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\title{Exercises}


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\section{First problem set}
\label{section-first}

\begin{exercise}
\label{exercise-kernel-ev}
Let $a = (a_1, \ldots, a_n) \in \mathbf{C}^n$.
Recall that $ev_a : \mathbf{C}[x_1, \ldots, x_n] \to \mathbf{C}$ is
the map $f \mapsto f(a)$. Carefully prove that
\begin{enumerate}
\item the ideal
$(x_1 - a_1, \ldots, x_n - a_n) \subset \mathbf{C}[x_1, \ldots, x_n]$
is a maximal ideal, and
\item we have $(x_1 - a_1, \ldots, x_n - a_n) = \text{Ker}(ev_a)$.
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exercise-existence-maximal-ideal}
Let $R$ be any ring. Let $I \subset R$ be an ideal.
Carefully prove that if $I \not = R$, then there exists 
a maximal ideal $\mathfrak m \subset R$ such that $I \subset \mathfrak m$.
\end{exercise}

\begin{exercise}
\label{exercise-find-equation-easy}
Consider the subset
$$
X = \{(t, 1/t) \mid t \in \mathbf{C}, t \not = 0\} \subset \mathbf{C}^2
$$
Show that this is an algebraic set by finding an ideal
$I \subset \mathbf{C}[x, y]$ such that $X = V(I)$.
\end{exercise}

\begin{exercise}
\label{exercise-find-equation}
Consider the subset
$$
X = \{(4t^2 - 4t + 1, t^3 - 1) \mid t \in \mathbf{C}\} \subset \mathbf{C}^2
$$
Show that this is an algebraic set by finding an ideal
$I \subset \mathbf{C}[x, y]$ such that $X = V(I)$. (Hint: Write $t$
in terms of $x$.)
\end{exercise}

\begin{definition}
\label{definition-affine-line}
Consider the affine space $\mathbf{C}^n$.
An {\it affine line} in $\mathbf{C}^n$
is a translate of a linear subspace of dimension $1$.
\end{definition}

\noindent
In other words, a line can be defined by a system of linear equations
$$
\left\{
\begin{matrix}
a_{11}x_1 + \ldots + a_{1n}x_n = b_1 \\
\ldots \\
a_{n - 11}x_1 + \ldots + a_{n- 1 n}x_n = b_{n - 1} \\
\end{matrix}
\right.
$$
where the rank of the matrix $(a_{ij})$ is $n - 1$.
This makes it clear that an affine line is an algebraic subset. Of course we
can also {\it parametrize} an affine line as
$$
l = \{ tv + w \mid t \in \mathbf{C}\}
$$
where $v, w \in \mathbf{C}^n$, and $v \not = 0$.

\begin{exercise}
\label{exercise-line-intersects-finite}
Let $X \subset \mathbf{C}^n$ be an algebraic subset.
Let $l \subset \mathbf{C}^n$ be an affine line.
Show that $X \cap l$ is either empty, or finite, or equal to $l$.
\end{exercise}

\begin{exercise}
\label{exercise-not-algebraic}
Show that the subset
$$
X = \{(t, e^t) \mid t \in \mathbf{C}\} \subset \mathbf{C}^2
$$
is a usual closed set, but not an algebraic set.
\end{exercise}

\begin{remark}
\label{remark-more-general}
More generally, if you have an algebraic set of the form
$\{(t, f(t)) \mid t \in \mathbf{C}\}$ where $f : \mathbf{C} \to \mathbf{C}$
is a function what can say about the function $f$?
\end{remark}









\section{Second problem set}
\label{section-second}

\begin{exercise}
\label{exercise-fibres}
Let $C \subset \mathbf{C}^2$ be a nonempty plane algebraic curve (remember
this just means that $C$ is a hypersurface in $\mathbf{C}^2$). Consider the
map
$$
C \longrightarrow \mathbf{C},\quad (x, y) \longmapsto x
$$
What can you say about the fibres of this map? More precisely, show the
following:
\begin{enumerate}
\item Show that if the fibre over $a \in \mathbf{C}$ is infinite,
then $\{a\} \times \mathbf{C} \subset C$.
\item Show that there exists an integer $d$ such that all but finitely
many fibres have cardinality $d$.
\item Show that, with $d$ as in (2), the other fibres have either $< d$
points, or are infinite.
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exercise-resultant}
For which primes $p$ do the polynomials
$x^2 + 2x + 3 \bmod p$ and $4x^2 + 5x + 6 \bmod p$
have a root in common?
(Hint: Compute the resultant of $x^2 + 2x + 3$ and $4x^2 + 5x + 6$. Please
state the result you are using.)
\end{exercise}

\noindent
The space of $n \times m$ (row $\times$ column) matrices is denoted
$\text{Mat}(n \times m, \mathbf{C})$. Of course we may think of this
as copy of affine $nm$-space, because we can use the coefficients of
the matrices to get a bijection
$$
\text{Mat}(n \times m, \mathbf{C}) \longrightarrow \mathbf{C}^{nm},\quad
A \longrightarrow (a_{11}, a_{12}, \ldots, a_{1m}, a_{21}, \ldots, a_{nm})
$$
In this way we can speak of algebraic sets in
$\text{Mat}(n \times m, \mathbf{C})$.

\begin{exercise}
\label{exercise-resultant-two}
Consider the subset
$$
X =
\{A \in \text{Mat}(2 \times 2, \mathbf{C}) \mid
A^2 =
\left(
\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}
\right)
\}
\subset
\text{Mat}(2 \times 2, \mathbf{C}).
$$
\begin{enumerate}
\item Show that $X$ is an algebraic set.
\item Consider the map
$$
\text{Char} :
\text{Mat}(2 \times 2, \mathbf{C}) \longrightarrow \mathbf{C}^2,
\quad
A \longmapsto (\text{Tr}(A), \det(A))
$$
What is $\text{Char}(X)$?
\item What does your answer to (2) mean for the topology of $X$?
(Just give the most obvious thing here.)
\end{enumerate}
\end{exercise}

\noindent
Since $\mathbf{C}^a \times \mathbf{C}^b = \mathbf{C}^{a + b}$
we can given $X \in \mathbf{C}^a$ and $Y \subset \mathbf{C}^b$
take their {\it product} $X \times Y$ in $\mathbf{C}^{a + b}$.
It turns out that if $X$ and $Y$ are algebraic sets, then so is
$X \times Y$. Namely, if $f_i \in \mathbf{C}[x_1, \ldots, x_n]$
define $X$ and if $g_j \in \mathbf{C}[y_1, \ldots, y_m]$
define $Y$, then
$$
X \times Y = V(f_i(x_1, \ldots, x_n), g_j(x_{n + 1}, \ldots, x_{n + m}))
$$
inside $\mathbf{C}^{a + b}$.
Then $X$, $Y$ and $X \times Y$ inherit the Zariski topology
from $\mathbf{C}^a$, $\mathbf{C}^b$ and $\mathbf{C}^{a + b}$.
Now, it is {\bf not} true that that the Zariski topology on $X \times Y$
as defined just now is the product topology! This is clear on considering
$X = \mathbf{C} \subset \mathbf{C}$, $X = \mathbf{C} \subset \mathbf{C}$
and $X \times Y = \mathbf{C}^2 \subset \mathbf{C}^2$ which clearly does
not have the product topology (it has many more closed sets than just
unions and intersections of products of closed subsets).
What is true is that if $X$ and $Y$ are irreducible
then $X \times Y$ is irreducible. Here are two exercises
whose combination implies this fact.

\begin{exercise}
\label{exercise-open-projection}
Let $X$, resp.\ $Y$ be an algebraic set in $\mathbf{C}^a$,
resp.\ $\mathbf{C}^b$.
\begin{enumerate}
\item For every $x_0 \in X$ show that the map
$Y \to X \times Y, y \mapsto (x_0, y)$ is continous in the Zariski
topology.
\item Show that the projection map
$$
X \times Y \longrightarrow Y,\quad
(x, y) \longmapsto y
$$
is continuous in the Zariski topology.
\item Combine (1) and (2) to show that the maps
$Y \to X \times Y$, $y \mapsto (x_0, y)$, and
$X \to X \times Y$, $x \mapsto (x, y_0)$
are homeomorphisms onto their image in the Zariski topology.
\item Show that the projection map
$$
X \times Y \longrightarrow Y,\quad
(x, y) \longmapsto y
$$
is open\footnote{A continuous map of topological spaces is open
if the image of an open set is open.} in the Zariski topology.
[Hint: If you have a continuous map
of topological spaces $f : Z \to Y$ such that for every $z \in Z$
there exists a continuous map $\sigma : Y \to Z$ with
$f \circ \sigma = \text{id}_Y$, then $f$ is open.]
\end{enumerate}
\end{exercise}

