Equivariant Iwasawa theory for elliptic curves

Abstract

We discuss abelian equivariant Iwasawa theory for elliptic curves over ${\mathbb {Q}}$ at good supersingular primes and non-anomalous good ordinary primes. Using Kobayashi’s method, we construct equivariant Coleman maps, which send the Beilinson–Kato element to the equivariant p-adic L-functions. Then we propose equivariant main conjectures and, under certain assumptions, prove one divisibility via Euler system machinery. As an application, we prove a case of a conjecture of Mazur–Tate.

Appendix: Properties of Fitting ideals

In this appendix, we collect auxiliary propositions about Fitting ideals, using the results in [18].

Let G be a finite abelian group and put $\Lambda = {\mathbb {Z}}_p[[T]], {\mathcal {R}}= {\mathbb {Z}}_p[G][[T]]$. We denote by ${\mathcal {M}}$ the category of finitely generated torsion ${\mathcal {R}}$-modules, where torsionness means being torsion as a $\Lambda$-module. Let ${\mathcal {P}}$ be the subcategory of ${\mathcal {M}}$ consisting of module with $\mathrm{pd}_{{\mathcal {R}}} \le 1$. Let ${\mathcal {C}}$ be the subcategory of ${\mathcal {M}}$ consisting of modules which does not have nonzero finite submodules. It is well-known (e.g. [31, Proposition (5.3.19)(i)]) that a finitely generated ${\mathcal {R}}$-module X does not contain a non-trivial finite submodule if and only if $\mathrm{pd}_{\Lambda }(X) \le 1$. Therefore ${\mathcal {P}}\subset {\mathcal {C}}$, meaning that any object of ${\mathcal {P}}$ is an object of ${\mathcal {C}}$.

Definition A.1

Let $\Omega$ be a commutative monoid.

• (1) A map ${\mathcal {F}}: {\mathcal {M}}\rightarrow \Omega$ is called a Fitting invariant if the following conditions hold:

  • If $P \in {\mathcal {P}}$, then ${\mathcal {F}}(P) \in \Omega$ is invertible.

  • If $P \in {\mathcal {P}}$ and $X, X' \in {\mathcal {M}}$ fit into an exact sequence $0 \rightarrow X' \rightarrow X \rightarrow P \rightarrow 0$, then ${\mathcal {F}}(X) = {\mathcal {F}}(P){\mathcal {F}}(X')$.

• (2) A map ${\mathcal {F}}: {\mathcal {C}}\rightarrow \Omega$ is called a quasi-Fitting invariant if the following conditions hold:

  • If $P \in {\mathcal {P}}$, then ${\mathcal {F}}(P) \in \Omega$ is invertible.

  • If $P \in {\mathcal {P}}$ and $X, X' \in {\mathcal {C}}$ fit into an exact sequence $0 \rightarrow X' \rightarrow X \rightarrow P \rightarrow 0$, then ${\mathcal {F}}(X) = {\mathcal {F}}(P){\mathcal {F}}(X')$.

  • If $P \in {\mathcal {P}}$ and $X, X' \in {\mathcal {C}}$ fit into an exact sequence $0 \rightarrow P \rightarrow X \rightarrow X' \rightarrow 0$, then ${\mathcal {F}}(X) = {\mathcal {F}}(P){\mathcal {F}}(X')$.

This definition of Fitting invariants (resp. quasi-Fitting invariants) is given in [18, Definition 2.4] (resp. [18, Definition 3.16]). A fundamental example is the following.

Proposition A.2

Let $\Omega$ be the monoid of fractional ideals of ${\mathcal {R}}$ and we put ${\mathcal {F}}(X) = {{\,\mathrm{Fitt}\,}}_{{\mathcal {R}}}(X)$, the Fitting ideal of X. Then ${\mathcal {F}}: {\mathcal {M}}\rightarrow \Omega$ is a Fitting invariant. Moreover, the restriction ${\mathcal {F}}|_{{\mathcal {C}}}: {\mathcal {C}}\rightarrow \Omega$ is a quasi-Fitting invariant.

Proof

The first assertion is proved in [18, Proposition 2.7]. Then the second follows, since [18, Proposition 3.17] proves that, in our present situation, any Fitting invariant gives rise to a quasi-Fitting invariant by restriction to ${\mathcal {C}}$. $\square$

Next we state the definition of shifts of (quasi-)Fitting invariants, introduced in [18].

