[18], Lecture IV. The theory of perfectoid fields, rings, and spaces has been described in numerous sources; instead of recapping this history here, see [81] (especially Remark 2.3.18).

In Section 7, we showed that a perfect prism \((A, I)\) can be recovered from the ring \(\overline{A} = A/I\text{.}\) Here, we study the rings of this form in more detail. These end up being closely related to perfectoid rings, which appear frequently in \(p\)-adic Hodge theory; however, we will not use too much of the existing theory of perfectoid rings, and in fact we will end up recovering some of it via a different approach.

Subsection8.1The category of lenses

Definition8.1.1.

A lens is a ring which occurs as the slice of some perfect prism. We define the category of lenses to be the full subcategory of \(\Ring\) consisting of lenses; by Theorem 7.3.5, the slice functor from perfect prisms to lenses is an equivalence of categories.

For \(\overline{A} = A/I\) a lens, we say that \(\overline{A}\) is an untilt of \(\overline{A}^\flat\text{.}\)

Example8.1.2.

For any perfect ring \(R\) of characteristic \(p\text{,}\) the pair \((W(R), (p))\) is a perfect prism with slice and tilt both equal to \(R\text{.}\) In particular, \(R\) is a lens.

Example8.1.3.

Let \(R\) be the \(t\)-adic completion of \(\FF_p[t^{p^{-\infty}}]\) and put \(A = W(R)\text{.}\) We can construct multiple perfect prisms \((A, I)\) with tilt \(R\text{,}\) such as the following.

Take \(I = (d)\) with \(d = \sum_{i=0}^{p-1} [t+1]^i\text{.}\) The lens \(A/I\) is isomorphic to the \(p\)-adic completion of \(\ZZ_p[\mu_{p^\infty}]\) via a map with \([t+1]^{p^{-n}} \mapsto \zeta_{p^n}\text{.}\) The prism \((A,I)\) is isomorphic to the prism from Example 5.1.6 and is the coperfection of the prism from Example 5.1.4.

Take \(I = (d)\) with \(d = p - [t]\text{.}\) The lens \(A/I\) is isomorphic to the \(p\)-adic completion of \(\ZZ_p[p^{p^{-\infty}}]\) via a map with \([t]^{p^{-n}} \mapsto p^{p^{-n}}\text{.}\) The prism \((A,I)\) is the coperfection of the prism from Example 5.1.5 in the special case \(K = \QQ_p, \pi = p\text{.}\)

Remark8.1.4.

While the terminology of tilting and untilting is now quite commonly used, our references to the category of lenses is highly nonstandard; in [22] and [25], objects of this category are called perfectoid rings. However, that usage is incompatible with most prior literature; in older terminology these would be integral perfectoid rings. To minimize confusion, we sidestep this issue by using a nonce terminology based on the metaphor of prisms.

Subsection8.2On the structure of lenses

Definition8.2.1.

A ring \(R\) of characteristic \(p\) is semiperfect if the Frobenius automorphism of \(R\) is surjective. Note that \(R\) is perfect if and only if it is both reduced and semiperfect.

Example8.2.2.

A basic example of a semiperfect ring that is not perfect is the ring \(\FF_p[t^{p^{-\infty}}]/(t)\text{.}\)

Lemma8.2.3.

Let \(R\) be a lens.

The ring \(R/p\) is semiperfect.

There exists an element \(\varpi \in R\) admitting a compatible system \(\varpi^{1/p^n}\) of \(p\)-power roots, such that \(\varpi = p u\) for some unit \(u \in R^\times\) and the kernel of the Frobenius map on \(R/p\) is generated by \(\varpi^{1/p}\text{.}\) (Note that \(\varpi = 0\) is possible.)

The ideal \(\sqrt{pR}\) is an increasing union of principal ideals and satisfies \((\sqrt{pR})^2 = \sqrt{pR}\text{.}\) (It is also flat as an \(R\)-module; see Lemma 8.2.4.)

Let \((A, I)\) be the perfect prism with \(A/I \cong R\) and let \(R^\flat\) be the tilt. Then \(R^\flat\) is perfect and \(R/p \cong R^\flat/p\text{,}\) so \(R/p\) is semiperfect. This proves (1).

