Dimension Theory

by iarnab_kundu, Jan 15, 2019, 12:01 PM

Here we talk about the subtle property of schemes called dimension. We first try to motivate the definition and then we make attempts to define the notion.

Our intuition is derived from the theory of Manifolds, where we define the dimension of a real manifolds at a point $x\in\mathbb{M}$ to be the number of free variables locally at $x$ , or equivalently the dimension of the tangent space at $x$ . Similarly given a complex manifold we can define the complex dimension at $x\in M$ to be the dimension of the tangent space at $x$ . Therefore if we define the dimension of a complex variety to the dimension of the tangent space, it does not work out because generally the complex variety has lot of singularities and therefore the tangent space defined does not realize our intuition. This is the first difficulty one faces. So we cannot go to tangent spaces, therefore we look at the stalk at $x\in M$ itself. Given a complex manifold we have a ring of (real or complex) functions on it. Therefore we can localize and define the local ring(or the stalk) at $x$ to the ring of germs of functions at $x\in M$ . This does indeed seem to perform better than the tangent space. This can be shown by a few examples. Let $C$ be the curve $y^2-x^3=0$ in the complex plane( $\mathbb{C}^2$ ). This curve has a singularity at the origin and the tangent space is two dimensional at that point. However the local ring at $(0,0)$ is $A_{(0,0)}$ where $A\coloneqq\mathbb{C}[X,Y]/(Y^2-X^3)$ . We can show that this ring is one dimensional and therefore captures our intuition better.

This post has been edited 1 time. Last edited by iarnab_kundu, Jan 15, 2019, 12:01 PM

algebraic geometry

Commutative Algeba

An important diagrama

by iarnab_kundu, Jan 2, 2019, 5:56 PM

All the algebras and rings considered are commutative. Let $A,B$ be a $C$ algebras. Suppose we have a morphism $F:A\to B$ which gives an $A$ -algebra structure on $B$ . Let $\Delta:A\otimes_C A\to A$ be a diagonal morphism, that is the map given by $a\otimes a'\mapsto aa'$ . Let $\psi:A\otimes A\to A\otimes B$ be the morphism given by $a\otimes a'\mapsto a\otimes f(a')$ . Suppose $\phi:A\otimes B\to B$ be a morphism $a\otimes b\mapsto f(a)b$ . Then the following diagram commutes, $B$ is the fiber product over....

Proof- Given a ring $R$ we consider

algebraic geometry

Commutative Algeba

Open Immersions

by iarnab_kundu, Dec 17, 2018, 5:27 PM

Definition- Let $U\subset X$ be an an open subset of a scheme $X$ . Then $U$ has a canonical sub-scheme structure given by restricting the structure sheaf of $X$ to $U$ , that is $\mathcal{O}_U=\mathcal{O}_X |_U$ . We denote this as $U\hookrightarrow X$
Definition- Given a map of schemes $V\to X$ , we say that it is an open immersion if, firstly on the topological spaces induces a homeomorphism into an open subset $U\subset X$ , and secondly it factors as $V\to U\hookrightarrow X$ .

Proposition- Suppose $f:Y\to X$ be a morphism of schemes, such that the image $f(Y)\subset U$ where $U$ is an open subset of $X$ . Then it factors via $f':Y\to U$ and $i:U\hookrightarrow X$ .

algebraic geometry

Points on Schemes

by iarnab_kundu, Dec 17, 2018, 11:28 AM

Definition- Let $S$ be a scheme, and $K$ be a field. Then a $K$ -point on $S$ is a morphism $x:\text{Spec}(K)\to S$ . The set of $K$ -points on $S$ shall be denoted by $S(K)$ .

Given a point $p\in S$ we denote the local ring of $S$ at $p$ to be $\mathcal{O}_{S,p}$ . The maximal ideal of the this local ring is denoted to be $p$ , excusing the abuse of notation. The residue field of $S$ at $p$ is denoted to be $\kappa(p)\coloneqq \mathcal{O}_{S,p}/p$ .

