Definition. Let V ⊆ A

2 Morphisms, co-ordinate rings, rational maps, and function fields

2.1 Morphisms and co-ordinate rings

The obvious functions to consider on an affine algebraic variety V ⊆ A

(K) are the polynomials. Two polynomials f , g define the same function V → K if and only if f − g ∈ I(V ), this motivates the following definition.

Definition. Let V ⊆ A

(K) be an affine algebraic variety. Its co-ordinate ring, denoted by K[V ], is defined to be K[V ] = K[x

, x

, . . . , x

]/I(V ).

Remark. K means the field over which the variety is defined, so if the field is C, then the notation is C[V ], if the field is called L, then L[V ].

Definition. Let V ⊆ A

(K) and W ⊆ A

(K) be affine algebraic varieties. A morphism ϕ : V → W is a function of the form ϕ(P ) = (ϕ

(P ), ϕ

(P ), . . . , ϕ

(P )) with ϕ

∈ K[V ] for each i, 1 ≤ i ≤ n.

Elements of K[V ] can be considered as morphisms V → A

.

The composition of two morphisms ϕ : V → W and ψ : W → X is a morphism ψ ◦ ϕ : V → X.

Definition. The morphism ϕ : V → W is called an isomorphism if and only if it has an inverse morphism, i. e., if there exists a morphism ψ : W → V such that ψ ◦ ϕ = I

and ϕ ◦ ψ = I

, where I

: X → X is the identity function on X for a set X. V and W are called isomorphic if and only if there exists an isomorphism between them.

Isomorphic affine algebraic varieties are the “same” for most purposes. One of the goals of algebraic geometry is to classify algebraic varieties up to isomorphism and to decide whether two varieties are isomorphic. One of the tools is to find properties of algebraic varieties which are preserved under isomorphism.

Examples:

1. Any affine map is a morphism. Invertible affine maps are isomorphisms.

Affine equivalent affine algebraic varieties are isomorphic.

2. Let K be an arbitrary infinite field.

Let t be the co-ordinate on A

and x, y the co-ordinates on A

. Let V = A

and W ⊂ A

be the parabola defined by the equation y − x

= 0. Then

I(V ) = {0}, I(W ) = hy − x

i (this is where we need K to be infinite), so

K[V ] = K[t] and K[W ] = K[x, y]/hy − x

i.

ϕ : V → W , t 7→ (t, t

) is a morphism. t and t

are elements of K[V ] and x = t, y = t

satisfy y − x

= 0 for any t. (This shows that ϕ(t) ∈ W for every t ∈ V = A

, so ϕ is a morphism V → W , not just V → A

.)

ψ : W → V , (x, y) 7→ x is also a morphism. x ∈ K[W ] (technically it should be x + I(W ), but this distinction is usually not made in practice) and there are no equations to check. (ψ ◦ϕ)(t) = ψ(t, t

) = t and (ϕ◦ψ)(x, y) = ϕ(x) = (x, x

) = (x, y) for (x, y) ∈ W , therefore ϕ and ψ are inverses of each other, so V and W are isomorphic.

3. Let K be an infinite field of characteristic other than 2, for example K = R or C. Let V = A

and let W ⊂ A

be the nodal cubic curve defined by the equation x

+ x

− y

= 0.

-1 1 2

-4 -2 2 4

ϕ : V → W , t 7→ (t

− 1, t(t

− 1)) is a morphism. t

− 1 and t(t

− 1) are elements of K[V ] and x = t

− 1, y = t(t

− 1) satisfy x

+ x

− y

= 0 for any t, so ϕ is a function V → W .

ϕ is not an isomorphism, since it is not injective, ϕ(1) = ϕ(−1) = (0, 0).

(We need the characteristic to be different from 2 to ensure that −1 6= 1.) In examples 4–5, K can be an arbitrary infinite field.

4. Let H ⊂ A

be the hyperbola defined by xy−1 = 0, and let ϕ : H → A

be the morphism (x, y) 7→ x. ϕ is not an isomorphism, since it is not surjective because 0 / ∈ im ϕ.

Warning: The image of a morphism is not necessarily an algebraic variety.

In this example im ϕ = A

\{0}, and this set is not a variety, since any proper

subvariety of A

is finite.

5. Let V = A

and let W = V(hy

− x

i) ⊂ A

be the cuspidal cubic curve.