\noindent
In the following exercise we use the convention that an irreducible
space is nonempty.

\begin{exercise}
\label{exercise-irreducible-topological}
Let $f : Z \to Y$ be a continuous map of topological spaces.
Assume that
\begin{enumerate}
\item $f$ is open,
\item $Y$ is irreducible,
\item for a dense set of points $y \in Y$ the fibre $f^{-1}(\{y\})$
is irreducible.
\end{enumerate}
Show that $Z$ is irreducible.
\end{exercise}









\section{Third problem set}
\label{section-third}

\noindent
For an $n \times n$ matrix $A$ over $\mathbf{C}$ we have the famous
Cayley-Hamilton which says that
$$
P(A) = 0
$$
where $P(x) \in \mathbf{C}[x]$ is the characteristic polynomial of $A$,
namely $P(x) = \det(x \mathbf{1}_{n \times n} - A)$. Here
$\mathbf{1}_{n \times n}$ indicates the identity $n \times n$ matrix.
Please use this in solving the exercise below.

\begin{exercise}
\label{exercise-Cayley-Hamilton}
Let $R$ be a ring. Let $A = (a_{ij})$ be an $n\times n$
matrix with coefficients in $R$. Let $P(x) \in R[x]$
be the characteristic polynomial of $A$ (defined
as $\det(x\text{id}_{n\times n} - A)$).
Then $P(A) = 0$ in $\text{Mat}(n\times n, R)$.
[Hints: Prove it for the ring $\mathbf{Z}[a_{ij}]$ where you
think of the $a_{ij}$ as variables and as entries of the matrix $A$.
Use that any polynomial ring $\mathbf{Z}[x_1, \ldots, x_N]$ is
isomorphic to a subring of $\mathbf{C}$.
Then use this to conclude for any $R$ and $A$.]
\end{exercise}

\noindent
The result of the preceding exercise says that if
$P(x) = x^n + a_{n - 1}x^n + a_{n - 2}x^{n - 2} + \ldots + a_0$
then the map
$$
\xymatrix{
R^{\oplus n} \ar[rrr]^{A^n + a_{n - 1}A^{n - 1} + \ldots + a_0} & & &
R^{\oplus n}
}
$$
is the zero map. Please use this in the exercise below.

\begin{exercise}
\label{exercise-apply-CH}
Suppose that $\varphi : R \to S$ is a ring map. Let $s_1, \ldots, s_n$
be elements of $S$ which generate $S$ as an $R$-module. Let $s \in S$
be an arbitrary element. By assumption for each $i$ we can choose
elements $a_{ij} \in R$ such that
$$
ss_i = \sum\nolimits_{j = 1, \ldots, n} \varphi(a_{ij})s_j
$$
Denote $A$ the $n \times n$ matrix over $R$ whose coefficients are the
elements $a_{ij}$. Let $P(x) \in R[x]$ be its characteristic polynomial.
Show that on writing
$$
P(x) = x^n + a_{n - 1}x^n + a_{n - 2}x^{n - 2} + \ldots + a_0,
\quad
a_i \in R
$$
we have
$$
s^n + \varphi(a_{n - 1})s^n + \varphi(a_{n - 2})s^{n - 2} + \ldots +
\varphi(a_0) = 0
$$
\end{exercise}

\noindent
This means that $s$ is integral over $R$, in other words this means
that $R \to S$ is integral.

\begin{exercise}
\label{exercise-irreducible}
Find the irreducible components of
$$
V(xy, xz, yz) \subset \mathbf{C}^3
$$
where we use $x, y, z$ as coordinates on $\mathbf{C}^3$ instead of the usual
$x_1, x_2, x_3$.
\end{exercise}

\begin{exercise}
\label{exercise-irreducible-again}
Find the irreducible components of
$$
V(xy - z^2, x^2 + y^2 + z^2) \subset \mathbf{C}^3
$$
where we use $x, y, z$ as coordinates on $\mathbf{C}^3$ instead of the usual
$x_1, x_2, x_3$.
\end{exercise}

\begin{exercise}
\label{exercise-irreducible-silly}
\footnote{In the actual homework there was a typo, namely
the second equation had a $z^3$ instead of $z^2$.}
Find the irreducible components of
$$
V(x^2 + y^2 + z^2 + 1, x^2 + y^2 + z^2 + 2) \subset \mathbf{C}^3
$$
where we use $x, y, z$ as coordinates on $\mathbf{C}^3$ instead of the usual
$x_1, x_2, x_3$.
\end{exercise}

\begin{exercise}
\label{exercise-irreducible-more}
\footnote{In the actual homework there was a typo, namely
the second equation had a $z^3$ instead of $z^2$.}
Find the irreducible components of
$$
V(x^2 + 2y^2 + 3z^2 + 2, x^2 + 3y^2 + 2z^2 + 1) \subset \mathbf{C}^3
$$
where we use $x, y, z$ as coordinates on $\mathbf{C}^3$ instead of the usual
$x_1, x_2, x_3$. (Hint: This one is pretty hard.
Try to eliminate a variable after changing coordinates.)
\end{exercise}








\section{Fourth problem set}
\label{section-fourth}

\noindent
Some questions about dimension.

\begin{exercise}
\label{exercise-dimension}
Let $X \subset Y \subset \mathbf{C}^n$ be affine algebraic subsets.
Show that $\dim(X) \leq \dim(Y)$.
\end{exercise}

\begin{exercise}
\label{exercise-dimension-linear-space}
Let $X \subset \mathbf{C}^n$ be a linear subspace.
Show that $X$ is a variety and that the dimension of $X$ as a variety is
the same as the dimension of $X$ as a linear space.
\end{exercise}

\begin{exercise}
\label{exercise-dimension-map}
Let $X \subset \mathbf{C}^n$ and $Y \subset \mathbf{C}^m$ be affine
varieties. Let $f_j(x_1, \ldots, x_n)$, $j = 1, \ldots, m$ be polynomials.
Consider the map
$$
f : \mathbf{C}^n \longrightarrow \mathbf{C}^m,\quad
(a_1, \ldots, a_n) \longmapsto
(f_1(a_1, \ldots, a_n), \ldots, f_m(a_1, \ldots, a_n))
$$
and assume
\begin{enumerate}
\item $f(X) \subset Y$, and
\item $f(X)$ is Zariski dense in $Y$.
\end{enumerate}
Prove that $\dim(X) \geq \dim(Y)$. (Hint: Consider the associated map
of coordinate rings $\Gamma(Y) \to \Gamma(X)$, see Fulton.)
\end{exercise}

\begin{exercise}
\label{exercise-when-finite}
Let $C = V(f) \subset \mathbf{C}^2$ be an irreducible affine plane curve.
Consider the projection
$$
\pi : C \longrightarrow \mathbf{C},\quad
(x, y) \longrightarrow x
$$
Prove the following are equivalent:
\begin{enumerate}
\item the corresponding ring map $\mathbf{C}[x] \to \Gamma(C)$
is finite,
\item the polynomial $f$ can be written as
$\lambda y^d + \sum_{i < d} a_i(x)y^i$ with $\lambda \in \mathbf{C}$ not
zero,
\item the map $\pi : C \to \mathbf{C}$ of usual topological spaces is
proper.
\end{enumerate}
Hints: In the course we proved that (1) implies (3). Prove (1) is equivalent
to (2) by doing some algebra. To prove that (3) implies (2) try to show that
if $f = a_d(x) y^d + \sum_{i < d} a_i(x)y^i$ and $z \in \mathbf{C}$ is a
zero of $a_d(x)$, then the solutions of $f(z', y) = 0$ are unbounded
as $z' \to z$.
\end{exercise}

\begin{definition}
\label{definition-finite}
Let $C \subset \mathbf{C}^2$ be an irreducible plane curve.
We say a (linear) projection $\pi : C \to \mathbf{C}$, i.e., a map
coming from a nonzero linear map $\mathbf{C}^2 \to \mathbf{C}$,
is {\it finite} if the equivalent conditions of
Exercise \ref{exercise-when-finite}
hold (after suitably changing coordinates for conditions (1) and (2)).
\end{definition}

