$\mathbb{A}^1$ has coordinate ring $k[\mathbb{A}^1] = k[x]$. A regular function on $\mathbb{A}^1$ is simply a polynomial $f(x) \in k[x]$. Rational functions like $1/x$ are not regular on all of $\mathbb{A}^1$ (they are undefined at $x = 0$), but $1/x$ is regular on the open set $D(x) = \mathbb{A}^1 \setminus \{0\}$.

Let $Y = V(y^2 - x^3 + x) \subseteq \mathbb{A}^2$ be the affine part of the elliptic curve $y^2 = x^3 - x$. Then

$k[Y] = k[x,y]/(y^2 - x^3 + x).$

Every element of $k[Y]$ can be written uniquely as $f(x) + g(x) y$ where $f, g \in k[x]$, since $y^2 = x^3 - x$ allows reducing the power of $y$. For instance, $y^3 = y \cdot y^2 = y(x^3 - x) = x^3 y - xy$ in $k[Y]$.

Consider $\mathbb{P}^1$. A rational function $g(x_0,x_1)/h(x_0,x_1)$ with $g, h$ homogeneous of the same degree $d$ defines a "function" $\mathbb{P}^1 \to k$, but it has poles where $h = 0$. The only way to avoid all poles on $\mathbb{P}^1$ is if $h$ divides $g$ in the homogeneous coordinate ring, giving a constant. For instance, $x_0/x_1$ is regular on $D(x_1)$ but has a pole at $[1:0]$. There is no nonconstant regular function defined on all of $\mathbb{P}^1$.

The morphism $\varphi: \mathbb{A}^1 \to Y = V(y - x^2) \subseteq \mathbb{A}^2$ given by $t \mapsto (t, t^2)$ corresponds to the $k$-algebra homomorphism

$\varphi^*: k[Y] = k[x,y]/(y - x^2) \to k[t], \quad x \mapsto t, \; y \mapsto t^2.$

This is an isomorphism of $k$-algebras (since $k[Y] \cong k[x]$ via $y \mapsto x^2$), confirming that $\varphi$ is an isomorphism of varieties.

Let $Y = V(y^2 - x^3 + x) \subseteq \mathbb{A}^2$ be the affine elliptic curve. The projection to the first coordinate

$\pi: Y \to \mathbb{A}^1, \quad (x, y) \mapsto x$

is a morphism. The corresponding pullback is $\pi^*: k[t] \to k[Y]$, $t \mapsto \bar{x}$. This is injective (so $\pi$ is dominant), and the extension $k[t] \hookrightarrow k[x,y]/(y^2 - x^3 + x)$ is integral of degree 2: generically, each fiber $\pi^{-1}(a) = \{(a, y) : y^2 = a^3 - a\}$ consists of 2 points (the two square roots), unless $a^3 - a = 0$ (i.e., $a = 0, \pm 1$), where the fiber is a single point.

Let $Y \subseteq X$ be a closed subvariety (e.g., $Y = V(y) \subseteq X = \mathbb{A}^2$). The inclusion $\iota: Y \hookrightarrow X$ is a morphism, and the pullback $\iota^*: k[X] \to k[Y]$ is the surjection $k[X] \twoheadrightarrow k[X]/I(Y) = k[Y]$. Closed immersions correspond to surjective $k$-algebra maps; open immersions correspond to localizations.

A morphism $\varphi: X \to \mathbb{A}^1$ is the same data as a single regular function $f = \varphi \in k[X]$: the pullback $\varphi^*: k[t] \to k[X]$ sends $t \mapsto f$. More generally, a morphism $X \to \mathbb{A}^n$ is exactly an $n$-tuple of regular functions $(f_1, \ldots, f_n) \in k[X]^n$, reproducing Definition 1.12.

The degree-$d$ Veronese embedding is the morphism

$\nu_d: \mathbb{P}^n \to \mathbb{P}^N, \quad [x_0:\cdots:x_n] \mapsto [\cdots : x_0^{i_0}\cdots x_n^{i_n} : \cdots]_{i_0+\cdots+i_n=d}$

where $N = \binom{n+d}{d} - 1$. All component polynomials are monomials of degree $d$, so they are homogeneous of the same degree. At any point $P \in \mathbb{P}^n$, at least one coordinate $x_j \neq 0$, so the monomial $x_j^d \neq 0$, ensuring the map is well-defined everywhere.

For $n = 1$, $d = 2$: $\nu_2: \mathbb{P}^1 \to \mathbb{P}^2$, $[s:t] \mapsto [s^2 : st : t^2]$. The image is the conic $V(xz - y^2)$, and $\nu_2$ is an isomorphism onto its image.

