# Solution to exercises

Exercise 1.1

Let ${\displaystyle A}$ be a finite generated ${\displaystyle \mathbb {C} }$-algebra and suppose ${\displaystyle {\mathfrak {m}}\subset A}$ is a maximal ideal. Then the natural injection ${\displaystyle \mathbb {C} \hookrightarrow A/{\mathfrak {m}}}$ is an isomorphism.

Proof

To begin notice that the polynomial ring ${\displaystyle \mathbb {C} }$ has countably infinite dimension over ${\displaystyle \mathbb {C} }$. Since such a polynomial ring surjects onto ${\displaystyle A/{\mathfrak {m}}}$ it follows that ${\displaystyle \dim _{\mathbb {C} }(A/{\mathfrak {m}})}$ is countably infinite. Next observe that, for any ${\displaystyle x\in A/{\mathfrak {m}}}$, a non-trivial ${\displaystyle \mathbb {C} }$-linear relation

${\displaystyle {\frac {b_{1}}{x-t_{1}}}+\ldots +{\frac {b_{r}}{x-t_{r}}}=0,\qquad b_{i}\in \mathbb {C} ,}$
among the set ${\displaystyle \lbrace {\frac {1}{x-t}}\mid t\in \mathbb {C} \rbrace }$ ensures that ${\displaystyle x}$ vanishes on some ${\displaystyle f(T)\in \mathbb {C} }$ (clear denominators). Since the set ${\displaystyle \lbrace {\frac {1}{x-t}}\mid t\in \mathbb {C} \rbrace }$ is uncountably infinite this set cannot be ${\displaystyle \mathbb {C} }$-linearly independent as this would contradict the fact that ${\displaystyle \dim _{\mathbb {C} }(A/{\mathfrak {m}})}$ is countably infinite. It follows that ${\displaystyle A/{\mathfrak {m}}}$ cannot contain any elements which are not algebraic over ${\displaystyle \mathbb {C} }$, i.e., which do not vanish on some polynomial ${\displaystyle f(T)\in \mathbb {C} }$. Since ${\displaystyle \mathbb {C} }$ is algebraically closed this forces ${\displaystyle A/{\mathfrak {m}}=\mathbb {C} }$.

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 ${\displaystyle X=Z(I)}$ be an affine variety in ${\displaystyle \mathbb {C} ^{n}}$, where ${\displaystyle I}$ is an ideal in ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]}$. Let ${\displaystyle G}$ be a finite group which acts on ${\displaystyle X}$. We can think of the elements of ${\displaystyle X}$ as maps ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]\rightarrow \mathbb {C} }$, 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 ${\displaystyle \mathbb {C} }$-algebra morphism. Given a morphism ${\displaystyle \phi :R\rightarrow S}$ of ${\displaystyle \mathbb {C} }$-algebras ${\displaystyle R,S}$, we can get a morphism of varieties in the opposite direction ${\displaystyle {\text{Spec}}(S)\rightarrow {\text{Spec}}(R)}$, using composition:

${\displaystyle f\in {\text{Spec}}(S)\mapsto f\circ \phi .}$

We want a variety which will be the quotient of ${\displaystyle X}$ 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 ${\displaystyle \pi :X\rightarrow Y}$ such that ${\displaystyle \pi (g\cdot x)=\pi (x)}$ for all ${\displaystyle x\in X}$ and ${\displaystyle g\in G}$, and which is somehow universal.

