Users Guide to Hartshorne Algebraic Geometry/Chapter 0

Categories of Commutative Rings and Algebras
The starting point for this section is the definition of a commutative ring: a unital ring with commutative multiplication. In this book you can assume that all rings are commutative, so we will omit the 'commutative' adjective. The most basic rings include


 * $$\mathbb{Z}$$
 * $$\mathbb{Z}/n$$
 * Fields $$\mathbb{F}$$
 * Polynomial rings $$R[x_1,\ldots,x_n]$$

We can relate rings to one another using a morphism of rings. A function $$\phi: R \to S$$ between rings is a morphism of rings if the following two axioms are satisfied we could have stated this succinctly as a function which respects the ring structure. It turns out that rings with ring morphisms form a category $$\textbf{CRing}$$. As an important technical note, there is no zero-ring given by a single element in our category. This category has an initial object given by the ring of integers because given a ring morphisms
 * 1) $$\phi(a + b) = \phi(a) + \phi(b)$$ (Additivity)
 * 2) $$\phi(ab) = \phi(a)\phi(b)$$ (Multiplicativity)
 * $$\phi:\mathbb{Z} \to R$$

the ring morphism axioms forces
 * $$\phi(1) = 1_R$$, $$\phi(-1) = -1_R$$, and $$\phi(n) = \phi\left(\sum_n 1\right) = \sum_n \phi(1)$$

Recall that the category of $$R$$-algebras has objects given by ring morphisms $$R \to S$$ and morphisms given by commutative diagrams $$\begin{matrix} && S \\ & \nearrow\\ R && \downarrow \phi \\ & \searrow \\ && S' \end{matrix}$$ If we consider only algebras, the category $$\mathbb{Z}-\textbf{CAlg}$$ is equivalent to the category $$\textbf{CRing}$$. Note that it is common to consider the categories $$\mathbb{F}_q-\textbf{CAlg}$$, $$\mathbb{C}-\textbf{CAlg}$$, $$\mathbb{Q}_p-\textbf{CAlg}$$. The motivation for why will be readily apparent when considering categories of schemes.

Ideals
One of the ways to construct new rings is by taking quotient rings. An ideal of a ring is a subset $$I \subset R$$ which is Then, we can take the quotient of abelian groups $$R/I$$ and use the multiplicative structure on $$R$$ to construct one on $$R/I$$. The second axiom of ideals guarantees that this is well-defined. This is called a quotient ring. Some typical examples of quotient rings are given by \frac{\mathbb{Z}[x_1,\ldots,x_n]}{(f_1,\ldots,f_k)} $$
 * 1) An abelian group under addition
 * 2) $$R\cdot I \subset R$$
 * $$\mathbb{Z}/(p)$$
 * $$\mathbb{Z}[x]/(x^2 - 5) \cong \mathbb{Z}[\sqrt{5}]$$
 * $$\mathbb{Q}[x]/(x^p-1)$$

Playing with Presentations
As we have seen, there are many ways to construct polynomial ring; but, another interesting technique for creating new polynomial rings is to attach variables which have relations between them. For example, consider $$\mathbb{Z}[x,x^2,x^3]$$. We can relabel the elements we've attached, so we consider the ring $$\mathbb{Z}[X,Y,Z]$$, but there are a couple relations between these variables:
 * $$X^2 - Y, XY - Z$$

note that these two relations can be used to show others such as $$XZ - Y^2$$ and $$X^3 - Z$$. Hence
 * $$\mathbb{Z}[x,x^2,x^3] \cong \frac{\mathbb{Z}[X,Y,Z]}{(X^2 - Y, XY - Z)}$$

Some other examples include
 * $$\mathbb{Z}[x,x^{3/2}] \cong \frac{\mathbb{Z}[X,Y]}{(Y^2 - X^3)} $$
 * $$\mathbb{Z}[x^2,xy,y^2] \cong \frac{\mathbb{Z}[X,Y,Z]}{(XZ - Y^2)}$$
 * $$\mathbb{Z}[x^3,x^2y,xy^2,y^3] \cong \frac{\mathbb{Z}[X,Y,Z,W]}{(XW - YZ, XZ - Y^2, YW - Z^2)}$$

