EverlastingDeadMan

Commutative Algebra

• Rings and hypersurfaces. Consider a ring $R$ and an ideal $I\subseteq R[t_{1},\cdots ,t_{n}]$, then we define the following set (set $\mathbb {A} _{R}^{n}=\mathbb {A} ^{n}$, the $n$-affine space)$$V(I)=\{x\in \mathbb {A} ^{n}:f(x)=0{\text{ for all }}f\in I\}$$which we call (affine) varieties. Finally, let ${\mathcal {O}}_{I}=R[t_{1},\cdots ,t_{n}]/I$ the space of functionals on $V(I)$.
• Spectrum of a ring. Given a ring $R$, we define the prime spectrum of a ring $${\textbf {Spec}}R=\{{\mathfrak {I}}{\text{ prime ideal of }}R\}$$We also define the maximal spectrum as the space $\mathbf {mSpec} R$ as the collection of all maximal ideals. Given a morphism $f:S\longrightarrow R$ and an ideal ${\mathfrak {p}}\in \mathbf {Spec} R$, the ideal $f^{-1}({\mathfrak {p}})$ is also prime, so $f$ induces a morphism $$f_{*}:\mathbf {Spec} R\longrightarrow \mathbf {Spec} S,\qquad f_{*}{\mathfrak {p}}=f^{-1}({\mathfrak {p}})$$In particular, the projection $R\twoheadrightarrow R/I$ onto the quotient by an ideal $I$, induces a map $\mathbf {Spec} R/I\rightarrow \mathbf {Spec} R$, whose image is the set of prime ideals containing $I$. Also, given multiplicative set $S$, the inclusion $j:R\longrightarrow S^{-1}R$ induces a map $$j_{*}:\mathbf {Spec} S^{-1}R\longrightarrow \mathbf {Spec} R$$(where $S^{-1}R$ is the localization of $R$ to $S$). The image of $j_{*}$ are those ideals that are disjoint from $S$.
• Radical and nilpotent elements. A nilpotent element $s\in R$ is one such that $s^{n}=0$ for some $n$. Let $${\mathfrak {rad}}R=\{s\in R{\text{ nilpotent element}}\}$$If $s\notin {\mathfrak {rad}}R$, the set $\{1,\cdots ,s^{n},\cdots \}$ is a multiplicative set, so using the result about prime ideals in $S^{-1}R$, we can conclude that ${\mathfrak {rad}}R$ is the intersection of all prime ideals in $R$.
• Local rings. A ring $R$ is a local ring if has a unique maximal ideal ${\mathfrak {m}}$. Let $$R/{\mathfrak {m}}$$be the residue field. It's easy to prove that every non-unit element $u\in R$ belongs to a maximal ideal (consider $uR$ and the Prime Ideal Theorem), therefore ${\mathfrak {m}}$ is the ideal of all non-unit elements and every $1-u$ is a unit, with $u\in {\mathfrak {m}}$.

Noetherian Rings

• Ascending chain condition. Given a ring $R$, a $R$-module $M$ is Noetherian if every family of submodules has a maximal element. Consider now a submodule $N\hookrightarrow M$ in a Noetherian module, then the family $$\{N\supseteq J{\text{ finitely-generated submodule of }}M\}$$has a maximal element $J$, which must be equal to $N$, therefore in a Noetherian module every submodule is finitely generated. On the other hand, a module where every submodule is finitely generated, is Noetherian.

Tensor product

• Exactness of Hom. Consider an exact sequence of modules $M\longrightarrow ^{p}N\twoheadrightarrow ^{q}Q$ and, for a general module $R$, the induced sequence $$0\longrightarrow \mathbf {Hom} (Q,R)\longrightarrow ^{q_{*}}\mathbf {Hom} (N,R)\longrightarrow ^{p_{*}}\mathbf {Hom} (M,R)$$first, it's immediate that since $q$ is onto, then $q_{*}$ is 1-1, so the first short sequence is exact. Then, $p_{*}q_{*}=(qp)_{*}=0$, so $\mathbf {im} q_{*}\subseteq \mathbf {ker} p_{*}$. On the other hand, consider $\phi \in \mathbf {ker} p_{*}$, which means that $\phi p=0\longrightarrow \mathbf {ker} q=\mathbf {im} p\subseteq \mathbf {ker} \phi$, so we have the following sequence of maps$$Q\longrightarrow ^{\simeq }N/\mathbf {ker} q=N/\mathbf {im} p\longrightarrow N/\mathbf {ker} \phi \longrightarrow R$$ (where the first map is the inverse of the canonical isomorphism $N/\mathbf {ker} q\longrightarrow ^{\simeq }q(N)=Q$), all the previous compositions define a morphism $\varphi :Q\longrightarrow R$ such that $q_{*}(\varphi )=\phi$, thus $\mathbf {ker} p_{*}\subseteq \mathbf {im} q_{*}$ and the sequence of Hom modules is exact. On the other hand, suppose that the Hom sequence is exact for all $R$, then take $R=Q/\mathbf {im} q$, it's easy to see that $$q_{*}(\pi )=0=q_{*}(0)$$(where $\pi$ is the canonical projection $Q\twoheadrightarrow Q/\mathbf {im} q$), thus by exactness of the Hom sequence (in particular, $q_{*}$ is 1-1), we conclude that $\pi =0\longrightarrow \mathbf {im} q=Q$ and $q$ is onto. Now, since $q$ is onto, we have an isomorphism $j:N/\mathbf {ker} q\longrightarrow Q$ such that $q=j\rho$ (where $\rho$ is the canonical projection), then $$\rho =j^{-1}q=q_{*}(j^{-1})\longrightarrow \rho \in \mathbf {im} q_{*}=\mathbf {ker} p_{*}$$so $\rho p=p_{*}(\rho )=0\longrightarrow \mathbf {im} p\subseteq \mathbf {ker} \rho =\mathbf {ker} q$. On the other hand, if $\rho$ is instead the canonical projection $N\longrightarrow N/\mathbf {im} p$, then $$\rho \in \mathbf {ker} p_{*}\longrightarrow \rho =kq,{\text{ for some }}k\longrightarrow \mathbf {ker} q\subseteq \mathbf {ker} \rho =\mathbf {im} p$$so the original sequence is exact if and only if the Hom sequence is exact for every $R$.
• Exactness of Tensor product. By definition, $\mathbf {Hom} (M\otimes N,R)\simeq \mathbf {Hom} (M,\mathbf {Hom} (N,R))$, so given a exact sequence $$M\longrightarrow ^{p}N\twoheadrightarrow ^{q}Q$$then for every pair $A,B$ of modules, the induced Hom sequence is exact$$0\longrightarrow \mathbf {Hom} (Q,\mathbf {Hom} (A,B))\longrightarrow ^{q_{*}}\mathbf {Hom} (N,\mathbf {Hom} (A,B))\longrightarrow ^{p_{*}}\mathbf {Hom} (M,\mathbf {Hom} (A,B))$$therefore the isomorphic sequence $$0\longrightarrow \mathbf {Hom} (Q\otimes A,B)\longrightarrow ^{(q\otimes 1)_{*}}\mathbf {Hom} (N\otimes A,B)\longrightarrow ^{(p\otimes 1)_{*}}\mathbf {Hom} (M\otimes A,B)$$is exact for every $B$, which implies (for the previous result), that the sequence $$M\otimes A\longrightarrow ^{p\otimes 1}N\otimes A\twoheadrightarrow ^{q\otimes 1}Q\otimes A$$is exact for every module $S$. In particular, consider the exact sequence $$I\longrightarrow R\twoheadrightarrow R/I$$for an ideal $I$, and a $R$-module $M$, then it's easy to prove that $IM\simeq I\otimes _{R}M$, so we have the following exact sequence $$IM\longrightarrow ^{\varphi }M\twoheadrightarrow M\otimes R/I$$which means $M\otimes R/I\simeq M/\mathbf {im} \varphi =M/IM$ (since $\varphi$ is the inclusion $IM\hookrightarrow M$).