Proposition- There is a natural bijection between the set of morphisms $\text{Spec}(K)\to S$ with the image as $p\in S$ and the set of embeddings $\kappa(p)\hookrightarrow K$ .
Proof- Suppose $x:\text{Spec}(K)\to S$ is a morphism with image $p$ . Then we get a map induces on local rings $\mathcal{O}_{S,p}\to\mathcal{O}_{\text{Spec}(K),q}$ , where $q$ is the unique point in $\text{Spec}(K)$ . But $\{q\}$ is open and therefore $\mathcal{O}_{\text{Spec}(K),q}=K$ . The morphism $\mathcal{O}_{S,p}\to K$ , induces a unique morphism $\kappa(p)\hookrightarrow K$ .

Now let $\kappa(p)\hookrightarrow K$ be an embedding. Suppose $p\in U=\text{Spec}(A)$ be an open affine sub-scheme of $S$ . Thus we have a morphism $A_p/p\hookrightarrow K$ which gives us a unique morphism $A\to K$ and this gives us a unique morphism $\text{Spec}(K)\to U=\text{Spec}(A)$ . We can compose with the open immersion $U\hookrightarrow S$ to get a morphism $\text{Spec}(K)\to S$ with image $p$ .

This post has been edited 1 time. Last edited by iarnab_kundu, Dec 17, 2018, 11:50 AM

algebraic geometry

Topology

by iarnab_kundu, Dec 12, 2018, 8:44 PM

For us here we consider $X,Y$ and $Z$ to be topological spaces, unless otherwise mentioned. Maps are NOT always assumed to be continuous unless otherwise specified.

Definition- We say $f:Y\to X$ is universally closed, if for every (Hausdorff?) base change $f\times\text{id}:Y\times Z\to X\to Z$ is a closed map.
Definition- We say that $f:Y\to X$ is a proper map, if pre-image of every quasi-compact subset of $Y$ is quasi-compact.

Proposition- (If $Y$ is Hausdorff?) $f$ (continuous?) is universally closed if and only if it is proper.

Definition- We say $f:Y\to X$ is universally open, if for every (Hausdorff?) base change $f\times\text{id}:Y\times Z\to X\to Z$ is an open map.

Proposition-

Direct-Sum

by iarnab_kundu, Dec 10, 2018, 2:14 PM

Proposition- Let $\mathcal{C}$ be an abelian category. Let $M_1,M_2$ be two objects. Then $F$ with four maps $i_1:M_2\to F, p_1:F\to M_1$ and $i_2:M_2\to F, p_2:F\to M_2$ is the direct sum of $M_1$ and $M_2$ if and only if we have that $i_1p_1+i_2p_2=\text{id}_F$ . Where the equation is viewed in $\text{Hom}(F,F)$ .
Proof- We have a morphism $(i_1+i_2):M_1\oplus M_2\to F$ and $(p_1,p_2):F\to M_1\oplus M_2$ . Then $(i_1+i_2)\circ (p_1,p_2)=(\text{id}_{M_1},\text{id}_{M_2})$ . Also $(p_1,p_2)\circ (i_1+i_2)=p_1\circ i_1+p_2\circ i_2=\text{id}_F$ .

This post has been edited 4 times. Last edited by iarnab_kundu, Dec 10, 2018, 2:25 PM

Flatness

by iarnab_kundu, Dec 9, 2018, 10:31 PM

Proposition- Let $M$ be an $A-$ module. Given an exact sequence of $A-$ modules $N'\to N\to N''\to 0$ , then the tensored sequence $N'\otimes M\to N\otimes M\to N''\otimes M\to 0$ is exact.

We say that tensoring is "right exact". Thus we can consider its left derived functors. We shall call them the $\text{Tor}$ functors. Given an $A-$ module $M$ , we consider its free(or projective) resolution $F_1\to F_0\to M\to 0$ . Then we define $\text{Tor}^A_n(N,M)$ to be the homology of the complex $\cdots\to N\otimes F_1\to N\otimes F_0\to 0$ . As we might check, we have that $\text{Tor}_0(N,M)=N\otimes M$ .

