Users Guide to Hartshorne Algebraic Geometry/Chapter 1

Affine Varieties

This section introduces the basic notions of algebraic varieties. It starts out by defining an algebraic subset as the vanishing locus, or zero set given by some subset $T$ of $A=k[x_1,\dots ,x_n]$; that is,

$Z(T):=\{p\in {𝔸}_k^n:f(p)=0\text{ for }f\in T\}$

In all examples we should consider $T$ as an ideal in $A$. Conversely, if we have a subset $Y\subseteq {𝔸}^n$ then it is called algebraic if it is defined as the zero set of some subset $A$.

(Non-)Examples of Vanishing Sets

• $Z((x^2-y))\subset {𝔸}_A^2$ is the parabola from elementary algebra where $(x^2-y)\subset A[x,y]$ and $A$ is any of $\bar{\mathbb{Q}},ℂ,\overline{{\mathbb{Q}}_p}$.
• $T=\{1,2,3,\dots \}\subset {𝔸}_{ℂ}^1$ is not an algebraic subset since there are no polynomials in $ℂ[x]$ which vanish at infinitely many points. This is a consequence of the fundamental theorem of algebra since any polynomial $ℂ[x]$ is the product of $(x-a_1)\cdots (x-a_n)$.
• $Z((xy))=Z((x)\cdot (y))\subset {𝔸}_A^2$ is another example where it is the union of the $y$-axis and the $x$-axis defined by the vanishing of $x$ and $y$ in $ℂ[x,y]$.
• The previous example encapsulates the idea of a whole family of examples, given subsets $T_1,T_2\subset A$ we can form the subset

$T_1\cdot T_2=\{t_1\cdot t_2:t_1\in T_1\text{ and }{t}_2\in T_2\}$

whose vanishing locus is the union of $Z(T_1)$ and $Z(T_2)$. The next proposition, 1.1, includes this example and the example of intersecting algebraic sets.

Zariski Topology and Basic Definitions

We can define a topology of ${𝔸}^n$ by defining the closed sets as the algebraic sets. Note that every open set is dense in this topology! Although it is very coarse, it is still useful for constructing invariants from algebraic topology, such as the cohomology of a space, for a large class of examples (for example On the de Rham cohomology of algebraic varieties). Later chapters in the book deal with the basic theory required for these constructions.

Irreducibility

A subset of a topological space is called irreducible if it is not the union of two proper subsets, each of which is closed. So for example, our space, $Z((xy))$ is reducible since it is the union of the components $Z((x))$ and $Z((y))$.

Affine Varieties and Quasi-Affine

He also defines an affine variety as an irreducible algebraic closed subset of some ${𝔸}^n$. Also, an open subset of an affine variety is called quasi-affine. Note that the topology of an affine variety is the induced topology. One interesting example of a quasi-affine variety which is not affine is origin removed from ${𝔸}^2$. This is the complement of $Z((x,y))\subset {𝔸}^2$. It is easy to show that this is not an affine variety using cohomological tools discussed later in the book.

Basic Properties

Proposition 1.2 gives some obvious properties of algebraic subsets, but it is worth noting that taking some power of a subset of $A$ will give the same algebraic set as the original subset. For example $(x,y)\subset ℂ[x,y]$ has the same vanishing locus as $(x,y{)}^2=(x^2,xy,y^2)$ and $(x^2,y^2)$. Note that in scheme theory, these ideals will correspond to distinct schemes. These examples are inspiration for the radical of an ideal $I$, or the nilradical, that is

$\sqrt{I}:=\{f\in A:f^k\in I\text{ for some }k\in \mathbb{N}\}$

These can be found in general by finding a generating set of $I$ whose elements are powers of some polynomial. For example,

$\begin{array}{cc}\sqrt{(x{y}^2,z^5)}& =(xy,z)\\ \sqrt{(x^2(x+y+z{)}^5)}& =(x(x+y+z))\\ \sqrt{(x^n+y^n+z^n)}& =(x^n+y^n+z^n)\end{array}$

Fundamental Theorem - Nullstellensatz

Although initially somewhat opaque, the Nullstellensatz is fundamental for thinking about algebras as spaces. Basically, it states that the vanishing locus of an algebraic set can always be found as the radical of some ideal. This is important because it let's us think geometrically about the ideals of the commutative ring $A$ but also lets us find an ideal for any algebraic set. Also, the corollary extends this out by noting the irreducible algebraic sets are always defined by a prime ideal.