\noindent
All the linear projections $\mathbf{C}^2 \to \mathbf{C}$ can be
parametrized (up to linear coordinate changes) by an element
$s \in \mathbf{P}^1 = \mathbf{C} \cup \{\infty\}$. Namely,
to a slope $s \in \mathbf{C}$ we associate the projection
$$
p_s : \mathbf{C}^2 \longrightarrow \mathbf{C},\quad
(x, y) \longmapsto sx - y
$$
and to the slope $s = \infty$ we associate the projection
$$
p_\infty : \mathbf{C}^2 \longrightarrow \mathbf{C},\quad
(x, y) \longmapsto x.
$$

\begin{exercise}
\label{exercise-which-projections}
Let $C = V(1 + x^3 + y^3) \subset \mathbf{C}^2$.
This is an irreducible affine plane curve (you can use this).
Which of the projections $p_s$ given above define a finite morphism
$$
\pi = p_s|_C : C \longrightarrow \mathbf{C}?
$$
Explain your answer.
More generally, if $C = V(f)$ which projections $p_s$ give a finite morphism
$C \to \mathbf{C}$? Only give an answer, no need to explain.
\end{exercise}






\section{Fith problem set}
\label{section-fifth}

\noindent
Let $R$ be a ring. Let $f \in R$ be an element. Any of the following
symbols
$$
R[1/f] = R_f = \{1, f, f^2, f^3, \ldots\}^{-1}R = R[x]/(fx - 1)
$$
will denote the ring constructed out of $R$ and $f$ in the following manner.
An element of $R[1/f]$ is a {\it fraction} $a/f^n$ with $n \geq 0$
and $a \in R$. We identify two fractions $a/f^n$ and $b/f^m$ if we can
find an integer $N$ such that $f^N(f^ma - f^nb) = 0$ in $R$. You can check that
this defines an equivalence relation. If $R$ is a domain (and $f \not = 0$)
then this is equivalent to asking $f^ma = f^nb$, which is the thing you
are used to. We add and multiply fractions by the rules
$$
a/f^n + b/f^m = (af^m + bf^n)/f^{nm}, \quad
a/f^n \cdot b/f^m = ab/f^{nm}.
$$
Then it is easy to verify that this is a ring with zero $0/1$ and unit $1/1$.
The map $R \to R_f$, $a \mapsto a/1$ is a ring map with the pleasing
property that $f$ maps to an invertible element. In fact the ring map
$R \to R_f$ is universal among all ring maps $R \to A$ which map $f$ to
an invertible element. The identification of $R_f$ with the ring
$R[x]/(fx - 1)$ uses the maps
$$
R_f \to R[x]/(xf - 1),\ a/f^n \mapsto ax^n,\quad
R[x]/(xf - 1) \to R_f,\ ax^n \mapsto a/f^n.
$$
Finally, if $R$ is a domain, then the ring $R_f$ is simply the subring of
the quotient field of $R$ consisting of all fractions whose denominator is
a power of $f$.

\begin{exercise}
\label{exercise-invert-one}
Let $f \in \mathbf{C}[x_1, \ldots, x_n]$ be nonzero.
Let
$$
X = \mathbf{C}^n \setminus V(f) = \{a \in \mathbf{C}^n \mid f(a) \not = 0\}
$$
Show that the ring of regular functions on $X$ can be described
as follows
$$
\mathcal{O}(X) = \mathbf{C}[x_1, \ldots, x_n][1/f] =
\mathbf{C}[x_1, \ldots, x_n]_f.
$$
(This may have been explained in the lectures, but please rewrite it anyway.)
\end{exercise}

\begin{exercise}
\label{exercise-hartog}
Let $f, g \in \mathbf{C}[x_1, \ldots, x_n]$ be nonzero.
Assume that $V(f)$ and $V(g)$ have no irreducible component in common.
Set
$$
X = \mathbf{C}^n \setminus V(f, g).
$$
Show that
$$
\mathcal{O}(X) = \mathbf{C}[x_1, \ldots, x_n].
$$
Hint: Use previous exercise.
\end{exercise}

\begin{exercise}
\label{exercise-take-out}
Consider the affine plane curve
$$
C = \{(x, y) \mid y^2 = x(x - 1)(x + 3) \}
$$
and the point $c = (-1, 2) \in C$. Let $U = C \setminus \{c\}$
which is a quasi-affine variety. Find an element of $\mathcal{O}(U)$
which is not an element of $\mathcal{O}(C) = \Gamma(C)$.
\end{exercise}

\begin{exercise}
\label{exercise-cusp-not-smooth}
Consider the cuspidal curve
$
C_{cusp} = \{(x, y) \in \mathbf{C}^2 \mid y^2 - x^3 = 0 \} \subset \mathbf{C}
$
and the affine line
$
C_{smooth} = \{t \in \mathbf{C} \mid 1 = 1\} = \mathbf{C}.
$
Consider the map
$$
\varphi : C_{smooth}  \longrightarrow C_{cusp},\ t \mapsto (t^2, t^3)
$$
Note that $\varphi$ is bijective. Show that its inverse
$\varphi^{-1} : C_{cusp} \to C_{smooth} = \mathbf{C}$ is not
a regular function.
\end{exercise}

\begin{exercise}
\label{exercise-conic}
Consider plane conics
$$
C = \{(x, y) \in \mathbf{C}^2 \mid a + bx + cy + dx^2 + exy + fy^2 = 0\}.
$$
over the complex numbers where we assume at least one of $d, e, f$ is nonzero.
For which $(a, b, c, d, e, f) \in \mathbf{C}^6$ does there exist a
``parametrization'' $\varphi : \mathbf{C} \to C$? Here ``parametrization''
means
\begin{enumerate}
\item $\varphi(t) = (P(t), Q(t))$ for some polynomials $P, Q$,
\item $\varphi$ is bijective, and
\item the inverse map $\varphi^{-1}$ is a regular function on $C$.
\end{enumerate}
Hint: Remember your classification of conics...!
\end{exercise}






\section{Sixth problem set}
\label{section-sixth}


\begin{exercise}
\label{exercise-principal-open-affine}
Let $X \subset \mathbf{C}^n$ be an affine variety.
Let $f \in \mathbf{C}[x_1, \ldots, x_n]$ be a polynomial
such that $X \not \subset V(f)$.
\begin{enumerate}
\item Show that
$$
X \setminus V(f) \longrightarrow \mathbf{C}^{n + 1},\quad
a \longmapsto (a_1, \ldots, a_n, 1/f(a))
$$
is a morphism.
\item Show that the image is a Zariski closed $Y \subset \mathbf{C}^{n + 1}$.
\item Show that the induced map $X \setminus V(f) \to Y$ is bijective.
\item Show that the inverse map $Y \to X \setminus V(f)$ is a morphism
too.
\end{enumerate}
Conclude that $X \setminus V(f)$ is an affine variety (as redefined in the
course).
\end{exercise}

\begin{exercise}
\label{exercise-affine-in-quasi-affine}
Let $X \subset \mathbf{C}^n$ be a quasi-affine variety
(as defined in the course). Let $a \in X$ be a point.
Show that there exists an open neighbhourhood $U \subset X$ of $a$
which is an affine variety (as redefined in the course).
(Hint: Use previous exercise.)
\end{exercise}

\noindent
Let $X \subset \mathbf{C}^n$ be an affine variety.
Let $a = (a_1, \ldots, a_n) \in X$ be a point.
Let $I = I(X)$ be the ideal of $X$.
Then we have $f(a) = 0$ for all $f \in I$.
But it is usually not the case that
$$
\frac{\partial f}{\partial x_i}(a) = 0
$$
for $f \in I$. Hence we get an interesting $\mathbf{C}$-linear map
\begin{equation}
\label{equation-derivative}
I \longrightarrow \mathbf{C}^n,\quad
f \longmapsto
\left(\frac{\partial f}{\partial x_1}(a), \ldots,
\frac{\partial f}{\partial x_n}(a)\right)
\end{equation}
We say $a$ is a {\it nonsingular} point of $X$ if and only if
$\text{rank}(\ref{equation-derivative}) = n - \dim(X)$.
We say $a$ is a {\it singular} point of $X$ if and only if
$\text{rank}(\ref{equation-derivative}) < n - \dim(X)$.
It is a theorem in commutative algebra that the rank is never
$> n - \dim(X)$, so that this covers all cases.