Prime Ideals
There are a special class of ideals called prime ideals: an ideal $$\mathfrak{p}$$ in a UFD $$R$$ is prime if
 * $$xy \in \mathfrak{p} \Leftrightarrow x \in \mathfrak{p} \text{ or } y \in \mathfrak{p}$$

For example, $$(p) \subset \mathbb{Z}$$ is the first known example of a prime ideal. It should be apparent that $$(6)$$ is not a prime ideal since $$ 2\cdot 3 \in (6)$$ but $$2,3 \not\in (6)$$. Now, given an irreducible polynomial $$f \in S[x_1,\ldots,x_n]$$ the ideal $$(f)$$ will be prime. A simple non-example of a prime ideal is given by $$(xy) \subset \mathbb{C}[x,y]$$. This can be generalized to $$(fg) \subset S[x_1,\ldots,x_n]$$. Some other examples of prime ideals include
 * $$(x^2 + 1) \subset \mathbb{Q}[x]$$
 * $$(x-\alpha) \subset \mathbb{C}[x]$$
 * $$(y^2 - x^3 + 1) \subset \mathbb{C}[x,y]$$

If you take the quotient ring of a prime ideal in a UFD $$R$$ you get an integral domain. This means your ring has the following multiplicative property:
 * $$xy = 0$$ if $$x=0$$ or $$y=0$$

For example, in
 * $$ \frac{\mathbb{C}[x,y]}{(y^2 - x^3)}$$

you will never be able to multiply two non-zero elements together to get zero. The two key non-examples of a ring being an integral domain are
 * $$\mathbb{C}[x,y]/(xy)$$ since $$xy = 0$$
 * $$\mathbb{C}[x]/(x^2)$$ since $$ x \cdot x = 0$$

In general, an ideal $$\mathfrak{p}$$ of a ring $$R$$ is called prime if $$R/\mathfrak{p}$$ is an integral domain. If $$R/\mathfrak{m}$$ is also a field, then we call $$\mathfrak{m}$$ a maximal ideal. One useful exercise is to check that for a morphism $$f:R \to S$$ and a prime ideal $$\mathfrak{p}\subset S$$ the inverse image $$f^{-1}(\mathfrak{p})$$ is a prime ideal. The second example motivates the operation of taking radicals of an ideal. Given an ideal $$I\subseteq R$$ we define its radical as"$\sqrt{I} = \{ f \in R : f^k \in I \text{ for some } r \in \mathbb{N} \}$"For example, the radical of the ideal $$((x^2 - y + z)^4(xyz - z^2)^5)\subset \mathbb{Z}[x,y,z]$$ is $$((x^2 - y + z)(xyz - z^2))$$. Given a quotient ring $$R/I$$ we call the ring $$R/\sqrt{I}$$ its reduction; sometimes this is denoted $$(R/I)_{\text{red}}$$. We define the nilradical of a ring $$R$$ as $$\sqrt{(0)}$$. The nonzero elements in the nilradical are called nilpotents.

Eisenstein's Criterion and Constructing Prime Ideals
There is a generalization of Eisenstein's criterion for integral domains: given a ring $$R$$ and a polynomial $$f(x) \in R[x]$$ which can be written as
 * $$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$$

then it cannot be written as a product of polynomials if the following conditions are satisfied: Suppose there exists a prime ideal $$\mathfrak{p}\subset R$$ such that

\begin{align} a_i \in \mathfrak{p} \text{ for } i \neq n \\ a_n \not\in \mathfrak{p} \\ a_0 \not\in \mathfrak{p}^2 \end{align} $$ then $$f(x)$$ cannot be written as a product of polynomials $$g_1(x)\cdots g_k(x)$$.