ϕ : V → W , t 7→ (t

, t

) is a morphism. t

and t

are elements of K[V ] and x = t

, y = t

satisfy y

− x

= 0 for any t, so ϕ is a function V → W .

-0.5 0.5 1.0 1.5 2.0 2.5

-4 -2 2 4

ϕ is bijective as a function, but it is not an isomorphim. It has an inverse as a function ψ : W → V , ψ(x, y) =

( y/x, if x 6= 0

0, if x = 0 , but this does not appear to be a morphism from that way it is defined. Let’s assume that ϑ : W → V is a morphism, which is the inverse of ϕ. The ϑ is represented by a polynomial f ∈ K[x, y] such that f (t

, t

) = t, but this is impossible.

We shall prove later (see the example after Corollary 2.2) that V and W are not even isomorphic, not just that this particular morphism is not an isomorphism.

Warning. Any isomorphism is bijective, since it has an inverse as a function, but as the above example shows, a bijective morphism is not necessarily an isomorphism.

6. Let x, y be the co-ordinates on A

(C) and u, v, w the co-ordinates on

A

(C). Let V = A

and W ⊂ A

be the variety defined by the equation

uw − v

= 0, it is the double cone shown below.

ϕ : V → W , (x, y) 7→ (x

, xy, y

)) is a morphism. x

, xy, y

are elements of K[V ] and u = x

, v = xy and w = y

satisfy uw − v

= x

y

− (xy)

= 0 for all x, y, therefore ϕ(V ) ⊆ W , so ϕ is indeed a morphism V → W , not just V → A

. ϕ is not an isomorphism because it is not injective, ϕ(x, y) = ϕ(−x, −y) for any (x, y) ∈ A

, but (x, y) 6= (−x, −y) unless (x, y) = (0, 0).

(This argument works as long as the characteristic of the field is not 2. In characteristic 2, (x, y) = (−x, −y) and ϕ is bijective if the field is algebraically closed, but it is still not an isomorphism.)

7. Let V = A

(R) and let W = V(hx

+ y

− 1i) ⊂ A

(R) be the unit circle.

ϕ : V → W , ϕ(t) = (cos t, sin t) is function from V to W , but it is not a morphism, as cos t and sin t are not elements of R[A

] ∼ = R[t].

Theorem 2.1 Let V ⊆ A

(K) and W ⊆ A

(K) be affine algebraic varieties.

Let y

, y

, . . . , y

be co-ordinates on A

. For a morphism ϕ : V → W , define ϕ

: K[W ] → K[V ], ϕ

(f ) = f ◦ ϕ.

: ϕ 7→ ϕ

is a bijection between morphisms of affine algebraic varieties ϕ : V → W and ring homomorphisms α : K[W ] → K[V ] preserving K.

Proof. Let f ∈ K[W ]. f = F + I(W ) for some F ∈ K[y

, y

, . . . , y

]. Now

ϕ

(f ) = f ◦ ϕ = F ◦ ϕ = F (ϕ

, ϕ

, . . . , ϕ

), where ϕ

, ϕ

, . . . , ϕ

∈ K[V ] are

the components of ϕ. As F is a polynomial, ϕ

(f ) = F (ϕ

, ϕ

, . . . , ϕ

) ∈

K[V ], too. If G ∈ K[y

, y

, . . . , y

] is another polynomial such that f =

G + I(W ), then F − G ∈ I(W ), so F (ϕ(P )) = G(ϕ(P )) for every P ∈ V

and therefore ϕ

(f ) is a well-defined element of K[V ]. This shows that ϕ

is indeed a function K[W ] → K[V ].

ϕ

(λ) = λ for any λ ∈ K and ϕ

(f +g) = ϕ

(f )+ϕ

(g), ϕ

(f g) = ϕ

(f )ϕ

(g) for all f , g ∈ K[W ] by the definition of ϕ

, therefore ϕ

is a homomorphism of rings which preserves K.

Given a homomorphism α : K[W ] → K[V ] with α(λ) = λ for every λ ∈ K, let ϕ

= α(y

+ I(W )) ∈ K[V ] for i = 1, 2, . . . , n. Then ϕ = (ϕ

, ϕ

, . . . , ϕ

) is clearly a morphism V → A

. We need to prove that ϕ(V ) ⊆ W so that ϕ is a morphism V → W .

