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.

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    $\begingroup$ I changed the title, but even clearer might be the plainer "The roots of unity in commutative rings". $\endgroup$
    – Todd Trimble
    Commented Oct 23, 2018 at 0:38
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    $\begingroup$ In the case of finite dimensional Q-algebras this seems false: the tensor square of $ K = \mathbb{Q}(\zeta_p)$ is the direct product of $(p-1)$ copies of $K$, so there are $(2p)^{p-1}$ roots of unity in it. So what happens in the case $A_1=A_2=\mathbb{Z}[\zeta_5]$? $\endgroup$ Commented Oct 23, 2018 at 8:45
  • $\begingroup$ @Servaes Limited experiment shows that there are $2p^2$ roots of unity in $\mathbb{Z}[\zeta_p]^{\otimes 2}$ for small $p$, so the map is surjective in this case. $\endgroup$ Commented Oct 23, 2018 at 17:55
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    $\begingroup$ @FrançoisBrunault We can prove this claim explicitly. First note that the only relevant roots of unity are of order $2p$. If $\alpha$ and $\beta$ are the two copies of $\zeta_p-1$ from the two rings, then in $\mathbb Z[\alpha, \beta]/ (p, \alpha^2, \alpha \beta, \beta^2)$, a element can only be a $2p$th root of unity if it is congruent to $\pm 1$ mod $(\alpha,\beta)$. There are $2p^2$ of these (and the map from the roots of unity of the two copies to them is surjective) so it suffices to show that each has a unique lift to a root of unity, or just that $1$ has a unique lift. $\endgroup$
    – Will Sawin
    Commented Oct 23, 2018 at 19:49
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    $\begingroup$ @FrançoisBrunault But if we look at the image in $\mathbb Q(\mathbb \zeta_p)^{p-1}$, or in one particularly copy, we see anything congruent to $1$ mod this ideal is congruent to $1$ mod $(\mathbb \zeta_p-1)^2$, and there is one root of unity of this type. $\endgroup$
    – Will Sawin
    Commented Oct 23, 2018 at 19:50

3 Answers 3


$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\Spec{\mathrm{Spec}}\def\fp{\mathfrak{p}}\def\fq{\mathfrak{q}}\def\FF{\mathbb{F}}$This is one of two answers recording my partial progress. The point of this answer is to show that it is enough to resolve the question for the case where $A_1=A_2$ and both are the ring of integers of a number field $K$, with $K$ Galois over $\QQ$ and ramified only at $p$ and $\infty$ (for some prime $p$). The other answer will contain my partial progress on that problem.

We begin with the following observation: Let $A_1 \subseteq B_1$ and $A_2 \subseteq B_2$ be free $\ZZ$-modules. Let $\zeta$ be an element of $A_1 \otimes A_2$ which factors as $b_1 \otimes b_2$ in $B_1 \otimes B_2$, and suppose that $b_1$ and $b_2$ are not divisible (in $B_1$ and $B_2$) by any integer $k \geq 2$. Then $b_j \in A_j$. Proof: Let $\pi_j$ be the projection $B_j \to B_j/A_j$. Since $\zeta \in A_1 \otimes A_2$, we have $(\pi_1 \otimes \mathrm{Id})(\zeta) =0$. So $\pi_1(b_1) \otimes b_2=0$ in $(B_1 / A_1) \otimes B_2$. Since $B_2$ is free over $\mathbb{Z}$ and $b_2$ is not divisible by any integer $\geq 2$, this means that $\pi_1(b_1)=0$, so $b_1 \in A_1$. Similarly, $b_2 \in A_2$. $\square$.

Also, notice that, in a ring finite over $\mathbb{Z}$, a root of unity can never have a nontrivial integer divisor, so when $B_1$ and $B_2$ are rings and $b_1$ and $b_2$ are roots of unity, that hypothesis is free. Thus, we deduce a key trick: If we can prove the theorem for a pair of rings $B_1$, $B_2$, we automatically deduce it for any subrings $A_1$, $A_2$.

Let $R_j$ be the radical of $A_j$ and let $S_j = A_j/R_j$. Then $S_j \otimes \QQ$ is a direct sum of fields. From the theory of Artin algebras over a field, we can split the exact sequence $0 \to (R_j \otimes \QQ) \to (A_j \otimes \QQ) \to (S_j \otimes \QQ) \to 0$ such that $S_j \otimes \QQ$ becomes a subring of $A_j \otimes \QQ$; write $\rho_j$ for the retraction $A_j \otimes \QQ \to R_j \otimes \QQ$. Let $\hat{R}_j$ be the $A_j$-submodule of $R_j \otimes \QQ$ generated by $\rho_j(A_j)$; note that this is still a finite $\ZZ$-module since $A_j$ and $\rho_j(A_j)$ are. So $A_j \subseteq \hat{R}_j \oplus S_j$ and it suffices to prove the result for $\hat{R}_j \oplus D_j$. We have reduced to the case that the radical of $S_j$ breaks off as a direct summand.