For $n = 1$, $d = 3$: $\nu_3: \mathbb{P}^1 \to \mathbb{P}^3$, $[s:t] \mapsto [s^3 : s^2 t : st^2 : t^3]$. The image is the twisted cubic curve in $\mathbb{P}^3$, defined by the $2 \times 2$ minors of $\begin{pmatrix} x_0 & x_1 & x_2 \\ x_1 & x_2 & x_3 \end{pmatrix}$.

The Segre embedding is the morphism

$\sigma: \mathbb{P}^m \times \mathbb{P}^n \to \mathbb{P}^{(m+1)(n+1)-1}, \quad ([x_i], [y_j]) \mapsto [x_i y_j]_{i,j}.$

For $m = n = 1$: $\sigma([s:t],[u:v]) = [su : sv : tu : tv]$. The image is $V(z_0 z_3 - z_1 z_2) \subseteq \mathbb{P}^3$, the smooth quadric surface.

The Segre map is well-defined: at least one $x_i \neq 0$ and one $y_j \neq 0$, so $x_i y_j \neq 0$. It is an isomorphism onto its image, making $\mathbb{P}^m \times \mathbb{P}^n$ a projective variety. This is how we give the product of projective varieties the structure of a projective variety.

Let $P = [1:0:\cdots:0] \in \mathbb{P}^n$ and let $H = V(x_0) \cong \mathbb{P}^{n-1}$. The projection from $P$ is the morphism

$\pi_P: \mathbb{P}^n \setminus \{P\} \to \mathbb{P}^{n-1}, \quad [x_0:x_1:\cdots:x_n] \mapsto [x_1:\cdots:x_n].$

This is well-defined: if $Q \neq P$, then not all of $x_1,\ldots,x_n$ are zero. Geometrically, $\pi_P$ sends $Q$ to the intersection of the line $\overline{PQ}$ with the hyperplane $H$.

If $X \subseteq \mathbb{P}^n$ is a projective variety with $P \notin X$, then $\pi_P|_X: X \to \mathbb{P}^{n-1}$ is a morphism. This is a fundamental tool: it can be used to show that every projective variety of dimension $d$ admits a finite surjective morphism to $\mathbb{P}^d$ (Noether normalization, geometric form).

Let $k$ be an algebraically closed field of characteristic $p > 0$ (e.g., $k = \overline{\mathbb{F}_p}$). The Frobenius endomorphism on $\mathbb{A}^n$ is

$\operatorname{Fr}: \mathbb{A}^n \to \mathbb{A}^n, \quad (a_1, \ldots, a_n) \mapsto (a_1^p, \ldots, a_n^p).$

This is a morphism (each component is a polynomial). The pullback is

$\operatorname{Fr}^*: k[x_1,\ldots,x_n] \to k[x_1,\ldots,x_n], \quad x_i \mapsto x_i^p.$

On a projective variety $X \subseteq \mathbb{P}^n$ defined over $\mathbb{F}_p$, the Frobenius acts as $[a_0:\cdots:a_n] \mapsto [a_0^p:\cdots:a_n^p]$. This is well-defined since $({\lambda a_i})^p = \lambda^p a_i^p$. The fixed points of $\operatorname{Fr}$ are precisely the $\mathbb{F}_p$-rational points: $X(\mathbb{F}_p) = X^{\operatorname{Fr}}$.

Key properties:

• $\operatorname{Fr}$ is a bijection on points (since $a \mapsto a^p$ is bijective on $k = \overline{\mathbb{F}_p}$).
• $\operatorname{Fr}$ is not an isomorphism: the pullback $x_i \mapsto x_i^p$ is injective but not surjective ($x_i$ is not in the image).
• $\operatorname{Fr}$ is a purely inseparable morphism of degree $p^n$.
• Counting fixed points of iterates $\operatorname{Fr}^r$ gives $|X(\mathbb{F}_{p^r})|$, which is the subject of the Weil conjectures.

The $d$-uple embedding (= Veronese embedding for general $n$) sends

$\rho_d: \mathbb{P}^n \to \mathbb{P}^N, \quad N = \binom{n+d}{d} - 1,$

mapping each point to its vector of all degree-$d$ monomials. For $n = 2$, $d = 2$ (the quadratic Veronese):

$\rho_2: \mathbb{P}^2 \to \mathbb{P}^5, \quad [x:y:z] \mapsto [x^2 : xy : xz : y^2 : yz : z^2].$

The image is the Veronese surface $V_2 \subseteq \mathbb{P}^5$. A line $V(ax + by + cz)$ in $\mathbb{P}^2$ corresponds to a conic on $V_2$ (a hyperplane section). The key property: $\rho_d$ converts degree-$d$ hypersurfaces to hyperplane sections.

As a morphism, $\rho_d$ is an isomorphism onto its image. The inverse is obtained by noting that the image lies in a linear subspace where the coordinate ratios recover the original coordinates.

