Any morphism $f : X \to Y$ determines a rational map by taking $U = X$. The rational map $\mathrm{id} : X \dashrightarrow X$ is the class of $(\mathrm{id}_X, X)$.

Let $X = V(y^2 - x^3) \subseteq \mathbb{A}^2$ (the cuspidal cubic). Define $\varphi : X \dashrightarrow \mathbb{A}^1$ by $\varphi(x, y) = y/x$. This is defined on the open set $U = X \setminus \{(0,0)\}$. On $U$, we have $y/x = x^2 \cdot x^{-1} \cdot y/y = \ldots$ --- more directly, $y/x$ is a well-defined regular function on $U$ since $x \neq 0$ there ($x = 0$ forces $y = 0$ on $X$). This gives a rational map $X \dashrightarrow \mathbb{A}^1$.

Define $\pi : \mathbb{P}^2 \dashrightarrow \mathbb{P}^1$ by

$\pi([x : y : z]) = [x : y].$

This is well-defined whenever $(x, y) \neq (0, 0)$, i.e., on $\mathbb{P}^2 \setminus \{[0:0:1]\}$. The point $P = [0:0:1]$ is the unique point of indeterminacy --- geometrically, $\pi$ is "projection away from $P$." Each fiber $\pi^{-1}([a:b])$ is the line through $P$ and the point $[a:b:0]$, minus $P$ itself.

The domain of definition is $\mathrm{dom}(\pi) = \mathbb{P}^2 \setminus \{P\}$, which has complement of codimension 2, consistent with the theorem above (since $\mathbb{P}^2$ is smooth).

Let $C = V(xz - y^2) \subseteq \mathbb{P}^2$ (a smooth conic) and let $P = [0:0:1] \in C$. The projection from $P$ restricts to a rational map

$\pi|_C : C \dashrightarrow \mathbb{P}^1, \quad [x:y:z] \mapsto [x:y].$

Since $C$ is a smooth curve, this extends to a morphism on all of $C$. To find $\pi(P)$: near $P = [0:0:1]$, use $y^2 = xz$. For points of $C$ near $P$ with $x \neq 0$, we get $[x:y] = [x:y]$. Taking the limit as $[x:y:z] \to [0:0:1]$ along $C$, we parametrize: $[t^2 : t : 1]$ gives $\pi = [t^2 : t] = [t:1]$, so $\pi(P) = \lim_{t \to 0} [t:1] = [0:1]$.

This gives an isomorphism $C \xrightarrow{\sim} \mathbb{P}^1$, confirming that every smooth conic is rational.

Consider the rational map $\varphi : \mathbb{A}^2 \dashrightarrow \mathbb{P}^1$ defined by $\varphi(x,y) = [x:y]$. The domain of definition is $\mathbb{A}^2 \setminus \{(0,0)\}$: the origin is the single point of indeterminacy. The "value" at the origin is ambiguous because lines through the origin have all possible slopes --- this is precisely the situation resolved by blowing up.

1. The inclusion $\mathbb{A}^1 \hookrightarrow \mathbb{A}^2$, $t \mapsto (t, 0)$, viewed as a rational map, is not dominant: its image is the $x$-axis, which is not dense in $\mathbb{A}^2$.

2. The projection $\mathbb{A}^2 \dashrightarrow \mathbb{A}^1$, $(x,y) \mapsto x$, is dominant (and in fact surjective).

3. The parametrization $\mathbb{A}^1 \dashrightarrow V(y^2 - x^3)$, $t \mapsto (t^2, t^3)$, is dominant: its image is the entire cuspidal cubic.

$\mathbb{A}^1 \setminus \{0\}$ and $\mathbb{A}^1$ are birational: both have function field $k(t)$. The inclusion $\mathbb{A}^1 \setminus \{0\} \hookrightarrow \mathbb{A}^1$ and the identity $\mathbb{A}^1 \to \mathbb{A}^1$ give an isomorphism on the open set $\mathbb{A}^1 \setminus \{0\}$.

