The roots of unity in a tensor product of commutative rings For $i\in\{1,2\}$ let $A_i$ be a commutative ring with unity whose additive group is free and finitely-generated. Assume that $A_i$ is connected in the sense that $0$ and $1$ are unique solutions of the equation $x^2=x$ in $A_i$. Denote by $\mu(A_i)$ the group of roots of unity of $A_i$, i.e., the elements of finite multiplicative order in $A_i$.

Problem. Is the map $$\mu(A_1)\times\mu(A_2)\ \longrightarrow\ \mu(A_1\otimes_{\mathbb Z}A_2),\;\;(u,v)\ \longmapsto\ u\otimes v,$$
  surjective?

(The problem is posed 9.07.2018 by Tristan Tilly from Leiden University on page 25 of Volume 2 of the Lviv Scottish Book). 
Prize for solution: An eternal welcome in Netherlands, and a bottle of liquor of your choice.
 A: $\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$My other answer is nice because it seems to follow a nice line of attack. Now, I start flailing around. To recall, we have reduced to the case of studying $p^k$ roots of unity in $A$, the ring of integers of a Galois extension $K$ of $\QQ$ ramified only over $p$, $\infty$. Recall the notation from the other answer: Let $G = \mathrm{Gal}(K/\QQ)$ and, for $g_1$, $g_2 \in G$ and $\omega = \sum a_j \otimes b_j \in A \otimes A$, set $\omega(g_1, g_2) = \sum g_1(a_j) g_2(b_j)$. The values $\omega(1, g)$ determine the others, since $\omega(g g_1, g g_2) = g \omega(g_1, g_2)$, and the values $\omega(1,g)$ exhibit $A \otimes A$ as a finite index subring of $A^{\oplus |G|}$. So $\zeta \in A \otimes A$ is a $p^k$ root of unity if and only if all $\zeta(1,g)$ are $p^k$-th roots of unity.
First, an observation which doesn't actually seem to help: For $p$ odd, it is enough to consider $p$-th roots of unity. For $p$ odd, let $p^a$ be the highest power of $p$ such that $A$ contains $p^a$-th roots of unity. Since $A \otimes A$ is a subring of $A^{\oplus |G|}$, all $p^k$-th roots of unity in $A \otimes A$ have $k \leq a$. So the group of roots of unity in $A \otimes A$ is $p^a$-torsion, and contains the subgroup $(\ZZ/p^a) \times (\ZZ/p^a)$ generated by the roots of unity in $A \otimes \ZZ$ and $\ZZ \otimes A$. A $p^a$-torsion abelian group containing a $(\ZZ/p^a) \times (\ZZ/p^a)$ subgroup is equal to that subgroup if and only if it only has $p^2$ $p$-torsion points. 
This argument doesn't work when $p=2$, because the $2^a$ roots of unity in $A \otimes \ZZ$ and $\ZZ \otimes A$ only generate the group $\ZZ/2^a \times \ZZ/2^{a-1}$. It isn't clear whether $\zeta_{2^a} \otimes \zeta_{2^a}$ could have a square root in $A \otimes A$.  For $A = \ZZ[\zeta_{2^k}]$, I have an ad hoc argument to rule this out, but not in general.
Now, let's try some special cases and fail. First of all, let $\zeta_p$ be a $p$-th root of unity. Will Sawin computes in comments above that the only $p$-th roots of unity in $\ZZ[\zeta_p] \otimes \ZZ[\zeta_p]$ are $\zeta_p^r \otimes \zeta_p^s$. Recalling that $G = (\ZZ/p)^{\times}$ in this case, this result can be restated that, if $\zeta$ is a $p$-th root of unity in $\ZZ[\zeta_p] \otimes \ZZ[\zeta_p]$ with $\zeta(1,g) = \zeta_p^{w(g)}$, then $w :  (\ZZ/p)^{\times} \to \ZZ/p$ is an affine linear function. This framing will be useful in the following.
Let's suppose that the prime $(1-\zeta_p)$ doesn't ramify in $A$. Choose a prime $\mathfrak{p}$ above $(1-\zeta_p)$ in $A$ and let $H$ be the ramification group of $\mathfrak{p}$. The composite $H \to G \to \mathrm{Gal}(\QQ(\zeta_p)/\QQ) = (\ZZ/p)^{\times}$ is an isomorphism, so $G \cong (\ZZ/p)^{\times} \ltimes V$ where $V = \mathrm{Gal}(K/\QQ(\zeta_p))$.
Let $w$ be a function $G \to \ZZ/p$. We can wonder whether there is some $\zeta \in A \otimes A$ with $\zeta(1,g) = \zeta_p^{w(g)}$. If there is, it is a $p$-th root of unity.
I don't want to write out the details of this, but I believe that such a $\zeta$ exists if and only if, for each coset $g_1^{-1} (\ZZ/p)^{\times} g_2$ in $G$, the function $w(g_1^{-1} c g_2)$ is an affine linear function of $c$. Of course, we may restrict attention to the case that $g_1$ and $g_2$ lie in $V$, absorbing any contribution from $(\ZZ/p)^{\times}$ into $c$. This $w$ will correspond to one of the obvious $p$-th roots of unity if and only if $w$ is the composite of the projection $G \to (\ZZ/p)^{\times}$ with an affine linear map.