Proposition- Given an exact sequence of $A-$ modules $0\to N'\to N\to N''\to 0$ we have a long exact sequence $\cdots\to\text{Tor}_1(N',M)\to\text{Tor}_1(N,M)\to\text{Tor}_0(N'',M)\to N'\otimes M\to N\otimes M\to N''\otimes M\to 0$ .

Some computations-
(a) Let $F$ be a free module. Then $0\to F\to F\to 0$ is a free resolution of $F$ . Thus $Tor_n(N,F)=0$ , for $n\ge 1$ . Thus free modules are flat, as one can prove otherwise.
(b) We can extend the above result for projective modules, and prove that they are flat.
(c) Let $M=M_1\times M_2$ . Then we have a

Examples-
(a) Let $f\in A$ be an (non-zero divisor?) element. Then $A_f$ is flat over $A$ .
Proof- $A_f=\text{colim}_n f^{-n}A$ . Thus it commutes with colimits, and therefore with kernels.
What I have in mind is this. Given an exact sequence $0\to M'\to M\to M''$ we have exact sequences $0\to f^{-n}M'\to f^{-n}M\to f^{-n}M''$ . Therefore we can take the colimit and preserve the kernel.
(b) Let $p$ be a prime ideal. Then $A_p$ is flat over $A$ .
Proof- Can we use similar idea as above?

Locality of Flatness-
Proposition- Let $M$ be an $A-$ module, then the following conditions are equivalent
(a) $M$ is flat over $A$
(b) $M_p$ is flat over $A_p$ for all prime ideals $p$ in $A$
(c) $M_P$ is flat over $A_m$ for all maximal ideals $p$ in $A$
Proof-
(a) $\implies$ (b)
We know $A_p$ is exact over $A$ and we know that $M_p=M\otimes A_p$ .

Proposition- Let $M$ be an $A-$ module, then the following conditions are equivalent
(a) $M$ is flat over $A$
(b) For all ideals $I$ in $A$ , the tensored map $I\otimes M\to A\otimes M$ is injective
(b) For all finitely generated ideals $I$ in $A$ , the tensored map $I\otimes M\to A\otimes M$ is injective.

Proposition- $$

This post has been edited 3 times. Last edited by iarnab_kundu, Dec 10, 2018, 1:56 PM

Commutative Algeba

Finite-Type Morphisms

by iarnab_kundu, Dec 9, 2018, 7:04 PM

Definition- Let $f:Y\to X$ be a morphism of schemes. Then for every open affine sub-scheme $U=\text{Spec}(A)$ of $X$ we have that the pre-image $f^{-1}(U)$ can be covered by affine opens $V_i=\text{Spec}(B_i)$ of $Y$ such that for each $i$ we have that $B_i$ is a finitely generated $A$ algebra via the induced morphism.

Example- Let $A$ be a ring. Then any basic open affine $\text{Spec}(A_f)\to\text{Spec}(A)$ is locally of finite type. In fact it is of finite-type.

