Lemma 15.9.1. Let $A$ be a ring, let $I \subset A$ be an ideal, let $\overline{u} \in A/I$ be an invertible element. There exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an invertible element $u' \in A'$ lifting $\overline{u}$.

## 15.9 Lifting

In this section we collection some lemmas concerning lifting statements of the following kind: If $A$ is a ring and $I \subset A$ is an ideal, and $\overline{\xi }$ is some kind of structure over $A/I$, then we can lift $\overline{\xi }$ to a similar kind of structure $\xi $ over $A$ or over some étale extension of $A$. Here are some types of structure for which we have already proved some results:

projective modules, see Algebra, Lemmas 10.76.4 and 10.76.5,

finite stably free modules, see Lemma 15.3.3,

ring maps, i.e., proving certain algebras are formally smooth, see Algebra, Lemma 10.136.4, Proposition 10.136.13, and Lemma 10.136.17,

syntomic ring maps, see Algebra, Lemma 10.134.18,

smooth ring maps, see Algebra, Lemma 10.135.19,

étale ring maps, see Algebra, Lemma 10.141.10,

factoring polynomials, see Algebra, Lemma 10.141.19, and

Algebra, Section 10.148 discusses henselian local rings.

The interested reader will find more results of this nature in Smoothing Ring Maps, Section 16.3 in particular Smoothing Ring Maps, Proposition 16.3.2.

Let $A$ be a ring and let $I \subset A$ be an ideal. Let $\overline{\xi }$ be some kind of structure over $A/I$. In the following lemmas we look for étale ring maps $A \to A'$ which induce isomorphisms $A/I \to A'/IA'$ and objects $\xi '$ over $A'$ lifting $\overline{\xi }$. A general remark is that given étale ring maps $A \to A' \to A''$ such that $A/I \cong A'/IA'$ and $A'/IA' \cong A''/IA''$ the composition $A \to A''$ is also étale (Algebra, Lemma 10.141.3) and also satisfies $A/I \cong A''/IA''$. We will frequently use this in the following lemmas without further mention. Here is a trivial example of the type of result we are looking for.

**Proof.**
Choose any lift $f \in A$ of $\overline{u}$ and set $A' = A_ f$ and $u$ the image of $f$ in $A'$.
$\square$

Lemma 15.9.2. Let $A$ be a ring, let $I \subset A$ be an ideal, let $\overline{e} \in A/I$ be an idempotent. There exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an idempotent $e' \in A'$ lifting $\overline{e}$.

**Proof.**
Choose any lift $x \in A$ of $\overline{e}$. Set

The ring map $A \to A'$ is étale because $(2t - 1)\text{d}t = 0$ and $(2t - 1)(2t - 1) = 1$ which is invertible. We have $A'/IA' = A/I[t]/(t^2 - t)[\frac{1}{t - 1 + \overline{e}}] \cong A/I$ the last map sending $t$ to $\overline{e}$ which works as $\overline{e}$ is a root of $t^2 - t$. This also shows that setting $e'$ equal to the class of $t$ in $A'$ works. $\square$

Lemma 15.9.3. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $\mathop{\mathrm{Spec}}(A/I) = \coprod _{j \in J} \overline{U}_ j$ be a finite disjoint open covering. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite disjoint open covering $\mathop{\mathrm{Spec}}(A') = \coprod _{j \in J} U'_ j$ lifting the given covering.

**Proof.**
This follows from Lemma 15.9.2 and the fact that open and closed subsets of Spectra correspond to idempotents, see Algebra, Lemma 10.20.3.
$\square$

Lemma 15.9.4. Let $A \to B$ be a ring map and $J \subset B$ an ideal. If $A \to B$ is étale at every prime of $V(J)$, then there exists a $g \in B$ mapping to an invertible element of $B/J$ such that $A' = B_ g$ is étale over $A$.

**Proof.**
The set of points of $\mathop{\mathrm{Spec}}(B)$ where $A \to B$ is not étale is a closed subset of $\mathop{\mathrm{Spec}}(B)$, see Algebra, Definition 10.141.1. Write this as $V(J')$ for some ideal $J' \subset B$. Then $V(J') \cap V(J) = \emptyset $ hence $J + J' = B$ by Algebra, Lemma 10.16.2. Write $1 = f + g$ with $f \in J$ and $g \in J'$. Then $g$ works.
$\square$

Next we have three lemmas saying we can lift factorizations of polynomials.