Localization

• Multiplicative set. Consider a ring $R$, a subset $S$ is multiplicative if $1\in S$ and $a,b\in S\longrightarrow ab\in S$. Thus a multiplicative set is simply a submonoid of $R$.
• Localization. Consider a ring $R$ and a multiplicative set $S$, then let $S^{-1}R$ be the the quotient of $R\times S$ modulo the equivalence $$(r,s)\sim (r',s')\longleftrightarrow {\text{there is }}u\in S{\text{ s.t. }}u(rs'-sr')=0$$let ${\textstyle {\frac {r}{s}}}$ be the equivalence class of the pair $(r,s)$, then we define the following operations $${\frac {r}{s}}{\frac {r'}{s'}}={\frac {rr'}{ss'}},\qquad {\frac {r}{s}}+{\frac {r'}{s'}}={\frac {rs'+sr'}{ss'}}$$and also define the map $j:R\longrightarrow S^{-1}R$ as $r\longmapsto {\tfrac {r}{1}}$. Also, notice that $j$ maps to zero those elements that are zero divisors of elements in $S$ (that is, $j(r)=0$, then $ur=0$ for some $u\in S$). Notice that if $R$ is an integral domain, then $S=R^{*}=R-\{0\}$ is a multiplicative set and $$S^{-1}R=R_{(0)}=\mathbf {Frac} R$$the field of fractions of $R$. In case $S=R-{\mathfrak {p}}$ is the complement of some prime ideal ${\mathfrak {p}}$, we denote $S^{-1}R$ by $R_{\mathfrak {p}}$.
• Properties of localization. Given a prime ideal ${\mathfrak {p}}\hookrightarrow R$ consider the localization $R_{\mathfrak {p}}$ and the ideal $$m_{\mathfrak {p}}=\{{\tfrac {r}{e}}|r\in {\mathfrak {p}},e\notin {\mathfrak {p}}\}$$this ideal is maximal (since every ideal properly greater than $m_{\mathfrak {p}}$ contains an element ${\tfrac {u}{e}}$, with $u\notin {\mathfrak {p}}$, which is a unit with inverse ${\tfrac {e}{u}}$) and the unique maximal ideal of $R_{\mathfrak {p}}$, thus $(R_{\mathfrak {p}},m_{\mathfrak {p}})$ is a local ring.
• Local property. A property of a ring is local if "true for the ring $A$" is equivalent to "true for $A_{\mathfrak {p}}$, for every prime ideal ${\mathfrak {p}}$".

Chain conditions and special elements

• Noetherian/Artinian ring. A ring $R$ is Noetherian if there is no strictly ascending infinite chain of ideals $$J_{1}\subset \cdots \subset J_{n}\subset \cdots$$The ring is called Artinian if there is no strictly descending infinite chain of ideals.
• Finite algebras and finite type. Given a ring $R$, an algebra $A$ over said ring is defined as $${\begin{cases}A{\text{ finite iff there is an onto }}R^{n}\longrightarrow A\\A{\text{ finite type iff there is an onto }}R[t_{1},\cdots ,t_{n}]\longrightarrow A\end{cases}}$$clearly $R^{n}$ is of finite type, thus every finite algebra is of finite type. Consider a Noetherian ring $R$ and an algebra $A$ of finite type, with a subalgebra $B\hookrightarrow A$ such that $A$ is finite as a $B$-algebra.
• Prime/Irreducible elements. An element $r\in R$ in a ring is prime if $(r)=rR$ is a prime ideal (which means that if $r|ab\longrightarrow r|a\vee r|b$). An element $r\in R$ is irreducible if $r\notin U$ (where $U$ is the group of units in $R$) and $r=ab$ implies that either $a\in U$ or $b\in U$.

Integral ring extensions

• Integral extension. A ring extension $R\hookrightarrow S$ is an integral extension if every element $s\in S$ is the zero of a monic polynomial, that is, in the form$$f=x^{n+1}+a_{n}x^{n}+\cdots +a_{0}\in R[x]$$
• Noether normalization theorem. Given a field $k$ and a $k$-algebra $A$ that is of finite type (that is, it is a finitely generated $k$-algebra), then there are elements $z_{1},\cdots ,z_{n}\in A$ such that the canonical morphism $$k[x_{1},\cdots ,x_{n}]\longrightarrow R=k[z_{1},\cdots ,z_{n}],\qquad x_{k}\longmapsto z_{k}$$is an isomorphism and the ring extension $R\hookrightarrow A$ is integral. In particular, if $A$ is a field (thus we have a field extension $k\hookrightarrow A$), then $R$ is a field, since every element $r\in R^{*}=R-\{0\}$ has an inverse $1/r\in A$ which satisfies a polynomial $${\frac {1}{r^{n+1}}}+{\frac {a_{n}}{r^{n}}}+\cdots a_{0}=0\longrightarrow 1+r(a_{n}+\cdots +a_{0}r^{n})=0$$so, the inverse of $r$ is $-(a_{n}+\cdots +a_{0}r^{n})\in R$ and $R$ is a field. But, a ring of polynomials can be a ring if and only if $n=0$ and thus the extension $k\hookrightarrow A$ is integral. Also, an integral $k$-algebra of finite type is finite as a $k$-vector space.
• Hilbert Nullstellensatz. Given a proper ideal $I\subset k[x_{1},\cdots ,x_{n}]$, there is a maximal ideal ${\mathfrak {m}}$ containing $I$, thus the field extension $$k\hookrightarrow L=k[x_{1},\cdots ,x_{n}]/{\mathfrak {m}}$$is a finite field extension and $(x_{1}+{\mathfrak {m}},\cdots ,x_{n}+{\mathfrak {m}})$ is an element in $L^{n}$ which is a zero of all polynomials in $I$. In particular, if $k$ is algebraically closed, then $L=k$ and every ideal $I$ has a zero in $k^{n}$. Now, every point $u\in k^{n}$ induces a ring homomorphism $$k[x_{1},\cdots ,x_{n}]\longrightarrow k,\qquad f(x)\longmapsto f(u)$$and its kernel ${\mathfrak {m}}_{u}$ is a maximal ideal. Thus we have a map (where $R=k[x_{1},\cdots ,x_{n}]$)$$k^{n}\longrightarrow \mathbf {mSpec} R=\{{\mathfrak {m}}\subseteq k[x_{1},\cdots ,x_{n}]{\text{ maximal ideal}}\}$$This map is 1-1, but not necessarely onto. If $k$ is algebraically closed, then every maximal ideal ${\mathfrak {m}}$ is contained in (an thus equal to) an ideal in the form ${\mathfrak {m}}_{u}$, for some $u\in k^{n}$ (and this ideal is necessarely unique), therefore we have an inverse morphism $$\mathbf {mSpec} R\longrightarrow k^{n}$$Now, consider a field extension $k\hookrightarrow L$ that is the algebraic closure of $k$. This map induces a map $k[x_{1},\cdots ,x_{n}]\rightarrow L[x_{1},\cdots ,x_{n}]$ and so, for every point $u\in L^{n}$ we have a map$$k[x_{1},\cdots ,x_{n}]\longrightarrow L[x_{1},\cdots ,x_{n}]\longrightarrow L$$thus, we defined a map from $L^{n}$ to the space of maximal ideals of $R=k[x_{1},\cdots ,x_{n}]$$$\lambda :L^{n}\longrightarrow \mathbf {mSpec} R$$Now, the space $L^{n}$ comes equipped with an action by the group $G=\mathbf {Gal} _{k}L$ of $k$-linear ring automorphisms of $L$. Then the action of $G$ extends to $L^{n}$ as $$\phi \cdot (u_{1},\cdots ,u_{n})=(\phi u_{1},\cdots ,\phi u_{n})$$Then notice that for every polynomial $f\in k[x_{1},\cdots ,x_{n}]$ and $u\in L^{n}$, we have $f(\phi \cdot u)=\phi \cdot f(u)$, so two elements in $L^{n}$ with the same orbit under $G$, have the same image under $\lambda$.

