20m
answered About the definition of norm in Clifford algebra
1h
answered Additive Order and Ring
11h
comment Help with a problem about Artinian rings from Christian Peskine's book
@user26857 Both tags seem bad. Since there were only 3 Artinian tags, I thought it was a good chance to nip the tag in the bud. The Noetherian tags are a little more numerous, and so I wouldn't bother removing them myself all at once.
14h
revised Help with a problem about Artinian rings from Christian Peskine's book
edited tags
14h
revised Total ring of fractions of a Noetherian reduced ring is artinian
edited tags
14h
revised Superfluous radical
edited tags
15h
comment Which algebraic intuition can be used in fields
@mvw you must be thinking of the result that the reals are the largest Archimedian ordered field.
1d
comment $Rad(I)$ is an ideal of $I$
Doh, I accidentally closed using the wrong link. This is certainly a duplicate.
1d
comment Is simple module over commutative ring always a field?
@Vladimir It does make it a field. I've said that at least three times. But the point is that this structure on $M$ is not very important and not very useful. Being isomorphic as modules is more natural and useful.
1d
comment Is simple module over commutative ring always a field?
@Vladimir Said another way, $M$ certainly isn't a ring until you define a multiplication on it. You can achieve that in the way I described, by declaring the module homomorphism to be a ring homomorphism by fiat. This is not very generally interesting, though.
1d
comment Is simple module over commutative ring always a field?
$R/I$ certainly has a ring structure. If $\phi:R/I \to M$ is any bijection, and $m,n\in M$, you can define $m+n:=\phi^{-1}(\phi(m)+\phi(n))$ and $mn:=\phi^{-1}(\phi(m)\phi(n))$, and these operations make $M$ into a ring.
1d
revised Is simple module over commutative ring always a field?
edited title
1d
answered Is simple module over commutative ring always a field?
1d
comment Ideal of $TV$ which trivially intersects $V$
You might consider checking out the left and right contraction operations in geometric algebra.
1d
comment Projective Geometry with Clifford Algebra - lost Inner Product
All of the models for projective geometry I know of in geometric algebra rely on null vectors, which necessitate indefinite metrics. So I just don't know why you think the metric does not come into play. It is apparently vital for supplying the behavior you need in the algebra. Or, if you prefer, the signature is definitely important, although the particular metric that makes the signature might not be important.
1d
comment Ideal of $TV$ which trivially intersects $V$
Yes, this is unclear. If $b$ were uniformly zero, then $I$ would indeed be homogeneous...
1d
comment Projective Geometry with Clifford Algebra - lost Inner Product
It depends on what you mean, I think. "metric" in the sense of angles and distances does not have any place in projective geometry: however, the choice of metric you use when building a Clifford algebra to do projective geometry in definitely affects the structure and properties of your model.
1d
comment What is the smallest positive integer that has never been mentioned?
And as soon as you have solved the problem, the answer is obsolete :(
Jan
29
comment Infinite rings with lots of zero divisors
What I mean is that nothing here says 0 isn't a zero divisor.
Jan
29
comment Infinite rings with lots of zero divisors
@user26857 I prefer that convention too, but it does not seem to make any difference to the statement of the question or answer here.