Example 27.21.3. The map $\text{Sym}^ n(\mathcal{E}) \to \pi _*(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(n))$ is an isomorphism if $\mathcal{E}$ is locally free, but in general need not be an isomorphism. In fact we will give an example where this map is not injective for $n = 1$. Set $S = \mathop{\mathrm{Spec}}(A)$ with
where $k$ is a field and
Denote $\overline{u}$ the class of $u$ in $A$ and similarly for the other variables. Let $M = (Ax \oplus Ay)/A(\overline{u}x + \overline{v}y)$ so that
where
In this case the projective bundle associated to the quasi-coherent sheaf $\mathcal{E} = \widetilde{M}$ on $S = \mathop{\mathrm{Spec}}(A)$ is the scheme
Note that this scheme as an affine open covering $P = D_{+}(x) \cup D_{+}(y)$. Consider the element $m \in M$ which is the image of the element $us_1x + vt_2y$. Note that
and
The first equation implies that $m$ maps to zero as a section of $\mathcal{O}_ P(1)$ on $D_{+}(x)$ and the second that it maps to zero as a section of $\mathcal{O}_ P(1)$ on $D_{+}(y)$. This shows that $m$ maps to zero in $\Gamma (P, \mathcal{O}_ P(1))$. On the other hand we claim that $m \not= 0$, so that $m$ gives an example of a nonzero global section of $\mathcal{E}$ mapping to zero in $\Gamma (P, \mathcal{O}_ P(1))$. Assume $m = 0$ to get a contradiction. In this case there exists an element $f \in k[u, v, s_1, s_2, t_1, t_2]$ such that
Since $I$ is generated by homogeneous polynomials of degree $2$ we may decompose $f$ into its homogeneous components and take the degree 1 component. In other words we may assume that
for some $a, b, \alpha _1, \alpha _2, \beta _1, \beta _2 \in k$. The resulting conditions are that
There are no terms $u^2, uv, v^2$ in the generators of $I$ and hence we see $a = b = 0$. Thus we get the relations
We may use the first generator of $I$ to replace any occurrence of $us_1$ by $vt_1 + ut_2$, the second generator of $I$ to replace any occurrence of $vs_1$ by $-us_2 + vt_2$, the third generator to remove occurrences of $vs_2$ and the third to remove occurrences of $ut_1$. Then we get the relations
This implies that $\alpha _1$ should be both $0$ and $1$ which is a contradiction as desired.
Comments (0)