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Let $X$ be a scheme over $\mathbb C$, such that $A=X_{\textrm{red}}$ is an abelian variety. Suppose the tangent spaces to $X$ are all of dimension $\dim X+1$. Then $X$ is non-reduced everywhere, but I cannot figure out what kind of non-reducedness can actually occur. So here is my question:

Q. Is $X$ isomorphic to $A\times \textrm{Spec }\mathbb C[\epsilon]$?

In general, one can produce non-trivial "ribbons" of, for instance, lines in $\mathbb P^3$. For smooth group schemes I would think there is only the trivial non-reduced structure, but I cannot prove it. Thank you for sharing any thought!

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  • $\begingroup$ Such things are classified by $H^1(A,T_A)$, with $T_A$ the sheaf of tangent vector fields. If $A$ is abelian then this is not zero, so there are non-trivial deformations. ($T_A$ is $\mathcal O^{\dim A}$, so you have to compute $H^1(A,\mathcal O_A)$; or see mathoverflow.net/questions/94150/…) $\endgroup$ Oct 1, 2015 at 7:42
  • $\begingroup$ @MarianoSuárez-Alvarez: Thanks! I am not asking for first-order deformations, I am asking for everywhere non-reduced structures. Isn't there a difference? $\endgroup$
    – Brenin
    Oct 1, 2015 at 7:57

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I believe an elliptic curve over $\mathbb{C}[\epsilon]$ whose $j$-invariant is not in $\mathbb{C}$ gives what you want.

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