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This may be a dumb question.

$\mathcal{M}(r,d)$ is a coarse moduli scheme for semistable pairs $(E,\phi:E \rightarrow K_X \otimes E)$ of rank $r$, degree $d$ on a smooth projective curve $X$ over $\mathbb{C}$.
Then,the Hitchin morphism from $\mathcal{M}(r,d)$ to $H^0(X,K_X) \times ...\times H^0(X,K_X^{\otimes r})$ is defined a pair $(E,\phi)$ to its characteristic polynomial. Explicitly, $(E,\phi) \mapsto (\operatorname{Tr}(\wedge^1\phi),...,\operatorname{Tr}(\wedge^r\phi))$
But strictly, this is not a scheme morphism in the sense of Hartshorne's algebraic geometry.

My idea:
Let $U_i=Spec(B_i)\subset{\mathcal{M}(r,d)}$ be affine cover. First, define $Sym(V^*)\rightarrow B_i$.
If it is possible to glue them, we get the morphism $\mathcal{M}(r,d) \rightarrow Spec(Sym(V^*))$ $(V:=H^0(X,K_X) \times ...\times H^0(X,K_X^{\otimes r}))$, using $Hom_{sch}(X,Spec(R))\simeq Hom_{ring}(R,\Gamma(X,O_X))$
But can we gule them? Or this is a completely wrong direction?

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    $\begingroup$ Yes, you are on a wrong track. All you have to prove is that $(E,\phi)\mapsto \operatorname{Tr}\wedge^{i}\phi $ is a regular function on $\mathscr{M}(r,d)$. This is clear on a fine moduli space (or stack), then use basic descent. $\endgroup$
    – abx
    Commented Sep 22, 2020 at 9:38
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    $\begingroup$ Another way of seeing it (perhaps a rewording of abx's comment?) is: the construction $(E,\phi)\mapsto\mathrm{tr}\wedge^i \phi$ works well in families cause it's done fiberwise. This gives a map from the moduli functor (i.e. the functor associating to $S$ the set of appropriate families on $S$) to (the functor of points of) your $SpecSym(V^*)$. By definition of coarse moduli space, this induces a unique map of schemes from the coarse moduli space to $SpecSym(V^*)$ which makes the diagram of the previous maps commute hence is the given map on points/objects. $\endgroup$
    – Qfwfq
    Commented Sep 22, 2020 at 13:03
  • $\begingroup$ Thanks for comments. But what do you mean by regular function? regular map in the category of algebraic variety? $\endgroup$
    – Aoki
    Commented Sep 22, 2020 at 16:44
  • $\begingroup$ @Aoki A regular function on $X$ is an element of $\mathcal O_X$, equivalently a map to $\mathbb A^1$. Since you're mapping to a vector space here, you want a tuple of regular functions I think. $\endgroup$
    – Tabes Bridges
    Commented Sep 22, 2020 at 19:34
  • $\begingroup$ @Tabes Bridges I think Hitchin map should be a morphism of the category of scheme. But $(E,\phi)\mapsto \operatorname{Tr}\wedge^{i}\phi$ is only defined at closed points and isn't a morphism between schemes.And, I don't know the usual definition of regular between schemes. $\endgroup$
    – Aoki
    Commented Sep 24, 2020 at 2:50

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