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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?

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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.

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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
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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.

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