Note that in general it is difficult to find generators for prime ideals and very easy to write down examples which are not prime. For example, the ideal $(y-x^2,(y-4{)}^5)$ corresponds to the two points $(2,2),(-2,2)\subset {𝔸}^2$. In all of the examples he gives, it is of an ideal defined by an irreducibly polynomial. It is useful to review Gauss' Lemma and Eisenstein's Criterion to determine if a polynomial is irreducible. For example, consider $f(x,y)=x^2-y^n\in ℂ[x,y]$. Then as a polynomial in $ℂ(x)[y])$ it can be written as

$g=1\cdot y^n-x^2$

so

$\frac{\partial g}{\partial y}=n{y}^{n-1}$

which has no common divisors with $g$ since its only solution is $0$. Another important class of polynomials are of the form

$y^2-x^3-ax-b\in ℂ[x,y]$

which define the vanishing locus of an elliptic curve. These are super important in mathematics because they are simple, define basic examples of abelian varieties, and can be used to study a lot of interesting phenomena in Arithmetic geometry.

Affine Coordinate Ring

One of the useful consequences of Nullstellensatz is we can find a unique ideal representing the geometry of any algebraic subset $Y$. Given any ideal $I$ whose vanishing locus is $Y$ this unique ideal is the radical of $I$, denoted $\sqrt{I}$. We can also associate a ring, called the affine coordinate ring of $Y$ defined as

$A(Y)=A/\sqrt{I}$

For example, the affine coordinate ring of $((y^3-x^3+1{)}^3)\subset ℂ[x,y]$ is

$A(Y)=\frac{ℂ[x,y]}{(y^3-x^3+1)}$

Basic Dimension Theory

In this section Hartshorne introduces the Krull dimension of a commutative ring as the supremum of heights of all prime ideals. That is, a prime ideal $𝔭\subset A$ is of height $n$ if there is a maximal chain of distinct prime ideals

${𝔭}_0\subset {𝔭}_1\subset \cdots \subset {𝔭}_n$

Examples of Heights and Dimensions

• For example, the height of the maximal ideal $(x,y,z)\subset ℂ[x,y,z]$ is $3$ since there is the (maximal) chain

$(0)\subset (x)\subset (x,y)\subset (x,y,z)$

of distinct prime ideals. Since all of the maximal ideals are of the form $(x-x_0,y-y_0,z-z_0)$ for some constants $x_0,y_0,z_0$ (by an application of corollary 1.4) we have that the dimension of $ℂ[x,y,z]$ is $3$.

• The prime ideals of the integers $\mathbb{Z}$ are $(0)$ or $(p)$. Since the $(p)$ are the maximal ideals, all of height $1$ we have that $\mathbb{Z}$ is one dimensional. At first, this seems odd, but once we get to flat families of schemes, we can see that this is "correct".

• Another interesting class of examples are infinite dimensional Noetherian rings whose prime ideals all have finite height. Check out this math.stackexchange discussion for a detailed overview.

• Finding an irreducible element in an integral domain $f\in R$ gives a height one prime ideal, since

$(0)\subset (f)$

is a maximal chain of prime ideals. This shows that an irreducible hypersurface of an integral domain always has dimension less than one.

• Krull dimension can also be used in local algebra to determine the dimension of a point on a variety. For example, consider the $x$-axis unioned with the $yz$-plane in ${𝔸}^3$. This has defining ideal

$I=(y,z)(x)=(xy,xz)$

If we localize $R=k[x,y,z]/I$ at the maximal ideal $(x,y,z)$, which is at the intersection of the axis with the plane, we want to know what the dimension of this point actually is. Krull dimension gives an answer to this! Consider the local ring

$(R,𝔪)=({\left(\frac{ℂ[x,y,z]}{(xy,xz)}\right)}_{(x,y,z)},(x,y,z))$

Since $R/𝔪\cong ℂ$ we have the height of the maximal ideal is equal to the dimension of the local ring. We have the two maximal chains in this local ring

$(0)\subset (x)\subset (x,y,z)$

$(0)\subset (y,z)\subset (x,y,z)$

giving the dimension of the ring, 2. This gives us an answer to our previous question.

