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.



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
    Posted May 2, 2008 at 2:43 pm | Permalink | Reply

    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
    Posted May 2, 2008 at 6:22 pm | Permalink | Reply

    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)

    which should have read

    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
    Posted May 2, 2008 at 11:46 pm | Permalink | Reply

    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) ?

    (INSTEAD of ‘x’)

    Which you have now explained (though perhaps accidentally). In the msg u posted, you wrote x instead of alpha by accident, which sort of made the question a bit strange. Either way, I have the explanation. Thanks.

  6. Nik
    Posted May 2, 2008 at 11:48 pm | Permalink | Reply

    That smiley face was unintenional. I meant to just close the bracket!

  7. Nik
    Posted May 2, 2008 at 11:49 pm | Permalink | Reply


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