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In the category of (commutative, unital) $\mathbb{C}$-algebras, let ${f : \mathbb{C}[t, t^{-1}] \rightarrow \mathbb{C}[t, t^{-1}, x]/(x^2 -t)}$ be the obvious map sending $t$ in the domain to $t$ in the codomain. Let $N$ be a nilpotent ideal in an algebra $C$ and let ${q : C \rightarrow C/N}$ be the resulting quotient. Let ${g : \mathbb{C}[t, t^{-1}] \rightarrow C}$ and ${h : \mathbb{C}[t, t^{-1}, x]/(x^2 -t) \rightarrow C/N}$ be such that ${h f = q g}$ (in other words, such that `the square commutes').

Why is it that there exists a unique ${k : \mathbb{C}[t, t^{-1}, x]/(x^2 -t) \rightarrow C}$ such that ${k f = g}$ and ${q k = h}$ (in other words, a unique $k$ making the inner triangles of the square commute)?

(Please, no appeal to schemes, references to EGA etc. I just want to see a direct proof using only the simplest commutative-algebraic facts. Thanks in advance.)

Boogie
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  • Could you share any ideas you have? Is there any properties of nilpotent ideals you think are important (I do not know how to solve the problem, btw, and I'm not too versed in AG so that is why I ask). Of course, the problem is equivalent to showing that $f(t)$ has a unique square root $c$ in $C$ such that $c+N=h(x)$. – Luiz Cordeiro Sep 27 '19 at 21:56
  • I am afraid I don't have useful ideas. Certainly, nilpotency has to play a role, and also the fact that $2x$ is invertible in the codomain of $f$, but I have not managed to make the thing work. (As you say, the problem can indeed be formulated without diagrams.) – Boogie Sep 27 '19 at 22:03
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    Are you only interested in how to show that $f$ is etale in your particular formulation? If it is acceptable to you to show that $f$ is flat and unramified, there is an easy and general approach. In particular, one can show that if $A$ is a commutative ring and $B = A[X]/\langle X^{n} - a\rangle$, then $B$ is an etale $A$-algebra if and only if $na \in A^{\times}$. – Alex Wertheim Sep 28 '19 at 06:40
  • @AlexWertheim, I needed to see the gory details; but thanks for the extra information. – Boogie Sep 28 '19 at 13:45
  • @Boogie: sure. I am happy to give the "gory" details in the approach I am suggesting - I was just wondering if you were asking for an answer which uses your particular definition of etale, or if mine would do. – Alex Wertheim Sep 28 '19 at 17:32
  • @AlexWertheim: I was looking for something along the lines of what diracdeltafunk provided, but I certainly would be interested to compare with an alternative. Please, though, do include your definitions of unramified and flat. (The definition of unramified I know is similar to the one I tacitly gave of étale except that 'there exists a unique' is replaced by 'there exists at most one'.) – Boogie Sep 28 '19 at 17:56
  • @AlexWertheim: I am still interested in the approach you were suggesting. – Boogie Oct 22 '19 at 07:58

2 Answers2

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Lemma 1 Let $A$ be a commutative ring in which no nonzero integer is a zero divisor (in particular, $A$ has characteristic $0$). Let $s \in A$ such that $s^2 = 1$. If $1-s$ is nilpotent, then $s = 1$.

Proof. Choose a positive integer $n$ such that $(1-s)^{2n+1} = 0$. By the binomial theorem, $$0 = \sum_{i=0}^{2n+1} \binom{2n+1}{i} (-1)^i s^i = \sum_{i=0}^n \binom{2n+1}{2i} - s \sum_{i=0}^n \binom{2n+1}{2i+1} = (1-s) \sum_{i=0}^n \binom{2n+1}{2i}.$$ Since $\sum_{i=0}^n \binom{2n+1}{2i}$ is not a zero divisor, $1-s = 0$, so $s = 1$. $\square$

Corollary If $k$ exists then it is unique.

Proof. Suppose there are two such maps, $k$ and $k'$. Since $kf = g = k'f$, we see that $k(x)^2 = k'(x)^2$ is a unit in $C$. In particular, $k(x)$ is a unit. Since $qk = h = qk'$, we see that $k(x) - k'(x)$ is nilpotent, so $1 - k'(x)/k(x)$ is nilpotent. By Lemma 1, $k'(x)/k(x) = 1$, i.e. $k(x) = k'(x)$. This shows that $k = k'$. $\square$

Existence of $k$ boils down to smoothness of the map $\operatorname{Spec} \mathbb{C}[t,t^{-1},x]/(x^2 - t) \to \operatorname{Spec} \mathbb{C}[t,t^{-1}]$, which we can unpack into straight algebra.

Lemma 2 If $N^2 = 0$ then there exists a map $k$ as desired.

