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Lukas Heger
Lukas Heger
05:27
@rschwieb shouldn't that that ring even be Bezout? It seems to me that every f.g. ideal is generated by $T^{1/n}$ for some $n$. And Bezout implies coherent
It suffices to show that every $2$-generated ideal is principal. Every element which is not a multiple of $T^{1/n}$ for some $T$ is a unit, thus we may focus on elements of the form $T^{1/n}$. Now $(T^{1/n},T^{1^k})$ contains $T^{1/\gcd(n,k)}$ (by using Bezout's lemma for $\Bbb Z$) and is in fact generated by it
Actually, looking at the page for Bezout rings in DaRT, I'm baffled by the assertion that Bezout does not pass to quotient rings. If $R/I$ is the quotient of the ring $R$ and $(\overline{f},\overline{g}) \leq R/I$ is a 2-generated ideal, then how doesn't the fact that the ideal $(f,g) \leq R$ is principal imply that $(\overline{f},\overline{g}) \leq R/I$ is also principal?

DaRT says that $\Bbb Z$ provides a counterexample, but clearly any quotient of $\Bbb Z$ is in fact Bezout (even PIR)
 
8 hours later…
rschwieb
rschwieb
13:35
@LukasHeger I'll check: maybe something wrong happened in automation.
@LukasHeger Oh, this looks like a manual entry. I suspect that a long time ago when I was entering the information for "Bezout domain does not pass to quotients" I used the wrong property (Bezout instead of bezout-domain) They are adjacent numbers.
Good find!
rschwieb
rschwieb
14:33
@LukasHeger Actually hang on: while I believe you that Bezout domains are coherent, it looks like we have examples of bezout rings that are not coherent: ringtheory.herokuapp.com/search/results/?H=67l&L=87l
The line of reasoning won't apply to the "harmonic series exponent polynomial ring"
I'm also not completely following why the argument for monomials will suffice for showing the same thing for ideals 2-generated by polynomials...
oh wait, right, the ring suggested above is a power series ring, not a polynomial ring :)
rschwieb
rschwieb
14:54
Since $\{T^{1/n}, T^{1/m}\}=\{T^{m/mn}, T^{n/mn}\}$, I can see how maybe $T^{gcd(m,n)/mn}=T^{1/lcm(m,n)}$ gets in there, but I'm not quite sure we have the negative exponents to make that happen. I might not be tracing the same line of thought you sketched though.
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2 hours later…
Lukas Heger
Lukas Heger
16:50
@rschwieb I don't really see it, tbh. Any prinicipal ideal is a finitely presented module, isn't it?
@rschwieb oh you're right, I'll have to rethink it
rschwieb
rschwieb
Right now the only logical links i have pointing to "coherent" are "semihereditary" and "Noetherian."
Lukas Heger
Lukas Heger
If $a \in R$, then $R \xrightarrow{\cdot a} R \to aR \to 0$ is a finite presentation
rschwieb
rschwieb
I guess I just have to work out how the kernel is f.g.
Lukas Heger
Lukas Heger
which definition of f.p. are you using?
The one I'm using just says that $M$ is $f.p.$ if there's an exact sequence $R^n\to R^m \to M \to 0$
rschwieb
rschwieb
And whether or not it's Bezout or not
Lukas Heger
Lukas Heger
16:58
oh no wait you're right
rschwieb
rschwieb
I think I might have misread what you wrote... I was reading the dot-a in this $R \xrightarrow{\cdot a} R \to aR \to 0$ as the multiplication map
Lukas Heger
Lukas Heger
I was talking nonsense
hmm
rschwieb
rschwieb
😅
this is why coherence and I don't get along
Lukas Heger
Lukas Heger
you're right, Bezout domain is sufficient, but Bezout might not be
hmm
@rschwieb I think you might add von-Neumann-regular
yeah I was talking nonsense, sorry
not every principal ideal is f.p. in general
actually I found a paper on rings in which principal ideals are f.p.
researchgate.net/publication/…
not sure if you want to add "p-coherent" as a property though
@rschwieb yeah coherence seems like a difficult property...
Lukas Heger
Lukas Heger
17:19
@rschwieb note that we divide out by $T$, that should help with negative exponents, since $T^{-k/n}=T^{(n-k)/n}$ in the quotient
thus it is indeed true that this quotient is Bezout
but as you correctly remarked, this is not sufficient for coherence
But in this case, it at least helps. The kernel of $\cdot T^{1/m}$ should be $T^{(m-1)/m}$, right?
and yeah I meant to say that $\langle T^{1/n},T^{1/m}\rangle=\langle T^{1/lcm(n,m)}\rangle$
rschwieb
rschwieb
17:40
@LukasHeger VNR -> semihereditary -> coherent already
@LukasHeger ooo
Lukas Heger
Lukas Heger
@rschwieb right

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