\begin{exercise}
\label{exercise-singular-points-silly}
Let $X = V(x_1^2) \subset \mathbf{C}^n$.
What are the singular points of $X$?
(This is a trick question. Think, don't compute.)
\end{exercise}

\begin{exercise}
\label{exercise-singular-points-ordinary-double-point}
Let
$$
X = V(x_1^2 + x_2^2 + \ldots + x_n^2) \subset \mathbf{C}^n.
$$
What are the singular points of $X$?
(Compute, don't think.)
\end{exercise}

\begin{exercise}
\label{exercise-singular-points-plane-curve}
Let $f \in \mathbf{C}[x, y]$ be irreducible, so
that $C = V(f) \subset \mathbf{C}^2$ is a plane curve
What equations define the singular points of $C$?
Can there be infinitely many singular points?
\end{exercise}

\begin{exercise}
\label{exercise-compute-singular-points-three-space}
Consider the affine curve $C \subset \mathbf{C}^3$
defined by the ideal
$$
I = ((x - 1)^2 + y^2 + z^2 - 1, (x + 1)^2 + 2*y^2 + z^2 - 1)
\subset
\mathbf{C}[x, y, z]
$$
Compute its singular points. (You may use that $I$ is a prime ideal
and that $C = V(I)$ is indeed a curve.)
\end{exercise}






\section{Seventh problem set}
\label{section-seventh}

\noindent
Please, please use any of the statements from the list in
Section \ref{section-review}.
If you need an extra general statement, then email it to me
and I'll add it to the list.

\begin{exercise}
\label{exercise-regular-functions-P1}
Compute $\mathcal{O}(\mathbf{P}^1)$.
\end{exercise}

\begin{exercise}
\label{exercise-morphism-from-P1}
Is there a nonconstant morphism $\mathbf{P}^1 \to \mathbf{C}^n$ for any $n$?
\end{exercise}

\begin{exercise}
\label{exercise-regular-functions-P2-line}
Compute $\mathcal{O}(\mathbf{P}^2 \setminus V_{+}(X_0 + X_1 + X_2))$.
\end{exercise}

\begin{exercise}
\label{exercise-regular-functions-P2-conic}
Compute $\mathcal{O}(\mathbf{P}^2 \setminus V_{+}(X_0^2 + X_1^2 + X_2^2))$.
\end{exercise}

\begin{exercise}
\label{exercise-map-from-P1}
Let $F_0, F_1, F_2 \in \mathbf{C}[X_0, X_1]$ be homogeneous of the
same degree. Assume that
$$
V_{+}(F_0) \cap V_{+}(F_1) \cap V_{+}(F_2) = \emptyset.
$$
Show that the map
$$
\mathbf{P}^1 \longrightarrow \mathbf{P}^2,\quad
[a_0 : a_1] \mapsto [F_0(a_0, a_1) : F_1(a_0, a_1) : F_2(a_0, a_1)]
$$
is a morphism of varieties. Generalize the statement to higher dimensional
projective spaces (but don't prove it).
\end{exercise}

\begin{exercise}
\label{exercise-conics-meet-in-one-point}
Find two (nondegenerate) conics in $\mathbf{P}^2$ which meet in one point.
Find two irreducible cubics in $\mathbf{P}^2$ which meet in one point.
\end{exercise}







\section{Eighth problem set}
\label{section-eighth}


\noindent
Review of holomorphic functions. Let $\Omega \subset C$ be an open subset.
A function $f : \Omega \longrightarrow \mathbf{C}$
is called {\it holomorphic} if for every $z_0 \in \Omega$ there exists a
complex number $f'(z_0)$, called the {\it derivative of $f$ at $z_0$}
such that
$$
\forall \epsilon > 0\ \exists \delta > 0 :
\left|\frac{f(z) - f(z_0)}{z - z_0} - f'(z_0)\right| < \epsilon
\ \forall z \in \Omega,\ |z - z_0| < \delta.
$$
These functions have the following properties:
\begin{enumerate}
\item Given complex numbers $a_0, a_1, a_2, \ldots$ such that
$|a_n| < C(1/R)^n$ for some $C, R > 0$ then the powerseries
$f(z) = \sum a_n(z - z_0)^n$ converges for all $z$, $|z - z_0| < R$
and is a holomorphic function on that disc.
\item Conversely, if $f : \Omega \to \mathbf{C}$ is holomorphic, and
a disc of radius $R$ around $z_0$ is contained in $\Omega$, then
$f$ is given by a convergent power series on the disc with radius
$R$ around $z_0$ as in (1).
\item If $f$ is holomorphic, then $f$ is continuous.
\item If $f$ is holomorphic, then the derivative $z \mapsto f'(z)$
is holomorphic too.
\item The derivative of $f(z) = \sum a_n(z - z_0)^n$ as in (1) is
$f'(z) = \sum na_n(z - z_0)^{n - 1}$ which converges on the same disc.
\item If $f : \Omega \to \mathbf{C}$ is holomorphic and nonconstant on
each connected component of $\Omega$, then $f$ is an open mapping.
\item If $f, g$ are holomorphic on $\Omega$ and $f(z_n) = g(z_n)$ for
some infinite set of points $\{z_n\} \subset \Omega$ which has a limit
point in $\Omega$ then $f = g$.
\item Let $f_i$, $i = 1, 2, 3, \ldots$ be holomorphic functions defined
on the open subset $\Omega \subset \mathbf{C}$. Assume the pointwise
limit $f(z) = \lim_i f_i(z)$ exists for all $z \in \Omega$. Assume moreover
that the convergence $\lim f_i = f$ is {\it uniform} on every compact subset
of $\Omega$. Then $f$ is holomorphic too.
\end{enumerate}

\begin{theorem}
\label{theorem-implicit}
Statement of the {\it implicit function theorem} as we proved it in the
lectures\footnote{I owe you the proof of the uniqueness of the solution.
Suppose that
$\varphi_1, \ldots, \varphi_{n - 1} \in \mathbf{C}[x_1, \ldots, x_n]$
are as in the lecture, i.e., have no linear or constant terms. Pick a
constant $C_1 > 0$ such that
$$
|\varphi_j(x_1 + y_1, \ldots, x_n + y_n) - \varphi(x_1, \ldots, x_n)|
<
C(\max\{|x_i|\}  + \max\{|y_j|\})\max\{|y_i|\}
$$
for all $x_i, y_i \in \mathbf{C}$ with $|x_i|, |y_i| \leq 1$. This constant
is slightly different from the constant in Lemma 1 of the lecture.
Let $|z| < \epsilon$. Suppose that we have
$y_1, \ldots, y_{n - 1}, y'_1, \ldots, y'_{n - 1} \in \mathbf{C}$
with $|y_j|, |y'_j| < \epsilon$ and
$y_j = \varphi_j(y_1, \ldots, y_{n - 1}, z)$, and
$y'_j = \varphi_j(y'_1, \ldots, y'_{n - 1}, z)$. Set
$\delta_j = y'_j - y_j$. Then
\begin{align*}
|\delta_j| = |y'_j - y_j| &
= |\varphi_j(y'_1, \ldots, y'_{n - 1}, z) -
\varphi_j(y_1, \ldots, y_{n - 1}, z)| \\
& = |\varphi_j(y_1 + \delta_1, \ldots, y_{n - 1} + \delta_{n - 1}, z) -
\varphi_j(y_1, \ldots, y_{n - 1}, z)| \\
& \leq C_1(\max\{|y_j|, |z|\} + \max\{|\delta_j|\}) \max\{|\delta_j|\}
\end{align*}
with $C_1$ as above. Now note that this is a contradiction
as soon as $\epsilon$ is small enough (Hint: $|\delta_j| \leq 2\epsilon$;
for example $4 \epsilon C_1 < 1$ is good enough). In this arguement
I did not have to impose a stronger condition on $|z|$ to make this work.
But in the statement of the theorem I do need to choose
$0 < \delta <\epsilon$ because in the theorem we are working
with a general coordinate system, and then the trick with
requiring $|z| < \delta$ is necessary, see
Exercise \ref{exercise-epsilon-delta}. The discrepancy happens in the
translation of the general result into the special coordinate system.
}. Suppose that $C \subset \mathbf{C}^n$ is a quasi-affine curve.
Let $p = (a_1, \ldots, a_n) \in C$ be a {\it nonsingular} point of $C$.
Then there exist
\begin{enumerate}
\item an $i \in \{1, \ldots, n\}$,
\item $\epsilon > \delta > 0$ , and
\item functions
$g_1, \ldots, g_n : \{z \in \mathbf{C} : |z| < \delta \} \to \mathbf{C}$
\end{enumerate}
such that the following conditions hold
\begin{enumerate}
\item $g_i(z) = z + a_i$,
\item $g_1, \ldots, g_n$ are holomorphic,
\item $|g_j(z) - a_j| < \epsilon$ for all $j$ and all $|z| < \delta$,
\item we have
$$
C \cap \{(z_1, \ldots, z_n) : |z_j - a_j| < \epsilon, |z_i - a_i| < \delta\}
=
\{(g_1(z), \ldots, g_n(z)) : |z| < \delta \}
$$
\end{enumerate}
\end{theorem}