Theorem A.3

([18, Theorem 2.6]) Let ${\mathcal {F}}: {\mathcal {M}}\rightarrow \Omega$ be a Fitting invariant. For $X \in {\mathcal {M}}$ and $n \ge 0$, the following ${\mathcal {F}}^{[n]}(X) \in \Omega$ is well-defined. Take an exact sequence $0 \rightarrow Y \rightarrow P_1 \rightarrow \dots \rightarrow P_n \rightarrow X \rightarrow 0$ in ${\mathcal {M}}$ with $P_i \in {\mathcal {P}}$ for $1 \le i \le n$. Then define

$$\begin{aligned} {\mathcal {F}}^{[n]}(X) = \left( \prod _{i=1}^n {\mathcal {F}}(P_i)^{(-1)^i} \right) {\mathcal {F}}(Y). \end{aligned}$$

Theorem A.4

([18, Corollary 3.21]) Let ${\mathcal {F}}: {\mathcal {C}}\rightarrow \Omega$ be a quasi-Fitting invariant. Then there exists a unique family $\{{\mathcal {F}}^{\langle n \rangle }: {\mathcal {M}}\rightarrow \Omega \}_{n \in {\mathbb {Z}}}$ of maps satisfying the following. Firstly, ${\mathcal {F}}^{\langle 0 \rangle }: {\mathcal {M}}\rightarrow \Omega$ is an extension of ${\mathcal {F}}: {\mathcal {C}}\rightarrow \Omega$. Secondly, if $0 \rightarrow Y \rightarrow P_1 \rightarrow \dots \rightarrow P_d \rightarrow X \rightarrow 0$ is an exact sequence in ${\mathcal {M}}$ with $P_i \in {\mathcal {P}}$ for $1 \le i \le d$, then

$$\begin{aligned} {\mathcal {F}}^{\langle n \rangle }(X) = \left( \prod _{i=1}^{d} {\mathcal {F}}(P_i)^{(-1)^{n-d+i}} \right) {\mathcal {F}}^{\langle n-d \rangle }(Y). \end{aligned}$$

The rest of this section is devoted to algebraic propositions which we use in Sect. 7.

Definition A.5

Let ${\mathcal {I}}, {\mathcal {J}}$ be fractional ideals of ${\mathcal {R}}$. If ${\mathcal {I}}\subset {\mathcal {J}}$ and moreover the quotient ${\mathcal {J}}/{\mathcal {I}}$ is finite, then we write ${\mathcal {I}}\subset _{{{\,\mathrm{fin}\,}}} {\mathcal {J}}$ and ${\mathcal {J}}\supset _{{{\,\mathrm{fin}\,}}} {\mathcal {I}}$.

Lemma A.6

Put ${\mathcal {F}}= {{\,\mathrm{Fitt}\,}}_{{\mathcal {R}}}$. For $X \in {\mathcal {M}}$ and a finite submodule $X'$ of X, we have ${\mathcal {F}}(X) \subset _{{{\,\mathrm{fin}\,}}} {\mathcal {F}}(X/X')$.

Proof

It is well-known that

$$\begin{aligned} {\mathcal {F}}(X/X') {\mathcal {F}}(X') \subset {\mathcal {F}}(X) \subset {\mathcal {F}}(X/X') \end{aligned}$$

in this case. Since ${\mathcal {F}}(X') \subset _{{{\,\mathrm{fin}\,}}} {\mathcal {R}}$, the lemma follows. $\square$

Definition A.7

For an ${\mathcal {R}}$-module X, we define $E^i(X) = {{\,\mathrm{Ext}\,}}_{{\mathcal {R}}}^i(X, {\mathcal {R}})$ for $i \ge 0$. Since ${{\,\mathrm{Hom}\,}}_{{\mathcal {R}}}(X, {\mathcal {R}})$ is an ${\mathcal {R}}$-module by $(af)(x) = a f(x)$ for $a \in {\mathcal {R}}, f \in {{\,\mathrm{Hom}\,}}_{{\mathcal {R}}}(X, {\mathcal {R}})$, and $x \in {\mathcal {R}}$, the derived functor $E^i(X)$ also admits an ${\mathcal {R}}$-module structure.

Proposition A.8

Put ${\mathcal {F}}= {{\,\mathrm{Fitt}\,}}_{{\mathcal {R}}}$. Let $0 \rightarrow X \rightarrow P_1 \rightarrow P_2 \rightarrow Y \rightarrow 0$ be an exact sequence in ${\mathcal {M}}$. Suppose that $P_i \in {\mathcal {P}}$ holds for $i = 1,2$. Then

$$\begin{aligned} {\mathcal {F}}(P_1){\mathcal {F}}(Y) \subset _{{{\,\mathrm{fin}\,}}} {\mathcal {F}}(P_2){\mathcal {F}}(E^1(X)). \end{aligned}$$

Proof

Note that this is an equality if $Y \in {\mathcal {C}}$ (see [2, Lemma 5] or [18, Proposition 4.7]). To treat the case where $Y \not \in {\mathcal {C}}$, we modify the argument of [18].

