Let $X$ be a reduced Noetherian scheme and $\mathcal{F}$ a coherent sheaf on $X$. What i am wondering is:
Is it true or not that the projectivization $$\mathbb{P}(\mathcal{F})=\underline{\mathrm{Proj}}(\underline{\mathrm{Sym}}(\mathcal{F}))$$ is reduced again?
If this has any influence on the question: In the case i want to consider, $X$ is actually a smooth, projective variety over an algebraically closed field.
EDIT: Removed a previous thought of mine including a mistake...
Finally one follow-up question: In case of a negative answer, are there easy conditions on $\mathcal{F}$ to assure reducedness of $\mathbb{P}(\mathcal{F})$?
Many thanks in advance!
The symmetric algebras of $R$-modules are precisely the ones which have a presentation (as a commutative $R$-algebra) in which every relation is a polynomial of degree $\leq 1$. An example is $\mathrm{Sym}_\mathbb{Z}(\mathbb{Z}/4)=\mathbb{Z}[x]/(4x)$. This is not reduced since $(2x)^2=0$. Notice that $\mathrm{Proj}\bigl(\mathrm{Sym}_\mathbb{Z}(\mathbb{Z}/4)\bigr)$ is actually affine, namely $\mathrm{Spec}(\mathbb{Z}[x]/(4x)_{(x)})$ where the subscript $(x)$ denotes the homogeneous localization at the element $x$ (in your answer you forgot to distinguish between the algebra and its homogeneous localizations; this is the big difference between $\mathrm{Spec}$ and $\mathrm{Proj}$). Notice that $\mathbb{Z}[x]/(4x)_{(x)} = \mathbb{Z}/4$ as rings and this is still not reduced. The same reasoning shows for arbitrary ideals $I \subseteq R$ that $\mathrm{Proj}\bigl(\mathrm{Sym}_R(R/I)\bigr)=\mathrm{Spec}(R/I)$ and this is reduced if and only if $I$ is a radical ideal.