For (2), apply Theorem 7.2.2 to write \(I = (d)\) with \(d \in A\) a distinguished element. By Lemma 7.1.2, we have \(d = [a_0] - pu\) for some \(a_0 \in R^\flat\) and some unit \(u \in A^\times\text{.}\) We may then take \(\varpi\) to be the image of \([a_0]\) to obtain (2).

To check (3), we show that \(\sqrt{pR} = \bigcup_n (\varpi^{p^{-n}})\text{.}\) Since the left side contains the right side, it suffices to observe that the quotient \(\overline{R} = R / \bigcup_n (\varpi^{p^{-n}})\) is itself perfect, and hence reduced.

To check (4), keep notation as above; it suffices to check that \(R[p]\) is killed by \([a_0^{p^{-n}}]\) for each \(n\) (since these elements generate \(\sqrt{pR}\)). To show that \(R[p]\) is killed by \([a_0^{p^{-n}}]\) for some particular \(n\text{,}\) note that neither \(p\) nor \(d\) is a zerodivisor in \(A\) (by Lemma 7.1.2 and Theorem 7.2.2), so we may write

We must check that for any \(M \in \Mod_R\text{,}\)\(\Tor_i^R(M, \sqrt{pR}) = 0\) for all \(i \gt 0\text{,}\) or equivalently \(\Tor_i^R(M, \overline{R}) = 0\) for all \(i \gt 1\text{.}\) By tensoring \(M\) with the exact sequence

\begin{equation*}
0 \to \Tor_1^{\ZZ_p}(M, \QQ_p) \to M \to M \otimes_{\ZZ_p} \QQ_p \to M \otimes_{\ZZ_p} \QQ_p/\ZZ_p \to 0
\end{equation*}

and using the fact that \(\overline{R} \otimes_{\ZZ_p} \QQ_p = 0\text{,}\) we may further reduce to the case of a module \(M\) which is \(p^\infty\)-power torsion. By the compatibility of tensor products with colimits, we may reduce to the case of a module which is killed by some power of \(p\text{;}\) by devissage, we may further reduce to the case where \(M\) is killed by \(p\text{.}\)

Since \(d\) is a non-zerodivisor in both \(W(R^\flat)\) and \(W(\overline{R})\) (Lemma 7.1.2),

If \(R = A/I\) is a lens, then all of the stated conditions follow directly from Theorem 7.2.2 and \(R/p\) is semiperfect by Lemma 8.2.3. Conversely, suppose that these conditions hold; we will show that for \(A = W(R^\flat), I = \ker(\theta_R)\text{,}\) the pair \((A,I)\) is a perfect prism with \(A/I \cong R\text{.}\) Since \(\theta_R\) induces a surjective map mod \(p\text{,}\) it is in fact surjective, so \(A/I \cong R\text{.}\) The ring \(A\) is classically \(p\)-complete; we may check that it is classically \((p,I)\)-complete by checking that \(A/p \cong R^\flat\) is classically \(I\)-complete, which is straightforward.

At this point, we must show that \(I\) admits a distinguished generator. To this end, choose \(x,v \in A\) lifting \(\varpi,u\) and put \(g = pv - x^p \in \ker(I)\text{.}\) The series expansion of \(x\) has zero coefficient of \(p\text{;}\) since \(v\) is a unit (because \(A\) is \(I\)-local), we deduce from Lemma 7.1.2 that \(g\) is distinguished. Since \(I\) is principal, we may apply Lemma 5.2.1 to deduce that \(g\) is in fact a generator.

In the \(p\)-torsion-free case we can make this description even simpler.

Proposition8.2.6.

A \(p\)-torsion-free commutative ring \(R\) is a lens if and only if the following conditions hold.

The ring \(R\) is classically \(p\)-complete and and \(R/p\) is semiperfect.

The ring \(R\) is \(p\)-normal: every \(x \in R[p^{-1}]\) with \(x^p \in R\) belongs to \(R\text{.}\)

There exists some \(\varpi \in R\) such that \(\varpi^p = pu\) for some unit \(u \in R\text{.}\)

Suppose first that \(R\) is a lens. In light of Proposition 8.2.5, we only need to check that \(R\) is \(p\)-normal. Take \(\varpi\) as in Lemma 8.2.3. Given \(x \in R[p^{-1}]\) with \(x^p \in R\text{,}\) let \(n\) be the smallest nonnegative integer such that \(\varpi^n x \in R\text{.}\) If \(n > 0\text{,}\) then by writing

\begin{equation*}
(\varpi^n x)^p = \varpi^{np} x^p \in \varpi^{np} R \subset \varpi^p R
\end{equation*}

and using that the Frobenius map \(R/\varpi \to R/\varpi^p\) is a bijection, we see that \(\varpi^n x \in \varpi R\) and so \(\varpi^{n-1} x \in R\text{,}\) a contradiction. Hence \(n=0\) and \(x \in R\text{,}\) as desired.