Ideals

• Radical of an ideal. Given an ideal $I\subseteq R$ in a commutative ring, we define the following $${\sqrt {I}}=\{f\in R:f^{n}\in I{\text{ for some }}n\}$$In particular, ${\sqrt {0}}$ is the ideal of nilpotent elements in $R$, so ${\sqrt {I}}$ is the pullback of ${\sqrt {0}}$ in $R/I$ under the projection $R\rightarrow R/I$. It can be proven that if $R=k[x_{1},\cdots ,x_{n}]$ for an algebraically closed field $k$, then $${\sqrt {I}}=\{f\in R:f(u)=0{\text{ for all }}u\in V(I)\}$$where $V(I)=\{u\in k^{n}:I\subseteq {\mathfrak {m}}_{u}\}$. Clearly ${\sqrt {\sqrt {I}}}={\sqrt {I}}$.
• Operations on ideals. Consider the ring $R=k[x_{1},\cdots ,x_{n}]$$$V(R)=0,\qquad V(0)=k^{n}$$also, given a family of ideals $\{I_{\alpha }\}_{\alpha }$, clearly $$V(\sum _{\alpha }I_{\alpha })=\bigcap _{\alpha }V(I_{\alpha })$$(since ${\textstyle I_{\beta }\subseteq \sum _{\alpha }I_{\alpha }}$, the inclusion $\subseteq$ is obvious, while a point that is a zero for every element of every $I_{\alpha }$, is also a zero of every linear combination of elements from these ideals and such combinations are all the elements in ${\textstyle \sum _{\alpha }I_{\alpha }}$). Now, given two ideals $I,J$ $$IJ=\{\sum _{j}^{k}f_{j}g_{j}|f_{j}\in I,g_{j}\in J,k\in \mathbb {N} \}\subseteq I,J$$then $V(I)\cup V(J)\subseteq V(IJ)$. On the other hand, if $u$ does not belong to $V(I)\cup V(J)$, there are $f\in I,g\in J$ such that $f(u),g(u)\neq 0$, thus $fg\in IJ$ and $(fg)(u)\neq 0$, so $$V(IJ)=V(I)\cup V(J)$$