\noindent
What this means geometrically is the following:
Consider the projection $\pi_i : X \to \mathbf{C}$ to the $i$th axis.
For a sufficiently small open neighbourhood $U \subset \mathbf{C}^n$
(above this is the ball of radius $\epsilon$)
of $p$, there exists a small open neighbourhood
$\Omega \subset \mathbf{C}$ (this is the disc of radius $\delta$)
of $a_i$ such that
$$
\pi_i : X \cap U \cap \pi_i^{-1}(\Omega) \longrightarrow \Omega
$$
is bijective, with inverse $\Phi : z \mapsto (g_1(z), \ldots, g_n(z))$
whose components are holomorphic. A reformulation which is easier to parse
is the following. (Here we reparametrize the disc $|z| < \delta$ to
convert it to the unit disc.)

\begin{theorem}
\label{theorem-reformulate}
Let $p \in C \subset \mathbf{C}^n$ a nonsingular point of an
algebraic curve. There exists a map
$$
\Phi :
\{z \in \mathbf{C} : |z| < 1\}
\longrightarrow
C,\quad
z \longmapsto (g_1(z), \ldots, g_n(z))
$$
such that: (a) each $g_i$ is holomorphic, (b) $\Phi(0) = p$, (c)
$g_i'(0) \not = 0$ for some $i$, and (d) $\Phi$ induces a homeomorphism
of the disc $\{z \in \mathbf{C} : |z| < 1\}$ with a usual open neighbourhood
of $p$ in $C$.
\end{theorem}

\begin{exercise}
\label{exercise-epsilon-delta}
Show that the statement of Theorem \ref{theorem-implicit} is false if you
try to take $\epsilon = \delta$ in it. Namely consider the curve
$$
C = \{(t + t^2, t - t^2), t \in \mathbf{C}\}
= \{(x, y) \mid 2y - 2x + x^2 + 2xy + y^2 = 0 \}
$$
Show that there does not exist an $\epsilon < 1$ such that
$$
\{(x, y) \in C : |x| < \epsilon \}
=
\{(x, y) \in C : |x| < \epsilon\text{ and }|y| < \epsilon \}
$$
and similarly that there does not exist an $\epsilon$ such that
$$
\{(x, y) \in C : |y| < \epsilon \}
=
\{(x, y) \in C : |x| < \epsilon\text{ and }|y| < \epsilon \}
$$
Hint: Look at real $t$!
\end{exercise}

\begin{exercise}
\label{exercise-node-parametrize}
Let $f \in \mathbf{C}[x, y]$. Assume that $f = xy + h.o.t.$, in other
words $f$ has no constant and linear terms and its quadratic term is $xy$.
(For example $f = xy + x^3 + y^3$).
Show that there exist holomorphic functions
$g_1, g_2 : \{z \in \mathbf{C} : |z| < 1\}$ such that
\begin{enumerate}
\item $f(g_1(z), g_2(z)) = 0$ for all $|z| < 1$,
\item $g_1'(0) \not = 0$.
\end{enumerate}
Hint: Apply
Theorem \ref{theorem-reformulate}
to the algebraic curve given by $x^{-2}f(x, xy) = 0$ and the point
$p = (0, 0)$.
\end{exercise}

\noindent
Consider a sphere of radius $\epsilon$ in $\mathbf{C}^2$. It is given by
$$
S^3_\epsilon =
\{(z_1, z_2) \in \mathbf{C}^2 :
z_1 \overline{z_1} + z_2 \overline{z_2} = \epsilon \}
=
\{(x_1, y_1, x_2, y_2) \in \mathbf{R}^4 :
x_1^2 + y_1^2 + x_2^2 + y_2^2 = \epsilon \}
$$
and we have a stereographic projection
$$
S^3_\epsilon \setminus \{north pole\}
\longrightarrow
\mathbf{R}^3,\quad
(x_1, y_1, x_2, y_2) \longmapsto
(\frac{x_1}{1 - y_2}, \frac{y_1}{1 - y_2}, \frac{x_2}{1 - y_2})
$$
Hence if we have a subset of the sphere then we may think of it as a
subset of $\mathbf{R}^3 \cup \{\infty\}$.

\begin{exercise}
\label{exercise-smooth-circle}
Consider an algebraic curve $C \subset \mathbf{C}^2$.
Assume that $(0, 0) \in C$ is a smooth point.
Convince yourself that for any small enough $\epsilon$
$$
C \cap S^3_\epsilon
$$
is (topologically) a single unknotted loop. Things you should do here:
(a) do a straightforward example, (b) do a less straightforward example, and
(c) give an idea of how you could use Theorem \ref{theorem-reformulate}
to start proving this (for example translate it into a question about
holomorphic functions which seems reasonable to you).
\end{exercise}

\begin{exercise}
\label{exercise-coordinates-linked}
Consider the algebraic set $X = V(xy) \subset \mathbf{C}^2$.
Show (by a calculation and picture) that for any small enough $\epsilon$
$$
X \cap S^3_\epsilon
$$
is (topologically) two circles and show that they are linked
(I mean that if they were made out of steel then you couldn't
separate them).
\end{exercise}

\begin{exercise}
\label{exercise-cusp-link}
Consider the algebraic curve $C = V(y^2 - x^3) \subset \mathbf{C}^3$.
Show (by a calculation and picture) that for any small enough $\epsilon$
$$
C \cap S^3_\epsilon
$$
is (topologically) a single loop which is knotted.
\end{exercise}

\begin{exercise}
\label{exercise-power-series-root}
Let $f = \sum_{i \geq 0} a_i z^i$ be a power series with complex coefficients.
Suppose that $a_0 \not = 0$. Let $n \geq 1$. Show that there exists a power
series $g = \sum_{i \geq 0} b_i z^i$ such that $f = g^n$.
\end{exercise}








\section{Ninth problem set}
\label{section-ninth}

\noindent
Suppose that $X \subset \mathbf{C}^n$ is an affine algebraic variety.
Let $\tilde f \in \mathbf{C}[x_1, \ldots, x_n]$ be an element such that
$X \not \subset V(\tilde f)$. Then in 
Exercise \ref{exercise-principal-open-affine}
we have seen that $U = X \setminus V(\tilde f)$ is an affine variety.
Let $f \in \mathcal{O}(X)$ denote the restriction of $\tilde f$ to $X$.
Then $f$ restricts to an invertible element of $\mathcal{O}(U)$. Hence
the restriction induces a canonical map
$$
\mathcal{O}(X)_f \longrightarrow \mathcal{O}(U).
$$
This map is an isomorphism. Actually, in the course of doing
Exercise \ref{exercise-principal-open-affine}
you most likely proved this along the way, but it wasn't
clearly stated as such. Also, in
Exercise \ref{exercise-invert-one}
you proved this when $X = \mathbf{C}^n$.