Let P ∈ V an arbitrary point. In order to prove ϕ(P ) ∈ W , we need to show g(ϕ(P )) = 0 for any g ∈ I(W ).

Let q : K[y

, y

, . . . , y

] → K[W ] = K[y

, y

, . . . , y

]/I(W ) be the quotient map q(f ) = f + I(W ). Then ϕ

= α(q(y

)), so

g(ϕ(P )) = g(α(q(y

))(P ), α(q(y

))(P ), . . . α(q(y

))(P )).

g is a polynomial, composed of its variables, scalars in K, addition and mul- tiplication, while α and q are ring homomorphisms preserving K, therefore

g(α(q(y

)), α(q(y

)), . . . , α(q(y

))) = α(q(g(y

, y

, . . . , y

))) = α(0) = 0 since g ∈ I(W ) = ker q. If we substitute the co-ordinates of P , we obtain g(ϕ(P )) = 0. As this holds for every g ∈ I(W ), we have ϕ(P ) ∈ W , so ϕ is indeed a morphism V → W .

If ϕ = (ϕ

, ϕ

, . . . , ϕ

) is a morphism of algebraic varieties V → W , then ϕ

= ϕ

(y

+ I(W )), so the ring homomorphism constructed from ϕ

is exactly ϕ. Conversely, if α is a ring homomorphism α : K[W ] → K[V ] and ϕ : V → W is the morphism of algebraic varieties constructed from α, then α = ϕ

.

This shows constructions associating a ring homomorphism to a homomor- phism of algebraic varieties and a homomorphism of algebraic varieties to a ring homomorphism are inverses of each other, therefore

: ϕ 7→ ϕ

is a bijection.

Note. ϕ 7→ ϕ

is contravariant, i. e., if ϕ : V → W and ψ : W → X are morphisms, then (ψ ◦ ϕ)

= ϕ

◦ ψ

: K[X] → K[V ].

The definition of the homomorphism ϕ

may seem complicated, but it can be written down very easily. In Example 6 before the theorem, V = A

, W = V(huw − v

i) ⊂ A

and ϕ : V → W is the morphism ϕ(x, y) = (x

, xy, y

).

ϕ

: K[W ] → K[V ] is defined by ϕ

(u + I(W )) = x

, ϕ

(v + I(W )) = xy,

ϕ

(w + I(W )) = y

and since it is ring homomorphism that preserves K, ϕ

(f ) can be calculated for any f ∈ K[W ].

Corollary 2.2 ϕ : V → W is an isomorphism of affine algebraic varieties if and only if ϕ

: K[W ] → K[V ] is an isomorphism of rings. V and W are isomorphic if and only if there is an isomorphism between K[V ] and K[W ] which preserves K.

Examples: In these examples K can be an arbitrary infinite field.

1. Let H = V(hxy − 1i) ⊂ A

be a hyperbola, we shall prove that it is not isomorphic to A

. It can be proved by direct calculation that I(H) = hxy−1i.

(Over an algebraically closed field this follows from the Nullstellensatz, as xy − 1 is an irreducible polynomial.) In K[H], (x + I(H))(y + I(H)) = 1 + I(H), i. e., there exist invertible elements which are not scalars, whereas in K[A

] the only invertible elements are the scalars. This shows that K[H]

and K[A

] are not isomorphic, therefore H is not isomorphic to A

either.

2. We shall show that the cuspidal cubic (Example 5 before Theorem 2.1) is not isomorphic to A

.

Let t be the co-ordinate on A

and x, y the co-ordinates on A

. Let V = A

, let W = V(hy

− x

i) ⊂ A

be the cuspidal cubic curve and let ϕ : V → W be the morphism t → (t

, t

). We already proved that ϕ is bijective, but it is not an isomorphism. Now we shall prove that W is not isomorphic to V by showing that K[W ] is not isomorphic to K[V ] ∼ = K[t].

It can be proved by direct calculation that I(W ) = hx

− y

i. (Over an algebraically closed field this follows from the Nullstellensatz, as x

− y

is an irreducible polynomial.)

In order to simplify notation, let f = f + I(W ) for f ∈ K[x, y]. ϕ

(x) = t

and ϕ

(y) = t

, so ϕ

(x

y

) = t

. This way we can obtain all non- negative powers of t except t itself. As the x

y

(a, b ∈ Z

) generate K[W ] as a vector space over K, the image of ϕ

is the subring S of K[t] consisting of polynomials with no degree 1 term.

