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

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

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

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

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

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

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

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

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

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

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

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