However, $\mathbb{A}^1 \setminus \{0\}$ is not isomorphic to $\mathbb{A}^1$ as a variety: the coordinate ring of $\mathbb{A}^1 \setminus \{0\}$ is $k[t, t^{-1}]$, which has a unit ($t$) that is not a scalar, while $k[\mathbb{A}^1] = k[t]$ does not.

$\mathbb{A}^n$ and $\mathbb{P}^n$ are birational: $\mathbb{A}^n$ embeds as the standard open set $U_0 = \{x_0 \neq 0\} \subseteq \mathbb{P}^n$, which is dense. Both have function field $k(x_1, \ldots, x_n)$.

Any smooth conic $C \subseteq \mathbb{P}^2$ is rational ($C \cong \mathbb{P}^1$). Given a point $P \in C$, projection from $P$ gives a birational map $C \dashrightarrow \mathbb{P}^1$ that is actually an isomorphism (see the projection example above). For instance, $V(x^2 + y^2 - z^2) \subseteq \mathbb{P}^2$ with $P = [0:1:1]$:

$[x:y:z] \mapsto [x : z - y]$

gives $\mathbb{P}^1 \to C$ by $[s:t] \mapsto [2st : s^2 - t^2 : s^2 + t^2]$.

The smooth quadric $Q = V(xw - yz) \subseteq \mathbb{P}^3$ satisfies $Q \cong \mathbb{P}^1 \times \mathbb{P}^1$ (via the Segre embedding). Since $\mathbb{P}^1 \times \mathbb{P}^1 \sim_{\mathrm{bir}} \mathbb{A}^2$ (the product of two copies of $\mathbb{A}^1$ is $\mathbb{A}^2$), we get $Q \sim_{\mathrm{bir}} \mathbb{P}^2$. So smooth quadric surfaces are rational.

Explicitly, $k(Q) = k(s, t)$ where $s, t$ parametrize the two rulings.

Let $Q = V(x_0^2 + x_1^2 + \cdots + x_n^2 - x_{n+1}^2) \subseteq \mathbb{P}^{n+1}$ be a smooth quadric hypersurface, and let $P = [0 : 0 : \cdots : 0 : 1 : 1] \in Q$. Projection from $P$ to the hyperplane $\{x_{n+1} = 0\} \cong \mathbb{P}^n$ gives a birational map $Q \dashrightarrow \mathbb{P}^n$.

Concrete case $n = 1$: $Q = V(x^2 + y^2 - z^2) \subseteq \mathbb{P}^2$, $P = [0:1:1]$. The line through $P$ and $[t:s:0] \in \mathbb{P}^1$ is parametrized by $[\lambda t : \lambda s + \mu : \mu]$. Substituting into $x^2 + y^2 = z^2$:

$\lambda^2 t^2 + (\lambda s + \mu)^2 = \mu^2$

$\lambda^2 t^2 + \lambda^2 s^2 + 2\lambda s \mu + \mu^2 = \mu^2$

$\lambda(t^2 + s^2) + 2s\mu = 0 \quad \Longrightarrow \quad \lambda/\mu = -2s/(t^2 + s^2).$

So the second intersection point (besides $P$) is:

$[x:y:z] = \left[-2st : t^2 + s^2 - 2s^2 : t^2 + s^2\right] \cdot \frac{1}{t^2+s^2} \sim [-2st : t^2 - s^2 : t^2 + s^2].$

This is the classical parametrization of Pythagorean triples: setting $s = 1$, $t \in k$:

$(x, y, z) \sim (-2t,\; t^2 - 1,\; t^2 + 1).$

Over $\mathbb{Q}$, this recovers all rational points on the unit circle $x^2 + y^2 = z^2$.

The standard Cremona transformation is the rational map $\sigma : \mathbb{P}^2 \dashrightarrow \mathbb{P}^2$ defined by

$\sigma([x : y : z]) = [yz : xz : xy] = [1/x : 1/y : 1/z].$

Domain of definition: $\sigma$ is defined wherever at least one of $yz, xz, xy$ is nonzero, i.e., wherever at most one coordinate vanishes. The indeterminacy locus consists of the three coordinate vertices:

$\mathrm{Ind}(\sigma) = \{[1:0:0],\; [0:1:0],\; [0:0:1]\}.$

Self-inverse: $\sigma \circ \sigma = \mathrm{id}$ as a rational map. Indeed:

$\sigma(\sigma([x:y:z])) = \sigma([yz:xz:xy]) = [(xz)(xy) : (yz)(xy) : (yz)(xz)] = [x^2yz : xy^2z : xyz^2] = [x:y:z].$

So $\sigma$ is a birational involution of $\mathbb{P}^2$.