Consider the set of ${\displaystyle G}$-invariant elements of ${\displaystyle R}$:

${\displaystyle R^{G}=\{f\in R:g\cdot x(f)=x(f)\forall x\in X,g\in G\}.}$
Because these an elements are ${\displaystyle G}$-invariant, we could think of an orbit of ${\displaystyle x\in {\text{Spec}}(R)}$ as giving a map ${\displaystyle R^{G}\rightarrow \mathbb {C} }$. If ${\displaystyle R^{G}}$ 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 ${\displaystyle i:R^{G}\rightarrow R}$ will give us a map in the other direction on the varieties, ${\displaystyle X\rightarrow {\text{Spec}}(R^{G})}$, which takes
${\displaystyle x\rightarrow x\circ i.}$
But this is ${\displaystyle G}$-invariant as we wanted, because by construction, for any ${\displaystyle f\in R^{G}}$,
${\displaystyle (g\cdot x)\circ i(f)=x\circ i(f).}$
In fact, ${\displaystyle Y={\text{Spec}}(R^{G})}$ has the right universal property as well. If ${\displaystyle \pi ':X\rightarrow Z}$ is another morphism of varieties which is ${\displaystyle G}$-invariant, then there is a corresponding morphism of ${\displaystyle \mathbb {C} }$-algebras ${\displaystyle S\rightarrow R}$. But because ${\displaystyle \pi '}$ is ${\displaystyle G}$-invariant, the image of ${\displaystyle S}$ has to lie in ${\displaystyle R^{G}}$, and so we should get a factoring ${\displaystyle S\rightarrow R^{G}\rightarrow R}$, which corresponds on the geometric side as a morphism of varieties ${\displaystyle X\rightarrow X/G\rightarrow Z}$.

So all we need to check is that ${\displaystyle R^{G}}$ is a finitely generated ${\displaystyle \mathbb {C} }$-algebra. Suppose ${\displaystyle R=\mathbb {C} [x_{1},\dots ,x_{n}]/I.}$ The elements of ${\displaystyle X}$ are a subset of ${\displaystyle \mathbb {C} ^{n}}$, so let ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]^{G}=\{f\in \mathbb {C} [x_{1},\dots ,x_{n}]:\forall x\in X,g\in G,f(g\cdot x)=f(x)\}}$. If ${\displaystyle R=\mathbb {C} [x_{1},\dots ,x_{n}]/I}$, then we have a morphism

${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]^{G}\rightarrow R^{G},f\mapsto f+I.}$
Suppose ${\displaystyle [f]\in R^{G}}$, then let
${\displaystyle g={\frac {1}{|G|}}\sum _{g\in G}(g\cdot x)(f).}$
Since ${\displaystyle [f]\in R^{G}}$, the image of ${\displaystyle g}$ under the projection is ${\displaystyle f}$.

So it is enough to show that ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]^{G}}$ is finitely generated. Writing ${\displaystyle R=\mathbb {C} [x_{1},\dots ,x_{n}]/I}$ 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 ${\displaystyle g\in G}$ acts on the generators as a permutation (of the indices). So we can assume that ${\displaystyle G}$ is a subgroup of ${\displaystyle S_{n}}$, the symmetric group on ${\displaystyle n}$-variables. Then we have inclusions

${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]^{S_{n}}\subset \mathbb {C} [x_{1},\dots ,x_{n}]^{G}\subset \mathbb {C} [x_{1},\dots ,x_{n}].}$
The algebra ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]^{S_{n}}}$ is generated by the ${\displaystyle n}$ elementary symmetric polynomials. So because ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]}$ is finitely generated over ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]^{S_{n}}}$, which is a finitely generated ${\displaystyle \mathbb {C} }$-algebra, and we have the triple inclusion above, ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]^{G}}$ must be finitely generated.