This gives another proof of the fact ϕ is not an isomorphism, since ϕ

is not an isomorphism, because it is not surjective.

We shall prove that K[W ] is not isomorphic to K[V ] ∼ = K[t] by showing that K[W ] cannot be generated by a single element and K as a ring. If there existed an element in z ∈ K[W ] which together with K generated K[W ] as a ring, then ϕ

(z) and K would generate S, the image of K[W ] under ϕ

. (In fact, ϕ

is an isomorphism between K[W ] and S.)

We shall prove that S cannot be generated by a single element and K as a

ring. Let’s assume to the contrary that there exists such an element u ∈ S.

We can write u as u = a

+ a

t

+ a

t

+ . . . + a

t

for some k ≥ 0 and for suitable coefficients a

, a

, a

, . . . , a

∈ K.

u and u − a

generate the same ring, so we may assume that a

= 0 and u = a

t

+ a

t

+ . . . + a

t

. The elements of the ring generated by u and K are of the form b

+ b

u + b

u

+ . . . b

u

, for some n ≥ 0 and b

, b

, b

, . . . , b

∈ K. If S is generated by K and u, then all elements of S, in particular t

and t

have to be able to written in this form with suitable coefficients b

, b

, b

, . . . , b

.

The degree 2 term in b

+ b

u + b

u

+ . . . b

u

is b

a

t

, the degree 3 term is b

a

t

. In order to be able to get t

with a non-zero coefficient, we need to have a

6= 0. Similarly, in order to get t

with a non-zero coefficient, we need a

6= 0. This, however, means that if b

= 0 then both t

and t

have coefficient 0 in b

+ b

u + b

u

+ . . . b

u

, while if b

6= 0 then both t

and t

have a non-zero coefficient, we are not able to obtain just t

or just t

. This contradiction shows that S cannot be generated by u and K.

K[t] ∼ = K[V ] is clearly generated by t and K, so we can conclude that K[V ] 6∼ = K[W ] and therefore V and W are not isomorphic either.

There are also other ways to prove that K[W ] and K[t] are not isomorphic:

K[t] is a principal ideal domain, but h¯ x, ¯ yi / K[W ] is not a principal ideal;

K[t] is a unique factorisation domain, while in K[W ], ¯ x

= ¯ y

and both ¯ x and ¯ y are irreducible.

The correspondence between varieties and ideals can be re-written in terms of varieties and co-ordinate rings.

Definition. An element r in a ring is called nilpotent if and only if there exists a positive integer n such that r

= 0.

Algebraic fact: An ideal I in a ring R is radical if and only if R/I has no nilpotent elements other than 0.

Co-ordinate rings of affine algebraic varieties over K are rings which contain K, have no nilpotent elements other than 0 and are generated as rings by K and finitely many other elements (the images of the co-ordinate functions).

Conversely, if we have such a ring R, we can take choose a set of elements

x

, x

, . . . , x

∈ K which together with K generate R. The fact that /

they and K generate R means that there is a surjective ring homomorphism

K[x

, x

, . . . , x

] → R, let J be its kernel, then R ∼ = K[x

, x

, . . . , x

]/J

by the First Ring Isomorphism Theorem. J is radical since the quotient

ring K[x

, x

, . . . , x

]/J has no nilpotent elements other than 0. If K is

an algebraically closed field, then J = I(V(J )) by the Nullstellensatz, so R ∼ = K[V(J )].

If K is algebraically closed, then V 7→ K[V ] gives a bijection between iso- morphism classes of affine algebraic varieties and isomorphism classes of rings which which contain K, have no nilpotent elements other than 0 and which can be generated by K and finitely many other elements.

An affine algebraic variety V is irreducible if and only if I(V ) is a prime ideal, which is equivalent to K[V ] being an integral domain, therefore in the above correspondence irreducible varieties correspond to integral domains which contain K and which can be generated by K and finitely many other elements.

If a property of affine algebraic varieties can be expressed purely in terms of algebraic properties of co-ordinate rings, then that property is preserved under isomorphism.

Being an integral domain an algebraic property preserved by isomorphism,

therefore irreducibility is preserved by isomorphism of affine algebraic vari-

eties.

