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Let $R$ be a commutative ring, and let $S$ be a finitely presented $R$-algebra of finite Tor-dimension over $R$. Can $R \to S$ be realized as the base change, along some ring map $R_0 \to R$, of a finitely presented ring map $R_0 \to S_0$ of finite Tor-dimension with $R_0$ Noetherian? Can we furthermore arrange that $R_0$ is actually of finite type over $\mathbf{Z}$?

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    $\begingroup$ Proposition 6.1.6.1 in Jacob Lurie's book Spectral Algebraic Geometry. $\endgroup$ – Riza Hawkeye Feb 17 at 13:36
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    $\begingroup$ Coming directly from the facebook page "Derived Memes for Spectral Schemes", reading that comment under a commutative algebra question was an absolutely fabulous moment. $\endgroup$ – Martin Brandenburg Feb 17 at 14:31
  • $\begingroup$ Thanks, Riza! That's exactly what I was looking for. $\endgroup$ – David Hansen Feb 17 at 17:26
  • $\begingroup$ I'm sorry for my incompetence but, could somebody translate that Proposition in Jacob Lurie's book to the language of commutative algebra (that is, what says that proposition for the case of commutative rings). $\endgroup$ – A.G Feb 18 at 20:54
  • $\begingroup$ @A.G Sure, I've posted an answer below. $\endgroup$ – Riza Hawkeye Feb 19 at 12:59
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As requested in the comments, here is Proposition 6.1.6.1 of Lurie's "Spectral Algebraic Geometry", specialized to the case of ordinary commutative rings:

Let $n\ge 0$ be an integer, let $A_0$ be a commutative ring and let $B_0$ be an $A_0$-algebra of finite presentation. Suppose we are given a diagram $\{A_\alpha\}_{\alpha\in I}$ of $A_0$-algebras indexed by a filtered partially ordered set $I$, and set $A = \varinjlim A_\alpha$. Set $B_\alpha = A_\alpha \otimes_{A_0} B_0$ and $B = A \otimes_{A_0} B_0 \simeq \varinjlim B_\alpha$. If $B$ has Tor-amplitude $\le n$, then there exists an index $\alpha$ such that $B_\alpha$ has Tor-amplitude $\le n$ over $A$.

So the answer to the question is: yes and yes.

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  • $\begingroup$ Thank you very much! $\endgroup$ – A.G Feb 19 at 19:35
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    $\begingroup$ I am a bit confused. This seems to be the case $m=0$ of proposition 6.1.6.1, but then you need $n=0$ as well, since there's the hypothesis $m\ge n$. $\endgroup$ – Denis Nardin Feb 19 at 20:31
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First of all, I strongly suspect that the answer to the question is: no.

Secondly, I want to give an example to show that the answer by Riza Hawkeye is incorrect as it stands (see also the comment by Denis Nardin for why the result is quoted incorrectly from Lurie). Namely, consider the system of rings $$ A_0 \to A_1 \to A_2 \to A_3 \to \ldots $$ where $A_i = \mathbf{Z}[x, z_i]/(xz_i)$ and where the maps send $x$ to $x$ and $z_i$ to zero. Then we see that the colimit of the system is $A = \mathbf{Z}[x]$. Now set $B_0 = A_0/xA_0$. Clearly, we see that $B_i = B_0 \otimes_{A_0} A_i$ is equal to $A_i/xA_i$ which has infinite tor dimension over $A_i$ for all $i$. On the other hand, we have $B = A/xA$ and this has tor dimension $1$ as a module over $A = \mathbf{Z}[x]$.

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  • $\begingroup$ @Thank you, Johan and Denis Nardin. I understood from the comment by Denis that the case n=0 holds. Is there any reference from that case ($B$ flat over $A$)? It is difficult for me to understand Lurie's proof without understanding the language of E$_\infty$-rings, mainly lemma 4.4.1.4. $\endgroup$ – A.G Feb 29 at 8:59

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