Exercise 1.3

Let ${\displaystyle X\subset \mathbb {C} ^{n}}$ 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 ${\displaystyle X}$ to be not empty and connected and let ${\displaystyle I}$ be the radical ideal of ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]}$ such that ${\displaystyle X=V(I)}$. Since ${\displaystyle X}$ is not empty, ${\displaystyle I}$ cannot be the entire polynomial ring. Let us consider any polynomial ${\displaystyle g\in \mathbb {C} [x_{1},\dots ,x_{n}]}$ which does not belong to ${\displaystyle I}$ and let us take the regular function ${\displaystyle g_{|X}\colon X\rightarrow \mathbb {C} }$ given by the restriction of the polynomial ${\displaystyle g}$ to the variety ${\displaystyle X}$. Since ${\displaystyle X}$ is a connected compact analytic space and ${\displaystyle g}$ is a holomorphic function, we deduce from the maximum principle that ${\displaystyle g|X}$ is equal to a constant ${\displaystyle c\in \mathbb {C} }$. We see that this constant ${\displaystyle c}$ is different from 0 since ${\displaystyle g}$ does not belong to the ideal ${\displaystyle I}$. However, ${\displaystyle g-c}$ vanishes on ${\displaystyle X}$ and thus it belongs to the radical idea ${\displaystyle I}$. Therefore we deduce that the ideal generated by ${\displaystyle I}$ and ${\displaystyle g}$ coincides with the whole ring ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]}$ since it contains ${\displaystyle c}$. To sum up, we have proven that for any elements ${\displaystyle g\notin I}$ the ideal generated by ${\displaystyle I}$ and ${\displaystyle g}$ coincide with the ring of polynomials ${\displaystyle \mathbb {C} [x_{1},\dots ,x_{n}]}$. This means that the ideal ${\displaystyle I}$ is maximal. Thus we conclude that ${\displaystyle X}$ is a point by the Nullstellensatz. If ${\displaystyle X}$ were not connected, it would be the union of a finite number of connected affine algebraic varieties ${\displaystyle \left\{X_{i}\right\}_{i=1}^{n}}$ and we can conclude by the above argument.

Exercise 1.4

A grading ${\displaystyle V=\bigoplus _{d\in \mathbb {Z} }V_{d}}$ on the vector space ${\displaystyle V}$ such that the ${\displaystyle V_{d}}$ are finite dimensional is "the same as" a ${\displaystyle \mathbb {C} ^{\ast }}$-action on ${\displaystyle V}$ with the property that each ${\displaystyle v\in V}$ is contained in a finite dimensional subspace ${\displaystyle W}$ of ${\displaystyle V}$ that is ${\displaystyle \mathbb {C} ^{\ast }}$-stable and such that the induced action on ${\displaystyle W}$ is algebraic.

Proof

Let us first find all one-parameter subgroups of ${\displaystyle \mathrm {GL} _{n}(\mathbb {C} )}$, i.e. all group homomorphisms