We thus have an interesting group theoretic problem. Let $V$ be a group with a $(\ZZ/p)^{\times}$ action and let $G = (\ZZ/p)^{\times} \ltimes V$. Let $w : G \to \ZZ/p$ be a function such that, for every $v_1$, $v_2 \in G$, the function $w(v_1^{-1} c g_2) :  (\ZZ/p)^{\times} \to \ZZ/p$ is affine linear. Does $w$ necessarily factor through the projection on $(\ZZ/p)^{\times}$?
I have found three cases where it does not: Let $V = \ZZ/p$ and let $c \in (\ZZ/p)^{\times}$ act on $V$ by $c^r$ for $r=-1$, $0$ or $1$. Explicitly, we want functions on $\{ (c,d) \in (\ZZ/p)^{\times}\times (\ZZ/p) \}$ such that $w(c, \ d_1+c^r d_2)$ is affine linear in $c$ for all $d_1$, $d_2$. For $r=1$, we can take $w(c,d) = d$; for $r=0$ we can take $w(c,d) = d$; and for $r=-1$, we can take $w(c,d) = cd$. I have not been able to find such a $w$ for other values of $r$, though I have not rigorously ruled it out.
Unfortunately, these values of $r$ cannot occur. If $r=0$, then $K$ is an extension of $\QQ$ with Galois group $(\ZZ/p) \times (\ZZ/p)^{\times}$, ramified only at $\{p, \infty \}$ with ramification group $(\ZZ/p)^{\times}$. But then the fixed field of $(\ZZ/p)^{\times}$ is a degree $p$ unramified extension of $\QQ$, violating Minkowski's theorem.
Proposition 6.16 in Washington's Cyclotomic Fields states that we can't have $r=1$ (this is the statement $A_1=0$); I didn't follow the proof of this.
And by Herbrand's theorem (Theorem 6.17 in Washington, I also don't follow the proof of this), if we had $r=-1$, then $p$ would divide the numerator of the Bernoulli number $B_2 = \tfrac{1}{6}$.
Can anyone make more progress on the group theoretic question? Or at least resolve the case of $A = \ZZ[\zeta_{p^k}]$?
A: $\newcommand\F{\mathbf{F}}$
$\newcommand\Z{\mathbf{Z}}$
$\newcommand\Q{\mathbf{Q}}$
$\newcommand\Gal{\mathrm{Gal}}$
$\newcommand\GL{\mathrm{GL}}$
$\newcommand\SL{\mathrm{SL}}$
Some more examples of groups with corresponding maps $w$:
Let $G = \GL_2(\F_p)$ and $V = \SL_2(\F_p)$, and consider $(\Z/p)^{\times} \subset G$ via the map
$$c \rightarrow \left( \begin{matrix} c & 0 \\ 0 & 1 \end{matrix} \right).$$
Then one can take $w: G \rightarrow \F_p$ to be (say) the $[1,1]$ entry of $G$, because then
$$w \left( \left( \begin{matrix} w & x \\ y & z \end{matrix} \right)
\left( \begin{matrix} c & 0 \\ 0 & 1 \end{matrix} \right)
\left( \begin{matrix} r & s \\ t & u \end{matrix} \right)\right)
= c p w + r x$$
is affine linear in $c$. Hence this gives an admissible $w$. Moreover, for $G$ to occur as a Galois group, we would want a representation:
$$\rho: \Gal(\overline{\Q}/\Q) \rightarrow \GL_2(\F_p)$$
which is unramified outside $p$ with cyclotomic determinant and so that the restriction to inertia was
$$\rho |_{I_p} = \left( \begin{matrix} \varepsilon & 0 \\ 0 & 1 \end{matrix} \right)$$
where $\varepsilon$ is the mod-$p$ cyclotomic character (which gives the canonical identification of $\Gal(\Q_p(\zeta_p)/\Q_p)$ with $(\Z/p)^{\times}$).
Speyer asked in another question whether one could prove that $V_{p-2} = 0$ using anything simpler than Herbrand --- perhaps with the idea that any simple direct negative answer to this question would give a new proof. This example is even worse --- to rule out the existence of such a representation $\rho$, I can't see any simpler argument than using the proof of Serre's conjecture by Khare-Wintenberger (the case $N(\rho) = 1$ and $k(\rho) = 2$). Even worse, if one replaces $\GL_2$ with $\GL_3$ and maps $(\Z/p)^{\times}$ to $G$ via $\mathrm{diag}(c,1,1)$, it becomes an open problem to show that a corresponding extension with Galois group $\GL_3(\F_p)$ and representation $\rho$ does not exist --- although various standard conjectures certainly imply that it does not. This strongly suggests that proving the answer to the original question is "no" will be very hard. So either one should