For example, consider the integral domain $$\mathbb{C}[x]$$ and the polynomial $$f \in \mathbb{C}[x][y]$$ given by
 * $$f(x,y) = 1\cdot y^3 + 0\cdot y^2 + 0\cdot y + x^3\cdot y^0 = y^3 + x^3$$

Using the prime ideal $$(x-1)\subset \mathbb{C}[x]$$ we have that $$a_1,a_2=0 \in (x-1)$$, $$a_3 = 1 \not\in (x-1)$$, and $$a_0 = x^3 \not\in (x-1)^2$$. Hence
 * $$(x^3 + y^3) \subset \mathbb{C}[x,y]$$

is a prime ideal. This example can be extended to show that
 * $$x_1^{k_1} + \cdots + x_n^{k_n} \in \mathbb{C}[x_1,\ldots,x_n]$$

generates a prime ideal.

Nullstellensatz
Now we are in the right place to discuss the foundational theorem of algebraic geometry: Hilbert's nullstellensatz. Here we fix $$k$$ as an algebraically closed field.

Theorem: The maximal ideals of $$\mathfrak{m}\subset k[x_1,\ldots,x_n]$$ are in bijection with the set $$k^n$$.

For example, the kernel of $$\text{ev}_{(1,2)}:\mathbb{C}[x,y] \to \mathbb{C}$$ is the ideal $$(x-1,y-2)$$. This allows one to interpret quotient rings give by ideals $$I \subset k[x_1,\ldots,x_n]$$ as algebraic subsets of $$k^n$$ because an evaluation morphism"$\text{ev}_{(\alpha_1,\ldots,\alpha_n)}:\mathbb{C}[x_1,\ldots,x_n]/I \to \mathbb{C}$"is well-defined only if $$(x_1-\alpha_1,\ldots,x_n-\alpha_n) \subset k[x_1,\ldots,x_n]/I$$ is a maximal ideal. For example, consider the following example and non-example: Now we can interpret rings which are not integral. For example, we saw that $$\mathbb{C}[x,y]/(xy)$$ is not an integral domain. Geometrically, this is the union of the $$x$$ and $$y$$ axes. The other main case of a non-integral ring is a non-reduced ring. For example, $$\mathbb{C}[x,y]/(y^3)$$ is the $$x$$-axis but there is extra algebraic information from the $$y,y^2$$ left over. The way you should interpret this ring as is a fat line.
 * $$\text{ev}_{(1,1)}:\mathbb{C}[x,y]/(y-x^2) \to \mathbb{C}$$ is a well defined morphism since$$(x-1,y-1,y^2 - x) = (x-1,y-1,1^2 - 1) = (x-1,y-1)$$This implies that $$(1,1) \in \{ (a,b) \in \mathbb{C}^2 : b^2 - a\}$$
 * $$\text{ev}_{(1,2)}:\mathbb{C}[x,y]/(y-x^2) \to \mathbb{C}$$ is not a well-defined morphism because $$(x-1,y-2,y^2-x) = (x-1,y-2,4-1) = (1)$$; there is no quotient ring $$R/(1)$$. Hence $$(1,2) \not\in \{ (a,b) \in \mathbb{C}^2 : b^2 - a\}$$

Basic Scheme Theory
We can now confidently define an affine scheme: it is a functor"$\text{Hom}_\mathbf{CRing}(R,-): \mathbf{CRing} \to \textbf{Sets}$"for some fixed commutative ring $$R$$.