Lemma 15.9.5. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $f \in A[x]$ be a monic polynomial. Let $\overline{f} = \overline{g} \overline{h}$ be a factorization of $f$ in $A/I[x]$ such that $\overline{g}$ and $\overline{h}$ are monic and generate the unit ideal in $A/I[x]$. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a factorization $f = g' h'$ in $A'[x]$ with $g'$, $h'$ monic lifting the given factorization over $A/I$.

**Proof.**
We will deduce this from results on the universal factorization proved earlier; however, we encourage the reader to find their own proof not using this trick. Say $\deg (\overline{g}) = n$ and $\deg (\overline{h}) = m$ so that $\deg (f) = n + m$. Write $f = x^{n + m} + \sum \alpha _ i x^{n + m - i}$ for some $\alpha _1, \ldots , \alpha _{n + m} \in A$. Consider the ring map

of Algebra, Example 10.141.12. Let $R \to A$ be the ring map which sends $a_ i$ to $\alpha _ i$. Set

By construction the image $f_ B$ of $f$ in $B[x]$ factors, say $f_ B = g_ B h_ B$ with $g_ B = x^ n + \sum (1 \otimes b_ i) x^{n - i}$ and similarly for $h_ B$. Write $\overline{g} = x^ n + \sum \overline{\beta }_ i x^{n - i}$ and $\overline{h} = x^ m + \sum \overline{\gamma }_ i x^{m - i}$. The $A$-algebra map

maps $g_ B$ and $h_ B$ to $\overline{g}$ and $\overline{h}$ in $A/I[x]$. The displayed map is surjective; denote $J \subset B$ its kernel. From the discussion in Algebra, Example 10.141.12 it is clear that $A \to B$ is etale at all points of $V(J) \subset \mathop{\mathrm{Spec}}(B)$. Choose $g \in B$ as in Lemma 15.9.4 and consider the $A$-algebra $B_ g$. Since $g$ maps to a unit in $B/J = A/I$ we obtain also a map $B_ g/I B_ g \to A/I$ of $A/I$-algebras. Since $A/I \to B_ g/I B_ g$ is étale, also $B_ g/IB_ g \to A/I$ is étale (Algebra, Lemma 10.141.8). Hence there exists an idempotent $e \in B_ g/I B_ g$ such that $A/I = (B_ g/I B_ g)_ e$ (Algebra, Lemma 10.141.9). Choose a lift $h \in B_ g$ of $e$. Then $A \to A' = (B_ g)_ h$ with factorization given by the image of the factorization $f_ B = g_ B h_ B$ in $A'$ is a solution to the problem posed by the lemma. $\square$

The assumption on the leading coefficient in the following lemma will be removed in Lemma 15.9.7.

Lemma 15.9.6. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $f \in A[x]$ be a monic polynomial. Let $\overline{f} = \overline{g} \overline{h}$ be a factorization of $f$ in $A/I[x]$ and assume

the leading coefficient of $\overline{g}$ is an invertible element of $A/I$, and

$\overline{g}$, $\overline{h}$ generate the unit ideal in $A/I[x]$.

Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a factorization $f = g' h'$ in $A'[x]$ lifting the given factorization over $A/I$.

**Proof.**
Applying Lemma 15.9.1 we may assume that the leading coefficient of $\overline{g}$ is the reduction of an invertible element $u \in A$. Then we may replace $\overline{g}$ by $\overline{u}^{-1}\overline{g}$ and $\overline{h}$ by $\overline{u}\overline{h}$. Thus we may assume that $\overline{g}$ is monic. Since $f$ is monic we conclude that $\overline{h}$ is monic too. In this case the result follows from Lemma 15.9.5.
$\square$

Lemma 15.9.7. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $f \in A[x]$ be a monic polynomial. Let $\overline{f} = \overline{g} \overline{h}$ be a factorization of $f$ in $A/I[x]$ and assume that $\overline{g}$, $\overline{h}$ generate the unit ideal in $A/I[x]$. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a factorization $f = g' h'$ in $A'[x]$ lifting the given factorization over $A/I$.