Image of lines: The line $V(ax + by + cz) \subseteq \mathbb{P}^2$ maps under $\sigma$ to

$a \cdot yz + b \cdot xz + c \cdot xy = 0 \quad \Longleftrightarrow \quad V(ayz + bxz + cxy),$

which is a conic passing through all three coordinate vertices. So $\sigma$ takes lines to conics and vice versa.

Image of conics through two base points: A conic $V(f)$ passing through two of the three coordinate vertices (say $[1:0:0]$ and $[0:1:0]$) maps to a line. For instance, $V(z^2 - xy)$ passes through $[1:0:0]$ and $[0:1:0]$, and its image under $\sigma$ is $V((xy)^2 - (yz)(xz)) = V(x^2 y^2 - xyz^2) = V(xy - z^2)$ ... wait, let us compute directly: substituting into $z^2 = xy$, we get $(xy)^2 = (yz)(xz)$, i.e., $x^2 y^2 = xyz^2$, giving $xy = z^2$. But the proper transform under $\sigma$ of the conic $V(z^2 - xy)$ is a line. To see this clearly, note that $\sigma^*$ sends degree-$d$ curves to degree-$2d$ curves, but passing through each base point with multiplicity $\geq d$ reduces the degree by the multiplicities.

The Cremona transformation is the simplest example of a birational automorphism of $\mathbb{P}^2$ that is not a linear automorphism ($\mathrm{PGL}_3$). In fact, the Noether--Castelnuovo theorem states that every birational automorphism of $\mathbb{P}^2$ (over an algebraically closed field) is a composition of linear automorphisms and the standard Cremona transformation.

Restricting $\sigma$ to the affine chart $z = 1$: $\sigma(x, y) = (1/x, 1/y)$ as a rational map $\mathbb{A}^2 \dashrightarrow \mathbb{A}^2$. This is defined on $\{xy \neq 0\}$ and maps the hyperbola $V(xy - 1)$ to the single point $(1, 1)$ (but only as a rational map --- the proper transform is more subtle).

On the torus $(\mathbb{A}^1 \setminus \{0\})^2 = \{xy \neq 0\}$, the map $(x,y) \mapsto (1/x, 1/y)$ is an automorphism (not just birational), reflecting the group inversion on the algebraic torus.

Let $C = V(y^2 - x^3) \subseteq \mathbb{A}^2$. Define $\varphi : \mathbb{A}^1 \to C$ by

$\varphi(t) = (t^2, t^3).$

Claim: $\varphi$ is a birational equivalence $\mathbb{A}^1 \dashrightarrow C$.

Forward map: $\varphi$ is a morphism (polynomial map), and $(t^2)^3 = t^6 = (t^3)^2$, so the image lies on $C$.

Inverse rational map: Define $\psi : C \dashrightarrow \mathbb{A}^1$ by $\psi(x, y) = y/x$. This is defined on $C \setminus \{(0,0)\}$. On $C$, we have $y^2 = x^3$, so $(y/x)^2 = x$, giving $(y/x)^3 = x \cdot (y/x) = y$. Thus

$\varphi(\psi(x,y)) = \left((y/x)^2, (y/x)^3\right) = (x, y).$

And $\psi(\varphi(t)) = t^3/t^2 = t$ for $t \neq 0$.

Key point: $\varphi$ is a bijection $\mathbb{A}^1 \to C$, but it is not an isomorphism of varieties. The inverse $\psi$ is only a rational map (undefined at the cusp). The coordinate rings are

$k[C] = k[t^2, t^3] \subsetneq k[t] = k[\mathbb{A}^1],$

and the inclusion $k[t^2, t^3] \hookrightarrow k[t]$ corresponds to $\varphi$. Since $k[t^2, t^3] \not\cong k[t]$, we have $C \not\cong \mathbb{A}^1$, even though $C \sim_{\mathrm{bir}} \mathbb{A}^1$.