2.2 Rational functions and maps, function fields

Definition. Let V be an irreducible affine algebraic variety. The function field of V , denoted by K(V ), is the set of fractions f /g f, g ∈ K[V ], g 6= 0, considering two such fractions f

/g

and f

/g

equal if and only if f

g

= f

g

in K[V ]. (K(V ) is indeed a field with the natural operations.)

The elements of K(V ) are called rational functions on V .

Just like in the case of co-ordinate rings, K means the field over which the variety is defined, so if the field is C, then the notation is C(V ).

A rational function ϕ on V is defined at P ∈ V if and only if it can be represented in the form f /g such that g(P ) 6= 0, and in that case the value ϕ(P ) is defined to be ϕ(P ) = f (P )/g(P ), and this value is independent of the choice of f and g.

Example: If V = A

, then K(A

) = K(x

, x

, . . . , x

), the field of rational functions in n variables.

Lemma 2.3 Let V be an irreducible affine algebraic variety. Let ϕ = f /g f, g ∈ K[V ], g 6= 0. If for some point P ∈ V , f (P ) 6= 0 and g(P ) = 0, then ϕ is not defined at P .

Proof. Let ϕ = f

/g

, f

, g

∈ K[V ], g

6= 0 be another representation of ϕ. By the definition of equality of rational functions, f

g = f g

in K[V ], so f

(P )g(P ) = f (P )g

(P ). g(P ) = 0, so the left-hand side is 0, but then f (P ) 6= 0 implies that on the right-hand side g

(P ) = 0, because K[V ] is an integral domain.

Remarks. 1. In general, all possible representations of a rational function ϕ need to be considered when deciding whether it is defined at a point. For example, let x, y, z, w be the co-ordinates on A

. Let V = V(hxy −zwi) ⊂ A

, and let ϕ(x, y, z, w) = x/z. x/z not defined at P = (0, 1, 0, 0), but x/z = w/y, since xy = zw ∈ K[V ]. w/y is defined at P , so ϕ(P ) = 0/1 = 0.

2. However, if K[V ] is a unique factorisation domain, then ϕ can be written as ϕ = f /g with f , g coprime. In this case it is sufficient to consider only this one representation of ϕ, because all others are of the form ϕ = (f h)/(gh) for some h ∈ K[V ], therefore ϕ is defined at a point P if and only if g(P ) 6= 0.

In particular, this applies if V = A

for some n.

Definition. Let V ⊆ A

(K) and W ⊆ A

(K) be affine algebraic varieties

and let V be irreducible. A rational map ϕ : V 99K W is a function defined

on a non-empty subset of V given by ϕ(P ) = (ϕ

(P ), ϕ

(P ), . . . , ϕ

(P )) with

ϕ

∈ K(V ) for each i, 1 ≤ i ≤ n, and such that if ϕ(P ) is defined at P , i. e.,

if ϕ

(P ) is defined for each i, 1 ≤ i ≤ n, then ϕ(P ) ∈ W .

Rational functions and rational maps are generally not functions on V , only partial functions defined on a non-empty subset of V . The set where a rational function or a rational map is not defined is a subvariety, the locus where the denominators of all the representations of the rational function or of the components of the rational map vanish. The set where a rational function or a rational map is defined is the complement of the previous set, therefore it is a Zariski open subset.

Definition. A rational map ϕ : V 99K W is called dominant if and only if its image is not contained in any proper subvariety of W .

Remark. Let V , W be an irreducible affine algebraic varieties, and let ϕ : V 99K W and ψ : W 99K X be rational maps. If the composite ψ ◦ ϕ : V 99K X is defined, it is also a rational map. If ϕ is dominant, ψ ◦ ϕ is always defined.

Definition. A rational map ϕ : V 99K W is called a birational equivalence if and only if there exists a rational map ψ : W 99K V such that ψ ◦ ϕ = I

and ϕ ◦ ψ = I

. (In this case ϕ and ψ are necessarily dominant.) V and W are birationally equivalent if and only if there exists a birational equivalence between them. A variety is rational if and only if it is birationally equivalent to A

for some n.

