Algebraic-Geometry

Let \(A\) be a ring. We have corresponding topological space \(X=\text{Spec}(A)\), the collection of all prime ideals of \(A\) with Zariski Topology. We now define a sheaf on \(X\) called the structure sheaf, denoted by \(\mathcal{O}_X\). This \(X\) with this structure sheaf \(\mathcal{O}_X\) is called an affine scheme, These affine schemes are building blocks of what is called an arbitrary scheme. To define a sheaf on \(X\) we need to associate a ring for each \(U\) open in \(X\). We do that as follows :where the condition \(\dagger\) says that given \(p\in U\) we have \(s(p)\in A_p\) and that \(s\) is locally a fraction i.e., given \(p\in U\) there exists an open set \(U(p)\subseteq U\) and \(a\in A, f\in A\) such that \(s(q)=\frac{a}{f}\in A_q\) for all \(q\in U(p)\). The verification that this gives a sheaf on \(X\) is same as that of the verification that sheafification of a sheaf is a sheaf. We can see the similarity between the definitions. More details can be found here about the similarity. So, \((X,\mathcal{O}_X)\) forms a ringed space, which we call an affine scheme. We will now see results about the global sections, stalks of structure sheaf and what does structure sheaf give on basic open subsets of \(X=\text{Spec}(A)\). Proposition : Let \(A\) be a ring, and \((\text{Spec A},\mathcal{O})\) its spectrum. For any \(\mathfrak{p}\in \text{Spec A}\), the stalk \(\mathcal{O}_{\mathfrak{p}}\) of the sheaf \(\mathcal{O}_{}\) is isomorphic to the local ring \(A_{\mathfrak{p}}\) i.e., \(\mathcal{O}_{\mathfrak{p}}\cong A_{\mathfrak{p}}\). For any element \(f\in A\), the ring \(\mathcal{O}(D(f))\) is isomorphic to the localized ring \(A_f\) i.e., \(\mathcal{O}(D(f))\cong A_f\). In particular, \(\Gamma(\text{Spec A}, \mathcal{O})\cong A\). Proof : Let \(\mathfrak{p}\in X\). We define a map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\) and show that this is a bijection. Defining the map - Let \([(U,s)]\in \mathcal{O}_{\mathfrak{p}}\) i.e., \(U\) is an open set in \(X\) containing \(p\) and \(s\in \mathcal{O}(U)\). By definition, \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\). To get an element in \(A_{\mathfrak{p}}\) given \(s\), its only natural to consider image of \(\mathfrak{p}\) under \(s\) namely \(s(\mathfrak{p})\in A\). Defining \(s\mapsto s(\mathfrak{p})\) gives a map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\). Showing that the map is well defined - Suppose \([(U,s)]=[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\) i.e., there is an open set \(W\subset U\cap V\) containing \(\mathfrak{p}\) such that \(s|_{W}=t|_W\). As \(\mathfrak{p}\in W\), we have in particular \(s(\mathfrak{p})=t(\mathfrak{p})\). So, there is a well defined map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\). Showing that the map is Injective - For \([(U,s)],[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\) with \(s(\mathfrak{p})=t(\mathfrak{p})\), we show that \([(U,s)]=[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\). As \(\mathfrak{p}\in U\), for \(s: U\rightarrow\bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) there exists open \(U(\mathfrak{p})\subset U\) containing \(\mathfrak{p}\) and \(a,f\in A\) such that \(s(\mathfrak{q})=\frac{a}{f}\) for all \(\mathfrak{q}\in