The cuspidal cubic $Y = V(y^2 - x^3) \subseteq \mathbb{A}^2$ has coordinate ring $k[Y] = k[t^2, t^3] \subset k[t]$. The normalization is the morphism

$\nu: \mathbb{A}^1 \to Y, \quad t \mapsto (t^2, t^3)$

corresponding to the inclusion $\nu^*: k[t^2, t^3] \hookrightarrow k[t]$. This is a bijection on points but not an isomorphism (the pullback is not surjective: $t \notin k[t^2, t^3]$).

The normalization "resolves" the cusp: $\mathbb{A}^1$ is smooth, and $\nu$ is a birational morphism that is an isomorphism away from the singular point. The preimage of the cusp $(0,0) \in Y$ is the single point $0 \in \mathbb{A}^1$.

The nodal cubic $Y = V(y^2 - x^2(x+1)) \subseteq \mathbb{A}^2$ has a node at the origin. The normalization is

$\nu: \mathbb{A}^1 \to Y, \quad t \mapsto (t^2 - 1, \; t(t^2 - 1)).$

This corresponds to $\nu^*: k[Y] = k[x,y]/(y^2 - x^2(x+1)) \hookrightarrow k[t]$ via $x \mapsto t^2 - 1$, $y \mapsto t^3 - t$.

Unlike the cusp, the normalization of the node is not bijective: the two points $t = 1$ and $t = -1$ both map to the node $(0, 0) \in Y$. The node has two distinct tangent directions, and the normalization "separates" them into two points on $\mathbb{A}^1$.

The rational normal curve of degree $d$ is the image of

$\varphi: \mathbb{P}^1 \to \mathbb{P}^d, \quad [s:t] \mapsto [s^d : s^{d-1}t : \cdots : st^{d-1} : t^d].$

This is the $d$-uple embedding of $\mathbb{P}^1$. For $d = 3$, the image $C \subseteq \mathbb{P}^3$ is the twisted cubic, defined by

$\operatorname{rank} \begin{pmatrix} x_0 & x_1 & x_2 \\ x_1 & x_2 & x_3 \end{pmatrix} \leq 1.$

The morphism $\varphi$ is an isomorphism onto $C$: the inverse on the chart $x_0 \neq 0$ is $[x_0:x_1:x_2:x_3] \mapsto [x_0:x_1]$, and on $x_3 \neq 0$ it is $[x_0:x_1:x_2:x_3] \mapsto [x_2:x_3]$.

The morphism $\varphi: \mathbb{A}^1 \to Y = V(y^2 - x^3)$, $t \mapsto (t^2, t^3)$ is:

• Bijective: given $(a, b) \in Y$ with $b^2 = a^3$, if $a = 0$ then $b = 0$ and $t = 0$; if $a \neq 0$ then $t = b/a$ is the unique preimage.
• A homeomorphism in the Zariski topology (since any bijective polynomial map between irreducible curves is a homeomorphism).
• Not an isomorphism: the pullback $\varphi^*: k[x,y]/(y^2 - x^3) \to k[t]$ sends $x \mapsto t^2$, $y \mapsto t^3$. The image is $k[t^2, t^3] \subsetneq k[t]$, so $\varphi^*$ is not surjective. Concretely, there is no polynomial in $(t^2, t^3)$ that equals $t$.

The "inverse" would need to send $(x,y) \mapsto y/x$, but $y/x$ is not a regular function at the origin (it is a rational function with an indeterminacy). The cusp singularity prevents the inverse from being a morphism.

Lesson: In characteristic 0, a bijective morphism $\varphi: X \to Y$ with $Y$ normal (e.g., smooth) is an isomorphism (by Zariski's Main Theorem). The failure for the cusp occurs because $Y$ is not normal.

In characteristic $p > 0$, the Frobenius $\operatorname{Fr}: \mathbb{A}^1 \to \mathbb{A}^1$, $a \mapsto a^p$ is bijective (since $k$ is algebraically closed), but the pullback $\operatorname{Fr}^*: k[x] \to k[x]$, $x \mapsto x^p$ is not surjective. So $\operatorname{Fr}$ is bijective but not an isomorphism.

This phenomenon is purely characteristic $p$: in characteristic 0, every bijective morphism $\mathbb{A}^1 \to \mathbb{A}^1$ is an isomorphism. Over $\mathbb{C}$, an endomorphism of $\mathbb{A}^1$ is given by $x \mapsto f(x)$, and bijectivity forces $\deg f = 1$, i.e., $f$ is linear.

An automorphism $\varphi: \mathbb{A}^1 \to \mathbb{A}^1$ corresponds to a $k$-algebra isomorphism $\varphi^*: k[x] \to k[x]$. Any such isomorphism sends $x \mapsto ax + b$ where $a \in k^*$, $b \in k$ (since $\varphi^*(x)$ must generate $k[x]$ and have degree 1). Therefore

$\operatorname{Aut}(\mathbb{A}^1) = \{x \mapsto ax + b \mid a \in k^*, b \in k\} \cong \operatorname{Aff}(1, k) = k^* \ltimes k.$

This is the affine group of the line. Higher-degree polynomial maps $x \mapsto x + x^2$, while injective over $\mathbb{C}$, are not surjective, hence not automorphisms.