Conversely, suppose that the given conditions hold. It will suffice to show that the kernel of \(\theta_R\) is principal, as then we can apply Proposition 8.2.5. We first show that the kernel of the (surjective) Frobenius map \(R/p \to R/p\) is generated by \(\varpi\text{.}\) Given \(x \in R\) with \(x^p \in pR\text{,}\) write \(x^p = \varpi^p y\) for some \(y \in R\text{.}\) Then \((x/\varpi)^p = y \in R\) and so \(x \in \varpi R\text{.}\)

Now the Frobenius on \(R/p\) factors as \(R/p \to R/\varpi \to R/p\) where the first map is the canonical projection and the second map is an isomorphism. Since this composite is surjective, the image of \(\varpi\) in \(R/p\) admits a compatible system of \(p\)-power roots \(\overline{\varpi}^{p^{-n}}\text{.}\) By induction on \(n\text{,}\) the kernel of \(\phi^n\) on \(R/p\) is generated by \(\overline{\varpi}^{p^{-n}}\text{.}\) Hence the kernel of \(\overline{\theta}_R\colon R^\flat \to R/p\) is generated by the element \(\overline{\varpi}\) of \(R^\flat\) corresponding to the coherent sequence \((\overline{\varpi}^{p^{-n}})_n\text{.}\) Since both \(W(R^\flat)\) and \(R\) are \(p\)-torsion-free and classically \(p\)-complete, the kernel of \(\theta_R\) is generated by any element of the kernel lifting \(\overline{\varpi}\text{,}\) and in particular is principal. (Compare [18], Lecture IV, Proposition 2.10.)

Remark8.2.7.

The example \(R = \ZZ_p\) shows that condition 3 of Proposition 8.2.5 does not follow from the others (and likewise for Proposition 8.2.6).

Crucially, Proposition 8.2.5 enables us to produce many lenses in cases where it is not so obvious how to give a direct construction of the corresponding perfect prism. In particular, we can make some prisms without specifying either a \(\delta\)-ring structure or a Frobenius lift!

Subsection8.3Perfectoid fields

Definition8.3.1.

A perfectoid field is a field \(K\) satisfying the following conditions.

The field \(K\) is complete for the topology induced by some nonarchimedean valuation with nondiscrete value group.

The valuation ring \(\frako_K\) of \(K\) has residue characteristic \(p\text{,}\) and the ring \(\frako_K/p\) is semiperfect.

By Lemma 8.3.3, the valuation ring of a perfectoid field is a lens. Its tilt is also the valuation ring of a perfectoid field (of characteristic \(p\)), denoted \(K^\flat\text{.}\)

Remark8.3.2.

We report an observation from [81], Remark 2.1.8: perfectoid fields are the same as the hyperperfect fields of [97].

Lemma8.3.3.

For any perfectoid field \(K\text{,}\) the valuation ring \(\frako_K\) is a lens.

If \(K\) is of characteristic \(p\text{,}\) then \(\frako_K/p=\frako_K\) is reduced and semiperfect, hence perfect. We thus assume hereafter that \(K\) is characteristic 0; we may then check the conditions of Proposition 8.2.6.

It is clear that \(\frako_K\) is classically \(p\)-complete and \(p\)-normal (since it is integrally closed), and by hypothesis \(\frako_K/p\) is semiperfect. Since \(K\) is not discretely valued, we can choose an element \(x \in \frako_K\) of positive valuation such that \(x^p\) divides \(p\text{.}\) Since \(\frako_K/p\) is semiperfect, there exists \(y \in \frako_K\) such that \(y^p \equiv p/x^p \pmod{p}\text{;}\) put \(\varpi = xy\text{.}\) Then \(\varpi^p/p \equiv 1 \pmod{x^p}\) and so \(u = \varpi^p/p\) is a unit in \(\frako_K\text{.}\)

The following result generalizes the field of norms isomorphism of Fontaine and Wintenberger [52]. We will later give an independent “prismatic” proof; see Remark 23.1.2.