Proposition- Let $f:Y\to X$ be a morphism of schemes. Suppose there is an open affine cover $U_i=\text{Spec}(A_i)$ such that for each $i$ we have that $f^{-1}(U_i)=\cup (V_{ij}=\text{Spec}(B_{ij}))$ where $B_{ij}$ is a finitely-generated $A_i$ algebra for each $j$ . Then $f$ is locally of finite type.
Lemma- Let $f:Y=\text{Spec}(B)\to X=\text{Spec}(A)$ be a morphism of affine schemes, such that there are $f_i\in A$ for which $B_{f(t_i)}$ is a finite-type algebra over $A_{t_i}$ . Then $B$ is of finite type over $A$ .
Proof- Since $X$ is quasi-compact, we extract a finite sub-cover $\{D(t_i)\}_{i=1\ldots n}$ of $X$ . Then $D(t_1,\ldots,t_n)=X$ implies that the radical of the ideal $(t_1,\ldots,t_n)$ equals $1$ .
Let $\{\alpha_{ij}\}_{j=1\ldots N_i}\in B$ be a finite generating set of the algebra $B_{f(t_i)}$ over $A_{t_i}$ . We consider the set $\{\alpha_{ij}\}_{i=1\ldots n, j=1\ldots N_i}$ , which we claim to be the generating set of $B$ over $A$ . Let $b\in B$ . By the definition of the generators, for each $i$ there exists $m_i$ such that $f^{m_i}(b-b_i)=0$ where $b_i=\sum_j\beta_i\alpha_{ij}$ . Since radical of the ideal $(t_1^{m_1},\ldots,t_n^{m_n})$ equals the radical of the ideal $(t_1,\ldots,t_n)$ . Thus there are elements $a_i\in A$ such that $\sum a_if_i^{m_i}=1$ , and therefore $b=\sum a_if_i^{m_i}b_i$ . Thus we are done.
Proof- Let $U=\text{Spec}(A)$ be an affine open of $X$ . Then $g$

Proposition- Let $f:Y\to X$ be a finite type morphism of schemes. Then for any open affine subset $U=\text{Spec}(A)$ of $X$ and $V=\text{Spec}(B)$ of $Y$ such that $f(V)\subset U$ we have that $B$ is an A-algebra of finite type.

Definition: Let $f:Y\to X$ be a morphism of schemes. Then $f$ is said to be of finite type if it is locally of finite type and quasi compact.

Proposition: Let $f:Y\to X$ be a morphism of schemes. Suppose there is an open affine cover $U_i=\text{Spec}(A_i)$ such that for each $i$ we have that $f^{-1}(U_i)=\cup_{j=1}^{N_i} (V_{ij}=\text{Spec}(B_{ij}))$ (pre-image covered by finitely many such affine opens) where $B_{ij}$ is a finitely-generated $A_i$ algebra for each $j$ . Then $f$ is of finite type.
Proof- Follows from the above proposition and the proposition about quasi-compact maps.

Proposition: Let $f:Y\to X$ be a finite type morphism of schemes. Then for any open affine subset $U=\text{Spec}(A)$ of $X$ and $V=\text{Spec}(B)$ of $Y$ such that $f(V)\subset U$ we have that $B$ is an A-algebra of finite type.
Proof- By the definition

This post has been edited 2 times. Last edited by iarnab_kundu, Dec 9, 2018, 9:35 PM

algebraic geometry

Quasi-Compact Morphisms

by iarnab_kundu, Dec 9, 2018, 6:07 PM

Def- Let $X,Y$ be schemes. A morphism $f:Y\to X$ is said to be quasi-compact if the pre-image of every open quasi-compact subset of $X$ is an open quasi-compact subset of $Y$ .
At this juncture it seems to me to be just a topological condition.

Proposition- $f:Y\to X$ is quasi compact iff the pre-image of every open affine sub-scheme of $X$ can be written as a finite union of open affine sub-schemes of $Y$ .
Proof- Every open quasi-compact subset can be written as a finite union of open affine subsets. Also note that any affine scheme is quasi-compact.

Lemma- Let $f:\text{Spec}(B)\to\text{Spec}(A)$ be a morphism of schemes, then $f^{-1}(D(r))=D(f(r))$ .
Corollary- A map of affine schemes is quasi-compact.

Proposition- $f:Y\to X$ is quasi-compact iff there is a finite open affine cover $U_i=\text{Spec}(A_i)$ of $X$ such that for each $i$ the pre-image $f^{-1}(U_i)$ can be written as a finite union of open affines of $X$ .
Proof- Suppose there is an open affine cover like the above and we want to show that $f$ is quasi-compact. Thus we take an open affine $U=\text{Spec}(A)$ of $X$ . Then we cover $U$ by open affines $W_j=\text{Spec}(C_j)$ such that for each $j$ there exists an $i$ such that $W_j=D(r_j)\subset U_i$ is a basic open set. Since $U$ is quasi-compact we extract a finite affine sub-cover from that. Let $f^{-1}(U_i)=\cup_k V_{ik}=\text{Spec}(B_{ik})$ , and $f_ik:A_i\to B_{ik}$ . Then $f^{-1}(V_j)=\cup_k D(f_{ik}(r_j))$ is a finite union of affine opens. Since there is a finite sub-collection of open affines $V_j$ we are done.

