Commit 504490b9 authored by Han-Miru Kim's avatar Han-Miru Kim
Browse files

comalg fix univ.prop quot ring, solex1, reformat exercises

parent 6a07d715
......@@ -279,6 +279,8 @@ Setup numbering system and define a function that creates environment.
#+BEGIN_SRC latex
\newcounter{theo}[section]
\renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}
\newcounter{exer}[subsection]
\renewcommand{\theexer}{\arabic{subsection}.\arabic{exer}}
\newcommand{\defboxenv}[3]{
\newenvironment{#1}[1][]{
......@@ -323,6 +325,36 @@ Setup numbering system and define a function that creates environment.
\defboxenv{prop}{Proposition}{orange!30}
\defboxenv{cor}{Corollary}{blue!30}
\defboxenv{xmp}{Example}{green!50!blue!30!white}
\defboxenv{exr}{Exercise}{black!30}
\defboxenv{rem}{Remark}{red!20}
\newenvironment{exr}[1][]{
\refstepcounter{exer}
\ifstrempty{#1} { % no Title
\mdfsetup{
frametitle={
\tikz[baseline=(current bounding box.east), outer sep=0pt]
\node[anchor=east,rectangle,fill=black!30]
{\strut Exercise~\theexer};
}
}
}{ % else: with Title
\mdfsetup{
frametitle={
\tikz[baseline=(current bounding box.east),outer sep=0pt]
\node[anchor=east,rectangle,fill=black!30]
{\strut Exercise~\theexer:~#1};
}
}
} % either case
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linecolor=black!30,
innertopmargin=0pt,
linewidth=2pt,
topline=true,
frametitleaboveskip=\dimexpr-\ht\strutbox\relax,
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\begin{mdframed}[]\relax
}{ % post env command
\end{mdframed}
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#+END_SRC
......@@ -8,7 +8,7 @@ A \textbf{natural transformation} $\Phi: F \to G$ is a rule that to every object
\begin{center}
\begin{tikzcd}[]
FD
FX
\arrow[]{r}{Ff}
\arrow[swap]{d}{\Phi(X)}
&
......
......@@ -14,7 +14,7 @@ We will look at the category of $R$-modules and describe the hom-functor for $R/
For any $R$-algebra $S$, there is a ``natural'' bijection
\begin{align*}
\Hom_{R\text{-alg}}(R/I,S) &\stackrel{\sim}{\to} \left\{
f: R \to S \big\vert \text{ring morphism with } \Ker f \subseteq I\right\}
f: R \to S \big\vert \text{ring morphism with } I \subseteq \Ker f \right\}
\\
g &\mapsto g \circ \pi_I
\end{align*}
......
......@@ -20,7 +20,7 @@
\begin{align*}
\bm{x} = \sum_{x \in X}' \lambda_x x \stackrel{Gf}{\mapsto}\sum_{x \in X}' \lambda_x f(x) \quad \text{for} \quad \lambda_x \in X
\end{align*}
where the sum goes over finitely many $x \in X$.
where only finitely many $\lambda_x$ are non-zero.
By linearity, it is sufficient to check the functor laws on basis elements.
Let $X$ be a set and $x \in X$.
......@@ -79,15 +79,26 @@
\end{exr}
\begin{proof}[Solution]
\phantom{a}
\begin{enumerate}
\item Since the forgetful functor maps every group homomorphism $G \stackrel{\phi}{\to}H$ to the underlying function, a bijection $\Hom_{\Grp}(G,S) \iso \Hom_{\Set}(G,S)$ would imply that every function is a group homomorphism, which is absurd.
\item Same argument as above: Not every function is continuous.
\item Since this is \emph{commutative} algebra, the category of rings only has commutative rings.
But not every group
\item False.
\item Yes, but note that the bijection is not a natural isomorphism because the isomorphism depends on the choice of basis.
\item Consider the rings $X = Y = \C[X]$. The unit groups are both isomorphic to $\C^{\times}$.