Define ${\mathcal {F}}^*: {\mathcal {C}}\rightarrow \Omega$ by ${\mathcal {F}}^*(X) = {\mathcal {F}}(E^1(X))$ for $X \in {\mathcal {C}}$. Then ${\mathcal {F}}^*$ is again a quasi-Fitting invariant by the duality properties of $E^1$ on ${\mathcal {C}}$ (see [18, Proposition 3.11]). Hence Theorem A.4 yields maps $({\mathcal {F}}^*)^{ \langle n \rangle }: {\mathcal {M}}\rightarrow \Omega$, which satisfies

$$\begin{aligned} ({\mathcal {F}}^*)^{ \langle 2 \rangle }(Y) = {\mathcal {F}}^*(P_2) {\mathcal {F}}^*(P_1)^{-1} ({\mathcal {F}}^*)^{\langle 0 \rangle }(X). \end{aligned}$$

Observe that ${\mathcal {F}}^*(P_i) = {\mathcal {F}}(E^1(P_i)) = {\mathcal {F}}(P_i)$ by [18, Lemma 4.6] and $({\mathcal {F}}^*)^{\langle 0 \rangle }(X) = {\mathcal {F}}^*(X) = {\mathcal {F}}(E^1(X))$ by $X \in {\mathcal {C}}$. Therefore we have

$$\begin{aligned} ({\mathcal {F}}^*)^{ \langle 2 \rangle }(Y) = {\mathcal {F}}(P_2){\mathcal {F}}(P_1)^{-1}{\mathcal {F}}(E^1(X)). \end{aligned}$$

(A.1)

Thus our goal is to show ${\mathcal {F}}(Y) \subset _{{{\,\mathrm{fin}\,}}} ({\mathcal {F}}^*)^{ \langle 2 \rangle }(Y)$ for $Y \in {\mathcal {M}}$.

Take an element $f \in \Lambda \setminus \{0\}$ which annihilates Y. By a presentation ${\mathcal {R}}^b \overset{h}{\rightarrow } {\mathcal {R}}^a \rightarrow Y \rightarrow 0$ of Y, construct an exact sequence

$$\begin{aligned} 0 \rightarrow X' \rightarrow ({\mathcal {R}}/f)^b \overset{{\overline{h}}}{\rightarrow } ({\mathcal {R}}/f)^a \rightarrow Y \rightarrow 0. \end{aligned}$$

(A.2)

Then by (A.1), we have $({\mathcal {F}}^*)^{ \langle 2 \rangle }(Y) = f^{a-b}{\mathcal {F}}(E^1(X'))$. Let Z be the image of ${\overline{h}}$ in the sequence (A.2), which is contained in ${\mathcal {C}}$. Then the two short exact sequences obtained by splitting (A.2) induce the following commutative diagram with exact row and column.

Here, we identify $E^1(R/f) \simeq R/f$ naturally and the superscript T denotes the transpose. This diagram induces an exact sequence

$$\begin{aligned} 0 \rightarrow E^2(Y) \rightarrow {{\,\mathrm{Cok}\,}}({\overline{h}}^T) \rightarrow E^1(X') \rightarrow 0. \end{aligned}$$

Since $E^2(Y)$ is a finite module, Lemma A.6 implies that ${\mathcal {F}}({{\,\mathrm{Cok}\,}}({\overline{h}}^T)) \subset _{{{\,\mathrm{fin}\,}}} {\mathcal {F}}(E^1(X'))$. Therefore

$$\begin{aligned} ({\mathcal {F}}^*)^{ \langle 2 \rangle }(Y) = f^{a-b} {\mathcal {F}}(E^1(X')) \supset _{{{\,\mathrm{fin}\,}}} f^{a-b} {\mathcal {F}}({{\,\mathrm{Cok}\,}}({\overline{h}}^T)) = {\mathcal {F}}({{\,\mathrm{Cok}\,}}({\overline{h}})) = {\mathcal {F}}(Y), \end{aligned}$$

which completes the proof. $\square$

For a prime ideal ${\mathfrak {q}}$ of $\Lambda$ of height one, let $\Lambda _{{\mathfrak {q}}}$ be the localization of $\Lambda$ at ${\mathfrak {q}}$ and put ${\mathcal {R}}_{{\mathfrak {q}}} = \Lambda _{{\mathfrak {q}}} \otimes _{\Lambda } {\mathcal {R}}$.

