Jeremy Rickard

A user here has posted four times in little more than a day, urging that this question on whether the equation a=a+1 is true for infinite cardinals without the axiom of choice should be closed/deleted. And they seem frustrated that members of the community are upvoting and voting to reopen. It's not the most elegantly written question, but it's perfectly reasonable, and has attracted a good quality answer.

Sorry, I accidentally posted the previous before I’d finished writing it, and can’t figure out how to delete or edit it. Please ignore, and I’ll try to start again!

A user here has posted four times in little more than a day urging for This on whether the equation a=a+1 is true for infinite cardinals without the axiom of choice,

@JyrkiLahtonen I can find examples of both “torsion” and “periodic” from the 1950s, so I don’t think it’s a modern coinage attempting to replace the well established term. Count yourself lucky: I found a 1950s use of “locally infinite” for “torsion-free”! ("Locally infinite-or-trivial” would be better terminology IMHO.)

@quid No. I’m just snooping!

@users [Sorry, the final paragraph of my answer had conflicting notation, with $n$ standing for two things, but I've fixed that now.]

@users (2) is the fact that $\aleph_n$ has uncountable cofinality: i.e., $\aleph_n$ is not the union of countably many ordinals that are strictly less than $\aleph_n$. If $d(\nu(s_k))<w_n$ then in the order isomorphism between $W$ and $\aleph_\omega$, $d(\nu(s_k))$ corresponds to some ordinal less than $\aleph_n$, so we can choose $s$ so that the ordinal corresponding to $d(\nu(s))$ is greater than the union of the ordinals corresponding to the $d(\nu(s_k))$ but still less than $w_n$.

@users (1) is just the definition of the lexicographic order on $\mathbb{Z}^W$. If $d(f)<d(g)$ and $f>0$ then $0=g(d(f))<f(d(f))$ and for $w<d(f)$, $f(w)=0=g(w)$. So $f>g$ in the lexicographic order. This also answers the second part of (2).

@users Sorry, for the uncountably generated ideal in a previous comment I meant the set of $r$ with $d(\nu(r))<w_1$, not $\nu(r)>f_1$.

@users The key fact is that there is no countable set of ordinals strictly less than $\aleph_1$ whose supremum is $\aleph_1$, but there is a countable set of ordinals strictly less than (the larger ordinal) $\aleph_\omega$ whose supremum is $\aleph_\omega$.

@users Sorry, “$f_n\leq n$” was a typo. I meant “$f_n\leq f$”. I’ve corrected it. The maximal ideal is generated by $\{r_n\}$ because for any $r$ in the maximal ideal, there is some $r_n$ with $\nu(r_n)<\nu(r)$, and so $r$ is in the ideal generated by $r_n$. The ideal consisting of all $r$ with $\nu(r)>f_1$ is not countably generated.

@users I think that exactly the same example works for $\alpha=\aleph_k$ for any $k\in\mathbb{N}$: Towards the end of the proof, where I choose $0<n$ with $w<w_n$, just choose $n\geq k$ with $w<w_n$ instead. And probably you can adjust the example for larger $\alpha$ by replacing $\aleph_\omega$ with a larger singular cardinal.

@users Sorry, I just mean that if $x$ is in there, and $x<y$, then $y$ is also in there. When I have a minute I’ll edit my answer to make that clearer.

@users I've added some explanations.

@users It is an ordinal. A cardinal is usually defined to be the smallest ordinal of a given cardinality, so $\aleph_\omega$ is the smallest ordinal that has larger cardinality than $\aleph_n$ for all $n\in\mathbb{N}$.

(where R is a subring of S)

If you define an "R-algebra" as an R-bimodule A with an associative multiplication A\otimes_RA\to A, then apart from the fact that this doesn't agree with the usual definition when R is commutative, I don't see any way in general of defining A\otimes_RS as an S-algebra. The obvious was of getting an S-bimodule is B=S\otimes_RA\otimes_RS, but then there's no obvious multiplication for B.

Going back to your original question, you could define an $R$-algebra as a ring $A$ together with a ring homomorphism $R\to A$, but that doesn't agree with the standard definition if $R$ is commutative (a ring $A$ with a ring homomorphism $R\to Z(A)$). And with this definition, $A\otimes_RS$ isn't an $S$-algebra in a natural way if $R$ is a subring of $S$.

That makes sense as a definition of $N\otimes M$ as a $Z(R)$-module, but then there are two different natural $R$-module structures: $r(x\otimes y)$ could be $(rx)\otimes y$ or $x\otimes (ry)$, and neither makes $_RMod$ a monoidal category (there isn't a unit).

But if $R$ isn't commutative then the left $R$-action and the right $R^{op}$-actions don't commute.

Sorry, just saw your edit. $M$ and $N$ are not $R^{op}\otimes R$-modules.

But then what does $_{R^{op}}\otimes_R$ mean? There isn't a natural monoidal structure on the category of (left, say) modules for a general non-commutative ring. E.g., if $R$ is the ring of $2\times2$ matrices over a field, and $V$ is the module of column vectors, then what is $V\otimes V$?

But $M\otimes_{R^{op}}N$ only makes sense if $M$ is a right $R^{op}$-module and $N$ is a left $R^{op}$-module.

But if $R$ is non-commutative, what is the tensor product in $_RMod$?

What's your definition of an $R$-algebra for a non-commutative ring $R$?