Then both maps
\begin{align*}
\id, \quad \text{and} \quad (f \mapsto f(0)) \in \Hom_{\Ring}(C[X],C[X])
\end{align*}
would get mapped to $\id_{C^{\times}}$ by the unit functor, so there would be an isomorphism of hom-sets.
\item If $V$ is infinitely dimensional, then $V^{\ast}$ has strictily larger dimension, so there cannot be a surjective map
\begin{align*}
\Hom_{\C}(V,W) \to \Hom_{\C}(W^{\ast},V^{\ast})
\end{align*}
\item See Linear Algebra II, the isomorphism is given by
\begin{align*}
\Phi: \Hom_\C(V,W) \to \Hom_{\C}(W^{\ast},V^{\ast})\\
f \mapsto (\beta \mapsto \beta \circ f: V \to \C)
\end{align*}
\end{enumerate}
\end{proof}
......@@ -98,15 +109,28 @@
\item The forgetful functor from groups to sets.
\item The forgetful functor from topological spaces to sets.
\item The unit functor from rings to groups.
\item The duality functor from $\C$-vector spaces to the opposite category of $\C$-vector spaces.
\item The duality functor from finite-dimensional $\C$-vector spaces to the opposite category of finite-dimensional $\C$-vector spaces.
\end{enumerate}
\end{exr}
\begin{proof}[Solution]
\begin{enumerate}
\item To prove this, we need to be able to put a group structure on any set.
\item As above, we can take any set $X$ and consider the discrete topology on $X$.
\item Since this is \emph{commutative} algebra, the category of rings only has commutative rings.
But not every group
\item No. There cannot be a group structure on the empty set.
However, if we only look at non-empty sets, then the statement is equivalent to the axiom of choice.
\item If we allow topological spaces to be empty, then consider the discrete topology on $X$. Then $F (X,\tau_{\text{disc}}) = X$.
\item (Since this is \emph{commutative} algebra, every ring is commutative, but not every group is abelian.)
This statement is also false if we consider non-commutative rings aswell.
Take for example the cyclic group $\Z/5\Z$.
For any ring $R$, either $\charac R = 2$, or $-1 \neq 1$.
But if $R^{\times} = \Z/5\Z$, then the group of units cannot have characteristic $2$.
Therefore, $R$ contains the field $\mathbb{F}_2$.
Moreover, it cannot contain any larger field or else it would contain a $2^{n}-1$-th root of unity, which only divides $5$ iff $n = 1$.
Missing rest of proof.
\item False, same reasioning as in the previous exercise.
\item Yes, but note that the bijection is not a natural isomorphism because the isomorphism depends on the choice of basis.
\end{enumerate}
\end{proof}
......@@ -120,9 +144,91 @@ Let $\textsf{C}$ be the category with objects the set $\N$ of natural numbers an
the set of matrices with $n$ rows and $m$ columns.
\begin{enumerate}
\item Check that the identity matrix $1_n$ with composition given by the product of matrices, $\textsf{C}$ does indeed form a cateogry.
\item Let $\textsf{fd-Vec}$ be the category of finitely-dimensional $\C$-vector spaces.
Show that $F: \textsf{C} \to \textsf{fd-Vec}$ defined by
\begin{align*}
F(n) = \C^{n}, \quad F(A) = (x \mapsto Ax)
\end{align*}
is a functor.
\item Show that $F$ is fully faithful and essentially surjective.
Such a functor is called an \textbf{equivalence of categories}.
It means that the categories behave in the same way, although the sets of objects and morphisms are very different.
\item Define a functor $G: \textsf{fd-Vec} \to \textsf{C}$ natural transformations $F \circ G \to \id_{\textsf{fd-Vec}}$ and $G \circ F \to \id_{\textsf{C}}$.
\emph{Hint: You can use the axiom of choice to obtain a basis of any fd-vector spaces.}
\end{enumerate}
\end{exr}
\begin{proof}[Solution]
\begin{enumerate}
\item Composition is well-defined as for any $A \in \Mat(n \times m, \C) = \Hom_{\textsf{C}}(m,n)$ and $B \in \Mat(l \times n,\C) = \Hom_{\textsf{C}(n,l)}$ we have
\begin{align*}
BA \in \Mat(l \times m,\C) = \Hom_{\textsf{C}}(m,l)
\end{align*}
It obviously also respects the identity.