Quasi-affine varieties and their dimensions

• Zariski topology on $k^{n}$. Consider an algebraically closed field $k$, then for every ideal $I$, we call $V(I)$ Zariski-closed (or z-closed). By the properties of the correspondence $I\longmapsto V(I)$ we proved before, the collection of z-closed subsets of $k^{n}$ form the closed sets of a topology on $k^{n}$, called Zariski topology. Since $k$ is algebraically closed, the correspondence $I\longmapsto V(I)$ is 1-1 between z-closed subspaces and ideals in $R$ such that ${\sqrt {I}}$.
• Noetherian spaces. A topological space $T$ is Noetherian if one of the following equivalent conditions is satisfied
  • There is no infinite descending chain of closed subspaces $T_{0}\supset T_{1}\supset \cdots \supset T_{n}\supset \cdots$
  • Every non-empty collection of closed subspaces has a $\subseteq$-minimal element
  • Every open subset is quasi-compact
• In particular, a sequence $T_{0}\supset T_{1}\supset \cdots \supset T_{n}\supset \cdots$ corresponds to an increasing sequence $I_{0}\subset I_{1}\subset \cdots \subset I_{n}\subset \cdots$, which cannot exists, since $k[x_{1},\cdots ,x_{n}]$ is a Noetherian ring, therefore $k^{n}$ with the Zariski topology is a Noetherian topologica space.
• Irreducible spaces. A topological space $T$ is irreducible if one of the following two conditions holds
  • If $T=A\cup B$, with $A,B$ closed subspaces, either $A=T$ or $B=T$
  • Every two non-empty open subspaces have non-empty intersection
  • Every non-empty open subspace is dense
• An irreducible subspace is a closed subspace which is irreducible in the subspace topology. Given an irreducible subspace $V$ and $$I=\{f\in R:f(u)=0{\text{ for all }}u\in V\}$$then, if $f,g$ are two polynomials such that $fg\in I$, then $(f)(g)=(fg)\subseteq I$ and thus $V\subseteq V(f)\cup V(g)$, therefore, by irreducibility, either $V=V(f)$ or, equivalently, $V=V(g)$, which implies that, assuming the first case $$f(u)=0{\text{ for all }}u\in V\longrightarrow f\in I$$therefore $I$ is a prime ideal. On the other hand, suppose $I$ is prime and that $V(I)=V(J)\cup V(K)=V(JK)$, then $JK\subseteq I$. If $J\not \subseteq I$ and $K\not \subseteq I$, then there are $f\in J,g\in K$ not in $I$ such that $fg\in I$, contrary to $I$ being prime. Therefore irreducible subspaces corresponds to prime ideals.
• Codimension. Given an irreducible subspace $A$ of a Noetherian space $X$, we define its codimension as $$\dim(A,X)=\sup\{k\in \mathbb {N} :\exists {\text{chain }}A=T_{1}\subset \cdots \subset T_{k}\subseteq X{\text{ of irreducible spaces}}\}$$The value of the codimension can also be infinite. The dimension of $X$ is $$\dim X=\sup\{\dim(A,X):A{\text{ irreducible subspace}}\}$$
• Quasi-affine varieties. An affine algebraic variety is an irreducible, z-closed subspace $Z\subseteq k^{n}$. A quasi-affine algebraic variety is a non-empty, z-open subset of an affine variety. So, a quasi-affine variety is one in the form $$A\cap Z\neq \emptyset$$with $A$ a z-open in $k^{n}$ and $Z$ is an affine variety. A function $\phi :Z\rightarrow k$, for $Z$ a quasi-affine variety, is regular at $x\in Z$ if there is a neighborhood $U\ni z$ in $Z$ and polynomials $f,g\in k[x_{1},\cdots ,x_{n}]$ such that $V(g)\cap U=\emptyset$ and $$\left.\phi \right|_{U}={\frac {f}{g}}$$If $\phi$ is regular at every point, we call $\phi$ regular. Let ${\mathcal {O}}(Z)$ be the ring of regular functions on $Z$. In particular, every polynomial $f\in k[x_{1},\cdots ,x_{n}]$ is regular and if $V(f)\cap Z=\emptyset$, then $1/f\in {\mathcal {O}}(Z)$. Now consider the case $Z=V({\mathfrak {p}})$, for a prime ideal ${\mathfrak {p}}$, and take the correspondence $$k[x_{1},\cdots ,x_{n}]=R\ni f\longmapsto f|_{Z}\in {\mathcal {O}}(Z)$$Clearly the kernel of this map is the ideal of polynomials that are zero over $Z$, which is, by what was proved before, to be ${\sqrt {\mathfrak {p}}}={\mathfrak {p}}$ (the radical of a prime ideal is the ideal itself). So, we have a 1-1 map $R/{\mathfrak {p}}\rightarrow {\mathcal {O}}(Z)$. This map is not only 1-1, but also onto, so $$R/{\mathfrak {p}}\simeq {\mathcal {O}}(Z)$$To prove it, consider $\phi \in {\mathcal {O}}(Z)$, then for every point $\zeta \in Z$ there is an open $U_{\zeta }$ and polynomials $f_{\zeta },g_{\zeta }$ such that $V(g_{\zeta })\cap U_{\zeta }=\emptyset$ and $$\left.\phi \right|_{U_{\zeta }}={\frac {f_{\zeta }}{g_{\zeta }}}$$Since $Z$ is a subspace of a Noether space, it is Noether space itself and it is compact, so there are $U_{1},\cdots ,U_{\kappa }$ such that $Z=U_{1}\cup \cdots \cup U_{\kappa }$. Now, by definition $${\frac {f_{i}}{g_{i}}}=\phi |_{U_{i}\cap U_{j}}={\frac {f_{j}}{g_{j}}}\longrightarrow f_{i}g_{j}=g_{i}f_{j}$$on the open subspace $U_{i}\cap U_{j}$. But, an open subspace of an irreducible space is dense, therefore $f_{i}g_{j}=g_{i}f_{j}$ on all $Z$. By definition, every point $z\in Z$ belongs to the complement of some $V(g_{j})$, therefore $$Z\cap \bigcap _{j}V(g_{j})=V({\mathfrak {p}}+(g_{1})+\cdots +(g_{\kappa }))=\emptyset$$But, by Hilbert's Nullstellensatz, if the ideal ${\mathfrak {p}}+(g_{1})+\cdots +(g_{\kappa })$ is proper, it would have a zero, contrary to the last equation, therefore it must be equal to $R=k[x_{1},\cdots ,x_{n}]$, so there is are $a_{1},\cdots ,a_{\kappa }\in R,h\in {\mathfrak {p}}$ such that $$a_{1}g_{1}+\cdots +a_{\kappa }g_{\kappa }+h=1$$which means that $a_{1}g_{1}+\cdots +a_{\kappa }g_{\kappa }=1$ over $Z$. Now, consider the polynomial $f=a_{1}f_{1}+\cdots +a_{\kappa }f_{\kappa }$, then for every $j$ and $x\in U_{j}$ $$g_{j}(x)\phi (x)=f_{j}(x)=\sum _{i}f_{j}(x)a_{i}(x)g_{i}(x)=\sum _{i}a_{i}(x)g_{j}(x)f_{i}(x)=g_{j}(x)\sum _{i}a_{i}(x)f_{i}(x)=g_{j}(x)f(x)$$so, since $g_{j}(x)\neq 0$, we conclude that $\phi (x)=f(x)$ and, since the same can be done for every $x\in Z$, we conclude that $\phi =f$ in $Z$, proving surjectivity.