${\displaystyle f\colon \mathbb {C} ^{\ast }\to \mathrm {GL} _{n}(\mathbb {C} )}$
which are given by Laurent polynomials. We will prove that there is ${\displaystyle D\in \mathrm {GL} _{n}(\mathbb {C} )}$ such that ${\displaystyle Df(\lambda )D^{-1}}$ is diagonal for all ${\displaystyle \lambda \in \mathbb {C} ^{\ast }.}$ Let ${\displaystyle U}$ be the set of all roots of unity in ${\displaystyle \mathbb {C} }$. Clearly, ${\displaystyle f(U)}$ is an abelian subgroup if ${\displaystyle \mathrm {GL} _{n}(\mathbb {C} )}$. Further, each element of ${\displaystyle f(U)}$ has finite order and is therefore diagonalisable. By linear algebra, this implies that ${\displaystyle f(U)}$ is simultaneously diagonalisable. This means that (after conjugating by ${\displaystyle D}$), all off-diagonal entries of ${\displaystyle f(\lambda )}$ (${\displaystyle \lambda }$ a non-zero complex number) are given by Laurent polynomials that vanish on the infinite set ${\displaystyle U}$. Hence the claim follows. Now we know that, after conjugating by ${\displaystyle D}$, ${\displaystyle f}$ is of the form
${\displaystyle \lambda \mapsto \mathrm {diag} (\psi _{1}(\lambda ),...,\psi _{n}(\lambda )),}$
with ${\displaystyle \psi _{j}\colon \mathbb {C} ^{\ast }\to \mathbb {C} ^{\ast }}$ a homomorphism given by a Laurent polynomial. Hence we can write ${\displaystyle \psi _{j}(T)=T^{-k}p(T)}$ for some nonnegative integer ${\displaystyle k}$ and some polynomial ${\displaystyle p}$. But ${\displaystyle \psi _{j}}$ cannot map any non-zero complex number to 0 (since the target of ${\displaystyle \psi _{j}}$ is ${\displaystyle \mathbb {C} ^{\ast })}$, so ${\displaystyle p(T)=T^{l}}$ for some nonnegative intereger ${\displaystyle l}$. This implies that, in fact, ${\displaystyle f}$ is of the form
${\displaystyle \lambda \mapsto \mathrm {diag} (\lambda ^{d_{1}},...,\lambda ^{d_{n}})}$
for suitable integers ${\displaystyle d_{j}}$, up to conjugation. Now suppose we are given a vector space ${\displaystyle V}$ with an action of ${\displaystyle \mathbb {C} ^{\ast }}$ satisfying the conditions from the exercise. Let ${\displaystyle v\in V}$. Pick a finite dimensional stable subspace ${\displaystyle W\subseteq V}$ containing ${\displaystyle v}$ and a basis ${\displaystyle w_{1},...,w_{n}}$ (${\displaystyle n=\dim W}$) of ${\displaystyle W}$ such that ${\displaystyle \lambda \in \mathbb {C} ^{\ast }}$ acts on ${\displaystyle w_{j}}$ by ${\displaystyle w_{j}\mapsto \lambda ^{d_{j}}w_{j}.}$ This is possible by our previous considerations. We can write
${\displaystyle v=\sum _{i=1}^{n}\alpha _{i}w_{i}}$
(${\displaystyle \alpha _{j}\in \mathbb {C} }$). Therefore, if we define
${\displaystyle V_{d}:=\{v\in V:\forall \lambda \in \mathbb {C^{\ast }} :\lambda \cdot v=\lambda ^{d}v\},}$
we obtain
${\displaystyle V=\bigoplus _{d\in \mathbb {Z} }V_{d}.}$
The other direction is trivial.

Exercise 1.5

Derive the degree-genus formula for a nonsingular curve in ${\displaystyle \mathbb {P} ^{2}}$ of degree ${\displaystyle d}$ by degenerating it to the union of ${\displaystyle d}$ lines in general position.

Proof

Let ${\displaystyle C=\{[x_{0}:x_{1}:x_{2}]\in \mathbb {P} ^{2}\colon f(x_{0},x_{1},x_{2})=0\}}$ be a nonsingular degree ${\displaystyle d}$ curve in ${\displaystyle \mathbb {P} ^{2}}$. 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

${\displaystyle g(x_{0},x_{1},x_{2})=(a_{10}x_{0}+a_{11}x_{1}+a_{12}x_{2})\ldots (a_{d0}x_{0}+a_{d1}x_{1}+a_{d2}x_{2})}$
which defines a singular curve in ${\displaystyle \mathbb {P} ^{2}}$ (a union of ${\displaystyle d}$ 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 ${\displaystyle d}$ curve in ${\displaystyle \mathbb {P} ^{2}}$ with a point in ${\displaystyle \mathbb {P} ^{{d+2 \choose 2}-1}}$. This is because a homogeneous polynomial of degree ${\displaystyle d}$ in three variables has ${\displaystyle {d+2 \choose 2}}$ coefficients and by the Nullstellensatz two such polynomials define the same curve in ${\displaystyle \mathbb {P} ^{2}}$ if and only if they are constant multiples of eachother. So we want our continuous deformation to be a path in ${\displaystyle \mathbb {P} ^{{d+2 \choose 2}-1}}$ connecting the point corresponding to ${\displaystyle f}$ to the one corresponding to ${\displaystyle g}$ such that each point on it, other than ${\displaystyle g}$, 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 ${\displaystyle g}$ is a union of ${\displaystyle d}$ projective lines, that is, ${\displaystyle d}$ Riemann spheres in ${\displaystyle \mathbb {P} ^{2}}$. But every two lines in ${\displaystyle \mathbb {P} ^{2}}$ intersect in a point. So we end up with ${\displaystyle d}$ spheres, every two of which touch, and the genus is the number of holes (right before we reach ${\displaystyle g}$ 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 ${\displaystyle d}$ vertices. The number of edges in the graph is ${\displaystyle {d \choose 2}}$ and a maximal tree has ${\displaystyle d-1}$ edges. Note that the maximal tree consists in fact of one root and ${\displaystyle d-1}$ 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