Transcendence degree and Dimension

Hartshorne gives another definition of dimension using transcendence degree in theorem 1.8.A — that the dimension of an integral domain $B$ can be defined as the transcendence degree of its fraction field $\text{Frac}(B)$ (which he denotes $K(B)$). He gives a quick sanity check in the next proposition that ${𝔸}^n$ has transcendence degree $n$, which follows from the fact that its fraction field is $k(x_1,\dots ,x_n)$.

Examples with Transcendence Degree

• $k(x_1,\dots ,x_n)$ has transcendence degree $n$
• Using the ring morphism

$k[x]\to k[x,y]\to \frac{k[x,y]}{(y^2-a{x}^3-b)}=R$

we can show that the transcendence degree of $R$ is $1$ since its fraction field is a degree two extension of $k(x)$. The set $\{x\}$ forms its transcendence basis and

$\text{Frac}(R)=\frac{k(x)[y]}{(y^2-(a{x}^3+b))}$ which is a degree two field extension.

• The technique in the previous example can be used to show the dimension of an irreducible hypersurface $X$ of ${𝔸}^n$ has dimension less than one. Although this requires knowledge of morphisms of varieties, this is a useful example to keep in mind. All you have to do is find a projection $\pi :X\to {𝔸}^{n-1}$ which generically has preimage a finite set of points over any point in ${𝔸}^{n-1}$. Taking the fraction fields of these two integral domains will give a finite field extension of $k(y_1,\dots ,y_n)$.
• For example, consider the integral domain

$R=\text{Frac}\left(\frac{ℂ[x,y,z]}{(z^3-x^3-y^3-1)}\right)$

If we take the inclusion $ℂ[x,y]\to ℂ[x,y,z]$ composed with the projection to $R$ we have such a projection $\pi :X\to {𝔸}^2$. This gives a field extension

$k(x,y)\to \frac{k(x,y)[z]}{(z^3-(x^3+y^3+1))}$

which is a degree $3$ field extension.

Proposition 1.10

This is a useful proposition which will extend to quasi-projective varieties (as defined in section I.2). Basically, the closure operation preserves dimension. For a simple intuitive example, an algebraic curve $C\subset {𝔸}^2$ could have a finite number of points removed. The closure of this open set is the original curve.

Proposition 1.12A

This is a useful proposition which has a simple counterexample from algebraic number theory if we take away the UFD hypothesis. Every Dedekind domain has dimension at most $1$, but the prime ideal

$(2,1+\sqrt{-5})\subset \mathbb{Z}[\sqrt{-5}]$

has two generators, so it is not principal.

Exercises

1.1

• (a) Notice that the ring $A(Y)=ℂ[x,y]/(y-x^2)$. If we rewrite it using the substitution $y\mapsto x^2$ then $A(Y)=ℂ[x,x^2]$. We can define a ring morphism $ℂ[t]\to ℂ[x,x^2]$ by $t\mapsto x$. Since it has no kernel and is surjective, it is an isomorphism.
• (b) We can rewrite the presentation of $A(Z)=ℂ[x,y]/(xy-1)$ as $ℂ[t,t^{-1}]$ where $t$ corresponds to $x$ and $t^{-1}$ corresponds to $y$. Using then Nullstellensatz it is easy to see there is no point in $Z$ corresponding to $0\in {𝔸}^1$ since the second presentation of $A(Z)$ corresponds to the variety ${𝔸}^1-\{0\}$
• (c) Check out this solution.

1.2

The trick here is to find all algebraic relations between the points in the set. Since $Y=\{(t,t^2,t^3)|t\in k\}$ we can write this as the quotient of $k[x,y,z]$ by the ideal $(x^2-y,x^3-z,y^3-z^2)$. The first generator corresponds to the face that $(t{)}^2=t^2$, the second $(t{)}^3=t^3$ and the third $(t^2{)}^3=(t^3{)}^2$.

1.3

Notice the second polynomial gives a variety which is the union of the $yz$-plane and the $xy$-plane shifted in the $z$-direction by $1$. We can then specialize to the solutions of $x^2-yz$ on each of the two planes to find all of the solutions. On the $yz$-plane we have $x=0$ so $x^2-yz=-yz$. This implies that the solutions are the union of the two coordinate axes on this plane. On the $xy$-plane, the equation reads $x^2-yz=x^2-y(1)=x^2-y$. This is the plane parabola on the shifted plane. There is an intersection point of components at $(0,0,1)$ of the parabola intersecting with the $z$-axis.