\item Clearly, $F(n) = \C^{n}$ is a vector space and for each $A \in \Hom(m,n)$, we have that
\begin{align*}
F(A) = (x \mapsto Ax): \C^{m} \to \C^{n}
\end{align*}
is well defined.
Now, let $A \in \Hom(m,n)$ and $B \in \Hom(n,l)$. Then
\begin{align*}
F(B \circ A) = (x \mapsto BAx) = F(B) \circ (x \mapsto Ax) = F(B) \circ F(A)
\end{align*}
\item $F$ fully faithful as, by definition:
\begin{align*}
\Hom(m,n) = \Mat(n \times m,\C) = \Hom_{\textsf{C}}(\C^{m},\C^{n})
\end{align*}
It is also essentially surjective, because if $V$ is a $\C$-vector space of dimension $n < \infty$, then
\begin{align*}
V \iso \C^{n} = F(n)
\end{align*}
\item First let's define $G$. For $V$ be a $\C$-vector space of dimension $n < \infty$ let $G(V) = n$.
If $V$ has dimension $n$ and $W$ has dimension $m$, then using the axiom of choice, we can fix a basis $v_{1}, \ldots, v_{n}$ of $V$ and $w_{1}, \ldots, w_{m}$ of $W$.
For $f \in \Hom(W,V)$ define $Gf \in \Hom_{\textsf{C}}(m,n)$ as follows:
\begin{align*}
Gf(f) = A = (a_{ij})_{i,j}
\quad \text{where} \quad
f(v_i) = \sum_{j=1}^{m}a_{ij} w_j
\end{align*}
Then define
\begin{align*}
\Phi: F \circ G \to \id_{\textsf{fd-Vec}} \quad \text{and} \quad
\Psi: G \circ F \to \id_{\textsf{C}}
\end{align*}
given by
\begin{align*}
\Phi_V : \C^{n} \to V, \quad e_i \mapsto v_i\\
\Psi_n = \id_n: n \to n
\end{align*}
Check for yourself that the following diagrams commute:
\begin{center}
\begin{tikzcd}[ ]
\C^{n} \arrow[]{d}{\Phi_V}
\arrow[]{r}{(F \circ G) f}
& \C^{m} \arrow[]{d}{}
\\
V \arrow[]{r}{f}
& W
\end{tikzcd}
\quad
\begin{tikzcd}[ ]
n \arrow[]{r}{A} \arrow[]{d}{\Psi_n}
& m \arrow[]{d}{\Psi_m}
\\
n \arrow[]{r}{A}
& m
\end{tikzcd}
\end{center}
\end{enumerate}
\end{proof}
......@@ -2,9 +2,39 @@
Let $p$ be a prime number and $R$ be the local ring $\Z_{p\Z}$.
Let $k$ be the residue field of $R$. Construct an isomorphism $\Z/p\Z \to k$.
\end{exr}
Recall the notation for $\Z_{p\Z} = (Z \setminus p\Z)^{-1}\Z$ and $k = R/m_R$.
Recall the notation for $\Z_{p\Z} = (Z \setminus p\Z)^{-1}\Z$ and $k = R/m_R$.
Denote by $\iota$ be inclusion mapping and by $\pi$ the projection mapping
\begin{align*}
\iota: \Z \hookrightarrow R, \quad a \mapsto \frac{a}{1}\\
\pi: \Z \mapsto \Z/p\Z, \quad a \mapsto a \mod p
\end{align*}
First, we find the maximal ideal $m_R$.
Since any ideal containing a unit must equal the whole ring and every element $a \in Z \setminus p\Z$ is invertible (in $\Z_{p\Z}$ when identified with $\iota(a) = \tfrac{a}{1}$), we claim that $m_R = \iota(p\Z)$.
