## Monthly Archives: April 2010

### 3704 Office hours

Because people haven’t been coming to my office hours, I thought I’d have the one on the coming Tuesday, 26 April, in the 5th floor common room. I’ll be there from 5 to 6.

### Algebraic number theory revision

There will be a revision session for algebraic number theory on Tuesday, 4 May, 2-4 pm in room 505. By this time, you should have read the notes several times and seriously attempted all practice questions.

### Algebraic geometry revision

There will be a further revision session for algebraic geometry this Thursday, 15 April, in my office. Check the directions on my webpage for getting to my office. I’ll come to open the door at 4 PM.

### Division of ideals

Hi,

I have a couple of questions concerning algebraic number theory. Why does $I$ contains $J$ imply $I$ divides $J$? Firstly assuming $I$ contains $J$ how do we know there exists a fractional ideal $I'$such that $I'I=J$ and then how do we get that $I'$ is actually an integral ideal?

Another problem I am having is in proving: if $I$ is ideal and $N(I)=p$ some prime $p$ then $I$ is prime ideal. So we assume $A,B$ are ideals such that $AB$ is contained in $I$. We want to show either $A$ is contained in $I$ or $B$ is. Assuming the above then we know $I$ divides $AB$. The proof then states that $I=ABC$ some ideal $C$. I thought $I$ divides $AB$ means $IC=AB$ for some ideal $C$ not other way round. I get that we could multiply by $C$ inverse to get $I$ on its own but then $C$ inverse is a fractional ideal.

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I’m presuming that you’ve read the course notes at least twice or so at this point. So for your first question, it’s already assumed that we’ve shown that the non-zero fractional ideals form a group. This gives the existence of the $I'$ you mention. But then
$I'I\subset J\subset I$, so each element of $I'$ stabilizes the ideal $I$. This implies that every element of $I'$ is an algebraic integer, i.e., that $I'$ is an ideal.

For the statement about prime ideals, note that we are assuming the factorization theorem. So any $I$ can be written

$I=\prod P_i$

for prime ideals $P_i$. But then

$N(I)=\prod N(P_i).$

Therefore, if $N(I)$ is prime, only one $P_i$ can occur.

### Algebraic Number Theory Sheet 1

Here are some solutions to the algebraic number theory coursework sheet 1. I am eventually bending to pressure and making them available. However, be warned that the material in sheet 1 is entirely of a preliminary nature that you should be able to do confidently, confident of the correctness of your answers. If that’s not the case, you can consult these solutions. But please be aware that this is an indication that you are still weak in the prerequisites, and may need to review earlier algebra courses.