1.4

Notice ${𝔸}^2-0$ is not an open in the product topology since its projection on each factor is ${𝔸}^1-0$, but the product of these two opens in ${𝔸}^2-Z((xy))$.

1.5

Nullstellensatz gives us the result almost immediately. We must have a surjection of some $k[x_1,\dots ,x_n]\to A(Y)$ and the defining ideal must not give nilpotents in $B$ since it is the radical or itself.

1.10

• (a) Since $Y$ has the induced topology, any chain of closed subsets in $Y$ comes from an intersection of closed subsets of $X$ with $Y$. Any such maximal chain may always be extended to a chain of closed subsets of $X$. Then, the dimension of $X$ is always greater than or equal to that of $Y$.
• (b)
• (c) With the set $X=\{a,b,c\}$ declare the following subsets as open $\{\varnothing ,X,\{a,b\}\}$. Then the smallest closed set containing $U=\{a,b\}$ is $X$, so it is a dense open subset. It has dimension zero since the only closed subset it $\varnothing$ but $X$ has the following maximal chain of closed subsets

$\varnothing \subset \{c\}\subset X$

so it is of dimension two.

• (c.1) If we remove the density hypothesis, take the set $X=\{a,b\}$ with the discete topology. Then the open subset $\{x\}$ has dimension one, but $X$ has dimension two.
• (d) Using the definition of dimension for Noetherian topological spaces, there is a maximal chain of closed subsets

$Y_0\subset Y_1\subset \cdots Y$

Since $Y\subset X$ is closed and proper, we could have extended this chain by one, giving a maximal chain of length greater than $\text{dim}(X)$, a contradiction. Hence $Y=X$.

• (e) Take the affine space ${𝔸}^{\mathrm{\infty }}$ and give it the topology where a subset $X$ is closed if it can be contained in some embedded ${𝔸}^n$ and is a closed subset when restricted to the embedded ${𝔸}^n$ with the Zariski topology.
• (e.1) Note that a space is locally Noetherian if every point has a neighborhood which is a Noetherian topological space. One example which is infinite dimensional, but has only finite dimension components is the infinite disjoint union ${𝔸}^1\coprod {𝔸}^2\coprod {𝔸}^3\coprod \cdots$ equipped with the Zariski topology on each affine space. Since we can only take a finite union of closed subvarieties $X_i\subset {𝔸}^{n_i}$, we have that this space is Noetherian.

1.11

The curve $Y\subset {𝔸}^3$ given parametrically by $x=t^3,y=t^4,z=t^5$ gives the following three relations among the generators $x,y,z$

$\begin{array}{c}{x}^4-y^3\\ x^5-z^3\\ y^5-z^4\end{array}$

hence its defining prime ideal is

$(x^4-y^3,x^5-z^3,y^5-z^4)$

which has three generators but has only height 2. We can use the dimension, height formula given in theorem 1.8A to should that this is the case. Since the dimension of $B$ is $3$ and the dimension of $B/𝔭$ is 1 (since it is isomorphic to ${𝔸}^1$), it has height 2.

1.12

The polynomial $y^2-x^3+7x-4$ defines the equation of an elliptic curve in the plane which has two components. This can be viewed using desmos. Playing around with different degrees and different parameters gives more solutions, just make sure the coefficients don't force the only solutions to be a finite set of points or all complex points. For example, the polynomial $f(x,y)=y^2-x^3+7x-i$ has no real solutions since any $f(a,b)$ cannot equal $i$ unless one of the $a,b$ are non-real.

Affine Algebraic Groups

Group Rings

An affine algebraic group is an algebraic set which is also has a group structure (and for later the group structure are morphisms of algebraic varieties). For example, the set

$\{(x,y,z,w)\in {𝔸}^4:xw-yz\ne 0\}$

defines the algebraic group $\text{GL}(2,k)$. We could have written this open set more suggestively as

$\{\left(\begin{array}{cc}x& y\\ z& w\end{array}\right)\in {𝔸}^4:\det \left(\begin{array}{cc}x& y\\ z& w\end{array}\right)=xw-yz\ne 0\}$

We can also define $\text{GL}(n,k)$ as the subset of ${𝔸}^{n^2}$ where the polynomial from the determinant is non-vanishing.