Localization
The next basic construction in commutative ring theory is localization. This defines a generalization of inverting the non-zero integers and getting the rational numbers. Let $$S\subset R$$ be a multiplicatively closed subset with unity, meaning $$1 \in S$$ and $$ s,s' \in S \Rightarrow ss' \in S$$. For example, for a fixed element $$f \in R$$ consider the subset $$\{1,s,s^2,s^3,\ldots\}$$. We define a commutative ring $$R[S^{-1}]$$ as follows. First, consider the set $$R\times S / \sim$$ where"$(r,s) \sim (r',s') \text{ if there exists } u \in S \text{ such that } u(rs' - r's) =0$"(don't worry, we will given a motivating example for this seemingly random $$u$$). It is an exercise to verify that this indeed defines an equivalence relation — it is standard to write these equivalence classes as $$r/s$$. These equivalence classes have a well-define commutative ring structure given by"$\frac{r}{s} \cdot \frac{r'}{s'} = \frac{rr'}{ss'} \text{ and } \frac{r}{s} + \frac{r'}{s'} = \frac{rs' + r's}{ss'}$"Some basic examples of localization include "$\mathbb{C}[x,y][S^{-1}] = \mathbb{C}[x,y]_{\mathfrak{p}} = \left\{ \frac{f(x,y)}{g(x,y)} : f,g \in \mathbb{C}[x,y] \text{ and } g(0,0) \neq 0 \right\}$|undefined"The last example is special because it motivates a definition: a ring is local if it has a unique maximal ideal. The pair $$(R_\mathfrak{p},\mathfrak{p})$$ is a local ring.
 * The subset $$S = \{1,p,p^2,p^3,\ldots\} \subset \mathbb{Z}$$ gives the ring $$\mathbb{Z}[S^{-1}] = \mathbb{Z}[1/p]$$. Notice that if we localized by the set $$T = \{1,p^3,p^6,p^9,\ldots\} \subset \mathbb{Z}$$ then this gives the ring $$\mathbb{Z}[1/p^3]$$. But, because we could write $$1/p$$ as $$p^2/p^3$$, these two rings are isomorphic. For brevity, we could just say that we localized $$\mathbb{Z}$$ by $$p$$. Try localizing by some other non-zero integers and see why you find.
 * An important geometric example is given by localizing by some non-zero polynomial $$f = f_1^{i_1}\cdots f_k^{i_k} \in \mathbb{C}[x_1,\ldots,x_n]$$.
 * Given an integral domain $$R$$, we can take the set $$S = R-\{0\}$$. Then, $$R[S^{-1}]$$ is called the field of fractions of the integral domain. (It is an exercise to check that this is a field)
 * Given a ring $$R$$ and a prime ideal $$\mathfrak{p}$$, we can consider the set $$S = R -\mathfrak{p}$$. This is multiplicatively closed because of the properties of primality of an ideal. The localization of $$R$$ by $$S$$ is typically denoted $$R_\mathfrak{p}$$. For example, consider $$(x,y) \subset \mathbb{C}[x,y]$$. The localization can be described as

Basic Module Theory
A $$R$$-module is defined as an abelian group $$M$$ with a fixed ring morphism $$\phi:R \to \text{End}_{\textbf{Ab}}(M)$$. We will use the notation
 * $$r\cdot m := \phi(r)(m)$$ where $$r\in R, m \in M$$

for the ring action on $$M$$. A morphism of $$R$$-modules $$\psi:M\to M'$$ is defined by a commutative diagram $$\begin{matrix} && \text{End}_{\textbf{Ab}}(M) \\ & \nearrow\\ R && \downarrow \psi \\ & \searrow \\ && \text{End}_{\textbf{Ab}}(M') \end{matrix}$$ We can use this construction to build a category of $$R$$-modules which is abelian. This means that it has a zero object, kernels and cokernels, products and coproducts, and images/co-images agree. Please note that we've had to enlarge our category of commutative rings to all rings since the endomorphism ring of an abelian group is generally non-commutative; This is one of the only cases where we use non-commutative unital rings in this book. Typical examples of $$R$$-modules includes Another useful technique for constructing new modules is taking the cokernel of a morphism $$\psi:R^n \to R^m$$. For example, the cokernel of"$\mathbb{C}[x,y,z]^{\oplus 2} \xrightarrow{\cdot (x^4 + y^4 + z^4 - 1)\oplus \cdot (x^4 - y^2 + z^2 + 1)} \mathbb{C}[x,y,z]$"is $$\mathbb{C}[x,y,z]/(x^4 + y^4 + z^4 - 1,x^4 - y^2 + z^2 + 1)$$. We can generalize this example using exact sequences. A sequence of objects in an abelian category $$ M_1\xrightarrow{\phi_1}M_2 \xrightarrow{\phi_2} \to \cdots \xrightarrow{\phi_{n-1}} M_n $$ is called exact if each
 * the zero object $$0$$
 * ideals $$I\subset R$$
 * direct sums, such as $$M_1\oplus \cdots \oplus M_k$$
 * a morphism $$\phi:R\to S$$ of rings gives the structure of an $$R$$-module on the underlying abelian group of $$S$$
 * $$\frac{\text{Ker}(\phi_i)}{\text{Ker}(\phi_{i-1})} \cong 0$$

