Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Nice question . – Georges Elencwajg Sep 4 '11 at 9:32
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.