**Proof.**
Say $f = x^ d + a_1 x^{d - 1} + \ldots + a_ d$ has degree $d$. Write $\overline{g} = \sum \overline{b}_ j x^ j$ and $\overline{h} = \sum \overline{c}_ j x^ j$. Then we see that $1 = \sum \overline{b}_ j \overline{c}_{d - j}$. It follows that $\mathop{\mathrm{Spec}}(A/I)$ is covered by the standard opens $D(\overline{b}_ j \overline{c}_{d - j})$. However, each point $\mathfrak p$ of $\mathop{\mathrm{Spec}}(A/I)$ is contained in at most one of these as by looking at the induced factorization of $f$ over the field $\kappa (\mathfrak p)$ we see that $\deg (\overline{g} \bmod \mathfrak p) + \deg (\overline{h} \bmod \mathfrak p) = d$. Hence our open covering is a disjoint open covering. Applying Lemma 15.9.3 (and replacing $A$ by $A'$) we see that we may assume there is a corresponding disjoint open covering of $\mathop{\mathrm{Spec}}(A)$. This disjoint open covering corresponds to a product decomposition of $A$, see Algebra, Lemma 10.23.3. It follows that

where the image of $\overline{g}$, resp. $\overline{h}$ in $A_ j/I_ j$ has degree $j$, resp. $d - j$ with invertible leading coefficient. Clearly, it suffices to prove the result for each factor $A_ j$ separatedly. Hence the lemma follows from Lemma 15.9.6. $\square$

Lemma 15.9.8. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal of $R$ and let $J \subset S$ be an ideal of $S$. If the closure of the image of $V(J)$ in $\mathop{\mathrm{Spec}}(R)$ is disjoint from $V(I)$, then there exists an element $f \in R$ which maps to $1$ in $R/I$ and to an element of $J$ in $S$.

**Proof.**
Let $I' \subset R$ be an ideal such that $V(I')$ is the closure of the image of $V(J)$. Then $V(I) \cap V(I') = \emptyset $ by assumption and hence $I + I' = R$ by Algebra, Lemma 10.16.2. Write $1 = g + f$ with $g \in I$ and $f \in I'$. We have $V(f') \supset V(J)$ where $f'$ is the image of $f$ in $S$. Hence $(f')^ n \in J$ for some $n$, see Algebra, Lemma 10.16.2. Replacing $f$ by $f^ n$ we win.
$\square$

Lemma 15.9.9. Let $I$ be an ideal of a ring $A$. Let $A \to B$ be an integral ring map. Let $b \in B$ map to an idempotent in $B/IB$. Then there exists a monic $f \in A[x]$ with $f(b) = 0$ and $f \bmod I = x^ d(x - 1)^ d$ for some $d \geq 1$.

**Proof.**
Observe that $z = b^2 - b$ is an element of $IB$. By Algebra, Lemma 10.37.4 there exist a monic polynomial $g(x) = x^ d + \sum a_ j x^ j$ of degree $d$ with $a_ j \in I$ such that $g(z) = 0$ in $B$. Hence $f(x) = g(x^2 - x) \in A[x]$ is a monic polynomial such that $f(x) \equiv x^ d(x - 1)^ d \bmod I$ and such that $f(b) = 0$ in $B$.
$\square$

Lemma 15.9.10. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $A \to B$ be an integral ring map. Let $\overline{e} \in B/IB$ be an idempotent. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an idempotent $e' \in B \otimes _ A A'$ lifting $\overline{e}$.

**Proof.**
Choose an element $y \in B$ lifting $\overline{e}$. Choose $f \in A[x]$ as in Lemma 15.9.9 for $y$. By Lemma 15.9.6 we can find an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and such that $f = gh$ in $A[x]$ with $g(x) = x^ d \bmod IA'$ and $h(x) = (x - 1)^ d \bmod IA'$. After replacing $A$ by $A'$ we may assume that the factorization is defined over $A$. In that case we see that $b_1 = g(y) \in B$ is a lift of $\overline{e}^ d = \overline{e}$ and $b_2 = h(y) \in B$ is a lift of $(\overline{e} - 1)^ d = (-1)^ d (1 - \overline{e})^ d = (-1)^ d(1 - \overline{e})$ and moreover $b_1b_2 = 0$. Thus $(b_1, b_2)B/IB = B/IB$ and $V(b_1, b_2) \subset \mathop{\mathrm{Spec}}(B)$ is disjoint from $V(IB)$. Since $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is closed (see Algebra, Lemmas 10.35.22 and 10.40.6) we can find an $a \in A$ which maps to an invertible element of $A/I$ whose image in $B$ lies in $(b_1, b_2)$, see Lemma 15.9.8. After replacing $A$ by the localization $A_ a$ we get that $(b_1, b_2) = B$. Then $\mathop{\mathrm{Spec}}(B) = D(b_1) \amalg D(b_2)$; disjoint union because $(b_1, b_2) = B$ and covers $\mathop{\mathrm{Spec}}(B)$ because $b_1b_2 = 0$. Let $e \in B$ be the idempotent corresponding to the open and closed subset $D(b_1)$, see Algebra, Lemma 10.20.3. Since $b_1$ is a lift of $\overline{e}$ and $b_2$ is a lift of $\pm (1 - \overline{e})$ we conclude that $e$ is a lift of $\overline{e}$ by the uniqueness statement in Algebra, Lemma 10.20.3.
$\square$

Lemma 15.9.11. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $\overline{P}$ be finite projective $A/I$-module. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite projective $A'$-module $P'$ lifting $\overline{P}$.