U(\mathfrak{p})\). Similarly, for \(t:V\rightarrow\bigsqcup_{\mathfrak{q}\in V}A_{\mathfrak{q}}\) there exists open \(V(\mathfrak{q})\subseteq V\) and \(b,g\in A\) such that \(t(\mathfrak{q})=\frac{b}{g}\) for all \(\mathfrak{q}\in V(\mathfrak{p})\). In particular, \(\frac{a} {f}=s(\mathfrak{p})=t(\mathfrak{p})=\frac{b}{g}\). Let \(\mathfrak{q}\in U(\mathfrak{p})\cap V(\mathfrak{p})\). Then, \(s(q)=\frac{a}{f}=\frac{b}{g}=t(q)\). Thus, we have \(s|_{U(\mathfrak{p})\cap V(\mathfrak{q})}=t|_{U(\mathfrak{p})\cap V(\mathfrak{q})}\). Thus, \([(U,s)]=[(V,t)]\). So, \(s\mapsto s(\mathfrak{p})\) is injective. Showing that the map is surjective - Let \(\frac{a}{f}\in A_{\mathfrak{p}}\), we want to choose an open set \(U\) containing \(\mathfrak{p}\) and \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) such that \(s(\mathfrak{p})=\frac{a}{f}\). One choice for \(s\) is sending \(q\) to image of \(\frac{a}{f}\) in \(A_{\mathfrak{q}}\). For this, we need \(f\notin\mathfrak{q}\) i.e., \(\mathfrak{q}\in D(f)\). Let \(U=D(f)\) and consider \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) sending \(\mathfrak{q}\) to image of \(\frac{a}{f}\) in \(A_{\mathfrak{q}}\). We then have \(s(\mathfrak{p})=\frac{a}{f}\in A_{\mathfrak{p}}\). Thus, the map is surjective. So, we have isomorphism \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\) given by \(s\mapsto s(\mathfrak{p})\). Let \(f\in A\). We define a map \(A_f\rightarrow \mathcal{O}(D(f))\) and show that this is a bijection. Defining the map - Given \(\frac{a}{f}\in A_f\) we assign \(s\in \mathcal{O}(D(f))\) where \(s:D(f)\rightarrow \bigsqcup_{q\in D(f)}A_q\). Let \(q\in D(f)\) then, \(f\notin q\). So, \(\frac{a}{f}\) is defined in \(A_q\). So, define \(s(q)\) to be the image of \(\frac{a}{f}\) in \(A_q\) for each \(q\in D(f)\). It is clearly a well defined function. Similarly we define for \(\frac{a}{f^n}\in A_f\) a map \(s:D(f^n)=D(f)\rightarrow \bigsqcup_{q\in D(f)}A_q\) as \(q\mapsto \frac{a}{f^n}\in A_q\). Showing that the map is injective - Suppose \(\frac{a}{f^n},\frac{b}{f^m}\in A_f\) is such that the corresponding maps \(s,t\) are equal i.e., \(\frac{a}{f^n}=\frac{b}{f^m}\in A_q~\forall q\in D(f)\) i.e., given \(q\in D(f)\) there exists \(t_q\notin q\) such that \(t_q(af^m-bf^n)=0\). Consider the case when \(D(f)=\{q\}\). As \(t\notin q\), we have \(q\in D(t)\) i.e., \(D(f)\subseteq D(t)\) i.e., \(V(t)\subseteq V(f)\) i.e., \(\sqrt{(f)}\subseteq \sqrt{(t)}\). As \(f\in \sqrt{(f)}\) we have \(f^l=td\) for some \(d\in A\). We have \(t(af^m-bf^n)=0\) which implies \(td(af^m-bf^n)=0\) i.e., \(f^l(af^m-bf^n)=0\) i.e., \(\frac{a}{f^n}=\frac{b}{f^m}\in A_f\) and we are done. Suppose \(D(f)=\{q_i\}_{i\in \Lambda}\). As \(t_i\notin q_i\) we have \(q_i\in D(t_i)\) i.e., \(D(f)\subseteq \bigcup_{i\in \Lambda} D(t_i)\). As in previous observation, this means \(f^l\) is in the ideal generated by \(\{t_i\}\) for some \(l\in \mathbb{N}\). So, we have (after rearranging indices in \(\Lambda\)) \(f^l=a_1t_1+\cdots+a_nt_n\) for some \(a_i\in A\). As \(t_i(af^m-bf^n)=0\), we have \(a_it_i(af^m-bf^n)=0\) for all \(i\). So, \(\sum_{i=1}^na_it_i(af^m-bf^n)=0\) i.e., \(f^l(af^m-bf^n)=0\). Thus, \(\frac{a}{f^n}=\frac{b}{f^m}\in A_f\). Thus, the map \(A_f\rightarrow \mathcal{O}(D(f))\) is injective.  