Definition A.9

For fractional ideals ${\mathcal {I}}, {\mathcal {J}}$ of ${\mathcal {R}}$, we say ${\mathcal {I}}$ and ${\mathcal {J}}$ are commensurable if both ${\mathcal {I}}\cap {\mathcal {J}}\subset _{{{\,\mathrm{fin}\,}}} {\mathcal {I}}$ and ${\mathcal {I}}\cap {\mathcal {J}}\subset _{{{\,\mathrm{fin}\,}}} {\mathcal {J}}$ hold. In that case we write ${\mathcal {I}}\sim _{{{\,\mathrm{fin}\,}}} {\mathcal {J}}$. Equivalently, ${\mathcal {I}}\sim _{{{\,\mathrm{fin}\,}}} {\mathcal {J}}$ if and only if ${\mathcal {I}}{\mathcal {R}}_{{\mathfrak {q}}} = {\mathcal {J}}{\mathcal {R}}_{{\mathfrak {q}}}$ for any prime ideal ${\mathfrak {q}}$ of $\Lambda$ of height 1. It follows that $\sim _{{{\,\mathrm{fin}\,}}}$ is an equivalence relation.

Lemma A.10

Let $f, g \in {\mathcal {R}}$ be non-zero-divisors and ${\mathcal {I}}\subset {\mathcal {R}}$ be an ideal containing a non-zero-divisor. Suppose that either ${\mathcal {I}}{\mathcal {R}}_{p\Lambda } = {\mathcal {R}}_{p\Lambda }$ or the order of G is relatively prime to p holds. Then $f{\mathcal {I}}\sim _{{{\,\mathrm{fin}\,}}} g{\mathcal {I}}$ implies $f{\mathcal {R}}= g{\mathcal {R}}$.

Proof

At first we suppose ${\mathcal {I}}= {\mathcal {R}}$ holds. Consider the natural injective map $f{\mathcal {R}}/(f{\mathcal {R}}\cap g{\mathcal {R}}) \hookrightarrow {\mathcal {R}}/g{\mathcal {R}}$. On the one hand, the assumption $f{\mathcal {R}}\sim _{{{\,\mathrm{fin}\,}}} g{\mathcal {R}}$ implies that $f{\mathcal {R}}/(f{\mathcal {R}}\cap g{\mathcal {R}})$ is finite. On the other hand, ${\mathcal {R}}/g{\mathcal {R}}$ does not contain non-trivial finite submodules since ${\mathcal {R}}/g{\mathcal {R}}\in {\mathcal {P}}\subset {\mathcal {C}}$. Therefore we obtain $f{\mathcal {R}}\subset g{\mathcal {R}}$. By symmetry we conclude $f{\mathcal {R}}= g{\mathcal {R}}$, which proves the lemma when ${\mathcal {I}}= {\mathcal {R}}$.

In general, for any prime ideal ${\mathfrak {q}}$ of $\Lambda$ of height 1, we claim that the ideal ${\mathcal {I}}{\mathcal {R}}_{{\mathfrak {q}}}$ is invertible. If ${\mathfrak {q}}\ne p\Lambda$ or the order of G is relatively prime to p, then this is clear since ${\mathcal {R}}_{{\mathfrak {q}}}$ is a product of principal ideal domains (and ${\mathcal {I}}$ contains a non-zero-divisor). Otherwise, our assumption ${\mathcal {I}}{\mathcal {R}}_{p\Lambda } = {\mathcal {R}}_{p\Lambda }$ implies the claim. Now suppose $f{\mathcal {I}}\sim _{{{\,\mathrm{fin}\,}}} g{\mathcal {I}}$ holds. Then, for any ${\mathfrak {q}}$, we have $f{\mathcal {I}}{\mathcal {R}}_{{\mathfrak {q}}} = g{\mathcal {I}}{\mathcal {R}}_{{\mathfrak {q}}}$, which implies $f {\mathcal {R}}_{{\mathfrak {q}}} = g {\mathcal {R}}_{{\mathfrak {q}}}$ by the above claim. Therefore $f{\mathcal {R}}\sim _{{{\,\mathrm{fin}\,}}} g{\mathcal {R}}$ and we conclude $f{\mathcal {R}}= g{\mathcal {R}}$ by the case ${\mathcal {I}}= {\mathcal {R}}$, which is already established. This completes the proof. $\square$

Remark A.11

The assumption ${\mathcal {I}}{\mathcal {R}}_{p\Lambda } = {\mathcal {R}}_{p\Lambda }$ is necessary when the order of G is divisible by p. For example, suppose G is a cyclic group of order p and consider ${\mathcal {I}}= (p, N_G)$, where $N_G$ is the norm element. Then we can easily verify $(p+N_G){\mathcal {I}}= p{\mathcal {I}}$ and $(p+N_G){\mathcal {R}}\ne p{\mathcal {R}}$.