Algebraic Geometry – Is Multiplication by $d$ on the Jacobian of a Nodal Curve Étale?

abelian-varietiesag.algebraic-geometrymoduli-spaces

Let $k$ be an alg.closed of char$k=0$ and let $A$ be an abelian variety over $k$. This
Lemma on stacks project states that $[d]\colon A\to A$ is étale. In particular, when $A$ is the Jacobian of a smooth curve $C$ over $k$, then multiplication by $d$ is étale.

Now, when $C$ is nodal, the Jacobian (parametrising multidegree $0$ line bundles on $C$) fails to be an abelian variety, although it is still a group scheme with the tensor product. Moreover, it fits into an exact sequence $$0\to \mathbb{G}_{m}^{t}\to J_{C}\to J_{C^v}\to 0$$ where $t=\#\text{nodes}-\#\text{components}+1$ and $v\colon C^v\to C$ is the normalization map. Hence, when $C$ is of compact type we still have that the multiplication by $d$ is étale since then, the Jacobian is the product of abelian varieties and multiplication will work component wise. My question is the following:

If $C$ has non-separting nodes is it still true that $[d]\colon J_C\to J_C$ is étale ?

Best Answer

Lemma. Let $G$ be a commutative group scheme of finite type over a field $k$, and let $d$ be a positive integer invertible in $k$. Then the multiplication by $d$ map $[d] \colon G \to G$ is finite étale.

Proof. Since multiplication $\mu \colon G \times G \to G$ (resp. the diagonal embedding $\Delta \colon G \hookrightarrow G \times G$) induces the diagonal embedding $\Omega_{G,e} \hookrightarrow \Omega_{G,e} \oplus \Omega_{G,e}$ (resp. the addition map $\Omega_{G,e} \oplus \Omega_{G,e} \to \Omega_{G,e}$) on cotangent spaces, the map $[d]_* \colon \Omega_{G,e} \to \Omega_{G,e}$ is just multiplication by $d$. We conclude that $[d] \colon G \to G$ is unramified at the identity (since $d$ is invertible in $k$), hence unramified since it is a group homomorphism.

On the other hand, the morphism of group schemes $[d]$ factors as $G \stackrel \pi\twoheadrightarrow H \stackrel\iota\hookrightarrow G$ where $\pi$ is faithfully flat and $\iota$ is a closed immersion. In particular, any component of $G$ has the same dimension as any component of $H$, since $\pi$ is quasi-finite. We conclude that $\iota$ is a clopen immersion picking out a subset of the connected components of $H$, and in particular $[d] \colon G \to G$ is flat. $\square$

I think the flatness of $[d]$ should actually hold without the assumption that $d$ is invertible in $k$, but I couldn't see a quick proof of this. On the other hand, the hypothesis that $G$ is commutative is crucial: otherwise $[d]$ is not a homomorphism and we probably can't say much.

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