Users Guide to Hartshorne Algebraic Geometry/Chapter 0

References

1. Algebraic Geometry - Hartshorne
2. Arithmetic - Serre
3. https://people.ucsc.edu/~weissman/Math222A/SerreAnn.pdf
4. http://web.mit.edu/18.705/www/12Nts-2up.pdf (for basic commutative algebra)
5. https://arxiv.org/pdf/1605.04832.pdf (for more advanced commutative algebra)
6. Algebraic Number Theory, A Computational Approach - Stein

Basic Commutative Algebra

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

1. $\phi (a+b)=\phi (a)+\phi (b)$ (Additivity)
2. $\phi (ab)=\phi (a)\phi (b)$ (Multiplicativity)

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

$\phi :\mathbb {Z} \to R$

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

1. An abelian group under addition
2. $R\cdot I\subset R$

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

• $\mathbb {Z} /(p)$
• $\mathbb {Z} [x]/(x^{2}-5)\cong \mathbb {Z} [{\sqrt {5}}]$
• $\mathbb {Q} [x]/(x^{p}-1)$
• ${\frac {\mathbb {Z} [x_{1},\ldots ,x_{n}]}{(f_{1},\ldots ,f_{k})}}$

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

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

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

$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}$

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

${\begin{aligned}a_{i}\in {\mathfrak {p}}{\text{ for }}i\neq n\\a_{n}\not \in {\mathfrak {p}}\\a_{0}\not \in {\mathfrak {p}}^{2}\end{aligned}}$

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:

• ${\text{ev}}_{(1,1)}:\mathbb {C} [x,y]/(y-x^{2})\to \mathbb {C}$ 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 $(1,2)\not \in \{(a,b)\in \mathbb {C} ^{2}:b^{2}-a\}$

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.

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

• 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

$\mathbb {C} [x,y][S^{-1}]=\mathbb {C} [x,y]_{\mathfrak {p}}=\left\{{\frac {f(x,y)}{g(x,y)}}:f,g\in \mathbb {C} [x,y]{\text{ and }}g(0,0)\neq 0\right\}$

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.

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

• 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 $S$

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

${\frac {{\text{Ker}}(\phi _{i})}{{\text{Ker}}(\phi _{i-1})}}\cong 0$

in the last example, we had the exact sequence

$R^{\oplus 2}\to R\to M\to 0$

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 $\cup _{i=1}^{\infty }M_{i}=R$

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

• Fields
• $\mathbb {Z}$
• Finite algebras over fields
• Quotients of Noetherian rings.

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

• $(\mathbb {C} [x]/(x^{5}),(x))$
• $(\mathbb {Q} [x,y]/((x-1)^{3},(x-1)^{2}(y-2),(y-2)^{5})),(x-1,y-2))$

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 $f(t)=t^{n}+a_{n-1}t^{n-1}+\cdot +a_{1}t+a_{0}\in R[t]$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

Chapter 0

Basic Commutative Algebra

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

Basic Differential/Complex Geometry

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