Let $C = V(y^2 - x^2(x + 1)) \subseteq \mathbb{A}^2$. The node at the origin has two tangent directions $y = \pm x$.

Define $\varphi : \mathbb{A}^1 \to C$ by

$\varphi(t) = (t^2 - 1,\; t(t^2 - 1)) = (t^2 - 1,\; t^3 - t).$

Verification: $(t^3 - t)^2 = t^2(t^2-1)^2$ and $(t^2-1)^2((t^2-1)+1) = (t^2-1)^2 \cdot t^2$. Check.

Inverse: $\psi(x, y) = y/x$, defined for $x \neq 0$. On $C$, $(y/x)^2 = x + 1$, so $t = y/x$ gives $x = t^2 - 1$, $y = t^3 - t$.

Failure of bijectivity: $\varphi(1) = (0, 0) = \varphi(-1)$. The map $\varphi$ identifies $t = 1$ and $t = -1$ to the node. So $\varphi$ is a bijection on $\mathbb{A}^1 \setminus \{1, -1\} \to C \setminus \{(0,0)\}$, but it is 2-to-1 at the node.

The normalization of $C$ is $\mathbb{A}^1$, and $\varphi$ is the normalization map. The nodal cubic is rational: $k(C) = k(t)$.

For each $d \geq 1$, the rational normal curve of degree $d$ is the image of

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

• $d = 1$: $\mathbb{P}^1 \xrightarrow{\sim} \mathbb{P}^1$ (the identity).
• $d = 2$: the smooth conic $V(x_0 x_2 - x_1^2) \subseteq \mathbb{P}^2$.
• $d = 3$: the twisted cubic $\subseteq \mathbb{P}^3$, cut out by $x_0 x_2 - x_1^2$, $x_0 x_3 - x_1 x_2$, $x_1 x_3 - x_2^2$.

All of these are rational: the Veronese map $\nu_d$ is an isomorphism onto its image.

The elliptic curve $E = V(y^2 z - x^3 + xz^2) \subseteq \mathbb{P}^2$ has genus 1, so $E \not\sim_{\mathrm{bir}} \mathbb{P}^1$. Equivalently, $k(E) \not\cong k(t)$: the function field $k(E) = k(x)[y]/(y^2 - x^3 + x)$ is a quadratic extension of $k(x)$, but it is not purely transcendental.

No rational parametrization of $E$ exists. This is the fundamental difference between genus 0 and genus $\geq 1$.

Let $S \subseteq \mathbb{P}^3$ be a smooth cubic surface over $k = \bar{k}$. A classical theorem of Cayley and Salmon states that $S$ contains exactly 27 lines. Choosing any one line $\ell \subseteq S$ and projecting from $\ell$ gives a birational map $S \dashrightarrow \mathbb{P}^2$. Thus every smooth cubic surface is rational.

For example, the Fermat cubic $x^3 + y^3 + z^3 + w^3 = 0$ contains the line $\{x + y = 0,\; z + w = 0\}$ (set $y = -x$, $w = -z$). Projection from this line gives $S \sim_{\mathrm{bir}} \mathbb{P}^2$.

The blowup of $\mathbb{A}^2$ at the origin is the variety

$\mathrm{Bl}_0 \mathbb{A}^2 = \{((x, y), [s : t]) \in \mathbb{A}^2 \times \mathbb{P}^1 \mid xt = ys\} \subseteq \mathbb{A}^2 \times \mathbb{P}^1.$

The projection $\pi : \mathrm{Bl}_0 \mathbb{A}^2 \to \mathbb{A}^2$ is a birational morphism: it is an isomorphism over $\mathbb{A}^2 \setminus \{(0,0)\}$, and the fiber over the origin is

$\pi^{-1}(0, 0) = \{(0, 0)\} \times \mathbb{P}^1 \cong \mathbb{P}^1,$

called the exceptional divisor $E$. The blowup "replaces the origin by the set of all tangent directions through it."