in the last example, we had the exact sequence
 * $$ R^{\oplus2} \to R \to M \to 0$$

In general, if there is an exact sequence
 * $$R^n \to R^m \to M \to 0$$

for finite integers $$m,n$$, then we say that the module is of finite-type. If there is just a sequence
 * $$R^n \to M \to 0$$

then we say that the module is finite. For example, the module
 * $$k[x_1,x_2,\ldots] \to k \to 0$$

is finite but not finite-type since the kernel of the non-trivial morphism is the ideal
 * $$(x_1,x_2,\ldots) \cong \bigoplus_{i=1}^\infty R\cdot x_i$$

Tensor Products

 * construct tensor products for modules
 * construct tensor products of algebras
 * show that tensor products of integral domains are integral
 * show that $$k[\underline{x}]/(f(\underline{x})\otimes_k k[\underline{y}]/(g(\underline{y})) \cong k[\underline{x},\underline{y}]/(f(\underline{x}, g(\underline{y}))$$
 * show $$k[\underline{x}]/(f(\underline{x}))\otimes_{k[\underline{x}]}k[\underline{x}]/(g(\underline{x})) \cong k[\underline{x}]/(f(\underline{x}),g(\underline{x}))$$

Finiteness, Chain Conditions
If we have an $$R$$-algebra $$R \to S$$ we say that $$S$$ it is a finite if it is finite as a module. We say that it is of finite-type if there exists a surjective morphism $$R[x_1,\ldots,x_n] \to S$$, implying that
 * $$S \cong \frac{R[x_1,\ldots,x_n]}{(f_1,\ldots,f_k)}$$

There are a couple other notions of "finiteness" which appear in commutative algebra called chain conditions. We say call a sequence of $$R$$-modules
 * $$M_1\subseteq M_2\subseteq \cdots$$

an ascending chain and
 * $$ N_1\supseteq N_2 \supseteq \cdots$$

a descending chain. They satisfy the ascending chain condition or descending chain condition if there is some $$k$$ such that $$M_k=M_{k+1}=\cdots$$, $$N_k=N_{k+1}=N_{k+2}=\cdots$$. If there exist chains
 * $$M_1\subseteq M_2 \subseteq \cdots$$ where $$ \cup^\infty_{i=1}M_i = R$$