Theorem8.3.4.Tilting correspondence for perfectoid fields.

Let \(K\) be a perfectoid field. Then for every finite extension \(L\) of \(K\text{,}\)\(L\) is a perfectoid field and \([L:K] = [L^\flat:K^\flat]\text{.}\) Consequently, the categories of finite etale algebras over \(K\) and \(K^\flat\) are canonically isomorphic; in particular, the absolute Galois groups of \(K\) and \(K^\flat\) are isomorphic.

Most familiar examples of lenses are either \(p\)-torsion-free or of characteristic \(p\text{.}\) We can prove a result that shows that this accounts for all possibilities up to a “glueing” construction.

Lemma8.4.1.

Let \(R\) be a perfect \(\FF_p\)-algebra. Let \(J\) be a radical ideal of \(R\) and let \(J' = R[J]\text{.}\) Then \(J'\) and \(J+J'\) are both radical ideals and the square in Figure 8.4.2 is both a pullback square and a pushout square of commutative rings.

We first check that \(J'\) is radical. If \(x \in R\) with \(x^p \in J'\text{,}\) then \(x^p J = 0\text{;}\) since \(R\) is perfect, it follows that \(xJ = 0\) and so \(x \in J'\text{.}\)

We next check that \(J+J'\) is radical. The ideal \(J+J'\) is the kernel of \(R \to R/J \otimes_R R/J'\text{;}\) the target is a colimit of perfect rings and hence is itself perfect.

The square in question is already a pushout square at the level of groups, hence also at the level of rings. To check that it is a pullback square, we must check that \(J \cap J' = 0\text{.}\) To this end, choose \(x \in J \cap J'\text{;}\) since \(x \in J'\) we have \(xJ = 0\text{,}\) but since \(x \in J\) this implies \(x^2 = 0\) and finally \(x = 0\) because \(R\) is perfect.

Proposition8.4.3.

Let \((A, I)\) be a perfect prism and put \(R = A/I\text{.}\) Put \(\overline{R} = R/\sqrt{pR}\text{,}\)\(S = R/R[\sqrt{pR}]\text{,}\)\(\overline{S} = S/\sqrt{pS}\text{.}\) Then \(\overline{R}, S, \overline{S}\) are all lenses and the square in Figure 8.4.4 is both a pullback square and a pushout square of commutative rings.

In addition, the following statements hold.

The ring \(S\) is \(p\)-torsion-free.

The ideal \(\sqrt{pR}\) maps isomorphically onto \(\sqrt{pS}\) (and hence is also \(p\)-torsion-free).

The map \(R \to \overline{R}\) induces an isomorphism \(R[\sqrt{pR}] \to \ker(\overline{R} \to \overline{S})\text{,}\) and thus \(x \mapsto x^p\) is bijective on \(R[\sqrt{pR}]\text{.}\)

We first show that the square is a pullback. By Theorem 7.2.2 we can write \(I = (d)\) with \(d\) distinguished. By Lemma 7.1.2 we can write \(d = [a_0] - pu\) with \(a_0 \in R^\flat\text{,}\)\(u \in A^\times\text{.}\) Consider the perfect ideals \(J = (a_0^{p^{-\infty}})\) and \(J' = R^\flat[J]\) of the perfect ring \(R^\flat\text{.}\) The square Figure 8.4.5 consists of \(p\)-torsion-free, \(p\)-adically complete rings and its reduction modulo \(p\) is the pullback square from Figure 8.4.2; hence by devissage it is a pullback square.

Since \(d\) is a non-zerodivisor in each of the rings in Figure 8.4.5 by Lemma 7.1.2, we may reduce modulo \(d\) to get another pullback square (Figure 8.4.6). Let \(S'\) be the top right and bottom right entry of the new square.