Explicit charts: In the chart $s \neq 0$ (set $s = 1$), we have $y = xt$, so the chart is $\mathbb{A}^2$ with coordinates $(x, t)$ and $\pi(x, t) = (x, xt)$. In the chart $t \neq 0$ (set $t = 1$), we have $x = ys$, giving coordinates $(s, y)$ and $\pi(s, y) = (sy, y)$.

The blowup is birational to $\mathbb{A}^2$: $\pi$ has an inverse rational map $\mathbb{A}^2 \dashrightarrow \mathrm{Bl}_0 \mathbb{A}^2$ given by $(x, y) \mapsto ((x, y), [x : y])$, defined for $(x, y) \neq (0, 0)$. So $\mathrm{Bl}_0 \mathbb{A}^2 \sim_{\mathrm{bir}} \mathbb{A}^2$.

Let $C = V(y^2 - x^2(x+1)) \subseteq \mathbb{A}^2$ be the nodal cubic. The strict transform of $C$ under the blowup $\pi : \mathrm{Bl}_0 \mathbb{A}^2 \to \mathbb{A}^2$ is obtained by taking $\pi^{-1}(C \setminus \{0\})$ and taking its closure.

In the chart $y = xt$: substituting into $y^2 = x^2(x+1)$ gives $(xt)^2 = x^2(x+1)$, so $x^2 t^2 = x^2(x+1)$. Canceling $x^2$ (this removes the exceptional divisor component): $t^2 = x + 1$, i.e., $x = t^2 - 1$. This is a smooth curve (check: $\partial/\partial t (t^2 - x - 1) = 2t$ and $\partial/\partial x = -1$ are never simultaneously zero).

The two branches at the node (slopes $t = 1$ and $t = -1$) are separated in the blowup: they correspond to two distinct points $(x, t) = (0, 1)$ and $(0, -1)$ on the strict transform. The blowup resolves the singularity.

For the cuspidal cubic $C = V(y^2 - x^3)$: in the chart $y = xt$, we get $x^2 t^2 = x^3$, so $t^2 = x$ (after canceling $x^2$). The strict transform is $V(t^2 - x)$, which is smooth --- but actually, we need to check the other chart as well. In the chart $x = ys$: $y^2 = (ys)^3 = y^3 s^3$, so $1 = ys^3$ (canceling $y^2$), giving a smooth curve.

So a single blowup resolves the cusp. The strict transform is isomorphic to $\mathbb{A}^1$ (parametrized by $t$, with $x = t^2$), confirming $C \sim_{\mathrm{bir}} \mathbb{A}^1$.

More generally, every singularity of a curve can be resolved by a finite sequence of blowups (Hironaka's theorem gives resolution of singularities in all dimensions in characteristic 0).

$C = V(y^2 - x^3) \subseteq \mathbb{A}^2$. The coordinate ring is $k[C] = k[x, y]/(y^2 - x^3) \cong k[t^2, t^3]$, where $t = y/x$. The function field is

$k(C) = \mathrm{Frac}(k[t^2, t^3]) = k(t),$

since $t = t^3/t^2 \in \mathrm{Frac}(k[t^2, t^3])$. So $k(C) = k(t)$, confirming $C \sim_{\mathrm{bir}} \mathbb{A}^1$.

$C = V(y^2 - x^2(x+1))$. Setting $t = y/x$, we get $t^2 = x + 1$, so $x = t^2 - 1$ and $y = t(t^2 - 1)$. The function field is

$k(C) = k(x, y) / (y^2 - x^2(x+1)) = k(t),$

again purely transcendental. So the nodal cubic is rational.

$E = V(y^2 - x^3 + x)$. The function field is

$k(E) = k(x)[y]/(y^2 - x^3 + x) = k(x)(\sqrt{x^3 - x}).$

This is a degree-2 extension of $k(x)$. Is it purely transcendental? No. The field $k(E)$ has genus 1 (as a function field of one variable over $k$), and by the Riemann--Roch theorem, $k(E) \not\cong k(t)$. This is what prevents $E$ from being rational.

Consider the subfield $L = k(t^2) \subseteq k(t)$. By Luroth's theorem, $L$ is purely transcendental over $k$, which is obvious: $L = k(s)$ where $s = t^2$.