Let $A$ be a finite free commutative $\ZZ$-algebra whose radical splits as a direct summand, say $A = S \oplus R$. We claim that all roots of unity of $A$ are in $S$. Proof: Recall that $R$ is nilpotent. Set $R(k) = \QQ R^k \cap R$, where the intersection takes place in $\QQ \otimes R$. So $R(k)=0$ for $k$ sufficiently large. Let $s+r$ be a root of unity of $A$ with $s \in S$ and $r \in R$. Let $(s+r)^n=1$. Since the projection $A \to D$ is a map of rings, we have $s^n=1$. We deduce that $(1+s^{-1} r)^n=1$ as well. But then we have $1+n s^{-1} r \equiv 1 \bmod r^2$ so $r \equiv 0 \bmod r^2$ in $A \otimes \QQ$. This shows by induction that $r \in R^(2^k)$ for every $k$, so $r \in \bigcap R(2^k) = \{ 0 \}$. $\square$

We now reduce to the case that $A_1$ and $A_2$ have no radical. Let $A_j = S_j \oplus R_j$. Then $A_1 \otimes A_2$ splits as $(S_1 \otimes S_2) \oplus (S_1 \otimes R_2 \oplus R_1 \otimes S_2 \oplus R_1 \otimes R_2)$, with $S_1 \otimes S_2$ semisimple and $S_1 \otimes R_2 \oplus R_1 \otimes S_2 \oplus R_1 \otimes R_2$ the radical. So the result of the previous paragraph shows that all roots of unity in $A_1 \otimes A_2$ are in $S_1 \otimes S_2$, so it is enough to prove the result for $S_1$ and $S_2$. We have reduced to the case that $A_1$ and $A_2$ have no radical.

So $A_1$ injects into a direct sum of domains, $A_1 \subseteq \bigoplus D_1^i$ and likewise $A_2 \subseteq \bigoplus D_2^k$. Our goal in the next several steps is to reduce to the case where $A_1$ and $A_2$ are each domains. So assume that we know the claim for each tensor product $D_1^i \otimes D_2^{k}$. For each case where $\Spec D_1^i$ and $\Spec D_1^j$ meet in $\Spec A_1$, choose a pair of primes $\fp_1^{ij} \in \Spec D_1^i$ and $\fp_1^{ji} \in \Spec D_1^j$ where they meet. I'll write an element of $\bigoplus D_1^i$ and $(a(1), a(2), \dots, a(r))$ with $a(i) \in D_1^i$, so we have $a(i) \bmod \fp_1^{ij} = a(j) \bmod \fp_1^{ji}$. We may and do replace $A_1$ by the larger subring of $\bigoplus D_1^i$ given by these congruences, and do the same thing for $A_2$.

We can write an element of $A_1 \otimes A_2$ as a matrix $\left[ z(i,k) \right]$ with $z(i,k) \in D_1^i \otimes D_2^k$. Note that we have $z(i,k) \bmod \fp_1^{ij} \otimes D_2^k = z(j,k) \bmod \fp_1^{ji} \otimes D_2^k$ and likewise for $z(i,k)$ and $z(i,\ell)$. This element will be a root of unity iff each $z(i,k)$ is a root of unity; by hypothesis, this occurs only if $z(i,k)$ is of the form $\alpha(i,k) \otimes \beta(k,i)$ for $\alpha(i,k)$ a root of unity in $D_1^i$ and $\beta(k,i)$ a root of unity in $D_2^k$. So $\alpha(i,k) \otimes \beta(k,i) \bmod \fp_1^{ij} \otimes D_2^k = \alpha(j,k) \otimes \beta(k,j) \bmod \fp_1^{ji} \otimes D_2^k$. This forces $\beta(k,i) = \pm \beta(k,j)$. Likewise, $\alpha(i,k) = \pm \alpha(i,\ell)$. So there are roots of unity $\alpha(i) \in D_1^i$ and $\beta(k) \in D_2^k$ such that $z(i,k) = \pm \alpha(i) \otimes \beta(k)$. Moreover $\alpha(i) \bmod \fp_1^{ij} = \alpha(j) \bmod \fp_1^{ji}$, so $\alpha:= (\alpha(1), \ldots,\alpha(r))$ is a root of unity in $A_1$. Likewise, we construct a root of unity $\beta = (\beta(1), \ldots, \beta(s))$ in $A_2$. Dividing by $\alpha \otimes \beta$, we are reduced to considering roots of unity of the form $z(i,k) = \pm 1$. We want to show such a matrix factors as a tensor product, meaning that it is rank $1$. It is enough to show that each $2 \times 2$ matrix $\left[ \begin{smallmatrix} z(i,k) & z(i,\ell) \\ z(j,k) & z(j, \ell) \\ \end{smallmatrix} \right]$ has rank $1$. If $D_1^i/\fp_1^{ij}$ has odd characteristic, then we deduce that $z(i,k) = z(j,k)$ and $z(i,\ell) = z(j,\ell)$ and we are done. We are similarly done if $D_2^k/\fp_2^{k\ell}$ has odd characteristic. So we may assume instead these quotient fields have even characteristic.