\begin{exercise}
\label{exercise-function-field}
Let $X$ be an affine algebraic variety. Let
$V \subset X$ be a nonempty Zariski open which is an affine variety also.
Show that the restriction mapping
$\mathcal{O}(X) \to \mathcal{O}(V)$
induces an isomorphism of fraction fields. Hint: Use
remarks above and a suitable choice of an open $U \subset V \subset X$.
\end{exercise}

\begin{exercise}
\label{exercise-valuations}
In the lectures we classified all discrete valuations on
$\mathbf{C}(x)/\mathbf{C}$ by a direct argument.
\begin{enumerate}
\item Let $K = \mathbf{C}(x)[y]/(y^2 - x(x - 1)(x - 2))$.
Classify all discrete valuations on $K/\mathbf{C}$ using a
direct argument.
\item Suppose $K = \mathbf{C}(x)[y]/(y^2 - x(x - 1)(x - 2)(x - 3))$.
What happens with the discrete valuations ``at $\infty$'' in this case?
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exercise-not-cyclic}
Give an example of a finite extension $\mathbf{C}(x) \subset K$
which is not cyclic; for example an extension which is not Galois or
an extension which is Galois but whose Galois group is not cyclic.
(This exercise is here to convince you that $\mathbf{C}(x)$ is very
different from $\mathbf{C}((x))$ which has only cyclic finite extensions.)
\end{exercise}

\begin{exercise}
\label{exercise-projection}
Let $P = [0 : 0 : 1] \in \mathbf{P}^2$. Consider projection from $P$
which is the morphism of quasi-projective varieties
$$
\pi : \mathbf{P}^2 \setminus \{P\} \longrightarrow \mathbf{P}^1, \quad
[a_0 : a_1 : a_2] \longmapsto [a_0 : a_1].
$$
Show that every fibre of $\pi$ is isomorphic (as a variety) to
the affine curve $\mathbf{C}$.
\end{exercise}

\begin{exercise}
\label{exercise-projection-on-curve}
Let $C = V_+(F) \subset \mathbf{P}^2$ be projective plane curve
of degree $d$. This means that $F \in \mathbf{C}[X_0, X_1, X_2]$
is irreducible and homogeneous of degree $d$. Assume that
$P = [0 : 0 : 1] \not \in C$. Consider the restriction of the
projection $\pi$ of
Exercise \ref{exercise-projection}
to $C$, which is a morphism $\pi|_C : C \to \mathbf{P}^1$.
Show that
\begin{enumerate}
\item $\pi|_C : C \to \mathbf{P}^1$ is a proper map on underlying usual
topological spaces, and
\item for all but finitely points in $\mathbf{P}^1$ the fibre of
$\pi|_C$ has exactly $d$ points.
\end{enumerate}
Hints: For (1) use results from Section \ref{section-fourth}.
For (2) use the result of Exercise \ref{exercise-fibres}.
\end{exercise}



\section{Tenth problem set}
\label{section-tenth}

\noindent
Let $A \to B$ be a ring map. The {\it integral closure} $B'$ of $A$ in $B$
is the subset
$$
B' = \{x \in B \mid x\text{ is integral over }A\}.
$$

\begin{exercise}
\label{exercise-integral-closure}
Show that $B'$ is an $A$-subalgebra of $B$.
Hint: The difficult step is to show $x, y \in B' \Rightarrow x + y, xy \in B'$.
To see this show that the $A$-subalgebra $B''$ of $B$ generated by $x$ and $y$
is finite over $A$, and apply the result of
Exercise \ref{exercise-apply-CH}.
\end{exercise}

\begin{exercise}
\label{exercise-computation}
Let $K \subset L$ be a finite separable field extension generated by
a single element. So $L = K[y]/(P(y))$ where $P \in K[T]$ is a monic
irreducible polynomial with $\text{gcd}(P, P') = 1$. Here
$P' = \text{d}P/\text{d}T$. Note that this implies $P'(y)$ is not
zero in $L$. Show that
$$
\text{Tr}_{L/K}\left(\frac{y^i}{P'(y)}\right)
=
\left\{
\begin{matrix}
0 & \text{if} & i = 0, \ldots, n - 2 \\
1 & \text{if} & i = n - 1
\end{matrix}
\right.
$$
where $n = \text{deg}_T(P) = [L : K]$. Use the following steps
(or if you have a different proof that would be great too):
\begin{enumerate}
\item Let $\overline{K}$ be an algebraic closure of $K$ and write
$P(T) = (T - \alpha_1)(T - \alpha_2) \ldots (T - \alpha_n)$ with
$\alpha_1, \ldots, \alpha_n \in \overline{K}$. The fact that $P$
is a separable polynomial means that $\alpha_i \not = \alpha_j$ for
$i \not = j$.
\item Let $\beta \in L$ be any element. Represent $\beta$ as the congruence
class of $Q(y)$ for some polynomial $Q(T) \in K[T]$. Show that
$$
\text{Tr}_{L/K}(\beta) = \sum\nolimits_{j = 1, \ldots, n} Q(\alpha_j).
$$
It is OK to find this in a book and refer to it.
\item Show that
$$
\text{Tr}_{L/K}\left(\frac{y^i}{P'(y)}\right)
= \sum\nolimits_{j = 1, \ldots, n} \frac{\alpha_j^i}{P'(\alpha_j)}
$$
\item Show that
$$
P'(\alpha_j) =
(\alpha_j - \alpha_1) \ldots
\widehat{(\alpha_j - \alpha_j)} \ldots (\alpha_j - \alpha_n)
$$
\item Show that
$$
\frac{1}{P(T)} =
\sum\nolimits_{j = 1, \ldots, n} \frac{1}{P'(\alpha_j)(T - \alpha_j)}
$$
\item Take the previous expression and do Taylor expansion in $1/T$
to conclude.
\end{enumerate}
\end{exercise}

\noindent
In the following exercises you may use the following fact that was proved in
the lecture by Jarod Alper: Suppose that we have
$$
A = \mathbf{C}[x] \subset B = \mathbf{C}[x, y]/(P)
$$
where $P$ is a polynomial in $x, y$ which is monic as a polynomial in $y$
and irreducible. Let $B'$ be the integral closure of $A$ in the fraction
field of the domain $B$. Then we have
$$
B \subset B' \subset \frac{1}{P'}B
$$
where $P' = \partial P / \partial y$.

\begin{exercise}
\label{exercise-hyperelliptic}
Let $f \in \mathbf{C}[x]$ be a nonconstant polynomial which is not a square.
This implies that $P = y^2 - f$ is irreducible. Let $A = \mathbf{C}[x]$
and $B = \mathbf{C}[x, y]/(P)$ as above.
Show the following:
\begin{enumerate}
\item The integral closure of $A$ in the fraction field of $B$ is
$\mathbf{C}[x, z]/(z^2 - g)$, where $g$ is the square free part\footnote{For
example the square free part of $f = (x - 1)^3(x - 2)^2$ is $g = (x - 1)$.}
of $f$.
\item The ring $B$ is integrally closed its fraction
field if and only if $f$ is square free.
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exercise-cyclic-degree-3}
Let $f = (x - 1)x^2(x + 1)^3$. Then $P = y^3 - f$ is irreducible.
Let $A = \mathbf{C}[x]$ and $B = \mathbf{C}[x, y]/(P)$ as above.
Compute the integral closure of $A$ in the fraction field of $B$.
\end{exercise}






\section{Eleventh problem set}
\label{section-eleventh}

\noindent
Catch up with homeworks you are behind on.