Theorem 2.4 Let V ⊆ A

(K) and W ⊆ A

(K) be irreducible affine al- gebraic varieties. There exists a bijection between dominant rational maps ϕ : V 99K W and field homomorphisms α : K(W ) → K(V ) preserving K given by the following constructions: for a rational map ϕ : V 99K W , define ϕ

: K(W ) → K(V ), ϕ

(f ) = f ◦ ϕ, and for a homomorphism α : K(W ) → K(V ), define a rational map by (α(y

), α(y

), . . . , α(y

)), where y

, y

, . . . , y

are the co-ordinates on A

.

Sketch of Proof: (Not examinable) We introduce an alternative view of ra- tional maps. Let g ∈ K[V ], g 6= 0. Let x

, x

, . . . , x

be the co-ordinates on A

. Let us consider A

∼ = A

× A

with a new co-ordinate z on A

and let V

= V (I(V ), zg − 1) ⊂ A

. (V

is the graph of z = 1/g over V .) Let K[V ][1/g] be the subring of K(V ) generated by K[V ] and 1/g, it consists of the elements of K(V ) that can be written as f /g

for some f ∈ K[V ] and some non-negative integer k. There is an isomorphism between ι

: K[V ][1/g] → K[V

] given by ι

(f /g

) = f z

. β

: V 99K V

, (x

, x

, . . . x

) → (x

, x

, . . . x

, 1/g(x

, x

, . . . , x

)) is a rational map, while its inverse γ

: V

→ V , (x

, x

, . . . x

, z) → (x

, x

, . . . x

) is a morphism.

For ρ ∈ K[V ][1/g], ι

(ρ) = ρ ◦ γ

and for σ ∈ K[V

], ι

(σ) = ρ ◦ β

. β

and

γ

give a bijection between the Zariski open subset {P ∈ V | g(P ) 6= 0} in V and V

.

Let now ϕ : V 99K W be a dominant rational map. We can assume that ϕ = (f

/g, f

/g, . . . , f

/g) for some g ∈ K[V ], if necessary we can multiply the numerator and denominator of each component by a suitable factor to get ϕ into this form. Now ϕ ◦ γ

: V

99K W is a rational map which is defined at every point of V

, therefore it is a morphism (homework sheet 6, question 2). By Theorem 2.1, there exists a corresponding homomorphism of rings (ϕ ◦ γ

)

: K[W ] → K[V

]. ϕ ◦ γ

is dominant as ϕ and γ are, therefore (ϕ ◦ γ

)

is injective (homework sheet 6, question 1 (b)). By composing it with the isomorphism ι

, we obtain an injective morphism ι

◦ (ϕ ◦ γ

)

: K[W ] → K[V ][1/g]. As it is injective, it can be extended to a morphism of function fields ϕ

: K(W ) → K(V ), and it can be checked that it has the property ϕ

(y

) = f

/g = y

◦ ϕ for each i, 1 ≤ i ≤ n, therefore ϕ

(f ) = f ◦ ϕ for any f ∈ K(W ).

Conversely, let α : K(W ) → K(V ) be a morphism of fields preserving K. Let g ∈ K[V ] be such that α(y

) = f

/g for for a suitable f

∈ K[V ] for each i, 1 ≤ i ≤ n. Then ϕ(K[W ]) ⊂ K[V ][1/g], so by Theorem 2.8, ι

◦ α : K[W ] → K[V

] determines a morphism ψ : V

→ W . Moreover, ψ is dominant since ι

◦α is injective (homework sheet 6, question 1 (b)), because α is injective, as it is a homorphism of fields. Therefore ϕ = ψ ◦ β

is a rational map V 99K W and ϕ

= α.

Corollary 2.5 ϕ : V 99K W is a birational equvalence if and only if ϕ

: K(W ) → K(V ) is an isomorphism of fields. V and W are birationally equivalent if and only if K(V ) and K(W ) are isomorphic as extensions of K.

A variety V is rational if and only if K(V ) ∼ = K(t

, t

, . . . , t

) for some k.

Birational equivalence is an equivalence relation on affine algebraic varieties, which is weaker than isomorphism. Birational equivalence classes of varieties are in bijection with the isomorphism classes of finitely generated extensions of K.

Birational equivalence of V and W means that there exist g ∈ K[V ] and h ∈ K[W ] such that V

∼ = W

, so in a sense, V and W have isomorphic non-empty Zariski open subsets.

Examples: Worked examples can be found in the separate handout at https:

//personalpages.manchester.ac.uk/staff/gabor.megyesi/teaching/MATH32062/

rational.pdf.