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I have a basic question concerning dual numbers and tangent vectors.

If I have a scheme $S$ over a field $k$ and a closed $k-$rational point $s\in S$, then one knows that to give a tangent vector in $s$ corresponds to a $k-$morphism $D\rightarrow S$, sending the unique point of $D$ to $s$, where $D$ denotes the scheme of dual numbers.

My question is simply: the morphism from $D$ to $S$ which you get is not a closed immersion, isn't it? Of course, the image is a closed point, namely $s$, but I don't see if the morphism on sheaves is surjective.

Thank you!

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Nice question . –  Georges Elencwajg Sep 4 '11 at 9:32

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The morphism $f:D\to S$ corresponding to the tangent vector $t\in T_S(s)$ is a closed immersion if and only if $t\neq0$.

Indeed, being a closed immersion here means that the induced morphism of local $k$-algebras
$f^\ast:\mathcal O_{S,s} \to \mathcal O_{D,d}=k[\epsilon ]$ is surjective. This will always be the case, unless $f^\ast {\frak m} _{S,s}=o \;$ i.e. unless the tangent vector $t$ is zero.

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