It's easy to see that this is a maximal ideal.
The isomorphism is given by
\begin{align*}
\Phi:
\Z/p\Z \to \Z_{p\Z}/\iota(p\Z), \quad
a + p\Z \mapsto \tfrac{a}{1} + \iota(p\Z)
\end{align*}
Now we check that it is well-defined.
If $a-b \in p\Z$, then $\tfrac{a}{1} - \tfrac{b}{1} = \tfrac{a-b}{1} = \iota(a-b) \in \iota(p\Z)$.
It is also clearly a ring morphism.
To show injectvivity, let $a,b \in \Z$. Then
\begin{align*}
(\Phi \circ \pi)(a) - (\Phi \circ \pi)(b) = 0 \iff \tfrac{a - b}{1} \in \iota(p\Z)
\iff \exists c \in p\Z:\ \tfrac{a - b}{1} = \tfrac{c}{1}
\\
\iff \exists t \in \Z \setminus p\Z: \quad t(a - b - c) \in p\Z \iff a - b \in p\Z
\end{align*}
For surjectivity, let $x + \iota(p\Z) \in k$. Then $x$ can be written as
\begin{align*}
x = \frac{a}{b} \quad \text{for some} \quad a \in \Z, p\not| a, b \in (\Z \setminus p\Z)
\end{align*}
With the prime number decomposition of $a = p_1^{s_1} p_2^{s_2} \cdots p_m^{s_m}$, we see that
We first find the maximal ideal $m_R \subseteq \Z_{p\Z}$.
\begin{exr}[]
......@@ -20,7 +50,7 @@ Let $R$ be a ring and $S \subseteq R$ a multiplicative subset.
\end{enumerate}
\end{exr}
\begin{proof}[Solution]
The intuition is that in $S^{-1}R$, elements of $S$ are made invertible. But if $S \subseteq R^{\times}$, they already are and we don't have to modify $R$ to obtain $S^{-1}R$.
The intuition is that in $S^{-1}R$ is created by modifying $R$ such that elements of $S$ are made invertible. But if $S \subseteq R^{\times}$, they already are and we don't have to make any modifications to $R$, which is why there should be such an isomorphism.
\begin{enumerate}
\item The canonical morphism $\phi_S$ is given by $r \mapsto \frac{r}{1}$.
We give the inverse map as follows
......@@ -75,10 +105,10 @@ Let $\psi: R \to B$ be the canonical morphism (composition of $R \to A \to A/I$)
We draw a diagram for an overview:
\begin{center}
\begin{tikzcd}[ ]
A
A
\arrow[]{r}{\pi}
&
A/I
B
\arrow[]{d}{f}
\\
R
......@@ -89,28 +119,45 @@ Let $\psi: R \to B$ be the canonical morphism (composition of $R \to A \to A/I$)
T
\end{tikzcd}
\end{center}
where $\iota: R \to R[(X_s)_{s \in S}$ is the inclusion mapping and $\pi: A \to A/I$ the projection.
where $\iota: R \to R[(X_s)_{s \in S}]$ is the inclusion mapping and $\pi: A \to A/I$ the projection.
\begin{enumerate}
\item
Let $s \in R$.
We claim that $(f \circ \pi)(X_s)$ is the inverse of $(f \circ \psi)(s)$ in $T$.
Indeed one finds
\begin{align*}
&1 - \iota(s) X_S = 1 - s X_s = Y_s \in I \implies (\iota(s) X_s) I = 1 I \\
\implies &\psi(s) \pi(X_s) = 1 \in B \implies (f \circ \psi)(s) \cdot (f \circ \pi)(X_s) = 1 \in T
&1 - \iota(s) X_S = 1 - s X_s = Y_s \in I \implies (\iota(s) X_s) + I = 1 + I \\
\implies &\psi(s) \pi(X_s) = \pi(\iota(s) X_s) = 1 \in B \implies (f \circ \psi)(s) \cdot (f \circ \pi)(X_s) = 1 \in T
\end{align*}
\item Let $g: R \to T$ as above.