Proposition-
(a) Closed immersions are quasi-compact.
(b) Composition of quasi-compact morphisms is quasi-compact.
(b) Base change of quasi-compact is quasi-compact.
Proof- (a) OMITTED

Warning! Open immersions are NOT, in general, quasi-compact.

This post has been edited 4 times. Last edited by iarnab_kundu, Dec 9, 2018, 10:06 PM

algebraic geometry

direct limit

by iarnab_kundu, Nov 26, 2018, 2:00 PM

We consider a locally small category $\mathcal{C}$ . Let $I$ be a small category, which is called the "diagram" category.
Given a covariant functor $\alpha:I\to\mathcal{C}$ we call it a "diagram" in $\mathcal{C}$ .
Example- We consider $I$ is the category of natural numbers, that is for each natural number $n\in\mathbb{N}$ there is an object in $I$ , and there is a unique morphism $n\to m$ if and only if $n\le m$ . Then a functor $\alpha:I\to\mathcal{C}$ is said to be a direct system in the category.
In Algebra we are interested in not only the diagrams in $\mathcal{C}$ but also in the "maps" between them. Let $\alpha,\beta:I\to\mathcal{C}$ be two diagrams. Then a map $m:\alpha\Rightarrow\beta$ is the data of a morphism $m(i):\alpha(i)\to\beta(i)$ for each $i\in I$ such that the (DIAGRAM BELOW) commutes.
Remark-Notice that this generalizes our notion of maps between directed systems.
Remark2- Notice that a map between $\alpha,\beta$ above is just a natural transformation between $\alpha,\beta$ .
Therefore we are naturally interested in studying the category $\mathcal{C}^I$ , the category of "diagrams in $\mathcal{C}$ ".
Notice that there are "constant diagrams". Given $A\in\mathcal{C}$ we have a diagram $C_A:I\in\mathcal{C}$ such that $C_A(i)=A$ for all $i\in I$ , and for each morphism $\phi:i\to j$ we have $C_A(\phi)=\text{id}_A$ . We note that we have a functor $C:\mathcal{C}\to\mathcal{C}^I$ given by $A\to C_A$ . In fact it is interesting to note that the map $\text{Hom}(A,B)\to\text{Hom}(C_A,C_B)$ is a bijection. Thus it is fully faithful. Now we fix a diagram $\alpha:I\to\mathcal{C}$ . We have a contravariant functor $f_{\alpha}:\mathcal{C}^{op}\to\text{Sets}$ given by $C\mapsto\text{Hom}(C_A,\alpha)$ .
Definition- An element $\mathfrak{A}$ is said to be the "limit" of $\alpha$ if it represents the functor $f_{\alpha}$ , that is $f_{\alpha}$ is isomorphic to $h_{\mathfrak{A}}$ .

category theory

Cracking the Nut Open

by iarnab_kundu, Jan 15, 2019, 12:01 PM

by iarnab_kundu, Jan 2, 2019, 5:56 PM

by iarnab_kundu, Dec 17, 2018, 5:27 PM

by iarnab_kundu, Dec 17, 2018, 11:28 AM

by iarnab_kundu, Dec 12, 2018, 8:44 PM

by iarnab_kundu, Dec 10, 2018, 2:14 PM

by iarnab_kundu, Dec 9, 2018, 10:31 PM

by iarnab_kundu, Dec 9, 2018, 7:04 PM

by iarnab_kundu, Dec 9, 2018, 6:07 PM

by iarnab_kundu, Nov 26, 2018, 2:00 PM