For $n = 2$, $\operatorname{Aut}(\mathbb{A}^2)$ is much larger. It contains:

• Affine maps: $(x, y) \mapsto (ax + by + e, cx + dy + f)$ with $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ invertible.
• Triangular (de Jonquieres) maps: $(x, y) \mapsto (ax + p(y), by + c)$ with $a, b \in k^*$, $p(y) \in k[y]$.

By the Jung--van der Kulk theorem, $\operatorname{Aut}(\mathbb{A}^2)$ is the amalgamated free product of the affine group and the triangular group over their intersection.

For $n \geq 3$, $\operatorname{Aut}(\mathbb{A}^n)$ is not fully understood. The Jacobian conjecture (open, one of Smale's problems) asserts: if $\varphi: \mathbb{A}^n \to \mathbb{A}^n$ has Jacobian determinant $\det(J_\varphi) \in k^*$ (a nonzero constant), then $\varphi$ is an automorphism.

An automorphism $\varphi: \mathbb{P}^1 \to \mathbb{P}^1$ is given by

$[x_0 : x_1] \mapsto [ax_0 + bx_1 : cx_0 + dx_1], \quad \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL(2, k).$

Two matrices give the same automorphism iff they differ by a scalar, so

$\operatorname{Aut}(\mathbb{P}^1) \cong PGL(2, k) = GL(2, k) / k^*.$

In the affine coordinate $t = x_1/x_0$, the automorphism becomes the Mobius transformation

$t \mapsto \frac{ct + d}{at + b} \quad \text{(or equivalently } t \mapsto \frac{\alpha t + \beta}{\gamma t + \delta}\text{)}.$

Key properties:

• $\dim PGL(2) = 3$, so automorphisms of $\mathbb{P}^1$ form a 3-dimensional group.
• $PGL(2)$ acts 3-transitively on $\mathbb{P}^1$: given any three distinct points $P_1, P_2, P_3$ and any other three distinct points $Q_1, Q_2, Q_3$, there is a unique $\varphi \in PGL(2)$ with $\varphi(P_i) = Q_i$.
• For $n \geq 2$: $\operatorname{Aut}(\mathbb{P}^n) = PGL(n+1, k)$, acting by linear changes of coordinates.

Let $E$ be an elliptic curve over $k$ with identity $O \in E$. An automorphism $\varphi: E \to E$ preserving $O$ (i.e., $\varphi(O) = O$) is an automorphism of $E$ as an abelian variety. The group of such automorphisms is:

• Generic case ($\operatorname{char} k \neq 2, 3$, and $j(E) \neq 0, 1728$): $\operatorname{Aut}(E, O) = \{\pm 1\} \cong \mathbb{Z}/2\mathbb{Z}$, generated by the negation map $P \mapsto -P$ (inversion in the group law).

• $j = 1728$ ($E: y^2 = x^3 - x$, char $k \neq 2$): $\operatorname{Aut}(E, O) \cong \mathbb{Z}/4\mathbb{Z}$, generated by $(x, y) \mapsto (-x, iy)$ where $i^2 = -1$. This is an order-4 automorphism.

• $j = 0$ ($E: y^2 = x^3 - 1$, char $k \neq 2, 3$): $\operatorname{Aut}(E, O) \cong \mathbb{Z}/6\mathbb{Z}$, generated by $(x, y) \mapsto (\zeta_3 x, -y)$ where $\zeta_3$ is a primitive cube root of unity. The order-3 element $(x,y) \mapsto (\zeta_3 x, y)$ is a "rotation" of the curve.

If we drop the condition $\varphi(O) = O$, then $\operatorname{Aut}(E) = E \rtimes \operatorname{Aut}(E, O)$, where $E$ acts on itself by translations $\tau_Q: P \mapsto P + Q$. Translations are automorphisms of $E$ as a variety (but not as a group).

Let $\varphi: Y = V(y^2 - x) \to \mathbb{A}^1$ be the projection $(x, y) \mapsto x$. The pullback is $\varphi^*: k[t] \to k[x,y]/(y^2 - x) \cong k[y]$, $t \mapsto y^2$. Is $k[y]$ a finite $k[y^2]$-module? Yes: $k[y] = k[y^2] \cdot 1 + k[y^2] \cdot y$, so $k[y]$ is free of rank 2 over $k[y^2]$. Therefore $\varphi$ is a finite morphism of degree 2. Geometrically, each fiber has at most 2 points: the two square roots of $x$.