\section{Twelth problem set}
\label{section-twelth}

\begin{exercise}
\label{exercise-krull-intersection-theorem}
(Krull intersection theorem.)
Let $R$ be a Noetherian domain.
Prove the following statements (please skip the ones that
you think are too easy):
\begin{enumerate}
\item If $I$, $J$ are ideals in $R$ and for every $x \in I$ there
exists an $n > 0$ such that $x^n \in J$, then $I^N \subset J$ for
some $N \geq 1$.
\item Let $I'$ be an ideal of $R$ and $x \in R$. Set
$I'_n = \{y \in R \mid yx^n \in I'\}$.
Show there exists a $k$ such that $I'_k = I'_m$ for all $m \geq k$.
\item Let $I$, $J$ be ideals of $R$. Consider the set of ideals
$I'$ of $R$ such that $I' \cap J \subset IJ$. Show that this set
ordered by inclusion has a maximal element. (Hint: Zorn's lemma.)
\item If $I$, $J$ are ideals of $R$ then $I^n \cap J \subset IJ$
for some $n$. [Hints: Let $I'$ be maximal with $I' \cap J \subset IJ$ as in
part (3). Show that $(I' + IJ) \cap J \subset IJ$ too, so
$I' = I' + IJ$ by maximality.
Hence $I' \cap J = IJ$. By part (1) it suffices to show that any $x \in I$
has a power which lies in $I'$. Let pick $k$ exactly as in (2).
Consider $I'' = I' + x^kR$. Check that $I'' \cap J \subset IJ$ by a
clever little argument. Hence $I' = I''$, hence $x^k \in I'$ as desired.]
\item Conclude that $\bigcap_{n \geq 0} I^n = 0$ if $I \not = R$.
[Hint: If $x \in \bigcap I^n$, then use (4) to show that
$x \in I^n \subset xI$ for some $n$ which gives $x(1 - f) = 0$ for
some $f \in I$.]
\end{enumerate}
This argument is from a paper by Karamzadeh. But I'm sure there
are lots of other even more elementary arguments. Actually I just found
one. It is an argument of H.~Perdy and you can find it in his paper
``An Elementary Proof of Krull's Intersection Theorem'' published in the
The American Mathematical Monthly,
Vol. 111, No. 4 (Apr., 2004), pp. 356-357.
If you want to look up his argument and explain it then that is
fine too (and it will probably save you quite a bit of time).
Note: it doesn't prove part (4) which is interesting in itself.
Part (4) is a special case of the Artin-Rees theorem.
\end{exercise}

\noindent
The rest of the exercises is a series of exercises aimed in some sense
at understanding the ``points at infinity''. While doing them you will
also be reviewing some of the material we've treated in the lectures.
This means that some of the questions are formulated in a somewhat
strange manner.

\begin{exercise}
\label{exercise-combined-growth-rate}
Consider a situation
$$
\xymatrix{
& \hat B \ar[r] & \hat L & \\
\mathbf{C}[[x]] \ar@{=}[r]& \hat A \ar[r] \ar[u] &
\hat K \ar@{=}[r] \ar[u] & \mathbf{C}((x))
}
$$
where $\hat K \subset \hat L$ is a finite ring extension, the ring
$\hat L$ is reduced, and $\hat B$ is the integral closure of $\hat A$ in
$\hat L$. Recall that this means we can find compatible isomorphisms
$$
\xymatrix{
\hat B \ar[r] & \hat L \\
\prod_{i = 1, \ldots, r} \mathbf{C}[[y_i]] \ar[r] \ar[u]^{\cong} &
\prod_{i = 1, \ldots, r} \mathbf{C}((y_i)) \ar[u]_{\cong}
}
$$
with moreover $x \mapsto (y_1^{e_1}, \ldots, y_r^{e_r})$. Prove the
following: Given $r + 1$ units
$$
u_1, \ldots, u_{r + 1} \in \hat L^*
$$
of $\hat L^*$ there exist integers
$m_1, \ldots, m_{r + 1} \in \mathbf{Z}$ not all zero
such that $u = u_1^{m_1} \ldots u_{r + 1}^{m_{r + 1}}$ is an element
of $\hat B^*$. (In other words, $u$ no longer has a ``pole'' at any
of the ``points'' lying over $x = 0$.)
\end{exercise}


\begin{exercise}
\label{exercise-at-infinity}
Let $C$ be a normal affine algebraic curve.
Recal that this means that the ring of regular functions $B = \mathcal{O}(C)$
is a normal domain. More precisely, if $L = \mathbf{C}(C) = f.f.(B)$
is the field of rational functions of $C$, then $B$ is integrally closed
in $L$. We have also seen that $C$ is a nonsingular curve.
Moreover, we proved that there exists a
ring map $\mathbf{C}[x] = A \to B$ such that $B$ is finite over $A$
(Noether Normalization -- works even for nonnormal affine curves).
Fix such a choice. Denote $K = \mathbf{C}(x) \subset L$. Diagram
$$
\xymatrix{
K \ar[r] & L \\
A \ar[u] \ar[r] & B \ar[u]
}
$$
OK, now let's introduce the ``variable'' $y = x^{-1}$.
Set $A' = \mathbf{C}[y] \subset K$.
Let $B'$ be the integral closure of $A'$ in $L$.
By the result of Jarod's lecture this is a finite ring extension of $A'$. 
We can also consider
$A'' = \mathbf{C}[x, y] = \mathbf{C}[x, x^{-1}] \subset K$ and
its integral closure $B'' \subset L$. This produces the following
diagram
$$
\xymatrix{
B' \ar[r] & B'' & B \ar[l] \\
A' \ar[u] \ar[r] & A'' \ar[u] & A \ar[l] \ar[u]
}
$$
Having said all of this prove that $B'' = B'_y = B_x$.
See Section \ref{section-fifth} for the notation $R_f$.
(You may use that the situation is symmetric in $x$ and
$y$ and hence that you only need to prove either $B'' = B'_y$
or that $B'' = B_x$.)
\end{exercise}


\begin{exercise}
\label{exercise-global-functions}
Notation as in Exercise \ref{exercise-at-infinity}.
Let $f \in L$ be an element which is contained in $B'$ and
in $B''$. Show that $f \in \mathbf{C}$ using the following
steps:
\begin{enumerate}
\item Show that $B \cong A^{\oplus d}$ as an $A$-module.
(Quote a theorem on modules over the polynomial ring $\mathbf{C}[x]$.)
\item Show that $B' \cong (A')^{\oplus d}$ as an $A'$-module.
(Quote a theorem on modules over polynomial ring $\mathbf{C}[y]$.
Yes, this is silly.)
\item Let $P(T) \in A[T]$ be the characteristic polynomial
of multiplication by $f$ on $B$ as an $A$-linear map.
Let $P'(T) \in A'[T]$ be the characteristic polynomial
of multiplication by $f$ on $B'$ as an $A'$-linear map.
Show that $P(T) = P'(T)$ in $A''[T]$.
[Hint: Char Pol independent of chosen basis.]
\item Conclude $P$ has constant coefficients.
\end{enumerate}
\end{exercise}



\begin{exercise}
\label{exercise-units-affine-normal-curve}
Notation as in Exercise \ref{exercise-at-infinity}.
Show that the group of units $B^*$ sits in a short exact
sequence
$$
0 \to \mathbf{C}^* \to B^* \to B^*/\mathbf{C}^* \to 0
$$
of abelian groups and that the group $B^*/\mathbf{C}^*$ on the right
is a finitely generated free abelian group.
[[[Hints: This exercise is a quite a bit harder...
First of all, by symmetry it is OK to switch the roles of $B$ and $B'$
(this is just a notational convenience).
Let $d = [L : K]$ as in Exercise \ref{exercise-global-functions}.
Suppose that $u_1, \ldots, u_{d + 1}$ are units of $B'$.
Try to find integers
$m_1, \ldots, m_r \in \mathbf{Z}$ not all zero
such that
$$
u = u_1^{m_1} \ldots u_{d + 1}^{m_{d + 1}}
$$
is also in $B$. Having found these then
by Exercise \ref{exercise-global-functions}
we see $u \in \mathbf{C}^*$. To find the $m_i$ use the result from
Exercise \ref{exercise-combined-growth-rate}
and the result from the lectures that says that $A \subset B$ matches
with $\hat A \subset \hat B$ in some sense, but try to be somewhat precise
here. If you do not remember the statement ask me.]]]
\end{exercise}