By the characteristic property of polynomial rings (Proposition \ref{prop:polynomial-ring}), $g$ uniquely determines a morphism of $R$-algebras
\item Let $g: R \to T$ with $g(S) \subseteq T^{\times}$.
By the characteristic property of polynomial rings (Proposition \ref{prop:polynomial-ring}), there exists a natural bijection
\begin{align*}
\Hom_{\Set}(S,T) \iso
\Hom_{R\Alg}(A,T)
\end{align*}
so $g$ uniquely determines a morphism of $R$-algebras
\begin{align*}
\tilde{g}: A \to T, \quad \text{given by} \quad \tilde{g}(X_{s_1}^{n_1} \dots X_{s_k}^{n_k}) = g(s_1)^{n_1} \dots g(s_k)^{n_k} \in T
\end{align*}
Now, we want to use the characteristic property of quotient rings (Proposition \ref{prop:quotient-ring}), but for this we first need to show that $\Ker \tilde{g} \subseteq I$.
Let
\item By (a), this map is well-defined.
By the characteristic property of quotient rings (Proposition \ref{prop:quotient-ring}), there also exists a natural bijection
\begin{align*}
\left\{
\text{ring morphism }f: A \to T \text{ with } I \subseteq \Ker f
\right\}
\iso
\Hom_{A\Alg}(A/I,T)
\end{align*}
so if we can show that $I \subseteq \Ker \tilde{g}$, there exists a ring morphism $f: A/I \to T$ such that the following diagram commutes:
\begin{center}
\begin{tikzcd}[ ]
A \arrow[]{r}{\tilde{g}} \arrow[]{d}{\pi}
& T
\\
B \arrow[]{ur}{f}
\end{tikzcd}
\end{center}
\item
\item Comparing the result of (c) with the description of $\Hom_{R\Alg}(S^{-1}R,-)$ in \ref{thm:localisation}, we see that they are isomorphic (in the sense of \ref{dfn:natural-transformation}).
By the Yoneda Lemma, (in particular Corollary \ref{cor:iso-with-hom}), the proof follows.
......
\newpage
\section{Exercises}
If you have any corrections or additions, please send me an email to \href{kimha@ethz.ch}{mailto:kimha@ethz.ch}.
\subsection{Sheet 1: Functors}
\input{./files/ex/e1}
\newpage
\subsection{Sheet 2: Localisation}
\input{./files/ex/e2}
\newpage
\subsection{Sheet 7: Transcendence and Integrality}
\input{./files/ex/e7}
......@@ -162,6 +162,8 @@
\newcounter{theo}[section]
\renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}
\newcounter{exer}[subsection]
\renewcommand{\theexer}{\arabic{subsection}.\arabic{exer}}
\newcommand{\defboxenv}[3]{
\newenvironment{#1}[1][]{
......@@ -203,5 +205,35 @@
\defboxenv{prop}{Proposition}{orange!30}
\defboxenv{cor}{Corollary}{blue!30}
\defboxenv{xmp}{Example}{green!50!blue!30!white}
\defboxenv{exr}{Exercise}{black!30}
\defboxenv{rem}{Remark}{red!20}
\newenvironment{exr}[1][]{
\refstepcounter{exer}
\ifstrempty{#1} { % no Title
\mdfsetup{
frametitle={
\tikz[baseline=(current bounding box.east), outer sep=0pt]
\node[anchor=east,rectangle,fill=black!30]
{\strut Exercise~\theexer};
}
}
}{ % else: with Title
\mdfsetup{
frametitle={
\tikz[baseline=(current bounding box.east),outer sep=0pt]
\node[anchor=east,rectangle,fill=black!30]
{\strut Exercise~\theexer:~#1};
}
}
} % either case
\mdfsetup{
linecolor=black!30,
innertopmargin=0pt,
linewidth=2pt,
topline=true,
frametitleaboveskip=\dimexpr-\ht\strutbox\relax,
}
\begin{mdframed}[]\relax
}{ % post env command
\end{mdframed}
}
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