**Proof.**
We can choose an integer $n$ and a direct sum decomposition $(A/I)^{\oplus n} = \overline{P} \oplus \overline{K}$ for some $R/I$-module $\overline{K}$. Choose a lift $\varphi : A^{\oplus n} \to A^{\oplus n}$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$. Let $f \in A[x]$ be the characteristic polynomial of $\varphi $. Set $B = A[x]/(f)$. By Cayley-Hamilton (Algebra, Lemma 10.15.1) there is a map $B \to \text{End}_ A(A^{\oplus n})$ mapping $x$ to $\varphi $. For every prime $\mathfrak p \supset I$ the image of $f$ in $\kappa (\mathfrak p)$ is $(x - 1)^ rx^{n - r}$ where $r$ is the dimension of $\overline{P} \otimes _{A/I} \kappa (\mathfrak p)$. Hence $(x - 1)^ nx^ n$ maps to zero in $B \otimes _ A \kappa (\mathfrak p)$ for all $\mathfrak p \supset I$. Thus $x(1 - x)$ is contained in every prime ideal of $B/IB$. Hence $x^ N(1 - x)^ N$ is contained in $IB$ for some $N \geq 1$. It follows that $x^ N + (1 - x)^ N$ is a unit in $B/IB$ and that

is an idempotent as both assertions hold in $\mathbf{Z}[x]/(x^ N(x - 1)^ N)$. The image of $\overline{e}$ in $\text{End}_{A/I}((A/I)^{\oplus n})$ is

as $\overline{p}$ is an idempotent. After replacing $A$ by an étale extension $A'$ as in the lemma, we may assume there exists an idempotent $e \in B$ which maps to $\overline{e}$ in $B/IB$, see Lemma 15.9.10. Then the image of $e$ under the map

is an idempotent element $p$ which lifts $\overline{p}$. Setting $P = \mathop{\mathrm{Im}}(p)$ we win. $\square$

Lemma 15.9.12. Let $A$ be a ring. Let $0 \to K \to A^{\oplus m} \to M \to 0$ be a sequence of $A$-modules. Consider the $A$-algebra $C = \text{Sym}^*_ A(M)$ with its presentation $\alpha : A[y_1, \ldots , y_ m] \to C$ coming from the surjection $A^{\oplus m} \to M$. Then

(see Algebra, Section 10.132) in particular $\Omega _{C/A} = M \otimes _ A C$.

**Proof.**
Let $J = \mathop{\mathrm{Ker}}(\alpha )$. The lemma asserts that $J/J^2 \cong K \otimes _ A C$. Note that $\alpha $ is a homomorphism of graded algebras. We will prove that in degree $d$ we have $(J/J^2)_ d = K \otimes _ A C_{d - 1}$. Note that

see Algebra, Lemma 10.12.2. It follows that $(J^2)_ d = \sum _{a + b = d} J_ a \cdot J_ b$ is the image of

The cokernel of the map $K \otimes _ A \text{Sym}^{d - 2}_ A(A^{\otimes m}) \to \text{Sym}^{d - 1}_ A(A^{\oplus m})$ is $\text{Sym}^{d - 1}_ A(M)$ by the lemma referenced above. Hence it is clear that $(J/J^2)_ d = J_ d/(J^2)_ d$ is equal to

as desired. $\square$

Lemma 15.9.13. Let $A$ be a ring. Let $M$ be an $A$-module. Then $C = \text{Sym}_ A^*(M)$ is smooth over $A$ if and only if $M$ is a finite projective $A$-module.

**Proof.**
Let $\sigma : C \to A$ be the projection onto the degree $0$ part of $C$. Then $J = \mathop{\mathrm{Ker}}(\sigma )$ is the part of degree $> 0$ and we see that $J/J^2 = M$ as an $A$-module. Hence if $A \to C$ is smooth then $M$ is a finite projective $A$-module by Algebra, Lemma 10.137.4.