${\displaystyle {d \choose 2}-(d-1)={\frac {(d-1)(d-2)}{2}}.}$

We now argue that the desired deformation always exists. We start with the two distinct points ${\displaystyle f}$ and ${\displaystyle g}$ in ${\displaystyle \mathbb {P} ^{{d+2 \choose 2}-1}}$. Through them passes a unique projective line ${\displaystyle L}$. Let ${\displaystyle Sing_{d}\subseteq \mathbb {P} ^{{d+2 \choose 2}-1}}$ be the set of points representing singular curves of degree ${\displaystyle d}$ in ${\displaystyle \mathbb {P} ^{2}}$. It is a Zariski closed subset of ${\displaystyle \mathbb {P} ^{{d+2 \choose 2}-1}}$ and hence ${\displaystyle Sing\cap L}$ is Zariski closed in ${\displaystyle L}$. But ${\displaystyle L=\mathbb {C} \cup \{\infty \}}$ and so Zariski closed subsets (i.e. zeros of polynomials) are only finite sets or the whole of ${\displaystyle L}$. But ${\displaystyle f\in L\setminus Sing_{d}}$, so ${\displaystyle L\cap Sing_{d}}$ is finite. Thus ${\displaystyle (L\setminus Sing_{d})\cup \{g\}}$ is a Riemann sphere with finitely many points removed and hence there is a smooth path from ${\displaystyle f}$ to ${\displaystyle g}$.

Exercise 1.6

Why the name "spectrum"? Given ${\displaystyle A\in M_{n\times n}(\mathbb {C} )}$ consider the (commutative!) ${\displaystyle \mathbb {C} }$-algebra generated by ${\displaystyle A}$. What is its Spec?

Proof

Using the linear algebra definition, the spectrum of the linear operator ${\displaystyle A}$ is the set of its eigenvalues: ${\displaystyle \sigma (A)=\operatorname {Spec} (A)=\left\{\lambda \in \mathbb {C} \;\;|\;\;\operatorname {Ker} (A-\lambda Id)\neq \left\{0\right\}\right\}}$. We want to see that this set is precisely the set of maximal ideal of the ring ${\displaystyle \mathbb {C} [A]}$ (so it's the same spectrum we defined in the lecture), i.e. the commutative ${\displaystyle \mathbb {C} }$-algebra generated by ${\displaystyle A}$. We have already seen that ${\displaystyle \left\{{\mathfrak {m}}\in \mathbb {C} [A]\;\;|\;\;{\mathfrak {m}}{\textit {ismaximal}}\right\}=\operatorname {Hom} _{\mathbb {C} -alg}(\mathbb {C} [A],\mathbb {C} ).}$

So we write down explicitly the maps between this two set and we show they are inverse maps. Let ${\displaystyle \lambda \in \sigma (A)}$ an eigenvalue, then

If ${\displaystyle A-\lambda Id\in \mathbb {C} [A]}$ is not a unit, then ${\displaystyle A-\lambda Id\in {\mathfrak {m}}_{\lambda }}$ for some maximal ideal ${\displaystyle m_{\lambda }}$. So we associate to ${\displaystyle \lambda }$ the homomophism ${\displaystyle \varphi }$ whose kernel is ${\displaystyle m_{\lambda }}$. In the other direction, fixed ${\displaystyle \varphi \in \operatorname {Hom} _{\mathbb {C} -alg}(\mathbb {C} [A],\mathbb {C} )}$, we have that ${\displaystyle A-\varphi (A)Id\in \operatorname {Ker} (\varphi )}$, so it's not invertible and then ${\displaystyle \varphi (A)\in \sigma (A)}$. It's straightforward to check that these maps are inverse to each other.