In Kedlayas lecture notes on prismatic cohomology it is an exercise (2.5.9) to prove that the category of rings equipped with a Frobenius lift (denotet $\mathbf{Ring}_\phi$, the morphisms are ringhoms compatible with the corresp. Frob. lift) admits equalizers. However in a lecture of James Borger on $\delta$-rings he claims that $\mathbf{Ring}_\phi$ doesn't have equalizers. Does someone know what is true and how to prove it or what would be a counterexample.
Link to Kedlayas work: https://kskedlaya.org/prismatic/sec_overview.html
Link to Borgers lecture: https://www.youtube.com/watch?v=_y2Tcu-iJV4&t=10s
Any help or advice would be great.
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3 Answers
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While it is true that Borger writes that the category does not have equalizers, what he actually explains is that the equalizer as computed in the category of rings equipped with an endomorphism (which, underlying, is the equalizer of rings) is not necessarily a Frobenius lift anymore, which is the same caveat as in Kedlaya's Execise 2.5.9(b).
It is indeed true that the category of rings equipped with a Frobenius lift admits equalizers. In fact, it is presentable: it sits in a pullback square $$ \require{AMScd} \begin{CD} \mathrm{Ring}_{\phi} @>>> \mathrm{Fun}(B\mathbb{N},\mathrm{Ring})\\ @VVV @VVV\\ \mathrm{Ring}^{\mathrm{char}=p} @>>> \mathrm{Fun}(B\mathbb{N},\mathrm{Ring}^{\mathrm{char}=p}). \end{CD} $$ The right vertical arrow takes a ring equipped with an endomorphism and mods out $p$. It is left adjoint to the obvious inclusion. The bottom horizontal arrow takes a characteristic $p$ ring and equips it with the Frobenius. It is an isofibration and left adjoint to the functor that takes a pair $(B,\varphi)$ consisting of a characteristic $p$ ring $B$ and a ring endomorphism $\varphi$ to the equalizer of $\varphi$ and the Frobenius. The categories in the cospan are all presentable, so the claim follows since $\mathrm{Pr}^L$ is closed under pseudo-limits in $\widehat{\mathrm{Cat}}$.
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$\begingroup$ Thanks for your answer! Unfortunately I am not familiar with the notion $\mathbf{Pr}^{\mathbf{L}}$. Could you explain this? $\endgroup$Mikkel– Mikkel2026-04-30 06:41:20 +00:00Commented yesterday
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1$\begingroup$ It is the (huge) category whose objects are presentable categories and whose morphisms are presentable functors. The inclusion thereof into the (huge) category of categories creates all small pseudo-limits (this is a slight correction of what I claimed in the original answer), a fact originally due to Bird. That $\mathrm{Ring}_{\varphi}$ is presentable is much stronger than what the question originally asked, though, so in light of @WillSawin's answer, it is fair to say that I have used a nuke to kill a fly. $\endgroup$Thorgott– Thorgott2026-04-30 08:33:34 +00:00Commented yesterday
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$\begingroup$ alright, thanks a lot $\endgroup$Mikkel– Mikkel2026-04-30 13:03:24 +00:00Commented 20 hours ago
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The equalizers do exist.
You can take the equalizer in the category of rings with a homomorphism to themselves and then transfinitely iterate the operation of taking the subring of elements $x$ such that $x^p\equiv \phi(x)$ mod $p$.
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Here is an example where the equalizer in rings with endomorphism is not an object of the category in question.
Let $p$ be a prime, and put $$ A=\mathbb Z[x,y],\qquad \phi(x)=x^p+py,\quad \phi(y)=y^p. $$ Let $$ B=A/(py), $$ with the induced endomorphism; this is possible since $$ \phi(py)=p y^p\in (py). $$ Let $$ q,r:A\rightrightarrows B $$ be given by taking $q$ to be the quotient map and $$ r(x)=x,\qquad r(y)=0. $$ These are maps preserving the Frobenius lifts.
Their equalizer in rings is $$ R=\{F\in\mathbb Z[x,y]:F(x,y)-F(x,0)\in (py)\} =\mathbb Z[x]+py\,\mathbb Z[x,y]. $$ This is stable under $\phi$. But the induced endomorphism of $R$ is not a Frobenius lift: $x\in R$, while $$ \phi(x)-x^p=py. $$ On the other hand $$ pR=p\mathbb Z[x]+p^2y\,\mathbb Z[x,y], $$ so $py\notin pR$.
Thus the forgetful functor to rings, or to rings with endomorphism, does not preserve this equalizer.