or
 * $$R\supseteq N_2 \supseteq \cdots$$

then we say $$R$$ is Noetherian or Artinian, respectively. One can show that every Artinian ring is Noetherian. The basic examples of Noetherian rings include A simple non-example is given by the ring $$k[x_1,x_2,x_3,\ldots]$$ where $$k$$ is a field. There is a fundamental theorem in algebra called Hilbert's Basis Theorem stating:"Theorem: If $R$ is Noetherian, then $R[x_1,\ldots,x_n]$ is Noetherian"Hence all rings of the form"$\frac{R[x_1,\ldots,x_n]}{(f_1,\ldots,f_k)}$"are Noetherian. Artinian rings are much simpler than Noetherian rings:"Theorem: Every Artin ring is a finite product of Artin local rings."All we have to analyze is the structure of an Artin local ring $$(R,\mathfrak{m})$$. Notice that we have a descending chain"$R \supseteq \mathfrak{m} \supseteq \mathfrak{m}^2 \supseteq \mathfrak{m}^3 \supseteq \cdots$"which eventually stabilizes at some $$\mathfrak{m}^k$$; this is the zero ideal $$(0)$$. We can use this to show the underlying $$R/\mathfrak{m}$$-vector space of $$R$$ is finite dimensional. Some examples of artin local rings are
 * Fields
 * $$\mathbb{Z}$$
 * Finite algebras over fields
 * Quotients of Noetherian rings.
 * $$(\mathbb{C}[x]/(x^5), (x))$$
 * $$(\mathbb{Q}[x,y]/((x-1)^3,(x-1)^2(y-2),(y-2)^5)), (x-1,y-2))$$

Integrality
Given a morphism of commutative rings $$R \to R'$$ we say an element $$x \in R' $$ is integral over $$R$$ if there is a monic polynomial $$f(t) = t^n + a_{n-1}t^{n-1} +\cdot + a_1t + a_0 \in R[t] $$and a morphism"$\frac{R[t]}{f(t)} \to R'$"sending $$t \mapsto x$$. For example, $$\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$$ is integral over $$\mathbb{Z}$$ since"$\frac{\mathbb{Z}[t]}{(t^2 + 5)} \cong \mathbb{Z}[\sqrt{-5}]$"Adjoining all of the integral elements $$\{x_i\}$$ $$R$$ is called the integral closure of $$R$$ in $$R'$$. An integral domain $$R$$ is called integrally closed if every element in its fraction field is integral over $$R$$. For example, we can compute the integral closure of"$R = \frac{\mathbb{C}[x,y]}{(x^2 - y^3)}$"fairly easily. Since it is isomorphic to the ring $$\mathbb{C}[x,x^{3/2}]$$ we should see immediately that $$x^{1/2}$$ is not contained in $$R$$. Adjoining this element to $$R$$ gives a ring isomorphic to $$\mathbb{C}[s]$$. As an exercise, try and unpack"$\frac{\mathbb{C}[x,y,z,w]}{(x^2 - y^5 - y^3, z^3 - w^4)}$"TODO:

- hyperelliptic curves

- quotient fields of curves

- https://math.stackexchange.com/questions/2304521/why-is-this-coordinate-ring-integral-over-kx

- rings of integers

Structures

 * tensor products of modules/algebras... localization of modules
 * categories of commutative algebras
 * support of modules
 * graded rings
 * colimits and stalks
 * limits, completions, p-adics
 * valuations - https://en.wikipedia.org/wiki/Valuation_(algebra)#P-adic_valuation_on_a_Dedekind_domain
 * dimension
 * transcendence degree
 * functor formalism
 * categorical structures such as pullbacks and pushforwards
 * kahler differentials/ basic differential algebra
 * smooth algebras/smooth morphisms
 * complete intersection/local complete intersection morphisms
 * etale algebras/etale morphisms
 * galois theory with useful terminology...
 * same w/ algebraic number theory terminology...
 * basic homological algebra, ext, tor
 * koszul complexes
 * derived categories
 * grothendieck group
 * hilbert polynomial
 * grobner bases/elimination theory - https://mathoverflow.net/questions/60957/how-to-determine-whether-an-ideal-is-prime-or-not-by-an-algorithm

Theorems

 * Eisenstein's Criterion
 * Primary Decomposition
 * Noether Normalization
 * Going up and down

Constructions

 * Smooth manifolds
 * Morphisms
 * Vector Bundles
 * Topological K-theory
 * de-Rham Cohomology
 * Complex manifolds and sheaves
 * hodge decomposition of complex manifolds

Theorems

 * Whitney embedding theorem
 * Submersion theorem
 * Sard's theorem