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Erratum for "Tropical superelliptic curves"

Erratum for "Tropical superelliptic curves" 1First ErratumIn [1] on page 532, in the proof of Proposition 3.12(2), the product of two algebras is assumed to give the base change of the two algebras. The authors cannot directly say whether this is true here. We note however that Proposition 3.12(2) is a consequence of the main theorems on strict Henselizations (or étale local rings) as described in [2, Section 0BSK]. Indeed, the strict Henselization Ash is a filtered colimit of étale algebras and the given algebra BH is étale at q ∩ BH, so we obtain an injection of the fraction field of BH into the fraction field of the strict Henselization Ksh. The statement then follows from infinite Galois theory and the fact that the Galois group of Ksh ⊂ Ksep is the absolute inertia group. More details can be found in Section 3.2Second ErratumOn page 538, in the statement of Lemma 4.2, the formula for the order of the decomposition group in this generality is wrong. It is correct however if the inertia group is trivial, which is exactly the case we need in the paper. The proof in that case is moreover as published.3Details for the First ErratumWe give some extra details here for the proof of Proposition 3.12(2). To prove the inclusion H ⊇ Iq, we use the material in [2, Section 0BSK] on strict Henselizations. We first note that we can localize A and assume that it is local with maximal ideal p = q ∩ A. The strict Henselization of the triple (A, p, k(p)) is then given by A s h := lim S , p S , α S, $\begin{equation} A^{\mathrm{sh}}:=\lim _{\left(S, \mathfrak{p}_S, \alpha\right)} S, \end{equation}$where the limit is taken over the filtered category of all triples (S, pS , α), where A → S is an étale ring map, pS is a prime ideal in S lying above p and α : k(pS)→ k(p)sep is an embedding of residue fields, see [2, Lemma 04GN]. Now consider the integral closure Asep of A in the separable closure Ksep. We choose a prime ideal qsep in Asep lying over q ∈ B. The localization A q s e p s e p $A_{q^{\mathrm{sep}}}^{\mathrm{sep}}$is then a strictly Henselian ring, and this induces an inclusion A sh  → A q sep  sep  , $A^{\text {sh }} \rightarrow A_{q^{\text {sep }}}^{\text {sep }},$see the beginning of [2, Section 0BSD]. As in the proof of [2, Lemma 0BSW], we then have(1)A sh  = A q sep  sep  I q q  sep  , $$ \begin{equation} A^{\text {sh }}=\left(A_{\mathfrak{q}^{\text {sep }}}^{\text {sep }}\right)^{I_{q^q} \text { sep }}, \end{equation}$$where I q sep  $I_{q^{\text {sep }}}$is the absolute inertia group.We now relate this material to the subgroups H ⊂ G. In terms of infinite Galois theory, H corresponds to a closed (and open) subgroup G L ⊂ H ¯ ⊂ G K $G_L \subset \bar{H} \subset G_K$such that H ¯ / G L = H . $\bar{H} / G_L=H.$Similarly, we have a natural surjective map(2)I q sep  → I q $$ \begin{equation} I_{q^{\text {sep }}} \rightarrow I_{\mathfrak{q}} \end{equation}$$obtained by restricting automorphisms to L, see [2, Lemma0BSX]. The kernel of this map is G L ∩ I q sep  , $G_L \cap I_{q^{\text {sep }}},$inducing the isomorphism I q = I q sep  / G L ∩ I q sep  . $I_{\mathrm{q}}=I_{\mathrm{q}^{\text {sep }}} /\left(G_L \cap I_{\mathrm{q}^{\text {sep }}}\right).$By our assumption on H, we have that BH ⊃ A is étale at qH := q ∩ BH. This means that there exists a g ∈ BH\qH such that the localization A → (BH)g is étale. After choosing an embedding k(qH)→ k(psep), we then obtain maps(3)A → B H g → A sh  = A q sep  sep  I q  sep  . $$A \rightarrow\left(B^H\right)_g \rightarrow A^{\text {sh }}=\left(A_{q^{\text {sep }}}^{\text {sep }}\right)^{I_{\mathrm{q}} \text { sep }}.$$These are easily seen to be injective. Since the fraction field of (BH)g is the invariant field of H ¯ , $\bar{H},$we have H ¯ ⊃ I q sep  $\bar{H} \supset I_{q^{\text {sep }}}$by infinite Galois theory. We then obtain an inclusion(4)I q = I q  sep  / G L ∩ I q sep  ⊆ H ¯ / G L = H , $$ \begin{equation} I_{\mathfrak{q}}=I_{\mathfrak{q}} \text { sep } /\left(G_L \cap I_{q^{\text {sep }}}\right) \subseteq \bar{H} / G_L=H, \end{equation}$$as desired. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Advances in Geometry de Gruyter

Erratum for "Tropical superelliptic curves"