Conversely, assume that $M$ is finite projective and choose a surjection $A^{\oplus n} \to M$ with kernel $K$. Of course the sequence $0 \to K \to A^{\oplus n} \to M \to 0$ is split as $M$ is projective. In particular we see that $K$ is a finite $A$-module and hence $C$ is of finite presentation over $A$ as $C$ is a quotient of $A[x_1, \ldots , x_ n]$ by the ideal generated by $K \subset \bigoplus Ax_ i$. The computation of Lemma 15.9.12 shows that $\mathop{N\! L}\nolimits _{C/A}$ is homotopy equivalent to $(K \to M) \otimes _ A C$. Hence $\mathop{N\! L}\nolimits _{C/A}$ is quasi-isomorphic to $C \otimes _ A M$ placed in degree $0$ which means that $C$ is smooth over $A$ by Algebra, Definition 10.135.1. $\square$

Lemma 15.9.14. Let $A$ be a ring, let $I \subset A$ be an ideal. Consider a commutative diagram

where $B$ is a smooth $A$-algebra. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an $A$-algebra map $B \to A'$ lifting the ring map $B \to A/I$.

**Proof.**
Let $J \subset B$ be the kernel of $B \to A/I$ so that $B/J = A/I$. By Algebra, Lemma 10.137.3 the sequence

is split exact. Thus $\overline{P} = J/(J^2 + IB) = \Omega _{B/A} \otimes _ B B/J$ is a finite projective $A/I$-module. Choose an integer $n$ and a direct sum decomposition $A/I^{\oplus n} = \overline{P} \oplus \overline{K}$. By Lemma 15.9.11 we can find an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite projective $A$-module $K$ which lifts $\overline{K}$. We may and do replace $A$ by $A'$. Set $B' = B \otimes _ A \text{Sym}_ A^*(K)$. Since $A \to \text{Sym}_ A^*(K)$ is smooth by Lemma 15.9.13 we see that $B \to B'$ is smooth which in turn implies that $A \to B'$ is smooth (see Algebra, Lemmas 10.135.4 and 10.135.13). Moreover the section $\text{Sym}^*_ A(K) \to A$ determines a section $B' \to B$ and we let $B' \to A/I$ be the composition $B' \to B \to A/I$. Let $J' \subset B'$ be the kernel of $B' \to A/I$. We have $JB' \subset J'$ and $B \otimes _ A K \subset J'$. These maps combine to give an isomorphism

Thus, after replacing $B$ by $B'$ we may assume that $J/(J^2 + IB) = \Omega _{B/A} \otimes _ B B/J$ is a free $A/I$-module of rank $n$.

In this case, choose $f_1, \ldots , f_ n \in J$ which map to a basis of $J/(J^2 + IB)$. Consider the finitely presented $A$-algebra $C = B/(f_1, \ldots , f_ n)$. Note that we have an exact sequence

see Algebra, Lemma 10.132.4 (note that $H_1(L_{B/A}) = 0$ and that $\Omega _{B/A}$ is finite projective, in particular flat so the Tor group vanishes). For any prime $\mathfrak q \supset J$ of $B$ the module $\Omega _{B/A, \mathfrak q}$ is free of rank $n$ because $\Omega _{B/A}$ is finite projective and because $\Omega _{B/A} \otimes _ B B/J$ is free of rank $n$ (see Algebra, Lemma 10.77.2). By our choice of $f_1, \ldots , f_ n$ the map

is surjective modulo $J$. Hence we see that this map of modules over the local ring $C_{\mathfrak q}$ has to be an isomorphism (this is because by Nakayama's Algebra, Lemma 10.19.1 the map is surjective and then for example by Algebra, Lemma 10.15.4 because $((f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2)_{\mathfrak q}$ is generated by $n$ elements the map is injective). Thus $H_1(L_{C/A})_{\mathfrak q} = 0$ and $\Omega _{C/A, \mathfrak q} = 0$. By Algebra, Lemma 10.135.12 we see that $A \to C$ is smooth at the prime $\overline{\mathfrak q}$ of $C$ corresponding to $\mathfrak q$. Since $\Omega _{C/A, \mathfrak q} = 0$ it is actually étale at $\overline{\mathfrak q}$. Thus $A \to C$ is étale at all primes of $C$ containing $JC$. By Lemma 15.9.4 we can find an $f \in C$ mapping to an invertible element of $C/JC$ such that $A \to C_ f$ is étale. By our choice of $f$ it is still true that $C_ f/JC_ f = A/I$. The map $C_ f/IC_ f \to A/I$ is surjective and étale by Algebra, Lemma 10.141.8. Hence $A/I$ is isomorphic to the localization of $C_ f/IC_ f$ at some element $g \in C$, see Algebra, Lemma 10.141.9. Set $A' = C_{fg}$ to conclude the proof. $\square$

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