\setcounter{section}{99}
\section{Review Course Material}
\label{section-review}

\noindent
Review of material in course.
\begin{enumerate}
\item Projective space
$\mathbf{P}^n = (\mathbf{C}^{n + 1} \setminus \{0\})/\mathbf{C}^*$
is defined as the set of nonzero vectors in $\mathbf{C}^{n + 1}$ up to scaling.
\item A point of $\mathbf{P}^n$ is denoted $[a_0 : a_1 : \ldots : a_n]$
which means that $(a_0, \ldots, a_n) \in \mathbf{C}^{n + 1}$ is a nonzero
vector and $[a_0 : a_1 : \ldots : a_n]$ is the corresponding point.
\item For $F \in \mathbf{C}[X_0, \ldots, X_n]$ we defined
$$
V_{+}(F) = \{
[a_0 : a_1 : \ldots : a_n] \in \mathbf{P}^n
\mid
F(a_0, \ldots, a_n) = 0\}
$$
and we showed that this is well defined.
\item The standard affine opens $U_i$, $i = 0, \ldots, n$. We have
$$
\mathbf{P}^n = U_0 \cup \ldots \cup U_n
$$
where $U_i = \mathbf{P}^n \setminus V_{+}(X_i)$ is the set of points
whose $i$th coordinate is nonzero. For each $i$ we have a bijection
$$
\Phi_i : U_i \longrightarrow \mathbf{C}^n,\quad
[a_0 : a_1 : \ldots : a_n] \longmapsto
\left(
\frac{a_0}{a_i}, \ldots, \widehat{\frac{a_0}{a_i}}, \ldots, \frac{a_n}{a_i}
\right)
$$
whose inverse is given by the map
$$
(c_1, \ldots, c_n)
\longmapsto
[c_1 : \ldots : c_{i-1} : 1 : c_i : \ldots : c_n].
$$
\item Let $\pi : \mathbf{C}^{n + 1} \setminus \{0\} \to \mathbf{P}^n$
denote the map $(a_0, \ldots, a_n) \mapsto [a_0 : a_1 : \ldots : a_n]$.
\item There is a topology on $\mathbf{P}^n$ defined by saying
$U \subset \mathbf{P}^n$ is open $\Leftrightarrow$
$\pi^{-1}(U) \subset \mathbf{C}^{n + 1} \setminus \{0\}$ is open in the
usual topology. This will be called the {\it usual topology}.
\item The map $\pi$ is continuous and open (in the usual topologies).
It follows that $\mathbf{P}^n$ is Hausdorff in the usual topology.
\item The maps $\Phi_i$ are homeomorphisms in the usual topologies
and the standard affine opens are open. (This determines the usual topology.)
\item The unit sphere $S^{2n + 2} \subset \mathbf{C}^{n + 1} \setminus \{0\}$
(the set of points $(a_0, \ldots a_n)$ such that $\sum |a_i|^2 = 1$)
surjects onto $\mathbf{P}^n$ we see that $\mathbf{P}^n$ is compact in
the usual topology.
\item The space $\mathbf{P}^n$ is a compact topological manifold in the
usual topology.
\item We defined a topology on $\mathbf{P}^n$ whose closed
subsets are
$$
Z = \bigcap\nolimits_{F \in E} V_{+}(F)
$$
where $E \subset \mathbf{C}[X_0, \ldots, X_n]$
is a subset consisting of homogeneous elements. This is the
{\it Zariski topology} on $\mathbf{P}^n$.
\item A Zariski closed subset of $\mathbf{P}^n$ is usual closed.
\item The maps $\Phi_i : U_i \to \mathbf{C}^n$ are homeomorphisms in
the Zariski topologies and the $U_i \subset \mathbf{P}^n$ are open in
the Zariski topology. (This determines the Zariski topology.)
\item The Zariski topological space $\mathbf{P}^n$ is Noetherian.
\item A {\it quasi-projective variety} is a Zariski irreducible locally closed
subset $X \subset \mathbf{P}^n$ for some $n$.
\item A {\it projective variety} is a Zariski irreducible closed
subset $X \subset \mathbf{P}^n$ for some $n$.
\item The maps
$$
\Phi_j \circ \Phi_i^{-1}|_{\Phi_i(U_i \cap U_j)} :
\Phi_i(U_i \cap U_j)
\longrightarrow 
\Phi_j(U_i \cap U_j)
$$
are isomorphisms of q-affine varieties.
\item\label{item-two-indices}
For any quasi-projective variety $X \subset \mathbf{P}^n$
with $X \subset U_i$ and $X \subset U_j$ the images $\Phi_i(X)$
and $\Phi_j(X)$ are quasi-affine varieties (by the above) and
$$
\Phi_j \circ \Phi_i^{-1}|_{\Phi_i(X)} :
\Phi_i(X)
\longrightarrow 
\Phi_j(X)
$$
is an isomorphisms of quasi-affine varieties.
\item\label{item-regular-in-standard-affine}
If $X \subset \mathbf{P}^n$ is a quasi-projective variety with
$X \subset U_i$ for some $i$ the we define the algebra of
{\it regular functions on $X$} by the rule
$$
\mathcal{O}(X) =
\{f : X \to \mathbf{C} \mid
\begin{matrix}
\text{the map }\Phi_i(X) \to \mathbf{C}, \Phi_i(x) \mapsto f(x)
\text{ is a regular} \\
\text{function on the quasi-affine variety }\Phi_i(X)
\end{matrix}
\}
$$
This is independent of the choice of $i$ such that $X \subset U_i$ by
(\ref{item-two-indices}).
\item If $X \subset \mathbf{P}^n$ is a quasi-projective variety we define
the algebra of {\it regular functions on $X$} by the rule
$$
\mathcal{O}(X) = 
\{f : X \to \mathbf{C} \mid
f|_{X \cap U_i} \in \mathcal{O}(X \cap U_i)
\text{ for }i = 0, \ldots, n\}
$$
This makes sense because we have defined $\mathcal{O}(X \cap U_i)$
in (\ref{item-regular-in-standard-affine}).
\item A regular function $f$ on a quasi-projective variety
$X \subset \mathbf{P}^n$
is continuous in both the Zariski and the usual topologies. This is true
because we have seen this holds for $f|_{X \cap U_i}$.
\item If $X$ is a quasi-projective variety and $Y \subset X$ is a
subvariety, then the restriction $f|_Y$ of a regular function on $X$
is a regular function on $Y$. (This is true because we've seen it holds on
$X \cap U_i$.)
\item Let $X \subset \mathbf{P}^n$ be a quasi-projective variety.
Let $f : X \to \mathbf{C}$ be a map of sets. The following are
equivalent
\begin{enumerate}
\item $f$ is a regular function, and
\item there exists an open covering $X = V_1 \cup V_2 \cup \ldots \cup V_m$
such that each restriction $f|_{V_j}$ is a regular function.
\end{enumerate}
\item A {\it morphism} $\varphi : X \to Y$ of quasi-projective varieties 
is a continuous map (in Zariski topology) such that for every open
$V \subset Y$ and any regular function $f \in \mathcal{O}(V)$ on $V$
the composition
$f \circ \varphi|_{\varphi^{-1}(V)} : \varphi^{-1}(V) \to \mathbf{C}$
is a regular function on $\varphi^{-1}(V)$.
\item If $X \subset \mathbf{P}^n$ and $Y \subset \mathbf{P}^m$ are
quasi-projective varieties and $\varphi : X \to Y$ is a map of sets
then the following are equivalent:
\begin{enumerate}
\item $\varphi$ is a morphism,
\item $\varphi$ composed with the map $Y \to \mathbf{P}^m$ is a morphism
from $X$ to $\mathbf{P}^m$,
\item there exists an open covering $X = V_1 \cup V_2 \cup \ldots \cup V_m$
such that each restriction $f|_{V_i}$ is a morphism,
\item for each $j \in \{0, \ldots, m\}$ we have that $\varphi^{-1}(U_j)$
is open in $X$, and the composition
$$
\xymatrix{
\varphi^{-1}(U_j) \ar[rr]_{\varphi|_{\varphi^{-1}(U_j)}} \ar@/^4ex/[rrrr] & &
U_j \ar[rr]_{\Phi_j} & &
\mathbf{C}^m
}
$$
is a morphism, and
\item there exists an open covering $X = \bigcup V_i$ such that for each
$i$ you have $\varphi(V_i) \subset U_{j(i)}$ and moreover the composition 
$$
\xymatrix{
V_i \ar[rr]_{\varphi|_{V_i}} \ar@/_4ex/[rrrr] & &
U_{j(i)} \ar[rr]_{\Phi_{j(i)}} & &
\mathbf{C}^m
}
$$
is a morphism.
\end{enumerate}
The easiest way to use these is to find $V_i$ as in the last condition
such that moreover each $V_i$ is {\it also} contained in a standard affine
open of $\mathbf{P}^n$, since in that case you reduced to checking that
the restriction is a morphism of quasi-affine varieties .
\end{enumerate}


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