• SL det = 1
• G_m = GL -> coordinate ring k[x,x^-1]
• Finite groups -> embed in some S_n -> embed with obvious action on A^n -> every finite group is an affine algebraic group
• Cyclic groups

Group Actions

• Group actions on sets
• Invariant polynomials
• Quotient varieties

Projective Varieties

Geometric Definitions

Hartshorne begins this section with the definition of projective space, it is the set of equivalence classes of tuples $(a_0,\dots ,a_n)$ such that

$(a_0,\dots ,a_n)\sim \lambda \cdot (a_0,\dots ,a_n)\sim (\lambda a_0,\dots ,\lambda a_n)$

Typically they are denoted as $[a_0:\cdots :a_n]$. To get used to these equivalence classes, consider the point $p\in {\mathbb{P}}^2$ where $p=[1:2:i]$. We have the following equivalent representations

$\begin{array}{cc}p& =[1:2:i]\\ & =2\cdot [1:2:i]\\ & =[2:4:i]\\ & =i\cdot [1:2:i]\\ & =[i:2i:-1]\end{array}0$

You can consider this as a generalization of ratios since any ratio $a:b$ can be represented as an element in $[a:b]\in {\mathbb{P}}^1$.

Another useful and equivalent way to think of ${\mathbb{P}}^n$ is as the quotient of ${𝔸}^n-\{0\}$ by the $k^{\ast }$-action $\lambda \cdot (a_0,\dots ,a_n)=(\lambda a_0,\dots ,\lambda a_n)$. It turns out this way to think about it makes more of the theory work. Check out these notes for more information.

In order to define functions on ${\mathbb{P}}^n$ to ${𝔸}^1$ we need to make sure they are well-defined with respect to the $k^{\ast }$-action. If we multiply by some $\lambda \in k^{\ast }$, the value needs to be the same, so

$f([a_0:\cdots :a_n])=f(\lambda \cdot [a_0:\cdots :a_n])=f([\lambda \cdot a_0:\cdots :\lambda \cdot a_n])=g(\lambda )f([a_0:\cdots :a_n])$

for some $g:{𝔸}^1-\{0\}\to {𝔸}^1-\{0\}$. But since every $p\in {𝔸}^1-\{0\}$ is equivalent from the $k^{\ast }$-action, and $0\in {𝔸}^1$ has a trivial $k^{\ast }$-action, we can only consider functions

$f:{\mathbb{P}}^n\to {𝔸}^1/k^{\ast }\cong \{0,1\}$

where $f(p)=1$ if $f$ is non-vanishing at $p$. It turns out that all such functions must be homogeneous polynomials. The linked noted above explain this further in detail, but Hartshorne goes on to describe the related algebra which explains what these polynomials are and how to find them.

Algebraic Definitions

Graded Rings and Ideals

Algebraically we can encode the structure of homogeneous elements using a grading. That is, we decompose a ring $S$ into abelian groups $S_d$ such that

$S=\underset{d\ge 0}{⨁}{S}_d$

and these groups behave well with respect to multiplication, that is, for $d,e\ge 0$ $S_d\cdot S_e\subseteq S_{d+e}$. We will denote the ring $S$ with its grading as $S_{\bullet }$ and call it a graded ring. For example, in the ring $S=k[x,y,z]$ we can define the $S_d$ as the abelian group of homogeneous polynomials of degree $d$ in $S$. Then we can decompose $S$ as

$S_{\bullet }=k\oplus k\cdot \{x,y,z\}\oplus k\cdot \{x^2,xy,xz,y^2,yz,z^2\}\oplus \cdots$

where $k\cdot \{x,y,z\}$ means the underlying abelian group of the $k$-vector space spanned by the three elements $x,y,z$. For example, $x{y}^2+x{z}^2+z^3\in S_3$. Graded rings of this form can be used to construct other graded rings. We can call an ideal $I\subset S$ graded if it is generated by homogeneous elements. That is, we can find $f_i\in S_{d_i}$ such that $(f_1,\dots ,f_k)=I$. Then, we can consider the subgroup of $S_d$ generated by the elements in $I$ as $I_d$ giving $I$ the structure of a grading. We can then form a graded ring

$(S/I{)}_{\bullet }=S_{\bullet }/I_{\bullet }=\frac{S_0}{I_0}\oplus \frac{S_1}{I_1}\oplus \frac{S_2}{I_2}\oplus \cdots$

For example, consider the ideal $I=(f_1,f_2)=(xy,x^3+y^3+z^3)\subset k[x,y,z]$. Since the generators are both homogeneous we can give the ideal $I$ the grading

