Reputation
24,466
Top tag
Next privilege 25,000 Rep.
Access to site analytics
Badges
8 24 112
Impact
~435k people reached

12h
comment Concrete balanced category
Related.
1d
asked If $P$ has all binary joins, and all chain-shaped joins, is $P$ necessarily a complete lattice?
1d
comment Regarding the axiom $2^\kappa = 2^{\kappa^+}$ for regular cardinals $\kappa$, and its relationship to a couple of other axioms.
Ahh yes, sorry. You're completely right, of course.
1d
comment Is there an adjective for rings whose every non-zero prime ideal is maximal?
@rschwieb, a simple "not now, I don't have access to a proper computer" would suffice. I shouldn't be expected to be able to read your mind, and you should be more cautious inferring "tone" from mere unspoken writing. The Wikipedia page does not give a proper definition. By the way, I think this is shameful. But my ire is reserved for those who wrote the relevant page.
1d
comment Is there an adjective for rings whose every non-zero prime ideal is maximal?
@rschwieb, can you be more precise? For starters, we're not taking suprema of chains, we're taking suprema of lengths of chains. So, we need to define the "length" of a chain. One way of doing this is to declare that the length of a chain is supremum among all cardinalities of proper subsets. This will give you $n-1$ for finite chains. If not, can you tell me, precisely, how you want to define "length"?
1d
comment Is there an adjective for rings whose every non-zero prime ideal is maximal?
@rschwieb, ah, I see. But what if the chain doesn't have links? E.g. what if our prime ideals form a poset isomorphic to $\mathbb{R}$?
1d
comment Is there an adjective for rings whose every non-zero prime ideal is maximal?
@Crostul, I don't get it. $\mathbb{Z}$ has Krull dimension $2$, right? $\{0\} \subseteq 2\mathbb{Z}$, for example. Or perhaps the zero ideal is omitted?
2d
revised Is there an adjective for rings whose every non-zero prime ideal is maximal?
edited title
2d
asked Is there an adjective for rings whose every non-zero prime ideal is maximal?
Apr
25
comment Regarding the axiom $2^\kappa = 2^{\kappa^+}$ for regular cardinals $\kappa$, and its relationship to a couple of other axioms.
Miha, it's clear that NJA proves $2^{<\kappa} = \kappa$. Do you know whether or not the converse holds?
Apr
25
answered Let $f\colon A\to B$ be a function
Apr
25
comment Find 10 commuting $2\times 2$ matrices of the same order
@MooS, thanks. $\;\!$
Apr
25
comment Find 10 commuting $2\times 2$ matrices of the same order
@MooS, what do you mean by $\varphi$ in this context?
Apr
25
comment Find 10 commuting $2\times 2$ matrices of the same order
This is a cool question, by the way.
Apr
25
answered Find 10 commuting $2\times 2$ matrices of the same order
Apr
25
comment Why is $v/\|v\|$ not a unit vector?
I agree. The statement isn't false; its ill-typed.
Apr
24
comment Object defined as a member of some category
Partial answer: You can certainly do this with $\mathbf{FinSet}$, which can be described as the cocartesian monoidal category freely generated by one object. It should be possible to do this with some other categories, too, but I can't quite see how at the moment.
Apr
24
comment Is there a set without a predicate?
@user254665, I disagree. If $X$ is a set, there's a canonical bijection between $\mathcal{P}(X)$ and $2^X$. Ergo, the number of subsets of $X$ equals the number of predicates on $X$. But, I think that by "predicate", what you really mean is "first-order formula in the language of set theory with one free variable." In that case, your statement is correct.
Apr
24
revised Is there a set without a predicate?
added 311 characters in body
Apr
24
answered Is there a set without a predicate?