• Morphism of varieties. Consider two quasi-affine varieties $X\subseteq k^{n},Y\subseteq k^{m}$, then a function $f:X\rightarrow Y$ is a morphism of varieties if there are $f_{1},\cdots ,f_{m}\in {\mathcal {O}}(X)$ such that $$f=(f_{1},\cdots ,f_{m})$$Consider now $U\subseteq Y$ and $\varphi \in {\mathcal {O}}(U)$ and define $(f^{*}\varphi )(x)=\varphi (f(x))$. Now, consider $x\in f^{-1}(U)$ and $y=f(x)$, then there is $V\subseteq U$ and polynomials $p,q$ such that $V(q)\cap V=\emptyset$ and $$\varphi |_{V}={\frac {p}{q}}\rightarrow f^{*}\varphi |_{f^{-1}(V)}={\frac {p(f)}{q(f)}}$$Now, there are open neighborhoods $D_{1},\cdots ,D_{n}$ of $x$ and polynomials $p_{1},\cdots p_{n},q_{1},\cdots ,q_{n}$ such that $V(q_{i})\cap D_{i}=\emptyset$ and $$f_{i}|_{D_{i}}={\frac {p_{i}}{q_{i}}}\rightarrow f=\left({\frac {p_{1}}{q_{1}}},\cdots ,{\frac {p_{n}}{q_{n}}}\right){\text{ in }}D=D_{1}\cap \cdots \cap D_{n}$$So, in the open neighborhood $f^{-1}(V)\cap D$ of $x$ $$f^{*}\varphi |_{f^{-1}(V)\cap D}={\frac {p(f_{1},\cdots ,f_{n})}{q(f_{1},\cdots ,f_{n})}}={\frac {p\left({\frac {p_{1}}{q_{1}}},\cdots ,{\frac {p_{n}}{q_{n}}}\right)}{q\left({\frac {p_{1}}{q_{1}}},\cdots ,{\frac {p_{n}}{q_{n}}}\right)}}$$Finally, we just need to collect all the $q_{i}$ at denominator, so that we are left with a quotient of polynomials multiplied by powers (eventually negative) of $q_{i}$'s. The result is that, Given a morphism $f$ of varieties, this induces a morphism between structure sheaves$$f^{*}:{\mathcal {O}}_{Y}\rightarrow {\mathcal {f}}^{\sharp }{\mathcal {O}}_{X},\qquad (f^{\sharp }{\mathcal {O}}_{X})(U)={\mathcal {O}}_{X}(f^{-1}(U)),\qquad f^{*}\varphi =\varphi \circ f$$On the other hand, a function $f:X\rightarrow Y$ with the previous property is a morphism of varieties, since$$f=(f^{*}\pi _{1},\cdots ,f^{*}\pi _{m})$$where $\pi _{i}:Y\rightarrow k$ is the $i$-th projection (which is equal to the polynomial $x_{i}$, therefore $\pi _{i}\in {\mathcal {O}}(X)$). In particular, a morphism $f:X\rightarrow k$ is regular if and only if it induces a map $f^{*}:{\mathcal {O}}_{Y}\rightarrow f^{\sharp }{\mathcal {O}}_{X}$. In particular, we have a morphism $$\mathbf {Hom} _{\text{Var}}(X,Y):=\{{\text{morphisms of varieties }}X\rightarrow Y\}\rightarrow \mathbf {Hom} _{k{\text{alg}}}({\mathcal {O}}(Y),{\mathcal {O}}(X))$$At the same time, to a morphism $\varphi :{\mathcal {O}}(Y)\rightarrow {\mathcal {O}}(X)$ we can associate the map $\varphi _{*}=(\varphi \pi _{1},\cdots ,\varphi \pi _{m})$ which is a map $X\rightarrow k^{m}$. Suppose $Y=V({\mathfrak {p}})$, for some prime ${\mathfrak {p}}$, (so $Y$ is an affine variety) and take $g\in {\mathfrak {p}}$, then given $u\in X$ $$g\varphi _{*}(u)=g(\varphi \pi _{1}(u),\cdots ,\varphi \pi _{m}(u))=\varphi (g(\pi _{1},\cdots ,\pi _{m}))(u)=0$$since $g(\pi _{1},\cdots ,\pi _{m})=0$ (because $g\in {\mathfrak {p}}$, thus $g|_{Y}=0$ by definition). Thus $\varphi _{*}$ is a morphism $X\rightarrow Y$. Clearly $(f^{*})_{*}=f$ (since the $i$-th coordinate of $(f^{*})_{*}$ is $f^{*}\pi _{i}=\pi _{i}f=f_{i}$). On the other hand, $$(\varphi _{*})^{*}\pi _{i}=\pi _{i}\varphi _{*}=\varphi \pi _{i}$$which means that $(\varphi _{*})^{*}=\varphi$, since $\pi _{i}$ generate ${\mathcal {O}}(Y)$ as a $k$-algebra. Thus Given a variety $Y$ and quasi-variety $X$, there is an isomorphism $$\mathbf {Hom} _{\text{Var}}(X,Y)\longrightarrow \mathbf {Hom} _{k{\text{alg}}}({\mathcal {O}}(Y),{\mathcal {O}}(X))$$Notice that every ${\mathcal {O}}(X)$ is an integral domain (since it is isomorphic to $R/{\mathfrak {p}}$, where $X=V({\mathfrak {p}})$ and ${\mathfrak {p}}$ is prime) and a $k$-algebra of finite type. On the other hand, given a $k$-algebra of finite type that is a domain $$k[x_{1},\cdots ,x_{n}]\rightarrow A$$the kernel ${\mathfrak {p}}$ is a prime ideal (since the quotient by the kernel is isomorphic to $A$, an integral domain), therefore $$A\simeq k[x_{1},\cdots ,x_{n}]/{\mathfrak {p}}\simeq {\mathcal {O}}(V({\mathfrak {p}}))$$ So Every $k$-algebra of finite type that is an integral domain is isomorphic to an algebra ${\mathcal {O}}(X)$, for some affine variety $X$. This proves (together with the previous result) that the category $k{\text{AffVar}}$ of affine varieties is equivalent to the category of domains of finite type over $k$.
• Isomorphism of affine and quasi-affine varieties. Consider an affine variety $X=V({\mathfrak {p}})\subseteq k^{n}$ and $f\in R=k[x_{1},\cdots ,x_{n}]$, then $X-V(f)$ is a quasi-affine variety defined as the locus of zeros of ${\mathfrak {p}}$ which are not zeros for $f$. Consider now the ideal $${\mathfrak {r}}\subseteq k[x_{1},\cdots ,x_{n},t]$$generated by ${\mathfrak {p}}$ and $1-tg$, where $g(x_{1},\cdots ,x_{n})$ is a polynomial such that $g\equiv f{\text{ mod}}{\mathfrak {p}}$. Consider now the two maps $$V({\mathfrak {r}})\rightarrow X-V(f),\qquad (x_{1},\cdots ,x_{n},t)\mapsto (x_{1},\cdots ,x_{n})$$and $$X-V(f)\rightarrow V({\mathfrak {r}}),\qquad (x_{1},\cdots ,x_{n})\mapsto \left(x_{1},\cdots ,x_{n},{\frac {1}{g(x_{1},\cdots ,x_{n})}}\right)$$Both maps are clearly morphisms of varieties (in particular, they are continuous) and are inverse of one another, therefore $V({\mathfrak {r}})\simeq X-V(f)$. Now, $X-V(f)$ is the open subspace of an irreducible space, therefore it is irreducible as well and so is $V({\mathfrak {r}})$, thus ${\mathfrak {r}}$ is prime and $V({\mathfrak {r}})$ is an affine variety. A quasi-affine variety $V({\mathfrak {p}})-V(f)$ is isomorphic to an affine variety $V({\mathfrak {r}})$
• In particular, consider a quasi-affine variety $X=V({\mathfrak {p}})-V({\mathfrak {r}})$ (by definition a quasi affine variety is the complement in an affine variety of an affine variety). For a point $x\in X$ there is a $f\in {\mathfrak {r}}$ such that $f(x)\neq 0$, so $$x\in V({\mathfrak {p}})-V(f)\subseteq X$$and $V({\mathfrak {p}})-V(f)$ is isomorphic to an affine variety, thus Every point of a quasi-affine variety has a neighborhood isomorphic to an affine variety.