Advances in Geometry , Volume 23 (1): 2 – Jan 1, 2023

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Publisher
de Gruyter
Copyright
© 2023 Madeline Brandt and Paul Alexander Helminck, published by De Gruyter
eISSN
1615-715X
DOI
10.1515/advgeom-2022-0029
Publisher site
See Article on Publisher Site

Abstract

1First ErratumIn [1] on page 532, in the proof of Proposition 3.12(2), the product of two algebras is assumed to give the base change of the two algebras. The authors cannot directly say whether this is true here. We note however that Proposition 3.12(2) is a consequence of the main theorems on strict Henselizations (or étale local rings) as described in [2, Section 0BSK]. Indeed, the strict Henselization Ash is a filtered colimit of étale algebras and the given algebra BH is étale at q ∩ BH, so we obtain an injection of the fraction field of BH into the fraction field of the strict Henselization Ksh. The statement then follows from infinite Galois theory and the fact that the Galois group of Ksh ⊂ Ksep is the absolute inertia group. More details can be found in Section 3.2Second ErratumOn page 538, in the statement of Lemma 4.2, the formula for the order of the decomposition group in this generality is wrong. It is correct however if the inertia group is trivial, which is exactly the case we need in the paper. The proof in that case is moreover as published.3Details for the First ErratumWe give some extra details here for the proof of Proposition 3.12(2). To prove the inclusion H ⊇ Iq, we use the material in [2, Section 0BSK] on strict Henselizations. We first note that we can localize A and assume that it is local with maximal ideal p = q ∩ A. The strict Henselization of the triple (A, p, k(p)) is then given by A s h := lim S , p S , α S, $\begin{equation} A^{\mathrm{sh}}:=\lim _{\left(S, \mathfrak{p}_S, \alpha\right)} S, \end{equation}$where the limit is taken over the filtered category of all triples (S, pS , α), where A → S is an étale ring map, pS is a prime ideal in S lying above p and α : k(pS)→ k(p)sep is an embedding of residue fields, see [2, Lemma 04GN]. Now consider the integral closure Asep of A in the separable closure Ksep. We choose a prime ideal qsep in Asep lying over q ∈ B. The localization A q s e p s e p $A_{q^{\mathrm{sep}}}^{\mathrm{sep}}$is then a strictly Henselian ring, and this induces an inclusion A sh  → A q sep  sep  , $A^{\text {sh }} \rightarrow A_{q^{\text {sep }}}^{\text {sep }},$see the beginning of [2, Section 0BSD]. As in the proof of [2, Lemma 0BSW], we then have(1)A sh  = A q sep  sep  I q q  sep  , $$ \begin{equation} A^{\text {sh }}=\left(A_{\mathfrak{q}^{\text {sep }}}^{\text {sep }}\right)^{I_{q^q} \text { sep }}, \end{equation}$$where I q sep  $I_{q^{\text {sep }}}$is the absolute inertia group.We now relate this material to the subgroups H ⊂ G. In terms of infinite Galois theory, H corresponds to a closed (and open) subgroup G L ⊂ H ¯ ⊂ G K $G_L \subset \bar{H} \subset G_K$such that H ¯ / G L = H . $\bar{H} / G_L=H.$Similarly, we have a natural surjective map(2)I q sep  → I q $$ \begin{equation} I_{q^{\text {sep }}} \rightarrow I_{\mathfrak{q}} \end{equation}$$obtained by restricting automorphisms to L, see [2, Lemma0BSX]. The kernel of this map is G L ∩ I q sep  , $G_L \cap I_{q^{\text {sep }}},$inducing the isomorphism I q = I q sep  / G L ∩ I q sep  . $I_{\mathrm{q}}=I_{\mathrm{q}^{\text {sep }}} /\left(G_L \cap I_{\mathrm{q}^{\text {sep }}}\right).$By our assumption on H, we have that BH ⊃ A is étale at qH := q ∩ BH. This means that there exists a g ∈ BH\qH such that the localization A → (BH)g is étale. After choosing an embedding k(qH)→ k(psep), we then obtain maps(3)A → B H g → A sh  = A q sep  sep  I q  sep  . $$A \rightarrow\left(B^H\right)_g \rightarrow A^{\text {sh }}=\left(A_{q^{\text {sep }}}^{\text {sep }}\right)^{I_{\mathrm{q}} \text { sep }}.$$These are easily seen to be injective. Since the fraction field of (BH)g is the invariant field of H ¯ , $\bar{H},$we have H ¯ ⊃ I q sep  $\bar{H} \supset I_{q^{\text {sep }}}$by infinite Galois theory. We then obtain an inclusion(4)I q = I q  sep  / G L ∩ I q sep  ⊆ H ¯ / G L = H , $$ \begin{equation} I_{\mathfrak{q}}=I_{\mathfrak{q}} \text { sep } /\left(G_L \cap I_{q^{\text {sep }}}\right) \subseteq \bar{H} / G_L=H, \end{equation}$$as desired.

Journal

Advances in Geometryde Gruyter

Published: Jan 1, 2023

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