This is an exercise from Hartshorne's Algebraic Geometry book. A part of this exercise is called QcQs lemma in Ravi Vakil's Foundations of Algebraic Geometry notes.  

Let \(A\) be a ring and let \((X,\mathcal{O}_X)\) be a scheme. Given a morphism \(f:X\rightarrow \text{Spec}(A)\) we have an associated map on sheaves \(f^{\#}:\mathcal{O}_{\text{Spec}(A)}\rightarrow f_* \mathcal{O}_X\). Taking global sections, we obtain a homomorphism \(A\rightarrow \mathcal{O}_X(X)\). Thus there is a natural map \(\alpha: \text{Hom}_{\text{Schemes}}(X,\text{Spec}(A))\rightarrow \text{Hom}_{\text{Rings}}(A,\mathcal{O}_X(X)).\) Then, \(\alpha\) is bijective. We will try to understand this adjointness of Global section functor and Spec functor. Suppose we are given a scheme \((X,\mathcal{O}_X)\) and a ring homomoprhism \(\varphi: A\rightarrow \mathcal{O}_X(X)\). We construct a morphism of schemes \((f,f^{\#}):(X,\mathcal{O}_X)\rightarrow (\text{Spec}(A),\mathcal{O}_{\text{Spec}(A)})\). We first define morphism of topological spaces \(f:X\rightarrow \text{Spec}(A)\). Let \(x\in X\), we want to assign a prime ideal \(P\) in \(A\). Let \(X=\text{Spec}(B)\) and \(x=\mathfrak{P}\in X=\text{Spec}(B)\), as we have a ring homomorphism \(\varphi : A\rightarrow \mathcal{O}_X(X)=O_{\text{Spec}(B)}(\text{Spec}(B))=B\) ...

Definition : Let \(S\) be a scheme. An \(S\) scheme is a scheme \(X\) together with a morphism \(p:X\rightarrow S\). A morphism of \(S\) schemes \((X,p:X\rightarrow S)\) and \((Y,q:Y\rightarrow S)\) is a morphism of schemes \(f:X\rightarrow Y\) such that \(q\circ f=p\).  

let \(\mathcal{F}\) be a presheaf. We define associated presheaf of \(\mathcal{F}\) to be the sheaf given by map \(U\mapsto \widetilde{F}(U)\) where where the condition \(^\dagger\) says that \(s(p)\in \mathcal{F}_p\) for every \(p\in U\) and there exists an open subset \(U(p)\subset U\) containing \(p\) and a section \(t\in \mathcal{F}(U(p))\) such that \(t_q=s(q)\) for every \(q\in U(p)\). Suppose \(\mathcal{F}\) is actually a sheaf then we will see that \(\widetilde{F}(U)\cong \mathcal{F}(U)\) for every open \(U\subseteq X\). Let \(s\in \widetilde{F}(U)\) i.e., \(s:U\rightarrow \bigsqcup_{i\in \Lambda}\mathcal{F}_{p_i}\) (this notation is just for my comfort, we have \(U=\bigcup_{i\in \Lambda}\{p_i\}\)) satisfying some conditions given above. Given such \(s\) we want to assign an element in \(\mathcal{F}(U)\). ...

I will add links of blogs of topics in sequence as in Hartshorne. Morphism of Sheaves – Morphism of Stalks Sheafification of a presheaf that is already a sheaf is itself – Reality check Structure sheaf on spectrum of a ring          

Let \(f:X\rightarrow Y\) be an affine morphism and \(\mathcal{F}\) is a quasi coherent sheaf of \(\mathcal{O}_X\) modules. Then, \(f_*\mathcal{F}\) is a quasi coherent sheaf of \(\mathcal{O}_Y\) modules. We have the following result : Let \(X\) be a scheme. Then, an \(\mathcal{O}_X\) module \(\mathcal{F}\) is quasi coherent iff for every open affine subset \(U=\text{Spec}(A)\) of \(X\), there is an \(A\) module \(M\) such that \(\mathcal{F}|_U=\tilde{M}\). Let \(U\subseteq Y\) be an open affine subset say \(U=\text{Spec}(A)\). As \(f\) is affine, \(f^{-1}(U)\) is affine open, say \(f^{-1}(U)=\text{Spec}(B)\subseteq X\). As \(\mathcal{F}\) is quasi coherent sheaf of \(\mathcal{O}_X\) modules and \(f^{-1}(U)=\text{Spec}(B)\) is open affine subset of \(X\), there exists a \(B\) module \(M\) such that \(\mathcal{F}|_{f^{-1}(U)}\cong \widetilde{M}\). As \(f^{-1}(U)=\text{Spec}(B)\) we have \(f:\text{Spec}(B)\rightarrow \text{Spec}(A)\) which also gives a ring morphsim \(A\rightarrow B\). The isomorphism \(\mathcal{F}|_{f^{-1}(U)}\cong \widetilde{M}\) implies \(f_*(\mathcal{F}|_{f^{-1}(U)})\cong f_*\widetilde{M}\). ...