A less trivial example: $L = k\!\left(\frac{t^2}{t+1}\right) \subseteq k(t)$. Set $u = \frac{t^2}{t+1}$. Then $ut + u = t^2$, so $t^2 - ut - u = 0$, giving $t = \frac{u \pm \sqrt{u^2 + 4u}}{2}$. Thus $[k(t) : L] = 2$ (generically). Luroth's theorem guarantees $L = k(u)$ is purely transcendental --- and indeed $u = t^2/(t+1)$ is a single rational function of $t$.

Let $C$ be a smooth projective curve and suppose there is a dominant rational map $\varphi : \mathbb{P}^1 \dashrightarrow C$. This gives an inclusion $k(C) \hookrightarrow k(\mathbb{P}^1) = k(t)$. Since $C$ is a curve, $\mathrm{tr.deg}_k \, k(C) = 1$, so Luroth's theorem says $k(C) \cong k(s)$, hence $C \sim_{\mathrm{bir}} \mathbb{P}^1$.

Consequence: A curve that can be parametrized by rational functions is necessarily rational. There is no distinction between "rational" and "unirational" for curves.

Let $S \subseteq \mathbb{P}^3$ be a smooth cubic surface. Then $\omega_S \cong \mathcal{O}_S(-1)$ (by adjunction: $\omega_S = \omega_{\mathbb{P}^3}(3)|_S = \mathcal{O}(-4+3)|_S = \mathcal{O}_S(-1)$). Since $\mathcal{O}_S(-1)$ has no global sections, $P_1 = 0$, and also $P_2 = h^0(\mathcal{O}_S(-2)) = 0$. Furthermore $q = h^1(\mathcal{O}_S) = 0$ (by the Lefschetz hyperplane theorem). So by Castelnuovo's criterion, $S$ is rational.

Let $X = V(x_0^3 + x_1^3 + x_2^3 + x_3^3 + x_4^3) \subseteq \mathbb{P}^4$ (the Fermat cubic threefold). This is:

• Unirational: Project from a line $\ell \subseteq X$ to get a conic bundle over $\mathbb{P}^2$. A unirational parametrization can be written down explicitly.
• Not rational: By Clemens--Griffiths. The intermediate Jacobian $J(X)$ is a 5-dimensional abelian variety that is not a Jacobian of a curve and not a product of lower-dimensional abelian varieties.

So $k(X) \hookrightarrow k(x_1, x_2, x_3)$ (unirational), but $k(X) \not\cong k(x_1, x_2, x_3)$ (not rational). This is the 3-dimensional failure of the Luroth problem.

A general smooth quartic threefold $X_4 \subseteq \mathbb{P}^4$ satisfies $\mathrm{Bir}(X_4) = \mathrm{Aut}(X_4)$. In particular, if $X_4$ has no nontrivial automorphisms (which is the general case), then $\mathrm{Bir}(X_4) = \{\mathrm{id}\}$.

This is called birational rigidity. It immediately implies $X_4$ is irrational, since $\mathrm{Bir}(\mathbb{P}^3)$ is enormous (it contains the Cremona group).

It is unknown whether a general smooth quartic threefold is unirational.

For surfaces, the MMP reduces to the classical theory of Castelnuovo contractions. A smooth projective surface $S$ contains a $(-1)$-curve $E \cong \mathbb{P}^1$ with $E^2 = -1$ if and only if $S$ is the blowup of another smooth surface $S'$ at a point, with $E$ the exceptional divisor. By repeatedly contracting $(-1)$-curves, we arrive at either:

• $\mathbb{P}^2$ or a Hirzebruch surface $\mathbb{F}_n$ ($\kappa = -\infty$), or
• A minimal surface with $K_S$ nef ($\kappa \geq 0$).

For example, the blowup of $\mathbb{P}^2$ at one point gives a surface $S$ with one $(-1)$-curve $E$. Contracting $E$ recovers $\mathbb{P}^2$. But $S$ is also isomorphic to $\mathbb{F}_1$ (the first Hirzebruch surface), so we see that $\mathbb{F}_1 \sim_{\mathrm{bir}} \mathbb{P}^2$ but $\mathbb{F}_1 \not\cong \mathbb{P}^2$.