Spectrum and localization

• Spectrum of a ring. Given a ring $R$, let $\mathbf {Spec} R$ be the collection of prime ideals in $R$, while let $\mathbf {mSpec} R$ be the collection of the maximal ideals. On $\mathbf {Spec} R$ we define the Zariski topology with closed spaces$$V(I)=\{{\mathfrak {p}}\in \mathbf {Spec} R:I\subseteq {\mathfrak {p}}\}$$for $I$ a generic ideal of $R$. Given a multiplicative set $S$ (i.e. $1\in S$ and $s,t\in S\rightarrow st\in S$), let $S^{-1}R$ be the localization of $R$ at $S$. We denote by $f^{-1}R$ the localization at $S=\{1,f,\cdots ,f^{n},\cdots \}$.
• Absolutely flat rings. Let $R$ be absolutely flat (that is, for every $x$ there is $a$ such that $x=ax^{2}$). If $R$ is local, then $$x(1-ax)=0$$implies that either $x$ is a unit or that $x$ belongs to the unique maximal ideal ${\mathfrak {m}}$ of $R$, which implies that $(1-ax)\notin {\mathfrak {m}}$, which means that $(1-ax)$ is a unit and $x=0$, so An absolutely flat, local ring is a field. In particular, if $S$ is a multiplicative subset, then $S^{-1}R$ is absolutely flat, so if we take $S=R-{\mathfrak {p}}$, we have that $R_{\mathfrak {p}}=S^{-1}R$ is an absolutely flat, local ring, therefore it is a field. So If $R$ is absolutely flat, then $R_{\mathfrak {p}}$ is a field, for every prime ideal ${\mathfrak {p}}$. On the other hand, $$((x)/(x^{2}))_{\mathfrak {p}}=(x)_{\mathfrak {p}}/(x^{2})_{\mathfrak {p}}=0$$where the first equality comes from the isomorphism $(M/N)_{\mathfrak {p}}\simeq M_{\mathfrak {p}}/N_{\mathfrak {p}}$, for $R$-modules $M,N$. In particular, $(x)_{\mathfrak {p}},(x^{2})_{\mathfrak {p}}$ are ideals in $R_{\mathfrak {p}}$, which is a field, so they are equal and their quotient zero. So $$((x)/(x^{2}))_{\mathfrak {p}}=0$$which implies $(x)/(x^{2})=0\rightarrow (x)=(x^{2})$. So $R$ is absolutely flat if and only if $R_{\mathfrak {p}}$ is a field, for every prime ${\mathfrak {p}}$. Consider the canonical map $j:R\longrightarrow S^{-1}R$, the induced map $$j_{*}:\mathbf {Spec} S^{-1}R\longrightarrow \{{\mathfrak {p}}\in \mathbf {Spec} R|{\mathfrak {p}}\cap S=\emptyset \}$$this map has an inverse, sending ${\mathfrak {p}}$ into $S^{-1}{\mathfrak {p}}=\{{\tfrac {r}{s}}|r\in {\mathfrak {p}},s\in S\}$ (which is a prime ideal). The two maps are inverses of each other, thus they are 1-1 onto. In particular, consider $R$ absolutely flat and $S=X-{\mathfrak {p}}$, with ${\mathfrak {p}}$ prime, then $$\{{\mathfrak {r}}\in \mathbf {Spec} R|{\mathfrak {r}}\cap S=\emptyset \}\simeq \mathbf {Spec} R_{\mathfrak {p}}=\{(0)\}$$clearly the condition ${\mathfrak {r}}\cap S=\emptyset$ is equivalent to ${\mathfrak {r\subseteq p}}$, so In an absolutely flat ring, there is no strict inclusion between prime ideals. Suppose that every prime ideal is maximal. Given a prime ideal ${\mathfrak {p}}\hookrightarrow R/{\mathfrak {N}}$, we have the following chain of morphisms $$\mathbf {Spec} (R/{\mathfrak {N}})_{\mathfrak {p}}\hookrightarrow \mathbf {Spec} R/{\mathfrak {N}}\longrightarrow ^{\pi ^{*}}\mathbf {Spec} R$$the second map is an homeomorphism (easy to prove). The first map has image the prime ideals in $R/{\mathfrak {N}}$ contained in ${\mathfrak {p}}$, which are in 1-1 correspondence with the prime ideals contained in $\pi ^{*}({\mathfrak {p}})$. But, by hypothesis, every prime ideal is maximal, thus $\mathbf {Spec} (R/{\mathfrak {N}})_{\mathfrak {p}}$ has a unique element $${\mathfrak {r}}=\left\{{\frac {r}{s}}|r\in {\mathfrak {p}},s\notin {\mathfrak {p}}\right\}$$But, this means that the nilradical of $(R/{\mathfrak {N}})_{\mathfrak {p}}$ is ${\mathfrak {r}}$ (since the nilradical is the intersection of all prime ideals), but the nilradical of $A_{\mathfrak {m}}$ is ${\mathfrak {M}}_{\mathfrak {m}}$, for ${\mathfrak {M}}$ the nilradical of $A$. In this case $$A=R/{\mathfrak {N}}\rightarrow {\mathfrak {M}}=0\rightarrow {\mathfrak {r}}={\mathfrak {M}}_{\mathfrak {p}}=0$$so $(0)$ is the unique prime ideal and $(R/{\mathfrak {N}})_{\mathfrak {p}}$ is a field, for every prime ${\mathfrak {p}}$, so $R/{\mathfrak {N}}$ is absolutely flat. Since $\mathbf {Spec} R\simeq \mathbf {Spec} (R/{\mathfrak {N}})$, we can assume that our ring has ${\mathfrak {N}}=0$ and is thus absolutely flat. Then for $f\in R$ there is $a\in R$ such that $$f(1-af)=0$$so $V(f)\cap V(1-af)=\emptyset$ and $V(f)\cup V({1-af})=\mathbf {Spec} R$, so if the quotient $R/{\mathfrak {N}}$ is absolutely flat, then $\mathbf {Spec} R$ is Hausdorff and totally disconnected.
• Torsion module. Consider a module $M$ and define $T(M)$, called torsion submodule, as the submodule of elements $v\in M$ such that ${\mathfrak {ann}}v$ is non zero. A module such that $T(M)=0$ is called torsion free and $M/T(M)$ is one such modules.
• Faithfully flat rings. Consider $\rho :A\rightarrow B$ such that $B$ is a flat $A$-algebra. Consider the following exact sequence $$0\longrightarrow \mathbf {ker} j\longrightarrow M\longrightarrow ^{j}B\otimes _{A}M$$Tensoring with $B$ we still have an exact sequence (since $B$ is flat) $$0\longrightarrow B\otimes \mathbf {ker} j\longrightarrow B\otimes M\longrightarrow ^{1\otimes j}B\otimes B\otimes M$$but $1\otimes j$ is 1-1, since it has a retraction $a\otimes b\otimes v\longmapsto ab\otimes v$, so $B\otimes \mathbf {ker} j\simeq \mathbf {ker} (1\otimes j)=0$. So, For a flat $A$-algebra $B$, we have that the map $j:M\rightarrow B\otimes _{A}M$ is 1-1 if and only if $B\otimes _{A}M=0$ implies $M=0$. On the other hand, suppose $M\rightarrow B\otimes M$ is 1-1, for every $A$-module $M$. Consider ${\mathfrak {p}}$ prime, then $$A\rightarrow ^{i}B/{\mathfrak {p}}B=A\rightarrow ^{\pi }A/{\mathfrak {p}}\rightarrow ^{j}B\otimes A/{\mathfrak {p}}\simeq B/{\mathfrak {p}}B$$But $j$ is 1-1, so the kernel of $i$ (which is equal to $\rho ^{-1}(B{\mathfrak {p}})$) is equal to the kernel of $\pi$, so $${\mathfrak {p}}=\rho ^{-1}(B{\mathfrak {p}})$$So if the map $j:M\rightarrow B\otimes _{A}M$ is 1-1, then ${\mathfrak {p}}=\rho ^{-1}(B{\mathfrak {p}})$, for every prime ideal ${\mathfrak {p}}\hookrightarrow A$. Now we show that ${\mathfrak {p}}=\rho ^{-1}(B{\mathfrak {p}})$ implies ${\mathfrak {p}}=\rho ^{-1}({\mathfrak {r}})$, for some ${\mathfrak {r}}\in \mathbf {Spec} B$: Take $S=\rho (A-{\mathfrak {p}})\hookrightarrow B$, then there is a prime ideal ${\mathfrak {r}}\hookrightarrow B$ disjoint from $\rho (A-{\mathfrak {p}})$ and containing $B{\mathfrak {p}}$, so ${\mathfrak {p}}\subseteq \rho ^{-1}({\mathfrak {r}})$ and $\rho ^{-1}({\mathfrak {r}})\cap (A-{\mathfrak {p}})=\emptyset$, so ${\mathfrak {p}}=\rho ^{-1}({\mathfrak {r}})$ and $$\rho ^{*}:\mathbf {Spec} B\rightarrow \mathbf {Spec} A$$is onto. Now, clearly this implies that for every maximal ideal ${\mathfrak {m}}$, we have $B{\mathfrak {m}}\subset B$ (otherwise, for a prime ideal ${\mathfrak {r}}\in (\rho ^{*})^{-1}({\mathfrak {m}})$, it would imply that $B=B{\mathfrak {m}}\subseteq {\mathfrak {r}}$), so if $\rho ^{*}$ is onto, then $B{\mathfrak {m}}\subset B$ for every maximal ideal ${\mathfrak {m}}$. Finally, suppose for a maximal ideal ${\mathfrak {m}}$, we have $B{\mathfrak {m}}\subset B$. Take a non zero $v\in M$, we have the following exact sequence $$0\longrightarrow {\mathfrak {ann}}v\longrightarrow {\mathfrak {m}}\longrightarrow M$$where ${\mathfrak {m}}$ is a maximal ideal containing ${\mathfrak {ann}}v$ and the last map is $a\mapsto av$. Tensoring with $B$ we still have the exact sequence $$0\longrightarrow B({\mathfrak {ann}}v)\longrightarrow B{\mathfrak {m}}\longrightarrow B\otimes M$$if $B\otimes M=0$, then $B({\mathfrak {ann}}v)=B{\mathfrak {m}}$. But, $B({\mathfrak {ann}}v)={\mathfrak {ann}}(1\otimes v)=B$ (since $1\otimes v=0$), implying $B{\mathfrak {m}}=B$, absurd, so $B\otimes _{A}M$ is non zero. Therefore, if $B{\mathfrak {m}}\subseteq B$ for every maximal ideal ${\mathfrak {m}}$, then $B\otimes _{A}M=0$ implies $M=0$, for every $A$-module $M$. If any of the previous equivalent conditions is verified by a flat $A$-algebra $B$, then we say that $B$ is a faithfully flat $A$-algebra.