At this point, an example may help. Let $R = \{ (a(1), a(2)) \subset \ZZ^{\oplus 2} : a(1) \equiv a(2) \bmod 2 \}$, and let's understand the square roots of $1$ in $R \otimes_{\ZZ} R$. So $R \otimes_{\ZZ} R$ can be identified with $2 \times 2$ matrices of integers $\left[ \begin{smallmatrix} z(1,1) & z(1,2) \\ z(2,1) & z(2,2) \\ \end{smallmatrix} \right]$, namely, those which are $\ZZ$-linear combinations of $\left[ \begin{smallmatrix} a(1) b(1) & a(1) b(2) \\ a(2) b(1) & a(2) b(2)\\ \end{smallmatrix} \right]$ with $a(1) \equiv a(2) \bmod 2$ and $b(1) \equiv b(2) \bmod 2$. We observe that every such matrix has $z(1,1) - z(1,2) - z(2,1) + z(2,2) \equiv 0 \bmod 4$. So $\left[ \begin{smallmatrix} 1 & 1 \\ 1 &-1 \\ \end{smallmatrix} \right]$ is not in $R \otimes_{\ZZ} R$. We want to generalize this computation to other rings.

As an abelian group, we have $D_1^i/\fp_1^{ij} \cong (\ZZ/2)^{\oplus r}$ for some $r$. So we can choose an integer basis $e_1$, $e_2$, ... , $e_s$ of $D_1^i$ such that $2 e_1$, $2 e_2$, ... $2 e_r$, $e_{r+1}$, ..., $e_s$ is an integer basis of $\fp_1^{ij}$; further more, we may take $e_1 = 1$. Let $\lambda_i : D_1^i \to \ZZ$ be projection onto the coefficient of $e_1$. So, for $a \in A_1$, we have $\lambda_i(a(i)) \equiv \lambda_j(a(j)) \bmod 2$. Likewise, we have maps $\mu_k : D_2^k \to \ZZ$ such that $\mu_k(b(k)) \equiv \mu_{\ell}(b(\ell)) \bmod 2$. Then, for any $a \in A_1$ and $b \in A_2$, we have $\lambda_i(a(i)) \mu_k(b(k)) - \lambda_i(a(i)) \mu_{\ell}(b(\ell)) - \lambda_j(a(j)) \mu_k(b(k)) + \lambda_j(a(j)) \mu_{\ell}(b(\ell)) = ( \lambda_i(a(i)) - \lambda_j(a(j)) ) ( \mu_k(b(k)) - \mu_{\ell}(b(\ell)) ) \equiv 0 \bmod 4$.

Thus, for any matrix $\left[ \begin{smallmatrix} z(1,1) & z(1,2) \\ z(2,1) & z(2,2) \\ \end{smallmatrix} \right]$ in $A_1 \otimes A_2$, we have $(\lambda_i \otimes \mu_k) z(i,k) - (\lambda_i \otimes \mu_{\ell}) z(i,\ell) - (\lambda_j \otimes \mu_k) z(j,k) + (\lambda_j \otimes \mu_{\ell}) z(j,\ell) \equiv 0 \bmod 4$. If all the $z(i,k)$ are $\pm 1$, then we deduce that $z(i,k) - z(j,k) - z(i,\ell) + z(k,\ell) \equiv 0 \bmod 4$. But by checking all $16$ matrices of $\pm 1$'s, this forces the matrix to be rank $1$. We have shown that, if the result holds for all $D_1^i \otimes D_2^k$, then it holds for $A_1 \otimes A_2$. We have reduced to the case that $A_1$ and $A_2$ are domains.

So, $A_1$ and $A_2$ are domains and, being finite $\ZZ$-algebras, they are orders in some number fields $K_1$ and $K_2$. Let $K$ be a Galois number field containing $K_1$ and $K_2$, and replace $A_1$ and $A_2$ by their rings of integers in $K$. Thus, we have reduced to the case that $A_1=A_2=:A$ is the ring of integers in some Galois number field $K$.