$\begin{array}{ccccccccccccccc}{I}_{\bullet }& =& I_0& \oplus & I_1& \oplus & I_2& \oplus & I_3& \oplus & I_4& \oplus & I_5& \oplus & \cdots \\ & =& 0& \oplus & 0& \oplus & k\cdot \{xy\}& \oplus & k\cdot \{x^3+y^3+z^3\}& \oplus & 0& \oplus & k\cdot \{xy(x^3+y^3+z^3)\}& \oplus & \cdots \end{array}$

Projective Varieties

Similarly to affine zero sets, we can construct projective algebraic sets. Given a homogeneous polynomial $f\in S_{\bullet }=k[x_0,\dots ,x_n]$ we can form the zero set

$Z(f)=\{p\in {\mathbb{P}}^n:f(p)=0\}$

Similarly, for a set $T$ of homogeneous elements in $S_{\bullet }$ we can form an algebraic set of all points which vanishing on each elements in $T$. Then, a projective algebraic variety is an irreducible algebraic set in ${\mathbb{P}}^n$ equipped with the induced topology. As you would expect, ${\mathbb{P}}^n$ has a Zariski topology whose closed sets are the algebraic sets.

With the previous section, we can talk about the associated graded coordinate rings of projective algebraic varieties. Given a graded ideal $I_{\bullet }$ we can find the projective algebraic set $Z(I_{\bullet })$ that is defined by the vanishing of every element in $I_{\bullet }$. For example, the vanishing set on the graded ideal

$I_{\bullet }=(x{y}^2,(x+y+z{)}^5)$

is the set

$\{[x_0:y_0:z_0]\in {\mathbb{P}}^n:x_0\cdot y_0=0\text{ and }{x}_0+y_0+z_0=0\}$

Notice that some of the degrees of the polynomials have changed! This is because of the analogue of the Nullstellensatz in the projective setting. One of the first exercises in this section is proving this theorem. Similarly to exercise I.1.3 we can analyze what the vanishing set looks like. In the first equation, notice that either $x_0$ or $y_0$ must equal $0$. If we take the first case, then this is the subset $[0:y_0:z_0]$ where $y_0+z_0=0$. This implies it is the point $[0:y_0:-y_0]=[0:1:-1]$. We can do a similar analysis for $y_0=0$ to get the point $[1:0:-1]$. These are all of the points in this algebraic set, but it is not a variety since it is a disjoint union of two irreducible components. If we want to find the corresponding ring, it is

${\left(\frac{S}{I}\right)}_{\bullet }=\frac{k[x,y,z]}{(xy,x+y+z)}$

since $(xy,x+y+z)=\sqrt{(x{y}^2,(x+y+z{)}^5)}$.

More Examples

Fermat Varieties

Another useful class of examples are the projective algebraic sets defined by the polynomials of the form

$x_0^k+\cdots +x_n^k=0$

These are called Fermat polynomials since they resemble the famous Fermat problem trying to find integer solutions to the equation $x^n+y^n=z^n$

Affine Stratification

If we consider the points $[x_0:\cdots :x_n]$ then the set where $x_0=0$ is

$\{[0:x_1:\cdots x_n]\in {\mathbb{P}}^n\}$

whose points can be identified with points in ${\mathbb{P}}^{n-1}$. The set ${\mathbb{P}}^n-{\mathbb{P}}^{n-1}$ is then the set of points

$[x_0:x_1:\cdots :x_n]\in {\mathbb{P}}^n:x_0\ne 0$

Since $x_0\ne 0$ we always have the equality

$[x_0:x_1:\cdots :x_n]=[1:\frac{x_1}{x_0}:\cdots :\frac{x_n}{x_0}]$

Notice we can identify a point $y\in {𝔸}^n$ with

$y\mapsto [1:y_1:\cdots :y_n]$

which is a unique point in ${\mathbb{P}}^n$ giving us the identification of ${𝔸}^n$ with ${\mathbb{P}}^n-{\mathbb{P}}^{n-1}$. This shows

$\begin{array}{cc}{\mathbb{P}}^n& \cong ({\mathbb{P}}^n-{\mathbb{P}}^{n-1})\cup {\mathbb{P}}^{n-1}\\ & \cong {𝔸}^n\cup {\mathbb{P}}^{n-1}\end{array}$

as sets. Applying this recursively shows

${\mathbb{P}}^n\cong {𝔸}^n\cup {𝔸}^{n-1}\cup \cdots \cup {𝔸}^1\cup {𝔸}^0$

On ${\mathbb{P}}^1$ we call the point $[0:1]$ the point at infinity since the set of all $[1:a]$ is the set ${𝔸}^1$ and ${\mathbb{P}}^1$ over $ℂ$ can be identified with $S^2$. Similarly, there is a line at infinity in ${\mathbb{P}}^2$ and a hyperplane at infinity in general.

