MAT00001M Algebraic Geometry 2019/20
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MAT00001M
MMath and MSc Examinations 2019/20
Algebraic Geometry
1 (of 4). Throughout this question, let U be an algebraically closed field and let V = An .
Recall that V has coordinate algebra A = U[X1 , . . . , Xn], the polynomial algebra over U in n indeterminates.
(a) Let S C A be any subset.
(i) Define the set v(S).
(ii) Recall that subsets of the form v(S) as in part (i) are called algebraic
sets. Show that the intersection of two algebraic sets is algebraic.
(b) Let U = C. Decide whether or not each of the following subsets of
affine space is algebraic, justifying your answer in each case. Write the statement of any result you use clearly:
(i) {(t, t2 - 1) l t e U} c A2 ;
(ii) k, the set of integers, considered as a subset of A1 ;
(c) What is projective n-space Zn ?
(d) What is meant by a homogeneous polynomial in A?
2 (of 4). Let U be an algebraically closed field, let V be a set and let A be a U-algebra
of U-valued functions on V . Recall that HomU −alg (A, U) denotes the set of all U-algebra homomorphisms from A to U.
(a) Given v e V , define the map evaluation at v , fv e HomU −alg (A, U) (you
do not need to prove any properties of this map).
(b) Write down what it means to say that the pair (V, A) is an affine variety
over U.
(c) Suppose (V, A) is an affine variety over U, let S C A and let U = v(S). Explain how to define an algebra B of functions on U so that (U, B) is an affine variety. Explain why the algebra you have defined is finitely generated.
(d) Let V = An with coordinate algebra A = U[X1 , . . . , Xn], and let W = Am with coordinate algebra B = U[Y1 , . . . , Ym]. Recall that a map φ : V → W is called a morphism if f o φ e A for all f e B. Determine whether the following maps are morphisms, justifying your answer in each case:
(i) V = A1 , W = A2 , φ : V → W given by φ(t) = (t + 3, t2 ), with U = C;
(ii) V = A1 = W , φ : A1 → A1 given by φ(t) = t (complex conjugation),
with U = C.
3 (of 4). Let (V, A) be an affine variety.
(a) What does it mean to say V is irreducible?
(b) Show that if V is irreducible and φ : V → W is a surjective morphism
of affine varieties, then W is irreducible.
(c) Now let U = C be the complex numbers, let V be affine 3-space A3 with coordinate algebra A = U[X, Y, Z], and let U be the following algebraic subset of V :
U := v(XY - Z2 , X - YZ).
(i) Show that U C v(X(Y2 - Z)). [Hint: try to write X(Y2 - Z) in terms of the defining polynomials for U.]
(ii) Show that v(XY - Z2 , X, X - YZ) = {(0, t, 0) l t e C}.
(iii) Show that v(X - YZ, XY - Z2 , Z - Y2 ) = {(t3 , t, t2 ) l t e C}.
(iv) Deduce that U has two irreducible components. Give a brief justifi-
cation why the subsets you have written down are irreducible.
4 (of 4).
(a) You are given that A is an integral domain. Explain how to construct the
field of fractions of the algebra A, and give a definition of the addition and multiplication rules in this field (you do not need to prove that anything you write down is well-defined).
(b) What does it mean to say that V is a rational curve?
(c) Let U = C, and let V = v(X3 + X2 - Y2 ) c A2 . You are given that V is irreducible. Giving full justification, show that V is a rational curve.
SOLUTIONS: MAT00001M
1. (a) (i) v(S) := {v e V l f(v) = 0 for all f e S}, where we view polynomi-
(ii) The result is that v(S) n v(T) = v(S u T). To see this, suppose
v e v(S) n v(T). Then f(v) = 0 for all f e S and f(v) = 0 for
all f e T; hence f(v) = 0 for all f e S u T. On the other hand,
/ '
(b) (i)
{(t, t2 - 1) l t e U} c A2 ; Clearly, {(t, t2 - 1) l t e U} c v(X2 -Y - 1). On the other hand, if (a, b) e v(X2 - Y - 1), then a2 = b + 1, and so
b = a2 - 1. Thus (a, b) = (a, a2 - 1) e {(t, t2 - 1) l t e U}. Therefore
/ '
(ii) k, the set of integers, is an infinite proper subset of A1 . As proper
algebraic subsets of A1 must be finite (as seen in the lect_(ur)e), k i、(s)
not an algebraic set. 7 Marks
(c) Zn (as we defined it) is the set of equivalence classes of points of Un+1 \
{(0, . . . , 0)} under the relation v ~ w if and only if v = λw for some/ '
(d) An element f e A is homogeneous of degree d > 0 if f can be written as a linear combination of monomials X1(i)1 X2(i)2 . . . Xn(i)n with .、
Remarks. Part (a)(i) is a standard definition and part (a)(ii) is book- work. Part (b) consists of examples very similar to those seen in class and in homeworks. Parts (c) and (d) are standard definitions.
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Total: 30 Marks
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2. (a) Given v e V , fv is defined by fv (f) = f(v) for all f e A. /5 Marks'
(b) We need A to be finitely-generated as a U-algebra, and we need the map
(c) We define B by restricting the domain of functions from A; that is/ '
A is finitely generated since (V, A) is an affine variety. Then B is gen- erated by the restriction to U of the generators of A, so B is finitely
generated.