• Fibers. Consider a morphism $f:A\rightarrow B$ and a prime ideal ${\mathfrak {p}}$, then let $B_{\mathfrak {p}}$ be the localization at $f(A-{\mathfrak {p}})$, so the map $f^{*}$ restricts to a map $$\mathbf {Spec} B_{\mathfrak {p}}\longrightarrow \mathbf {Spec} A_{\mathfrak {p}}$$(where we can identify $\mathbf {Spec} A_{\mathfrak {p}}$ with the ideals in $\mathbf {Spec} A$ contained in ${\mathfrak {p}}$). Now, given a morphism $f:A\rightarrow B$ and an ideal $I$, then the morphism $f^{*}$ restricts to $$\mathbf {Spec} B/BI\longrightarrow \mathbf {Spec} A/I$$(where we can identify $\mathbf {Spec} A/I$ with the ideals in $\mathbf {Spec} A$ containing ${\mathfrak {p}}$). Therefore, the map $f^{*}:A\rightarrow B$ restricts to a map $$\mathbf {Spec} B_{\mathfrak {p}}/{\mathfrak {p}}B_{\mathfrak {p}}\longrightarrow \mathbf {Spec} A_{\mathfrak {p}}/{\mathfrak {p}}A_{\mathfrak {p}}$$therefore the fiber $(f^{*})^{-1}({\mathfrak {p}})$ is homeomorphic to $\mathbf {Spec}$ of $$B_{\mathfrak {p}}/{\mathfrak {p}}B_{\mathfrak {p}}\simeq (B_{\mathfrak {p}})\otimes _{A_{\mathfrak {p}}}k({\mathfrak {p}})\simeq (B\otimes _{A}A_{\mathfrak {p}})\otimes _{A_{\mathfrak {p}}}k({\mathfrak {p}})\simeq B\otimes _{A}k({\mathfrak {p}})$$(where $k({\mathfrak {p}})$ is the residue field of $A_{\mathfrak {p}}$). Therefore, given a morphism $A\rightarrow B$, we call $\mathbf {Spec} (B\otimes _{A}k({\mathfrak {p}}))$ the fiber of $f^{*}$ over ${\mathfrak {p}}$.

Grassmanian Manifold

• Grassmanian coordinates. Let $\mathbb {G} _{n,k}$ be the collection of $k$-dimensional subspaces of $\mathbb {P} ^{n}$. From $V\in \mathbb {G} _{n,k}$ we take $k+1$ points $x^{j}=[x^{j}{}_{i}]$ which generate the subspace, then the matrix $${\begin{pmatrix}x^{1}\\\vdots \\x^{n}\end{pmatrix}}$$has rank $k+1$, meaning that the minors of order $k+1$ don't all have zero determinant$$\mathbb {G} _{n,k}\longrightarrow \mathbb {P} ^{{\binom {n+1}{k+1}}-1},\qquad X\longmapsto [p_{i_{0},\cdots ,i_{k}}]$$where $p_{i_{0},\cdots ,i_{k}}$ is the minor of the previous matrix from the columns $i_{0}<\cdots <i_{k}$ (picking different points only changes the matrix by a constant factor, resulting in the same point in the projective space).

