Commutative Ring - Algebraic Curves - Solved Exam | Exams Algebra

ALGEBRAIC CURVES (MA40188): SOLUTIONS TO 2007 EXAM

1. (a): an ideal is a nonempty subset I⊂Rsuch that if a, b ∈Iand r∈Rthen a+b∈I

and ra ∈I. It is a prime ideal if rs ∈Iimplies r∈Ior s∈I.

(b): an affine variety Vis a subset of Ansuch that there exists an ideal Iof K[t1, . . . , tn]

for which V=V(I), i.e. V={P∈An|f(P) = 0 for f∈I}. It is irreducible if it is not

the union of two proper subvarieties.

The first part of (c) is also bookwork. A polynomial map f:V→Wis a collection

f= (f1, . . . , fm) of elements fi∈K[t1, . . . , tn] such that if P∈V⊂Anthen f(P)∈

W⊂Am. It induces f∗by composition: an element of K[W] is a map g:W→A1and

f∗(g) = g◦f. For the last part of (c), notice that Wis irreducible if and only if K[W] is a

domain. Suppose K[W] is not a domain, so we have 0 6=a, b ∈K[W] with ab = 0. Then

0 = f∗(ab) = f∗(a)f∗(b) and since f∗is injective f(a)6= 0 and f(b)6= 0. So K[V] is not a

domain, i.e. Vis reducible.

For (d), the last two equations give y=z2and x=z3if (x, y, z)∈W, so (x, y , z)7→ z

is a morphism W→A1with inverse z7→ (z3, z2, z). But the equations in (e) also allow

x=y= 0, z=anything, i.e. W0=W∪ {(0,0, z)}which is reducible.

2. (a) The function field of a projective variety Vis the field of fractions of the integral

domain K[t0, . . . , tn]/I(V).

(b) A rational map φ:V99K W, where V⊆Pn,W⊆Pmare irreducible projective

varieties, is an equivalence class of m+ 1 polynomials f0, . . . , fmin n+ 1 variables, all of

the same degree, where {fi}and {gi}are equivalent if figj−fjgi∈I(V) for all i, j, such

that the fican be chosen so that they do not all vanish simultaneously on the whole of V

and such that if P∈Vand (f0(P) : . . . :fm(P)) ∈Pmthen (f0(P) : . . . :fm(P)) ∈W.

φis regular at Pif there is a representation φ= (f0:. . . :fm) such that fi(P) are not

all zero. It is dominating if the image of φis Zariski dense in W, i.e. is not contained in

any proper subvariety of W. It is birational if it is dominating and there is a dominating

rational map ψ:W99K Vsuch that φ◦ψand ψ◦φare both the identity on the (dense)

sets where they are defined.

P1(or A1). curves.

For (d), consider the plane y=λx, whose points are (t, λt, s) with s, t ∈K. This meets C0

when st =λt and t=s2(s−1), i.e. at (λ2(λ−1), λ3(λ−1), λ) so the map A1→C0given

by λ7→ (λ2(λ−1), λ3(λ−1), λ) has rational inverse (x, y, z )7→ z. Hence C0is rational.

For (e), consider the projection map p:A3→A2given by p: (x, y, z)7→ (x, y). This maps

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ALGEBRAIC CURVES (MA40188): SOLUTIONS TO 2007 EXAM

