if $P$ is any prime ideal in $R$, then what I can say about $PS$? I meant if $P$ is any prime ideal in the factorisation of $I$ , then is $PS\subset I$???

How to show $J=IS\cap R$?

Is there any website or book explain these concepts in more details and examples.

I really need to understand the concepts of this question.

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    $\begingroup$ Who is Marcus ? Don't believe everybody has been introduced to this nice man... You should at least give the title of his book... $\endgroup$ – Jean Marie Apr 10 '16 at 19:57
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    $\begingroup$ @JeanMarie Marcus is one of the standard algebraic number theory textbooks. Almost anyone able to answer this question will understand what the OP is referring to $\endgroup$ – Mathmo123 Apr 10 '16 at 19:58
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    $\begingroup$ @Mathmo123 Permit the beotian I am not to agree: it's good to give the students the habit to cite their sources in a proper way. $\endgroup$ – Jean Marie Apr 10 '16 at 20:05
  • $\begingroup$ I edited the title of the post.. $\endgroup$ – Leonardo Apr 10 '16 at 20:08
  • $\begingroup$ To answer your first question (and perhaps the root of your confusion), multiplying ideals makes them smaller not bigger. We have $PS \supset IS$ not the other way round. Convince yourself this is the case by considering what happens in $\mathbb Z$. $\endgroup$ – Mathmo123 Apr 10 '16 at 20:15

It's not terrible, just prove ($2$) and then note $IS|JS \iff JS\subseteq IS$, now intersect both sides with $R$ and you get $J= R\cap SJ\subseteq R\cap IS = I$.

For ($2$) you get the intersection properties come to you since you can do it for prime powers, and then using the fact that distinct prime powers' intersections are their product.

For the last one, it's easy to see based on how primes split in the extension, so if $I=\prod_{i=1}^r\mathfrak{P}_i^{e_i}$ where if $\{\mathfrak{p}_j\}=\left\{\mathfrak{P}_i\cap R\right\}_{i=1}^r$ has that


for every $s$ where $ne'_{i_k}=e_{i_k}$ for all $k$.

  • $\begingroup$ that is really helpful. I really appreciate it . Many thanks dude. $\endgroup$ – Leonardo Apr 15 '16 at 16:05

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