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4 Fixed more mathematical syntax using p tags

Let $\mathcal{F}$ be a locally free sheaf on $X$. For any $x$ in $X$ there exists $x \in U \subset_{open} X$ such that

$\mathcal{F}|_U \cong \mathcal{O}_X|_U^{(I)}$ $\ \ \ \ (\star)$.

In particular, for each $y$ in this particular $U$, one has $\mathcal{F}y \mathcal{F}_y \cong \mathcal{O}{X,y}^{(I)}$ mathcal{O}_{X,y}^{(I)}$(which is given by the isomorphism above!!!). Suppose now$X$is connected and$\mathcal{F}$is locally free (we need this). Fix an indexing set$I$(and I think I need to take this$I$to be one of the indexing sets from$(\star)$above). The properties of$\mathcal{F}$show that the set$S_I = \left(x \in X : \mathcal{F}_x \cong \mathcal{O}_{X,x}^{(I)}\right)$is both closed and open in$X$. We know that there exists$x$in$X$with$\mathcal{F}_x \cong \mathcal{O}_{X,x}^{(I)}$, mathcal{O}_{X,x}^{(I)}$,

we have $S_I = X$.

In particular, $\text{rank}_{\mathcal{O}_{X,x}}(\mathcal{F}_x)$ is constant as $x$ varies in $X$.

Let $\mathcal{F}$ be a locally free sheaf on $X$. For any $x$ in $X$ there exists $x \in U \subset_{open} X$ such that

$\mathcal{F}|_U \cong \mathcal{O}_X|_U^{(I)}$ $\ \ \ \ (\star)$.

In particular, for each $y$ in this particular $U$, one has $\mathcal{F}y \cong \mathcal{O}{X,y}^{(I)}$ (which is given by the isomorphism above!!!).

Suppose now $X$ is connected and $\mathcal{F}$ is locally free (we need this). Fix an indexing set $I$ (and I think I need to take this $I$ to be one of the indexing sets from $(\star)$ above). The properties of $\mathcal{F}$ show that the set

$S_I = \left(x \in X : \mathcal{F}x mathcal{F}_x \cong \mathcal{O}{X,x}^{(I)}\right)$ mathcal{O}_{X,x}^{(I)}\right)$ is both closed and open in$X$. We know that there exists$x$in$X$with $\mathcal{F}x \mathcal{F}_x \cong \mathcal{O}{X,x}^{(I)}$, mathcal{O}_{X,x}^{(I)}$,

we have $S_I = X$.

In particular, $\text{rank}{\mathcal{O}{X,x}}(\mathcal{F}_x)$ \text{rank}_{\mathcal{O}_{X,x}}(\mathcal{F}_x)$ is constant as$x$varies in$X$. 2 added 1 characters in body Let$\mathcal{F}$be a locally free sheaf on$X$. For any$x$in$X$there exists$x \in U \subset_{open} X $such that$\mathcal{F}|_U \cong \mathcal{O}_X|_U^{(I)} \ \ \ \ (\star)$. In particular, for each$y$in this particular$U$, one has$\mathcal{F}y \cong \mathcal{O}{X,y}^{(I)}$(which is given by the isomorphism above!!!). Suppose now$X$is connected and$\mathcal{F}$is locally free (we need this). Fix an indexing set$I$(and I think I need to take this$I$to be one of the indexing sets from$(\star)$above). The properties of$\mathcal{F}$show that the set$S_I = \left(x \in X : \mathcal{F}x \cong \mathcal{O}{X,x}^{(I)}\right)$is both closed and open in$X$. We know that there exists$x \x$in$X$with$\mathcal{F}x \cong \mathcal{O}{X,x}^{(I)}$, we have$S_I = X$. In particular,$\text{rank}{\mathcal{O}{X,x}}(\mathcal{F}_x)$is constant as$x$varies in$X\$.

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