Charts on Projective Space

• compute transition functions

Projective Closure

We can use the previous stratification of ${\mathbb{P}}^n$ to turn an affine variety in ${𝔸}^n$ into a projective variety in ${\mathbb{P}}^n$. It should be more clear after reading about the charts on projective space. If we have a polynomial, say

$f=xyz+x^3+y^3+z^3\in k[x,y,z]$

then on the open set $U_x=\{[x:y:z]:x\ne 0\}$ the vanishing locus $Z(f)\cap U_x$ is

$\{[x:y:z]:f(x,y,z)=0\text{ and }x\ne 0\}$

Since we can divide by $x$ in this open set, we get the equality

$f(1,y/x,z/x)=\frac{yz}{x^2}+1+\frac{y^3}{x^3}+\frac{z^3}{x^3}=\frac{f}{x^3}$

If we rewrite

$\begin{array}{cc}\frac{y}{x}& =Y\\ \frac{z}{x}& =Z\end{array}$

this equation reads as

$f(1,Y,Z)=YZ+1+Y^3+Z^3$

giving the equation of an affine variety. This gives us a way to associate an affine variety to a projective variety. All we have to do is go the other direction! Let's try this in degree two first so we can get the general picture. Consider the polynomial

$G(X,Y)=X^{4}Y+XY+Y^3+X$

and notice the degree is $5$. If we want to turn this into a homogeneous polynomial in three variables (since we need three for ${\mathbb{P}}^2$ we can multiply each of the terms by some $z^k$ to get a degree $5$ homogeneous polynomial. So

$g(x,y,z)=x^{4}y+xy{z}^3+y^3{z}^2+x{z}^4$

is such a polynomial, and on the set $\{[x:y:z]:z\ne 0\}\subset {\mathbb{P}}^2$ we can identify $g(1,x,y)$ with $G(X,Y)$.

Quasi-Projective Varieties

You may be able to convince yourself that ${\mathbb{P}}^n$ for $n\ge 2$ is an example of a quasi-projective variety which is not affine.

• Graded Ring + graded ideals + graded generators
• Homogeneous polynomials
• https://math.stackexchange.com/questions/969894/quotient-ring-of-a-graded-algebra-with-respect-to-a-graded-ideal
• https://mathoverflow.net/questions/126722/why-does-the-naive-choice-of-homogeneous-coordinate-ring-of-a-product-of-project

More Geometry

• Projective varieties
• Stratification
• Point and line at infinity ${\mathbb{P}}^1$ and ${\mathbb{P}}^2$
• Patches
• Affine coverings

Gradings Continued, Multi-Graded, Weighted Homogeneous

• Product spaces and double grading
• Weighted homogeneous polynomials
• Graded ring again
• Weighted projective space (https://arxiv.org/pdf/1604.02441.pdf)
• Maybe GIT basics? http://www-fourier.univ-grenoble-alpes.fr/~mbrion/lin_rev.pdf

Morphisms

• definition
• automorphisms -> group actions, GL, PGLd
• Coverings
• Ramified covers
• Quotients -> weighted projective space

Talk about ramified coverings

$\frac{ℂ[x,y,z]}{(x^3+y^3+z^3-1)}$

should have transcendence degree 2.

Just take the composition of the include

$ℂ[x,y]\to ℂ[x,y,z]$

with the projection. This gives a dominant morphism. Looking at the fibers gives the answer.

Generically, this will be a three-sheeted cover, but along the locus

$x^3+y^3-1$ there will be a degeneration.

Nonsingular Varieties

https://math.stackexchange.com/questions/678419/normalization-of-a-quotient-ring-of-polynomial-rings-reid-exercise-4-6/681926#681926 Include p-derivations + Witt vectors https://www.uvm.edu/~tdupuy/notes/dupuy-ttu-slides.pdf