Algebraic geometry, Sheets 1, 2: selected solutions
For a polynomial f , write [f ]d for the degree d homogeneous component of f . So
[f + g]d = [f ]d + [g]d
and
[f g]d =
X
[f ]i [g]j .
i+j=d
We will frequently be making use of the fact that the polynomial ring k[x1 , x2 , . . . , xn ] over a
field k is a unique factorization domain. In particular, if a polynomial f is divisible by two distinct
irreducible polynomials p and q, then it is divisible by pq. Also, p is irreducible if and only if (p) is a
prime ideal if and only if k[x1 , x2 , . . . , xn ]/(p) is an integral domain.
(1) Prove that for any ideal J, the radical
√
J is also an ideal.
√
This result holds in any commutative ring R. Let a, b ∈ J. Then an , bm ∈ J for some n, m ∈ N,
so that ai ∈ J and bj ∈ J for any i ≥ n and j ≥ m. Let N = 2max{n, m}. Then
X µN ¶
(a + b)N =
ai bN −i .
i
i
For each summand ai bN −i , we clearly
N/2 ≥ n or N√− i ≥ N/2 ≥ m, so that
√ have either i ≥ √
(a + b)N ∈ J. Therefore, a + b√∈ J. Obviously, 0 ∈ J and −a ∈ J. Finally, if r ∈ R, then
(ra)n = rn an ∈ J, so that ra ∈ J.
(2) Let V := V (I) ⊂ A3 be the algebraic set corresponding to the ideal I = (x2 − yz, xz − x).
Decompose V into its irreducible components.
Clearly the equations
are equivalent to
or
That is,
But
x2 − yz = 0; xz − x = 0
x2 − yz = 0; x = 0
x2 − yz = 0; z − 1 = 0.
V (I) = V (x2 − yz, x) ∪ V (x2 − yz, z − 1).
k[x, y, z]/(x2 − yz, z − 1) ' k[x, y]/(x2 − y) ' k[x],
which is an integral domain. Therefore, (x2 − yz, z − 1) is a prime ideal, and V (x2 − yz, z − 1) is
irreducible. On the other hand,
V (x2 − yz, x) = V (x, y) ∪ V (x, z)
and k[x, y, z]/(x, y) ' k[z], k[x, y, z]/(x, z) ' k[y]. So both V (x, y) and V (x, z) are irreducible.
Therefore,
V (I) = V (x, y) ∪ V (x, z) ∪ V (x2 − yz, z − 1)
is the decomposition of V (I) into its irreducible components.
(3) In this exercise, we investigate the relation between ideals and varieties for the following ideals
in C[x, y, z]:
I1 := (xy + y 2 , xz + yz);
1
I2 := (xy + y 2 , xz + yz + xyz + y 2 z);
I3 := (xy 2 + y 3 , xz + yz).
(a) Does Ik = Il for some k 6= l?
Note that I1 ⊃ I2 and I1 ⊃ I3 . But xz +yz +xyz +y 2 z = xz +yz +z(xy +y 2 ), so that xz +yz ∈ I2 .
Hence, I2 = I1 . But we claim that xy + y 2 ∈
/ I3 . To see this, assume
xy + y 2 = a(xy 2 + y 3 ) + b(xz + yz)
for a, b ∈ C[x, y, z]. Then
xy + y 2 = [xy + y 2 ]2 = [a(xy 2 + y 3 )]2 + [b(xz + yz)]2 = [b(xz + yz)]2 = [b]0 (xz + yz).
Since [b]0 is constant, this cannot hold for any b.
(b) Does V (Il ) = V (Ik ) for some i 6= k?
Of course V (I1 ) = V (I2 ). The points in V (I3 ) satisfy xz + yz = 0 and xy 2 + y 3 = 0. This is the
same as [x + y = 0 or z = 0] and [x + y = 0 or y = 0]. Clearly, this is the same zero set as V (I1 ).
Therefore, all zero sets are the same.
(6) Show that the hyperbola {(x, y) ∈ A2 | xy = 1} is not isomorphic to A1 .
Let
φ : k[x, y]/(xy − 1)→k[t]
be a homomorphism of k-algebras. Then φ(x), φ(y) ∈ k[t] must be invertible. Therefore, they must
both be constant. Hence, the coordinate rings of the two varieties are not isomorphic as k-algebras.
So the affine varieties are not isomorphic.
(9) Give an example to show that the image of a polynomial map
f : Cn →Cm
is not necessarily an algebraic set.
Consider the map f : C2 →C2 given by
(x, y) 7→ (x, xy).
If the image of this map were contained in the zero set of any polynomial p(s, t), then the polynomial
p(x, xy) would be identically zero. Writing
X
p(s, t) =
ai (s)ti ,
i
we get
X
ai (x)xi y i = 0
i
i
so ai (x)x is identically zero for each i. Hence, ai (x) = 0 for each i, and p(s, t) = 0. Therefore, the
image of f is not contained in any proper Zariski closed subset of C2 . On the other hand any (0, t)
for t 6= 0 is not in the image of φ. Therefore, the image of φ is not the whole space C2 . Therefore, the
image cannot be Zariski closed.
You might try to show that this example is in some sense ‘the simplest possible.’ That is, try to
show that that for polynomial maps C→C, C2 →C, and C→C2 , the image is always Zariski closed.
(12) Show that the Zariski topology on A2 = A1 × A1 is not the product topology.
2
Notice that a basis for the product topology are sets of the form U × V where U and V are open
in the first and second A1 -factors.
But since the diagonal ∆ := V (x − y) is a closed set, its complement A2 \ ∆ is an open set. This
set cannot contain any non-empty U × V .
(13)
Let X = V (x1 , x2 ) and Y = V (x3 , x4 ) in A4 . Show that I(X ∪ Y ) cannot be generated by two
elements.
In fact, I(X ∪ Y ) cannot even be generated by three elements. Suppose f ∈ I(X ∪ Y ). Then
f ∈ I(X) ∩ I(Y ), so that f can be written f = ax1 + bx2 as well as f = cx3 + dx4 . Therefore, every
monomial occurring in f is divisible by x1 or x2 and is divisible by x3 or x4 . Hence, every monomial
in f is divisible by x1 x3 or x1 x4 or x2 x3 or x2 x4 . Therefore,
I(X ∪ Y ) = (x1 x3 , x1 x4 , x2 x3 , x2 x4 ).
Now, let {vj }j∈J be any set of generators for I(X ∪ Y ). Then there would be polynomials ckl
j such
P kl
that xk xl =
cj vj for each of the monomials xk xl generating the ideal. Since each vj is in I(X ∪ Y ),
and hence, can be written in terms of the xk xl , they have no homogeneous components of degree less
than 2. Therefore, we find
X
xk xl =
[ckl ]0 [vj ]2
for each k, l. That is, x1 x3 , x1 x4 , x2 x3 , x2 x4 will be linear combinations of the [vj ]2 with constant
coefficients. Since these four monomials are linearly independent over C, there must be at least four
[vj ]2 .
3