*

*Try to prove the answer is "conditionally no" by using conjectures in number theory, after more fully understanding the group theoretic condition.

*Try to prove the answer is "yes".

earlier: This answer is basically a response to Speyer (although he omitted this part of the argument, so hopefully it is correct). edit: This doesn't seem to work.
Suppose that $G = \Gal(K/\Q)$, where $K$ is unramified over $\Q(\zeta_p)$. Let $V = \Gal(\Q(\zeta_p)/\Q)$. As noted by Speyer, byconsidering the inertia group, we have that that $G$ is a semi-direct product, so $G = (\Z/p \Z)^{\times} \ltimes V$, and elements of $G$ have the form $(c,g):=(c,0)(0,g)$. Since the genus field of $\Q(\zeta_p)$ is trivial, we certainly have $V = [G,G]$. Let us now assume that
$$H = [G,[G,G]] = [G,V] \ne V.$$
Then we can define a function $w: G \rightarrow \Z/p$ on $G$  as follows:
$$\psi(c,g) = \begin{cases}  c, & g \in H, \\ 0, & g \notin H. \end{cases}$$
For fixed $x$ and $y$ in $V$, 
$$\psi((0,x)(c,0)(0,y^{-1})) = \psi(c,c^{-1} xc y^{-1}) = \psi(c,[c^{-1},x] x y^{-1}).$$
Now $[c^{-1},x] \in [G,V] = H$, and hence
$$\psi((0,x)(c,0)(0,y^{-1}))  = \begin{cases}  c, & xH = yH, \\ 0, & \text{otherwise}.\end{cases}$$
If I understand correctly, this satisfies the properties of $w$ desired by Speyer. 
edit:I guess this doesn't work --- or rather not in an interesting way --- the conditions are never satisfied! They force $G/H$ to be abelian and then $G/H = G/V$.
