Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $k$ be a field and $D:=\operatorname{Spec}(k[t]/(t^2)$ the scheme of dual numbers over $k$.

Then what is the fibre product $D \times_k D$ with itself over $k$? In other words, what is $\operatorname{Spec}(k[t]/(t^2) \otimes_k k[t]/(t^2)$ And how do line bundles over this scheme look like?

share|cite|improve this question
up vote 5 down vote accepted

The tensor product $k[t]/(t^2)\otimes k[t]/(t^2)$ is equal to $k[x,y]/(x^2,y^2)$, where $k[x,y]$ is the polynomial ring in the two variables $x,y$ over $k$, together with the natural maps $f,g: k[t]/(t^2)\rightarrow k[x,y]/(x^2,y^2)$, $f(t+(t^2))=x+(x^2)$, $g(t+(t^2))=y+(y^2)$.

Proving the universal property is a bit lengthy but straightforward as far as I see.

share|cite|improve this answer
It is not lengthy, $R[x]/(p) \otimes_R R[y]/(q) = R[x,y]/(p,q)$ is just abstract nonsense (compare hom-functors and use universal properties). – Martin Brandenburg Nov 11 '11 at 9:06

As Hagen explained, the scheme $X=D\times_{k} D$ is the affine scheme associated to the ring $R=k[x,y]/(x^2,y^2)$.
Since that ring is local, $all$ vector bundles of any rank on $X$ are trivial, not only line bundles .
Indeed vector bundles (or equivalently locally free sheaves $\mathcal F$) on $X$ correspond to finitely generated projective $R$-modules $P$ ( $\mathcal F \leftrightarrow \tilde P$ ) and all projective modules on a local ring are trivial.

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.