We now show that Figure 8.4.6 coincides with Figure 8.4.4. Inside \(W(R^\flat/J)\) we have \((d) = (p)\) since \(a_0 \in I\text{,}\) so by Lemma 8.2.3,

Since both \(d\) and \(p\) are non-zerodivisors on \(S'\text{,}\) by Exercise 8.5.3 we have \((S'/(d))[p] \cong (S'/(p))[d] = (R^\flat/J')[d]\text{.}\) The latter vanishes because the element \(d = a_0\) of \(R^\flat/J'\) is a non-zerodivisor (by Lemma 6.4.2). We deduce that \(S'/(d)\) is \(p\)-torsion-free, and so the surjection \(R \to S'\) from the top row factors through \(R/R[p^\infty] = S\text{.}\) As in the previous paragraph, we may identify the bottom right entry with \(S'/\sqrt{pS'}\text{.}\)

Let \(K\) be the kernel of \(R \to S'\text{.}\) Since Figure 8.4.6 is a pullback square, \(K\) embeds into \(W(R^\flat)/J)/d = \overline{R}\) and hence is \(p\)-torsion. Hence \(K \subseteq R[p^\infty]\) and so the induced map \(R/R[p^\infty] = S \to S'\) is injective. Since it is also surjective (because \(R \to S'\) is) it is an isomorphism; this proves that Figure 8.4.4 and Figure 8.4.6 are the same square, and hence the former is a pullback square.

To conclude, note that the first numbered assertion is included in Lemma 8.2.3; the second and third assertions follow from the fact that Figure 8.4.4 is now a pullback square; and these in turn imply that the square is a pushout. (Compare [18], Lecture IV, Proposition 3.2.)

By Proposition 8.4.3, we may reduce to the cases of a perfect ring of characteristic \(p\) and of a \(p\)-torsion-free untilted ring. In the former case, it is evident that any perfect ring is reduced. in the latter case, let \(R\) be the lens in question. Apply Lemma 8.2.3 to choose an element \(\varpi \in R\) such that \(\varpi^p = pu\) for some unit \(u \in R^\times\text{.}\) It will suffice to check that any \(x \in R\) with \(x^p = 0\) vanishes, or (because \(R\) is \(p\)-adically separated) that any such \(x\) is divisible by \(\varpi^n\) for any positive integer \(n\text{.}\) We prove this by induction starting with the base case \(n=0\text{.}\) Given that \(x = \pi^n y\) for some nonnegative integer \(n\) and some \(y \in R\text{,}\) we have \(\varpi^{np} y^p = 0\) and hence \(y^p = 0\) because \(R\) is \(p\)-torsion-free. By Lemma 8.2.3 again, the kernel of the Frobenius on \(R/p\) is generated by \(\varpi\text{;}\) hence \(y \in \varpi R\) and \(x \in \varpi^{n+1} R\text{.}\) (Compare [18], Lecture IV, Corollary 3.3.)

The following argument makes a mild use of derived categories; see Section 10.

Proposition8.4.8.

Let \(A \to B\text{,}\)\(A \to C\) be morphisms of lenses. Then the derived \(p\)-completion of \(B \otimes_A^L C\) has cohomology only in degree \(0\text{,}\) which is a lens.

where \(\widehat{\otimes}^L\) denotes the derived \(p\)-completion of the derived tensor product. Write \(A \cong W(A^\flat)/d\) with \(d \in W(A^\flat)\) distinguished (Theorem 7.2.2). Since \(d\) is a non-zerodivisor in \(W(A^\flat), W(B^\flat), W(C^\flat)\) (Lemma 7.1.2), we get an isomorphism

\begin{equation*}
W(R)/(d) \cong B \widehat{\otimes}^L_A C.
\end{equation*}

This proves the claim. (Compare [18], Lecture IV, Proposition 2.11.)

Exercises8.5Exercises

1.

Show that the category of lenses is closed under arbitrary colimits and products in the category of all derived \(p\)-complete rings. However, this does not imply closure under arbitrary limits; see Exercise 8.5.2.

2.

Show that the category of lenses is not closed under formation of equalizers in the category of rings.

One approach is to use the theorem of Ax-Sen-Tate (see [11]); this implies for example that if \(K\) is a (possibly infinite) Galois algebraic extension of \(\QQ_p\) with Galois group \(G\text{,}\) then the invariant subfield of the completion of \(K\) under the action of \(G\) is equal to \(\QQ_p\text{.}\)

3.

For \(A \in \Ring\) and \(x,y \in A\) two non-zerodivisors, prove that the \(A\)-modules \((A/x)[y]\) and \((A/y)[x]\) are isomorphic.