Locally Ringed Space
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Locally Ringed Space
Let \( X \) be a topological space and let \( \mathscr{C}_{X} \) be the sheaf of continuous function on \( X \). Show that \( \left( X , \mathscr{C}_{X} \right) \) is a locally ringed space and describe (for each \( x \in X \)) the maximal ideal \( \mathfrak{m}_{x} \) of \( \mathscr{C}_{X,x} \).
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- Community Team
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Re: Locally Ringed Space
Let \(\displaystyle{U}\) be an open subset of \(\displaystyle{X}\) and \(\displaystyle{O_{X}(U)}\) the
set of all \(\displaystyle{\mathbb{K}}\)-valued and continuous functions on \(\displaystyle{U}\).
1. Obviously, \(\displaystyle{O_{X}(U)}\) is a \(\displaystyle{\mathbb{K}}\) - subalgebra of the \(\displaystyle{\mathbb{K}}\)
algebra \(\displaystyle{C(U,\mathbb{K})}\).
2. If \(\displaystyle{f\in O_{X}(U)}\) and \(\displaystyle{U'}\) is an open subset of \(\displaystyle{U}\),
then \(\displaystyle{f|_{U'}\in O_{X}(U')}\).
3. Let \(\displaystyle{f\in O_{X}(U)}\). Then, for each \(\displaystyle{p\in U}\), if \(\displaystyle{U_{p}}\)
is an open neighbourhood of \(\displaystyle{p}\), then, \(\displaystyle{f|_{U_{p}}\in O_{X}(U_{p})}\).
Conversely, if \(\displaystyle{U=\bigcup_{i\in I}U_i}\) is an open subcovering of \(\displaystyle{U}\)
then, \(\displaystyle{f|_{U_i}\in O_{X}(U_i)\,,\forall\,i\in I}\) and then, \(\displaystyle{f\in O_{X}(U)}\).
According to 1,2,3 , \(\displaystyle{\left(X,C_{X}\right)}\) is a locally \(\displaystyle{\mathbb{K}}\) - ringed space.
Now, let \(\displaystyle{x\in X}\). If \(\displaystyle{U\,,U'}\) are open neighbourhoods of \(\displaystyle{x}\)
and \(\displaystyle{f\in O_{X}(U)\,,f'\in O_{X}(U'}\), then,
\(\displaystyle{(U,f)\equiv (U',f')}\) iff \(\displaystyle{f=f'}\) on some neighbourhood of \(\displaystyle{x\in U\cap U'}\).
Therefore,
\(\displaystyle{m_{x}=\left\{f\in C_{X}: f(x)=0\right\}}\).
set of all \(\displaystyle{\mathbb{K}}\)-valued and continuous functions on \(\displaystyle{U}\).
1. Obviously, \(\displaystyle{O_{X}(U)}\) is a \(\displaystyle{\mathbb{K}}\) - subalgebra of the \(\displaystyle{\mathbb{K}}\)
algebra \(\displaystyle{C(U,\mathbb{K})}\).
2. If \(\displaystyle{f\in O_{X}(U)}\) and \(\displaystyle{U'}\) is an open subset of \(\displaystyle{U}\),
then \(\displaystyle{f|_{U'}\in O_{X}(U')}\).
3. Let \(\displaystyle{f\in O_{X}(U)}\). Then, for each \(\displaystyle{p\in U}\), if \(\displaystyle{U_{p}}\)
is an open neighbourhood of \(\displaystyle{p}\), then, \(\displaystyle{f|_{U_{p}}\in O_{X}(U_{p})}\).
Conversely, if \(\displaystyle{U=\bigcup_{i\in I}U_i}\) is an open subcovering of \(\displaystyle{U}\)
then, \(\displaystyle{f|_{U_i}\in O_{X}(U_i)\,,\forall\,i\in I}\) and then, \(\displaystyle{f\in O_{X}(U)}\).
According to 1,2,3 , \(\displaystyle{\left(X,C_{X}\right)}\) is a locally \(\displaystyle{\mathbb{K}}\) - ringed space.
Now, let \(\displaystyle{x\in X}\). If \(\displaystyle{U\,,U'}\) are open neighbourhoods of \(\displaystyle{x}\)
and \(\displaystyle{f\in O_{X}(U)\,,f'\in O_{X}(U'}\), then,
\(\displaystyle{(U,f)\equiv (U',f')}\) iff \(\displaystyle{f=f'}\) on some neighbourhood of \(\displaystyle{x\in U\cap U'}\).
Therefore,
\(\displaystyle{m_{x}=\left\{f\in C_{X}: f(x)=0\right\}}\).
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