9
votes
Do all Euclidean domains admit a Euclidean function that is "weakly multiplicative"
Throughout this answer, I'm using the following definition of a Euclidean norm: A map $\varphi\colon R-\{0\}\to {\rm Ord}$, such that for any elements $a,b\in R$ with $b\neq 0$ and $b\nmid a$, there ...
- 18.9k
9
votes
Accepted
Classifying associators for the tensor product of graded modules
Looks like you're describing a $3$-cocycle on $\mathbb Z$ with coefficients in $R^\times$ (with trivial action on $R^\times$).
But $\mathbb Z$ has no $H^3$ (with any coefficients), so a $3$-cocycle is ...
- 16.6k
7
votes
Accepted
The classification of finitely generated modules over the ring of (Laurent polynomials in multiple variables)
There will be a finitely generated abelian group $B$, finitely generated abelian groups $U_1,\dots, U_n$, and monomorphisms $f_i,g_i \colon U_i \to B$ such that $M$ is isomorphic to the pushout of a ...
- 151k
5
votes
A local ring with the property $r^p=r$
(From the comments):
It is ambiguous whether $p$ is allowed to depend on $r$ or
not. Here is an argument that covers both cases. (If $p$ is
fixed one can be slightly more efficient.)
Suppose that $r$ ...
- 798
5
votes
Proof of Quillen's dimension formula?
The equality has no meaning unless $D_n(k/R,k)=0$ for $n>>0$. This hypothesis is included in Quillen statement. As it was conjetured by Quillen, we now know that this hypothesis is equivalent to ...
- 804
4
votes
Accepted
Equality of functions on Tannakian fundamental group
Yes. We can take $X$ to be a quotient of $X_1 +X_2$.
The right way to think of these things is that, after constructing the group$ G_\omega$ and checking the equivalence of $T$ with the category of ...
- 151k
4
votes
Alternative versions of a very basic fact
Let $D$ be an integral domain. Then $d \in D$ is a gcd of $a, b \in D$, written $\gcd(a,b) = d$, if $(d)$ is the smallest principal ideal of $D$ containing $(a,b)$. Any gcd of $a$ and $b$ in $D$ is ...
4
votes
Accepted
Existence of a regular sequence with degrees prescribed by the Hilbert series
No, this is not at all true. The easiest example is $A=k[x,y]/(x^2)$ where $|x|=1$ and $|y|=2$. Then $f_A(q)=1/(1-q)$ but there is no regular element of degree one.
If you want an integral domain, ...
- 16.9k
3
votes
Accepted
"Nonarchimedean" Euclidean domains
The claim in Question 2 is true if $\sigma$ satisfies property $(*)$, also known as submultiplicativity. For what follows, denote by $A^*$ the set of nonzero elements, and $A^\times$ the set of units.
...
- 970
2
votes
Do all Euclidean domains admit a Euclidean function that is "weakly multiplicative"
$\def\ZZ{\mathbb{Z}}$There are two conditions here:
(WM) If $\sigma(a) < \sigma(b)$ then $\sigma(ax) < \sigma(bx)$ for $x \neq 0$.
($\ast$) $\sigma(a) \leq \sigma(ab)$.
Given a Euclidean ...
- 157k
2
votes
Accepted
Multiplicative closure of $ax^2+bxy+cy^2$ with discriminant $d$ and class number $h(d)=3m?$
Courtesy of Jianing Song's code,
for(d=1, 217627, if(isfundamental(-d) && quadclassunit(-d)[2] == [12,3], print1(d, ", ")))
in the comments and the online PARI/GP calculator, then ...
- 12.9k
1
vote
Subsets of the integers which are closed under multiplication
Part I.
In his #5, the OP talks about the well-known case of two squares $x^2+y^2$. In general, we ask what quadratic forms $F(x,y) = ax^2+bxy+cy^2$ have multiplicative closure and which is still an ...
- 12.9k
1
vote
Exotic principal ideal domains
While being a PID is certainly a strong property, some PID can still be in some sense pathological, for example by not being excellent or even Nagata.
This doesn't happen in characteristic zero: Every ...
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