## More on ideals

Hi Sir.

Couple of small things in proofs.

*In Dedekind’s PFT

there is a step in the printed notes where we go

$O/P_i \simeq Z[x]/(m,p,m_i) \simeq F_p[x]/(m,m_i) \simeq F_p[x]/(m_i)$

From my understanding, when we went from 3rd to 4th we modded out by ‘m’. Should the last part, therefore, not be $F_p[x]/(m_i)$?
(where $m$ is min poly of $\alpha$, i think)

*In the proof of norms of ideals being multiplicative we choose ‘a’ from I\Ip and consider an isomorphism from O -> I/Ip.

I just wanted to make sure a is, indeed, from I (without) Ip (as opposed to I/Ip) and what our thinking was behind constructing this isomorphism? The rest of the proof is understood.

Thanks.

The answer is yes to both questions. Except the reason for the first is that
$(m,m_i)=(m_i)$ inside $F_p[x]$ because $m_i$ divides $m$. That is, in the sequence of isomorphisms, we use the notation $m, m_i$ for polynomials in $F_p[x]$ and also for their lifts to $Z[x]$. In $Z[x]$ there is no divisibility (there can’t be since $m$ is irreducible) but in $F_p[x]$, $m_i$ is, by definition, a factor of $m$.

For the question on multiplicativity of norms, I’m afraid I can’t think of much enlightening to say beyond what’s written in the proof. That is, one needs to show that $|I/IP|=|O/P|$.

1. Nik

I think there should have been Fp[alpha] in the question instead of x.

2. Posted May 2, 2008 at 3:01 pm | Permalink | Reply

No, it should have been x. Make sure you understand the distinction in this context.

3. Nik

I think I worded my comment wrong, which has caused confusion.

I meant, the question I asked (which you went over to display the correct symbols) should have read “Should the last part be … Fp[alpha]?”.
I understand what you are saying about the answer (that it should read ‘x’), but it might confuse people reading the entry, since it doesn’t make sense if corrected in the initial question (which i think was an accident).

4. Posted May 2, 2008 at 8:48 pm | Permalink | Reply

Oh, I see what you mean. I thought the original email, which I’ve unfortunately erased, pointed out a misprint like

$F_p[x]/(m,m_i) \simeq F_p[x]/(m)$

$F_p[x]/(m,m_i) \simeq F_p[x]/(m_i)$

But now I’m no longer sure, because I just checked the course notes and it’s written correctly there. If you can repeat the original question, which I appear to have mangled, I can try to clarify.

5. Nik

No worries. I was just asking for the benefit of other readers.
The question was actually:

Why is it not

Fp[x]/(m,mi) -> Fp[alpha]/(mi) ?