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This is the Solved Exam of Algebraic Curves which includes Nullstellensatz, Radical, Algebraically, Variety, General, Ideal, Characteristic, Non Singular etc. Key important points are: Nullstellensatz, Radical, Algebraically, Variety, General, Ideal, Characteristic, Non Singular, Coordinates, Group Law
Typology: Exams
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Full marks will be given for correct answers to THREE questions. Only the best three answers will contribute towards the assessment.
I = {a โ R | an^ โ I for some n โ N}. It is an ideal because if an^ โ I and r โ R then (ra)n^ โ I and if an, bm^ โ I then
(a + b)n+m^ =
โ (n + m r
arbn+mโr^ โ I
since either r โฅ n or n + m โ r โฅ m. [4, unseen but hint given in lectures.] (c) V (I) = {(a 1 ,... , an) โ An^ | f (a 1 ,... , an) = 0 for all f โ I}, and I(V ) = {f โ K[x 1 ,... , xn] | f (a 1 ,... , an) for all a 1 ,... an โ V }. If n = 1 and I = ใx^2 ใ then V (I) = { 0 } โ A^1 but I(V (I)) = ใxใ. [3, bookwork] (d) Suppose f โ A = K[x 1 ,... , xn]. Consider the ring B = A[y] = K[x 1 ,... , xn, y], and the ideal I+^ = IB + (yf โ 1)B of B. Notice that Q โ An+1^ is in V (I+) if and only if the point P โ An^ got by taking the first n coordinates of Q is in V (I) and, in addition, the last coordinate of Q is 1 /f (P ) (in particular f (P ) 6 = 0). The set (f 6 = 0) โ V (I) is empty when f = 0 everywhere on V (I), i.e. when f โ I(V (I)). So suppose f (P ) = 0 for all P โ V (I): that means that V (I+) = โ . By the Nullstellensatz, that implies that 1 โ I+, and because I+^ is generated by I and yf โ 1 we can find polynomials g 0 , g 1 ,... gk โ B such that g 0 (yf โ 1) + g 1 f 1 + ยท ยท ยท + gkfk = 1, where f 1 ,... , fk are generators for the ideal I. This equation is an identity, so writing 1/f instead of y we have
โ^ k
i=
gi
x 1 ,... , xn, 1 /f (x 1 ,... , xn)
fi(x 1 ,... , xn) = 1.
The left-hand side is a rational function with denominator f N^ where N is the maximum of the degrees of the gi in y), so
gi
x 1 ,... , xn, 1 /f (x 1 ,... , xn)
= hi(x 1 ,... , xn)/
f (x 1 ,... , xn)
for some polynomials hi. If we multiply through by f N^ we get
โ^ k
i=
hi(x 1 ,... , xn, 1)fi(x 1 ,... , xn) = f (x 1 ,... , xn)N
so f โ
I as claimed. [10, bookwork]
Question 2 continued...
(d) First, P โ E because b = 19 = โ4 and 1^3 + 9 ร 1 โ 4 = 6 = 11^2 (mod 23). By the formula, the x-coordinate of โ 2 P is
(1^2 โ 9)^2 โ 8 ร 1 ร (โ4) 24
= 4 mod 23.
So the x-coordinate of 4P is (4^2 โ 9)^2 โ 8 ร 4 ร (โ4) 4 ร (4^3 + 9 ร 4 โ 4)
Therefore 4P = ยฑP , but if 4P = P then 2P = โP ; but we have already seen that P and 2P have different x-coordinates, whereas P and โP have the same x-coordinate. So 4P = โP , so 5P = 0. [7, unseen]
f = x^2 (x โ y)(x + y)z + x^5 + 3y^5 = 0
are found by setting z = 1 and fx = fy = 0 (writing fx for โfโx ), and similarly for y and z. It is easiest to begin with y = 1. Then f = x^4 z โ x^2 z + x^5 + 3, so fz = x^4 โ x^2 and fx = 4x^3 z โ 2 x + 5x^4. The equation fz = 0 gives x = 0, x = 1 or x = โ1: but none of these satisfy both f = 0 and fx = 0. If y 6 = 1 then y = 0 and on that line the equation is x^4 z + x^5 = 0, so x = 0 or x = โz, i.e. the points (0 : 0 : 1) and (โ1 : 0 : 1). So we can check these on the z = 1 part, where we have f = x^4 โ x^2 y^2 + x^5 + 3y^5 , fx = 4x^3 โ 2 xy^2 + 5x^4 and fy = โ 2 x^2 y + 15y^4. At the point (โ1 : 0 : 1), fx does not vanish so that is not a singular point, but all three vanish at (0 : 0 : 1) which is thus the only singular point of C. [6, unseen] (f) Projecting from the singular point gives a birational map ฯ : C 99K P^1. We may do this on the part z = 1, since the line z = 0 is not contained in C. Then the line of slope t has y = tx and passes through C where x^4 (1 โ t^2 ) + x^5 (1 + 3t^5 ) = 0, so the unique nonzero point is at x = t (^2) โ 1 3 t^5 +1 and this gives a birational map^ P
inverse to ฯ. [5, unseen]
