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 12h comment Concrete balanced category 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?