At this point, it is convenient to set up some notation. Let $G = \mathrm{Gal}(K/\QQ)$. For $\omega = \sum a_j \otimes b_j \in A \otimes A$ and $g_1$, $g_2 \in G$, put $\omega(g_1, g_2) = \sum g_1(a_j) g_2(b_j)$. It is well known that $\omega$ is determined by the $|G|^2$ values $\omega(g_1, g_2)$ and, since $\omega(g g_1, g g_2) = g \omega(g_1, g_2)$, it is in fact determined by the $|G|$ values $\omega(1,g_2)$. In this manner, $A \otimes A$ is identified with a subring of $A^{\oplus |G|}$.

Now, the set of roots of unity in $A \otimes A$ is a finite abelian group. If $\Gamma_1 \subseteq \Gamma_2$ are finite abelian groups, and their $p$-parts are equal for all $p$, then $\Gamma_1 = \Gamma_2$. So it is enough to show (for all $p$) that the $p^k$ roots of unity in $A \otimes A$ are of the form $\zeta_1 \otimes \zeta_2$ for $\zeta_1$ and $\zeta_2$ $p^k$-th roots of unity in $A$. Fix such a $p$.

We will now reduce to the case that $p$ is the only finite prime of $\QQ$ ramified in $K$. Suppose that $q$ is some other ramified prime, let $\mathfrak{q}$ be a prime of $A$ above $q$ and let $H$ be the ramification group. For any $\omega \in A \otimes A$, we have $\omega(h g_1, g_2) \equiv \omega(g_1, g_2) \bmod \mathfrak{q}$. Now, suppose that $\zeta$ is a $p^k$-root of unity in $A \otimes A$, so every $\zeta(g_1, g_2)$ is a $p^k$-root of unity. The $p^k$ roots of unity stay distinct modulo $\mathfrak{q}$. So we deduce that $\zeta(h_1 g_1, h_2 g_2) = \zeta(g_1, g_2)$ for all $h_1$, $h_2 \in H$. Since $\zeta$ is determined by the $\zeta(g_1, g_2)$, we deduce that $\zeta$ is invariant under $H \times H$. For a free $\ZZ$-module $A$ with an action of a finite group $H$, we have $(A \otimes A)^{H \times H} = (A^H) \otimes (A^H)$. So we may replace $K$ by the subfield $K^H$ and replace $A$ by the ring of integers in $K^H$. Repeating this argument for each of the primes above $q$, we shrink $K$ down to the minimal subextension unramified at $q$. Running the same argument for every finite prime $q \neq p$, we have reduced to the case that $K$ is only ramified above $\{ p, \infty \}$.


$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$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 $A = \ZZ[\zeta_{p^k}]$?

  • 2
    $\begingroup$ I have now confused myself concerning your condition in the case when $V$ is not abelian. Does one require that $w(gch)$ is affine linear for all $(g,h)$ and ONE section from $(\mathbf{Z}/p)^{\times}$ to $G$ or for ALL sections? I am thinking of the case $G = \mathrm{GL}_2(\mathbf{F}_p)$ and $V = \mathrm{SL}_2(\mathbf{F}_p)$ and $w$ is the just (say) the first entry. Then this works for the section which takes $c \in (\mathbf{Z}/p)^{\times}$ to $\mathrm{diag}(c,1)$ but not $\mathrm{diag}(c^2,c^{-1})$. $\endgroup$
    – user131093
    Commented Nov 19, 2018 at 3:10
  • $\begingroup$ @user131093 Good point. It should be the splitting coming from the inertia subgroup. $\endgroup$ Commented Nov 19, 2018 at 17:17

$\newcommand\F{\mathbf{F}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\Gal{\mathrm{Gal}}$ $\newcommand\GL{\mathrm{GL}}$ $\newcommand\SL{\mathrm{SL}}$

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

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

  2. 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 $G/H = G/V$.

  • $\begingroup$ Wow! That is quite a computation. I'll try to get my argument for this part written up tonight or tomorrow night. $\endgroup$ Commented Nov 18, 2018 at 3:38
  • $\begingroup$ I am confused. I see that $[V,V]$, in your example, is the order 3 subgroup you describe. But it looks to me like $[G,V] = V$, which makes your example trivial. $\endgroup$ Commented Nov 18, 2018 at 4:14
  • $\begingroup$ I agree that in this case, $[G,G]=[G,[G,G]]$ (a normal subgroup of order 27). $\endgroup$
    – YCor
    Commented Nov 18, 2018 at 4:32
  • $\begingroup$ You are (both) completely correct --- my computations on scribble paper on the bus were nonsense $\endgroup$
    – user131093
    Commented Nov 18, 2018 at 5:23

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