Sheave and Schemes

• Sheaves. Given a functor $p:C\rightarrow D$, this induces a map $p_{*}:[D,\mathbf {Set} ]\rightarrow [C,\mathbf {Set} ]$ given by precomposition with $p$. Now, given an object $d\in D$ we define the category $p/d$ with objects the pairs $$(c,p(c)\rightarrow d)$$with a morphism $(c,\phi )\rightarrow (e,\chi )$ is a morphism $f:c\rightarrow e$ such that $\chi \circ p(f)=\phi$. Now, given a copresheaf ${\mathcal {F}}:C\rightarrow \mathbf {Set}$, we define the following diagram $${\mathcal {F}}\circ \pi :p/d\rightarrow C\rightarrow \mathbf {Set}$$(where $\pi$ is the projection $p/d\rightarrow C$ onto the first component). Since $\mathbf {Set}$ is complete, assuming that the categories involved are small enough, the following definition makes sense$$(p^{*}{\mathcal {F}})(d)={\text{colim}}{\mathcal {F}}\circ \pi$$The functor $p^{*}$ is called inverse image functor. Now, given a natural transformation $\delta :{\mathcal {G}}\rightarrow p_{*}{\mathcal {F}}$ we can define for each $d\in D$ a cone ${\mathcal {G}}\circ \pi \rightarrow {\mathcal {F}}(d)$ by sending $(c,\phi )$ to the map $${\mathcal {F}}(\phi )\circ \delta _{c}:{\mathcal {G}}(c)\rightarrow {\mathcal {F}}(p(c))\rightarrow {\mathcal {F}}(d)$$it easy to prove from the definition of morphism in $p/d$ that this is a cone, therefore there is a unique lift to a map $(p^{*}{\mathcal {G}})(d)\rightarrow {\mathcal {F}}(d)$, with this is easy to prove that $p^{*}:[C,\mathbf {Set} ]\rightarrow [D,\mathbf {Set} ]$ extend to a functor which is a left adjoint to $p_{*}$.On the same vein, define the category $d/p$ and $$(p_{!}{\mathcal {F}})(d)={\text{lim}}{\mathcal {F}}\circ \pi$$where $\pi :d/p\rightarrow C$ is the projection like the one above. The functor $p_{!}$ is right adjoint to $p_{*}$, so that we have the following triple of adjunction $$p^{*}\dashv p_{*}\dashv p_{!}:[C,\mathbf {Set} ]\rightarrow [D,\mathbf {Set} ]$$we call this an effective geometric morphism between $[C,\mathbf {Set} ]$ and $[D,\mathbf {Set} ]$.
• Geometric morphism for presheaves on topological spaces. Consider the case of a continuous map $f:X\rightarrow Y$ and set $$[X,\mathbf {Set} ]:=[O(X)^{op},\mathbf {Set} ]$$(with $O(X)$ the topology of the space $X$). The map $f$ induces a functor $f^{-1}:O(Y)\rightarrow O(X)$ between the categories of open subsets. This, in turn, define first a functor $$f_{*}:[X,\mathbf {Set} ]\rightarrow [Y,\mathbf {Set} ]$$and then, by the above construction (and the fact that $O(X)$ is always small, for every space), we other two functors, namely $f_{!},f^{*}$. By unraveling the definition above, the two functors are defined as $$f^{*}{\mathcal {F}}:U\mapsto {\text{colim}}_{f(U)\hookrightarrow V}{\mathcal {F}}(V)$$In particular, we can induce another adjunction $\mathbf {Sh} (X)\rightarrow \mathbf {Sh} (Y)$ by restricting $f_{*}$ to $\mathbf {Sh} (X)$ ($f_{*}$ preserve sheaves), while restricting $f^{*}$ to $\mathbf {Sh} (Y)$ and the map $f^{*}{\mathcal {F}}$ to its associated sheaf. In particular, given a point $x:*\rightarrow X$, we call $x_{*}:\mathbf {Set} \rightarrow \mathbf {Sh} (X)$ the skyscraper functor, while we call the functor $x^{*}:\mathbf {Sh} (X)\rightarrow \mathbf {Set}$ is called the stalk functor at $x$, we also indicate ${\mathcal {F}}_{x}:=x^{*}({\mathcal {F}})$.
• Scheme. Consider a commutative ring $A$ and the corresponding spectrum $\mathbf {Spec} A$. Let $A_{\mathfrak {p}}$, for a point ${\mathfrak {p}}\in \mathbf {Spec} A$, the localization at the complement of ${\mathfrak {p}}$. Given $U\subseteq \mathbf {Spec} A$ open, we define ${\mathcal {O}}(U)$ as the space of the sections $s$ of the map $$\coprod _{{\mathfrak {p}}\in U}A_{\mathfrak {p}}\rightarrow U$$sending $A_{\mathfrak {p}}$ to ${\mathfrak {p}}$, such that for each there is a cover $\{U_{i}\hookrightarrow U\}_{i}$ and $a_{i}\in A$ and $f_{i}\in A$ such that $f_{i}\notin {\mathfrak {p}}$ for every ${\mathfrak {p}}\in U_{i}$ and $$s:{\mathfrak {p}}\longmapsto {\frac {a_{i}}{f_{i}}}\in A_{\mathfrak {p}}$$We call the pair $(\mathbf {Spec} A,{\mathcal {O}})$ a spectrum. A ringed space is a pair $(X,{\mathcal {O}})$ made of a topological space $X$ and a sheaf of rings ${\mathcal {O}}\in \mathbf {Sh} (X)$, then every affine scheme is a ringed space. A morphism of ringed spaces $(X,{\mathcal {O}}_{X})\rightarrow (Y,{\mathcal {O}}_{Y})$ is a continuous function $f:X\rightarrow Y$ and a sheaf homomorphism $$f^{\sharp }:{\mathcal {O}}_{Y}\rightarrow f_{*}{\mathcal {O}}_{X}$$The space $(X,{\mathcal {O}})$ is called locally ringed space if the stalk ${\mathcal {O}}_{x}$ at every $x\in X$ is a local ring. A morphism of locally ringed spaces is a morphism of ringed spaces, such that, for each $x\in X$, the induced map on stalks $${\mathcal {O}}_{Y,f(x)}\rightarrow {\mathcal {O}}_{X,x}$$is a map of local rings (that is, counterimage of the maximal ideal in ${\mathcal {O}}_{X,x}$ is the maximal ideal in ${\mathcal {O}}_{Y,f(x)}$). A ringed space $(X,{\mathcal {O}})$ is an affine scheme if $$(X,{\mathcal {O}})\simeq (\mathbf {Spec} A,{\mathcal {O}}_{\mathbf {Spec} A})$$for some ring $A$. We call ${\mathcal {O}}$ the structure sheaf. Given a spectrum $(\mathbf {Spec} A,{\mathcal {O}})$, notice there is a map, for every $U$ containing ${\mathfrak {p}}$$${\mathcal {O}}(U)\rightarrow A_{\mathfrak {p}},\qquad s\mapsto s({\mathfrak {p}})$$this map induces a morphism ${\mathcal {O}}_{\mathfrak {p}}\rightarrow A_{\mathfrak {p}}$ which is onto: For every $a/f\in A_{\mathfrak {p}}$, with $f\notin {\mathfrak {p}}$, take $U=D(f)$, then $f\notin {\mathfrak {q}}$ for every ${\mathfrak {q}}\in U$, by definition, so define $s\in {\mathcal {O}}(U)$ as constant $=a/f\in A_{\mathfrak {q}}$, then $s({\mathfrak {p}})=a/f$, proving surjectivity. As per injectivity, if $s({\mathfrak {p}})=a/f=0$, for $s\in {\mathcal {O}}(U)$, then there is $g\notin {\mathfrak {p}}$ such that $gf=0$, but then take $V=D(g)\cap U'$ (since $g\notin {\mathfrak {p}}\rightarrow {\mathfrak {p}}\in D(g)$, so ${\mathfrak {p}}\in V$), where $U'$ is an open neighborhood of ${\mathfrak {p}}$ such that $s=a/f$ in $U'$, thus $g\notin {\mathfrak {q}}$ for every ${\mathfrak {q}}\in V$, and so $a/f=0$ in every $A_{\mathfrak {q}}$. So ${\mathcal {O}}_{\mathfrak {p}}\simeq A_{\mathfrak {p}}$ In particular, notice that $A_{\mathfrak {p}}$ is a local ring, therefore ${\mathcal {O}}_{\mathfrak {p}}$ is a local ring and $(\mathbf {Spec} A,{\mathcal {O}})$ is a locally ringed space. A scheme is a locally ringed space $(X,{\mathcal {O}})$ for which there is a cover $\{U_{i}\hookrightarrow X\}_{i}$ such that $(U,{\mathcal {O}}|_{U})$ is an affine scheme.