ERRATA FOR “CANONICAL SUBGROUPS VIABREUIL-KISIN MODULES”
SHIN HATTORIThe proof of [1, Proposition 4.3 (2)] is incorrect. In page 950 line 1–2,
the author claims that the assertion (2) of the proposition is deduced from
“The uniqueness of the canonical subgroup in Theorem 3.1 (1)”. However,
the truncated Barsotti-Tate group p−1 D/D has Hodge weight p−1 w, while
the subgroup scheme G[p]/D coincides with the Frobenius kernel only onmodulo m⩾1−w
and we cannot apply the uniqueness assertion of Theorem
K3.1 (1). Here we give a correct proof of Proposition 4.3 (2).
Lemma 0.1. Let K/Qp be a finite extension. Let G be a truncated BarsottiTate group over OK of level one, height h, dimension d with 0 < d < h andHodge height w. Let H be a finite flat closed subgroup scheme of G over OKof height d. Put f = deg(G/H). Assume f < 1/2. Then Hdg(G) = f and
H is the canonical subgroup of G.
Proof. Let e be the absolute ramification index of K. Let L, M and N bethe Breuil-Kisin modules associated to G/H, G and H, respectively. We∨e ∨denote the dual of M by M∨ . The k[[u]]-module M∨
1 = M /u M has a∨natural structure of a φ-module induced from that of M and put∨
Fil1 M∨: φ ∗ M∨
1 = Im(1 ⊗ φM∨
1 → M1 ).
1Since G ∨ is of dimension h − d, this is a free k[[u]]/(ue )-module of rank d,and also a direct summand of the k[[u]]-module M∨
1 [1, §2.3]. We write vu
for the u-adic valuation on k[[u]] normalized as vu (u) = 1. By [1, Lemma
2.5 (3)], we have f = e−1 vu (det(φL )). By the definition of the dual L∨ , theimage φL∨ (L∨ ) is contained in ue(1−f ) L∨ .
We have an exact sequence of φ-modules over k[[u]]/(ue(1−f ) )0/ N∨ /ue(1−f ) N∨

/ M∨ /ue(1−f ) M∨

/ L∨ /ue(1−f ) L∨

/0and the Frobenius map of the φ-module L∨ /ue(1−f ) L∨ is zero. Thus theFrobenius map of M∨ /ue(1−f ) M∨ factors through N∨ /ue(1−f ) N∨ , and we
obtain the inclusione(1−f )∨e(1−f ) ∨Fil1 M∨
Fil1 M∨N1 .
1 /u
1 ⊆ N1 /u

Since the both sides are of length de(1 − f ) as k[[u]]-modules, the equalitye(1−f )∨e(1−f ) ∨Fil1 M∨
Fil1 M∨
N11 /u1 = N1 /u

Date: May 2, 2015.
Supported by JSPS KAKENHI Grant Number 26400016.
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2SHIN HATTORI

holds, from which we obtainmin{1 − f, e−1 vu (det(φFil1 M∨ ))} = min{1 − f, e−1 vu (det(φN∨ ))}.
1On the other hand, we havevu (det(φN∨ )) = e deg(H∨ ) = e(d − deg(H)) = e(d − (d − f )) = ef.
By the assumption f < 1/2, we have f < 1 − f and the above equalityimplies
w = Hdg(G) = Hdg(G ∨ ) = e−1 vu (det(φFil1 M∨ )) = f < 1/2.
1

By [1, Theorem 3.1], the truncated Barsotti-Tate groups G and G ∨ have thecanonical subgroups of level one. Note that the canonical subgroup of G ∨is constructed in [1, Theorem 3.1] as the finite flat closed subgroup scheme
C ′ of G ∨ such that the quotient G ∨ /C ′ is associated to the unique lift ofe(1−w) Fil1 M∨ in M∨ as a Breuil-Kisin submodule. This implies
Fil1 M∨
11 /uthat (G/H)∨ is the canonical subgroup of G ∨ . By [1, Theorem 3.1 (b)], wesee that H is the canonical subgroup of G.
□Now we prove [1, Proposition 4.3 (2)]. Put w = Hdg(G). We are assumingw < 1/2. Since we have a generic isomorphism C → G[p]/D, [1, Theorem
3.1 (a)] impliesd − w = deg(C) ≤ deg(G[p]/D) = d − deg(D)and thusdeg((p−1 D/D)/(G[p]/D)) = deg(D) ≤ w < 1/2.
Therefore we can apply Lemma 0.1 and we see that G[p]/D is the canonicalsubgroup of p−1 D/D.
References[1] S. Hattoti: Canonical subgroups via Breuil-Kisin modules, Math. Z. 274 (2013), 933–
953.
E-mail address: shin-h@math.kyushu-u.ac.jp
Faculty of Mathematics, Kyushu University