Proof. Since $q$ is surjective, choose some $y \in C$ such that $q(y) = h(x)$. Now $q(y^2 - g(t)) = 0$, so $y^2 - g(t) \in N$. Since $q(2y) = h(2x)$ is a unit in $C/N$, $2y$ is a unit in $C$. So let $z = (g(t) - y^2)/(2y)$ (whence $z \in N$). Let $k_0 : \mathbb{C}[t,t^{-1},x] \to C$ be defined by sending $t \mapsto g(t)$ and $x \mapsto y+z$. Then $k_0(x^2 - t) = (y+z)^2 - g(t) = y^2 + (2y)z + z^2 - g(t) = y^2 - g(t) + (2y)z = 0$ since $N^2 = 0$. Thus, $k_0$ descends to give a map $k : \mathbb{C}[t,t^{-1},x]/(x^2 - t) \to C$ where $k(t) = g(t)$ and $k(x) = y + z$. It's easy to check that $kf = g$ and $qk = h$. $\square$

Now we can lift from $C/N$ to $C/N^2$, to $C/N^4$, ..., until we reach $C/N^m = C$. Formally, as an inductive proof:

Corollary There exists a map $k$ as desired.

Proof. Let $n$ be the smallest positive integer such that $N^n = 0$. We will use induction on $n$. If $n = 1$ then $q$ is an isomorphism and we are done. For the inductive step, suppose $n > 1$ and that the result holds for all smaller values of $n$. Factor $q$ as the composition of the quotients $C \to C/N^2 \to (C/N^2)/(N/N^2) \cong C/N$. Since $(N/N^2)^2 = 0$, Lemma 2 says that $h$ lifts up the projection $C/N^2 \to C/N$ to give a map $\ell : \mathbb{C}[t,t^{-1},x]/(x^2 - t) \to C/N^2$ making everything commute. By inductive hypothesis, $\ell$ lifts up to $k : \mathbb{C}[t,t^{-1},x]/(x^2 - t) \to C$. $\square$

diracdeltafunk
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    +1 let's catch up soon! – Joshua Mundinger Sep 28 '19 at 04:12
  • Great. This is the kind of stuff I was looking for, @diracdeltafunk. Am I correct in guessing that essentially the same proof would show that standard-étale implies formally-étale? By the way, you say that existence of $k$ boils down smoothness but would that simplify the proof? Finally, could you recommend an item in the bibliography treating étale maps in this purely commutative-algebraic fashion? – Boogie Sep 28 '19 at 13:40
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    Honestly this is not my area of expertise; I don't know the proof that an étale map is formally étale. I was just having fun bashing out the algebra in this specific case, but I think @AlexWertheim may know where you should look.

    The smoothness appears in the argument where we invert $2y$ (which was possible because $2x = \partial_x (x^2 - t)$ was a unit and units lift modulo nilpotents). You could replicate a very similar proof for any smooth extension of $\mathbb{C}$-algebras $A \to A[x]/(f)$!

     – diracdeltafunk Sep 28 '19 at 20:04
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    Nice answer. @Boogie: indeed, a very similar approach shows that étale implies formally-étale. See Milne's "Etale Cohomology", page 30 for details. – Alex Wertheim Sep 29 '19 at 22:38
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Let me give a bare-hands answer.

Consider the Taylor expansion for $(1+u)^{1/2}$: $$ s(u) = \sum_{n\geq 0} \binom{1/2}{n} u^n.$$ This power series converges in a small disk around $1$ in $\mathbb C$, so at the level of formal power series with coefficients in $\mathbb Z[1/2]$ as well we have $s(u)^2 = 1+ u$. Thus, for any nilpotent element $v$ of a ring where $2$ is a unit, we have $s(v)^2 = 1+v$.

Now suppose we are in your situation. We are given $g(t) \in C$ a unit and some $y \in C$ such that $y^2 - g(t) \in N$. We wish to find a square root of $g(t)$ in $C$. But $ g(t)/y^2 - 1 \in N$, so $$s( (g(t)/y^2 - 1))^2 = 1 + (g(t)/y^2 - 1) = g(t)/y^2$$ and thus $$v = y s(g(t)/y^2 - 1)$$ is a square root of $g(t)$ in $C$. Uniqueness follows since if $v$ and $w$ are square roots, then $w^2/y^2 - 1 = g(t)/y^2 - 1 \in N$, so $w/y = s(1 + ((w/y)^2 - 1)) = s(1 + ((v/y)^2 - 1)) = v/y$ as we have an equality $s(1 + (u^2 -1 )) = u$ of formal functions.

This method seems mysterious on the face of it, relying on the Taylor expansion of the square root. I suspect it is more general.

Joshua Mundinger
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