Solution to exercises

Exercise 1.1

Let be a finite generated -algebra and suppose is a maximal ideal. Then the natural injection is an isomorphism.

 
Proof

To begin notice that the polynomial ring has countably infinite dimension over . Since such a polynomial ring surjects onto it follows that is countably infinite. Next observe that, for any , a non-trivial -linear relation

among the set ensures that vanishes on some (clear denominators). Since the set is uncountably infinite this set cannot be -linearly independent as this would contradict the fact that is countably infinite. It follows that cannot contain any elements which are not algebraic over , i.e., which do not vanish on some polynomial . Since is algebraically closed this forces .

 


Exercise 1.2

What is the interpretation of the quotient of an affine variety by the action of a finite group in terms of coordinate rings?

 
Proof

Let be an affine variety in , where is an ideal in . Let be a finite group which acts on . We can think of the elements of as maps , because the kernel of such a map is a maximal ideal, or equivalently, a maximal ideal gives such a map by the projection. This is convenient because it gives an easy way to define the Spec of a -algebra morphism. Given a morphism of -algebras , we can get a morphism of varieties in the opposite direction , using composition:

We want a variety which will be the quotient of by this action, a variety where we can think of the elements as orbits over the group action. That is, we want a surjective morphism of varieties such that for all and , and which is somehow universal.

Consider the set of -invariant elements of :

Because these an elements are -invariant, we could think of an orbit of as giving a map . If is a finitely generated sub-algebra, then we can take the Spec of it, and in fact, it will be precisely what we want. That is, the inclusion will give us a map in the other direction on the varieties, , which takes
But this is -invariant as we wanted, because by construction, for any ,
In fact, has the right universal property as well. If is another morphism of varieties which is -invariant, then there is a corresponding morphism of -algebras . But because is -invariant, the image of has to lie in , and so we should get a factoring , which corresponds on the geometric side as a morphism of varieties .

So all we need to check is that is a finitely generated -algebra. Suppose The elements of are a subset of , so let . If , then we have a morphism

Suppose , then let
Since , the image of under the projection is .

So it is enough to show that is finitely generated. Writing is like choosing a basis for a vector space, and in fact, we can choose the generators and the ideal in such a way that each acts on the generators as a permutation (of the indices). So we can assume that is a subgroup of , the symmetric group on -variables. Then we have inclusions

The algebra is generated by the elementary symmetric polynomials. So because is finitely generated over , which is a finitely generated -algebra, and we have the triple inclusion above, must be finitely generated.

 
Exercise 1.3

Let be an affine complex algebraic variety. It is compact in the euclidean topology if and only if it is a finite collection of points.

 
Proof

Let us suppose to be not empty and connected and let be the radical ideal of such that . Since is not empty, cannot be the entire polynomial ring. Let us consider any polynomial which does not belong to and let us take the regular function given by the restriction of the polynomial to the variety . Since is a connected compact analytic space and is a holomorphic function, we deduce from the maximum principle that is equal to a constant . We see that this constant is different from 0 since does not belong to the ideal . However, vanishes on and thus it belongs to the radical idea . Therefore we deduce that the ideal generated by and coincides with the whole ring since it contains . To sum up, we have proven that for any elements the ideal generated by and coincide with the ring of polynomials . This means that the ideal is maximal. Thus we conclude that is a point by the Nullstellensatz. If were not connected, it would be the union of a finite number of connected affine algebraic varieties and we can conclude by the above argument.

 
Exercise 1.4

A grading on the vector space such that the are finite dimensional is "the same as" a -action on with the property that each is contained in a finite dimensional subspace of that is -stable and such that the induced action on is algebraic.

 


Proof

Let us first find all one-parameter subgroups of , i.e. all group homomorphisms

which are given by Laurent polynomials. We will prove that there is such that is diagonal for all Let be the set of all roots of unity in . Clearly, is an abelian subgroup if . Further, each element of has finite order and is therefore diagonalisable. By linear algebra, this implies that is simultaneously diagonalisable. This means that (after conjugating by ), all off-diagonal entries of ( a non-zero complex number) are given by Laurent polynomials that vanish on the infinite set . Hence the claim follows. Now we know that, after conjugating by , is of the form
with a homomorphism given by a Laurent polynomial. Hence we can write for some nonnegative integer and some polynomial . But cannot map any non-zero complex number to 0 (since the target of is , so for some nonnegative intereger . This implies that, in fact, is of the form
for suitable integers , up to conjugation. Now suppose we are given a vector space with an action of satisfying the conditions from the exercise. Let . Pick a finite dimensional stable subspace containing and a basis () of such that acts on by This is possible by our previous considerations. We can write
(). Therefore, if we define
we obtain
The other direction is trivial.

 
Exercise 1.5

Derive the degree-genus formula for a nonsingular curve in of degree by degenerating it to the union of lines in general position.

 


Proof

Let be a nonsingular degree curve in . We know that such a curve is a compact and connected Riemann surface, so in particular it is a closed, connected, orientable manifold of real dimension 2 and hence it is topologically determined by its genus. We would like to "continuously deform" our given polynomial to one of the form

which defines a singular curve in (a union of projective lines) from which we can "read off" the genus (in fact we'll want the lines to lie in general position, that is, no three of them passing through the same point).

To make this a little more precise, note that we can identify each degree curve in with a point in . This is because a homogeneous polynomial of degree in three variables has coefficients and by the Nullstellensatz two such polynomials define the same curve in if and only if they are constant multiples of eachother. So we want our continuous deformation to be a path in connecting the point corresponding to to the one corresponding to such that each point on it, other than , corresponds to a nonsingular curve (so that its genus is well-defined).

Suppose we have achieved such a deformation (a word on why this is possible to follow). Then the curve defined by is a union of projective lines, that is, Riemann spheres in . But every two lines in intersect in a point. So we end up with spheres, every two of which touch, and the genus is the number of holes (right before we reach on our path of deformation we have a curve which looks like the touching spheres but every two of them are connected by a thin tube). To count the holes imagine shrinking the spheres to points, while the common point of every two spheres stretches to become a line segment connecting the two points. We thus obtain a complete graph with vertices. The number of edges in the graph is and a maximal tree has edges. Note that the maximal tree consists in fact of one root and leaves. From then on, every added edge creates a hole---the triangle with vertices the endpoints of that edge and the root. So the number of holes is

We now argue that the desired deformation always exists. We start with the two distinct points and in . Through them passes a unique projective line . Let be the set of points representing singular curves of degree in . It is a Zariski closed subset of and hence is Zariski closed in . But and so Zariski closed subsets (i.e. zeros of polynomials) are only finite sets or the whole of . But , so is finite. Thus is a Riemann sphere with finitely many points removed and hence there is a smooth path from to .

 


Exercise 1.6

Why the name "spectrum"? Given consider the (commutative!) -algebra generated by . What is its Spec?

 
Proof

Using the linear algebra definition, the spectrum of the linear operator is the set of its eigenvalues: . We want to see that this set is precisely the set of maximal ideal of the ring (so it's the same spectrum we defined in the lecture), i.e. the commutative -algebra generated by . We have already seen that

So we write down explicitly the maps between this two set and we show they are inverse maps. Let an eigenvalue, then

If is not a unit, then for some maximal ideal . So we associate to the homomophism whose kernel is . In the other direction, fixed , we have that , so it's not invertible and then . It's straightforward to check that these maps are inverse to each other.

 
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