(a): an ideal is a nonempty subset I ⊂ R such that if a, b ∈ I and r ∈ R then a + b ∈ I and ra ∈ I. It is a prime ideal if rs ∈ I implies r ∈ I or s ∈ I. (b): an affine variety V is a subset of An^ such that there exists an ideal I of K[t 1 ,... , tn] for which V = V (I), i.e. V = {P ∈ An^ | f (P ) = 0 for f ∈ I}. It is irreducible if it is not the union of two proper subvarieties. The first part of (c) is also bookwork. A polynomial map f : V → W is a collection f = (f 1 ,... , fm) of elements fi ∈ K[t 1 ,... , tn] such that if P ∈ V ⊂ An^ then f (P ) ∈ W ⊂ Am. It induces f ∗ by composition: an element of K[W ] is a map g: W → A^1 and f ∗(g) = g ◦ f. For the last part of (c), notice that W is irreducible if and only if K[W ] is a domain. Suppose K[W ] is not a domain, so we have 0 6 = a, b ∈ K[W ] with ab = 0. Then 0 = f ∗(ab) = f ∗(a)f ∗(b) and since f ∗^ is injective f (a) 6 = 0 and f (b) 6 = 0. So K[V ] is not a domain, i.e. V is reducible. For (d), the last two equations give y = z^2 and x = z^3 if (x, y, z) ∈ W , so (x, y, z) 7 → z is a morphism W → A^1 with inverse z 7 → (z^3 , z^2 , z). But the equations in (e) also allow x = y = 0, z =anything, i.e. W ′^ = W ∪ {(0, 0 , z)} which is reducible.
(a) The function field of a projective variety V is the field of fractions of the integral domain K[t 0 ,... , tn]/I(V ). (b) A rational map φ: V 99K W , where V ⊆ Pn, W ⊆ Pm^ are irreducible projective varieties, is an equivalence class of m + 1 polynomials f 0 ,... , fm in n + 1 variables, all of the same degree, where {fi} and {gi} are equivalent if figj − fj gi ∈ I(V ) for all i, j, such that the fi can be chosen so that they do not all vanish simultaneously on the whole of V and such that if P ∈ V and (f 0 (P ) :... : fm(P )) ∈ Pm^ then (f 0 (P ) :... : fm(P )) ∈ W. φ is regular at P if there is a representation φ = (f 0 :... : fm) such that fi(P ) are not all zero. It is dominating if the image of φ is Zariski dense in W , i.e. is not contained in any proper subvariety of W. It is birational if it is dominating and there is a dominating rational map ψ: W 99K V such that φ ◦ ψ and ψ ◦ φ are both the identity on the (dense) sets where they are defined. (c) A curve C, projective or affine, is rational if there is a birational rational map φ: C 99K P^1 (or A^1 ). curves. For (d), consider the plane y = λx, whose points are (t, λt, s) with s, t ∈ K. This meets C 0 when st = λt and t = s^2 (s − 1), i.e. at (λ^2 (λ − 1), λ^3 (λ − 1), λ) so the map A^1 → C 0 given by λ 7 → (λ^2 (λ − 1), λ^3 (λ − 1), λ) has rational inverse (x, y, z) 7 → z. Hence C 0 is rational. For (e), consider the projection map p: A^3 → A^2 given by p: (x, y, z) 7 → (x, y). This maps

C 0 to C 1 (if (x, y, z) ∈ C 0 then (x, y) ∈ C 1 ) and it has rational inverse (x, y) 7 → (x, y, y/x) defined away from (x, y) = (0, 0) (the only point of C 1 where x = 0 is (0, 0)).

(a) An^ is simply Kn. Pn^ is Kn+1/K∗, where K∗^ acts by coordinatewise multiplication in K. (b) A polynomial is said to be homogeneous of degree d if it is a sum of degree d mono- mials. And ideal I of K[t 0 ,... , tn] is called a homogeneous ideal if it can be generated by homogeneous polynomials (possibly of different degrees). (c) If I is a homogeneous ideal generated by f 1 ,... , fk homogeneous of degrees di then the projective variety V (I) is the image in Pn^ of the variety f 1 = · · · = fk = 0 in An+1, which is also {(x 0 :... : xn) | fi(x 0 ,... , xn) = 0, i = 1,... , k}. Note that this makes sense because if fi(x 0 ,... , xn) = 0 then fi(λx 0 ,... , λxn) = λdi^ fi(x 0 ,... , xn) = 0 also. (d) If V ⊂ An^ is an affine variety given by fi(t 1 ,... , tn) = 0 its projective closure is the variety V¯ ⊂ Pn^ given by f¯I (t 0 , t 1 ,... , tn) = 0, where f¯ is the polynomial obtained from f by homogenising with respect to t 0 : that is, any monomial of degree d < deg f is multiplied by tdeg 0 f^ −d. The points at infinity are the solutions of f¯i(0, t 1 ,... , tn) = 0. (e) If V is a projective variety and P ∈ V , we say that P is singular if the dimension of the tangent space to V at P is greater that the dimension of the tangent space to V at some other point Q ∈ V. Only (f) is not bookwork. V is given by homogenising: x^4 +x^2 y^2 − 3 x^2 z^2 +3y^2 z^2 +8yz^3 +6z^4. To find the points at infinity, put z = 0: we get x^4 + x^2 y^2 = 0 so x = 0 or x = ±y, so the points are (0 : 1 : 0), (1 : 1 : 0) and (1 : −1 : 0). To find the singular points, differentiate the equation given and set both partial derivatives equal to zero. That gives

2 x(2x^2 + y^2 − 6) = 2y(x^2 + 3) + 8 = 0.

Since x = 0 gives y = −4 from the second equation and (0, −4) is not on the curve, we may divide out the x from the first equation and substitute y = − 8 /(x^2 + 3) from the second (x = ±

−3 does not give a point of the curve either). That gives the equation in the hint; dividing out the x^2 − 1 leaves 2x^4 + 11x^2 + 11. But we don’t need to solve that: taking x = 1 gives y = −1 and taking x = −1 gives y = 1 also. So two of the singular points are (1 : 1 : 1) and (−1 : 1 : 1), and (0 : 1 : 0) is singular also so that is the lot as the question says there are only three.

Main points are: Smooth plane cubic curves are not rational;

Commutative Ring - Algebraic Curves - Solved Exam, Exams of Algebra

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ALGEBRAIC CURVES (MA40188): SOLUTIONS TO 2007 EXAM