conjecture: if p is prime, the no. of rings of order p^2 is 11
Christian G. Bower
bowerc at usa.net
Thu Apr 17 20:03:12 CEST 2008
suppose that for some element a, a^2 is not a multiple of a.
Now if Q isn't 0, we can divide by it, to get an equation
4) the quadratic satisfied by a is irreducible. In this
case what we have is a quadratic extension of the field of
order p, which must be the (unique) field of order p^2.
5) This quadratic is the product of two distinct linear factors.
In this case, we can (by replacing a by a suitable ra+s)
make these be x and x-1, so that what we have is the
field of order p extended by an idempotent a^2 = a.
6) It's a square - but then we can make it x^2, and what
we have is the field of order p extended by a quadratic
unipotent a with a^2 = 1.
If Q IS 0, then either
7) P isn't 0 - in this case we can replace a by b = a/P,
which satisfies b^3 = b, and so the ring is that
generated by an "idemcubic" ...
8) or it is, and the ring is that generated by a cubic nilpotent.
Eight rings so far - not too bad!
In the remaining cases, every element satisfies an equation
9) the zero (non-cyclic) ring, in which a^2 is always 0.
send this letter off before considering these cases.
John Conway
In all remaining cases, every element was either a quadratic
Let's suppose there are two independent nilpotents a and b.
ab = Pa + Qb ba = Ra + Sb , say.
Premultiplying the first of these by a, we see that Qab = 0,
so that either Q = 0 or ab = 0, in which case also Q = 0. So
So in any new case, all nilpotents are multiples of a given one,
ab = Pa, ba = Ra
ab = Pab, ba = Rba
showing that P and R must each be 0 or 1. This seemingly
0 0 0 a 0 0 0 a
0 b 0 b a b a b.
Obviously there must be either some isomorphisms or some nonrings
showing that either Q=1 and P=0 or ab is a multiple of a,
showing that either P=1 and Q=0 or P=0. So they're either
ab = 0 or a or b, and similarly ba = 0 or a or b.
Summary: any ring not in my previous letter has either
aa = 0, bb = b, ab = 0/a ba = 0/a
aa = a, bb = b, ab = 0/a/b, ba = 0/a/b.
Enough for now! John Conway
Let me attack those 9 last cases
aa = a, bb = b, ab = 0/a/b, ba = 0/a/b
ab = ba = 0; ab = a, ba = b; ab = b, ba = a;
4a + 0 + 0 + b
4a + 2a + 2b + b
10) aa = ab = a, bb = ba = b.
Now let me look at the four cases
aa = 0, bb = b, ab = 0/a, ba = 0/a
showing that we have either
11) aa = ab = 0, ba = a, ab = b
All that remains to be done is to check these rings 10) and
For 10), we find
(Pa + Qb)(Ra + Sb) = (Pa+Qb)(R+S)
showing that the product of any two elements is just a multiple
For 11), we find
(Pa + Qb)(Ra + Sb) = Q(Ra+Sb)
This makes it clear that the extra two rings are in fact
NEW 10): xy = h(x).y
NEW 11): xy = x.h(y)
There remains one tiny point: the case
aa = bb = a, ab = ba = 0
(Pa + Qb)(Ra + Sb) = PRa + QSb.
0 0 0 0
0 a 0 a
0 0 b b
0 a b c, where c = a+b.
Oh! I see that if we'd put c into our base, this would have
My next (and maybe last?) letter will summarize the
John Conway
Additive group cyclic on a:
three cases aa = a, aa = pa, aa = 0.
Non-cyclic on 1,a (1 a unit):
three cases aa = 0, aa = a, field of order p^2.
No unit, but generated by a single element:
two cases aaa = a, aaa = 0.
No unit, needs two generators:
three cases xy = 0, xy = h(x).y, xy = x.h(y)
While preparing this list I noticed an error in its previous
should be amended to read "a^2 = 0" in the last line.
I quite enjoyed doing that! Apologies to those who didn't want
John Conway
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Date: Thu, 17 Apr 2008 11:06:22 -0700
From: "Christian G. Bower" <bowerc at usa.net>
To: "seqfan" <seqfan at ext.jussieu.fr>
Subject: Re: conjecture: if p is prime, the no. of rings of order p^2 is 11
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From: "David Wilson" <davidwwilson at comcast.net>
To: "seqfan" <seqfan at ext.jussieu.fr>
Subject: Re: conjecture: if p is prime, the no. of rings of order p^2 is 11
showed a(8)=52, a(p^3)=3p+50 if p is odd prime.
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Date: Thu, 17 Apr 2008 14:30:56 -0400
From: "Maximilian Hasler" <Maximilian.Hasler at martinique.univ-ag.fr>
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You are absolutely right, the word "sorted" is missing in my comment,
and the name of the second PARI procedure is misleading, it should be
replaced by
%o A037916 (PARI) primesig(n)=sum(i=1,#n=vecsort(factor(n)[,2]),10^(#n-i)*n[i])
\\ up to the order of digits, the same result up to 2^10
But in any case simple concatenation becomes ambiguous from 2^11 on at
least, and multiplying with powers of 10 doesn't really give the pime
signature, either.
Maximilian
On Thu, Apr 17, 2008 at 2:10 PM, <franktaw at netscape.net> wrote:
> It's a subtle distinction, but I have generally seen prime signature used to
> mean
> the multiset of exponents in the prime factorization, rather than the
> sequence of
> same. This agrees with the Wikipedia definition -- note the words "sorted
> by size"
> in that definition.
>
> Franklin T. Adams-Watters
>
>
>
> -----Original Message-----
> From: Maximilian Hasler <Maximilian.Hasler at martinique.univ-ag.fr>
>
> %I A037916
> %S A037916
> 0,1,1,2,1,11,1,3,2,11,1,21,1,11,11,4,1,12,1,21,11,11,1,31,2,11,3,21,1
> %C A037916 The sequence of (nonzero) exponents in the prime
> factorization of a number is called its prime signature.
> %H A037916 Wikipedia, <a
> href="http://en.wikipedia.org/wiki/Prime_signature">
> Prime signature</a>.
> %o A037916 (PARI) A037916(n)=if( n>1,
> eval(concat(concat([""],factor(n)[,2]~))))
> %o A037916 (PARI) primesig(n)=sum(i=1,#n=factor(n)[,2],10^(#n-i)*n[i])
> \\ the same result up to 2^10
> %K A037916 nonn,easy
> %O A037916 1,4
> %A A037916 M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 17 2008
>
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