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4 Marks
/ '
(d) |
(i) |
This is a morphism. In this case A = U[X], say, and B = U[X1 , X2], say. It suffices to check the condition for the generators X1 and X2 of B. We have, for each t e A1 , (X1 o φ)(t) = X1 (t + 3, t2 ) = t + 3, so X1 o φ = X + 3 e U[X]. Similarly, X2 o φ = X2 e U[X], so we’re / ' |
(ii) Not a morphism. We need to show that there is no polynomial
f e U[X] with f(t) = t for each t e C. Any such f can only have 0 as a root, so is a power of X. But no power of X fixes all real
numbers, unlike complex conjugation. So no such f exis_(ts), and 、(φ)
cannot be a morphism. 5 Marks
Remarks. Part (a) and (b) are standard definitions. Part (c) is a stan- dard definition together with bookwork. Part (d)(i) is new but similar to seen examples. Part (d)(ii) is an example from a problem sheet.
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Total: 25 Marks
3. (a) V is irreducible if V cannot be written as the union of two proper closed
(b) We prove this by contrapositive: suppose φ : V → W is surjective and
W = C1 u C2 is reducible (so C1 , C2 are proper closed subsets of W). Then φ is a continuous map, so φ − 1 (C1 ) and φ − 1 (C2 ) are closed in V , and V = φ − 1 (C1 ) u φ − 1 (C2 ). Since C1 and C2 are proper and φ is surjective, φ − 1 (C1 ) and φ − 1 (C2 ) are proper too (if V = φ − 1 (Ci ) for some i, then
W = φ(V) = φ(φ− 1 (Ci )) = Ci ). Thus V is reducible too. This proves/ '
(c) (i) We write Y (XY - Z2 ) - Z(X - YZ) = XY2 - XZ = X(Y2 - Z).
Hence any common solution of XY - Z2 = 0 and X - YZ_= 0 als、(o)
satisfies X(Y2 - Z) = 0. 3 Marks
(ii) Suppose (a, b, c) is killed by the three given polynomials. Then a = 0
because we have to be killed by X, but then XY - Z2 gives rise to
c = 0 as well. Now note that any point (0, t, 0) does satisfy the three/ '
(iii) Suppose (a, b, c) satisfies the three given polynomials. Then the third
polynomial gives c = b2 , and then the first gives a = b3 . Hence ev-
ery point satisfying these polynomials has the given form. On the
other hand, every point of the given form clearly satisfies _(t)he thre、(e)
polynomials. 3 Marks
(iv) Let the algebraic set from (b) be denoted C1 and the set from (c)
be denoted C2 . Then C1 and C2 are the images of A1 under sur- jective morphisms t ,→ (t, 0, 0) and t ,→ (t, t5 , t2 ), respectively. A1 is irreducible, since C[X], its coordinate algebra, is an integral do-
main. The image of an irreducible set is irreducible, so _(C1) and C、(2)
are irreducible. 2 Marks
Now, the inclusions (XY - Z2 , X, X - YZ)2 (XY - Z2 , X - YZ)
and (X - YZ, XY - Z2 , Z - Y2)2 (XY - Z2 , X - YZ)imply that
C1 and C2 lie inside U. On the other hand, part (a) shows that every
point of U also satisfies the polynomial X(Y2 - Z) and hence lies in
C1 or C2 (since X and Y2 - Z are the extra generators for the ideals
(XY - Z2 , X, X - YZ) and (X - YZ, XY - Z2 , Z - Y2), respec-
tively). Thus U = C1 u C2 and since there are no mutual inclusions/ '
Remarks. Part (a) is a standard definition and part (b) is bookwork. Part (c) is an unseen example, similar to ones seen in lectures and on problem sheets.
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Total: 25 Marks
4. (a) We define an equivalence relation on Ax(A\{0}) by (a, b) ~ (c, d) if and
only if ad = bc. Then write the equivalence class of a pair (a, b) under
this relation as . The addition rule is + = and multiplication
is . = . /(_)6 Marks
(b) V is a rational curve if V is irreducible and the field of fractions of the/ '
(c) In this case, if we let X0 = XlV and Y0 = Y lV , then we have the coordinate algebra of V is A = U[X0 , Y0]. We’re given that A is an integral domain, so we can define T in the field of fractions by T = , so that Y0 = TX0 . Since, in A, we have X0(3) + X0(2) - Y02 = 0, we can substitute in to get
X0(3) + X0(2) - T2X0(2) = X0(2)(X0 + 1 - T2 ) = 0.
Since X0 is not identically 0, we conclude that X0 = T2 - 1 and hence Y0 = T3 - T.
This shows that the field of fractions of A is generated by the single element T, and since the reverse inclusion is obvious, we have that the
field of fractions of A is equal to U(T).
Thus V is a rational curve.
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10 Marks
/ '
Remarks. Part (a) and (b) are standard definitions. Part (c) is similar to examples seen in lectures and on problem sheets, but it is made harder by the fact that no prompts are given.
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Total: 20 Marks
2022-01-11