Let \(X=\text{Spec}(R)\) be an affine scheme. Let \(X'\) be an affine \(X\) scheme i.e., \(X'=\text{Spec}(R')\) for some ring \(R'\) with a morphism of schemes \(\pi: X'\rightarrow X\). This \(\pi\) comes with morphism of global sections \(R=\Gamma(X,\mathcal{O}_X)\rightarrow \Gamma(X',\mathcal{O}_X')=R'\) giving \(R'\), structure of an \(R\) algebra. So, any affine scheme over \(X=\text{Spec}(R)\) is simply the specturm of an \(R\) algebra. Conversely, given an \(R\) algebra say \(R'\), we have an affine scheme \(X'=\text{Spec}(R')\) over \(X\) with morphism \(\pi:X'\rightarrow X\). Being a morphism of affine schemes, \(\pi: X\rightarrow X\) is an affine morphism. Let \(X\) be an arbitrary scheme. We want to associate an \(X\) scheme \(X'\) such that the structure morphism \(\pi:X'\rightarrow X\) is an affine morphism. To do this in case of \(X=\text{Spec}(R)\) we have fixed an \(R\) algebra and then associated an affine scheme for this. In case of an arbitrary scheme \(X\) unlike the case of affine scheme \(X=\text{Spec}(R)\) there is no single ring that has all information about the scheme \(X\). It is only natural to consider the collection \(\{\mathcal{O}_X(U): U\subseteq X\}\) varying over all open subsets of \(X\) to get information about the scheme \(X\). Choosing an \(\mathcal{O}_X(U)\) algebra \(\mathcal{F}(U)\) for each open \(U\subseteq X\) we associate an \(X\) scheme \(X'\) for this collection \(\{\mathcal{F}(U)\}\) of \(\mathcal{O}_X(U)\) algebras. It is only natural to put a condition that this collection \(\{\mathcal{F}(U)\}\) to be compatible with structure sheaf \(\mathcal{O}_X\) i.e., we want \(U\mapsto \mathcal{F}(U)\) to give a structure of sheaf of \(\mathcal{O}_X\) algebras on \(X\). So, given an arbitrary scheme \(X\) and a sheaf \(\mathcal{F}\) of \(\mathcal{O}_X\) algebras we associate an \(X\) scheme \(X'\) such that the structure map \(X'\rightarrow X\) is an affine morphism. It is not obvious at this point but we also want \(\mathcal{F}\) to be a quasicoherent sheaf of \(\mathcal{O}_X\) modules. We call this \(X'\), Global spec or Relative spec of sheaf \(\mathcal{F}\) of \(\mathcal{O}_X\) algebras over \(X\) denoted by \(\textbf{Spec} (\mathcal{F})\). Here we make two important remarks : The \(\textbf{Spec}\) construction gives an important way to understand affine morphisms. Note that \(\textbf{Spec}(\mathcal{F})\rightarrow X\) is an affine morphism. The converse is also true. If \(f:X\rightarrow Y\) is an affine morphism then \(\mathcal{A}=f_*\mathcal{O}_X\) is a quasi coherent sheaf of \(\mathcal{O}_Y\) algebras and \(X\cong \text{Spec} (\mathcal{A})\). The \(\textbf{Spec}\) construction is used to assign a geometric vector bundle on a scheme \(Y\) to each locally free sheaf \(\mathcal{E}\) of rank \(n\) on a scheme \(Y\) which gives a bijection between isomorphism classes of locall free sheaves of rank \(n\) on \(Y\), and isomorphism classes of vector bundles of rank \(n\) on \(Y\). To define \(\textbf{Spec}(\mathcal{F})\) we do not need \(\mathcal{F}\) to be quasi coherent, but \(\mathcal{F}\) quasi coherent implies the structure map \(\textbf{Spec}(\mathcal{F})\rightarrow X\) is an affine morphism. Now, we try to construct \(\textbf{Spec}(\mathcal{F})\) and structure map \(\textbf{Spec}(\mathcal{F})\rightarrow X\). One way to do this is gluing the schemes \(\text{Spec}(\mathcal{F}(U))\) over all open subsets \(U\subseteq X\). Another way is to use universal property of \(\text{Spec}\) of a ring. We have following result : Let \(A\) be a ring and let \((X,\mathcal{O}_X)\) be a scheme. Given a morphism \(f:X\rightarrow \text{Spec}(A)\) we have an associated map on sheaves \(f^{\#}:\mathcal{O}_{\text{Spec}(A)}\rightarrow f_*\mathcal{O}_X\). Taking global sections, we obtain a ring homomorphism \(A\rightarrow \mathcal{O}_X(X)\). Thus there is a natural map \(\alpha: \text{Hom}_{\text{Schemes}}(X,\text{Spec}(A))\rightarrow\text{Hom}_{\text{Rings}}(A,\mathcal{O}_X(X)).\) ...

Here I will add links of pages of interesting questions/answers from Math Stack Exchange and Math over flow. Learning Algebraic Geometry 1 transitive Lie groupoid is Morita equivalent to the isotropy group