Commit 504490b9 by Han-Miru Kim

### 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 \mdfsetup{ linecolor=black!30, innertopmargin=0pt, linewidth=2pt, topline=true, frametitleaboveskip=\dimexpr-\ht\strutbox\relax, } \begin{mdframed}[]\relax }{ % post env command \end{mdframed} } #+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 manyx \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$F: \textsf{C} \to \textsf{fd-Vec}defined by \begin{align*} F(n) = \C^{n}, \quad F(A) = (x \mapsto Ax) \end{align*} is a functor. \item Show thatF$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$A \in \Hom(m,n), we have that \begin{align*} F(A) = (x \mapsto Ax): \C^{m} \to \C^{n} \end{align*} is well defined. Now, letA \in \Hom(m,n)$and$B \in \Hom(n,l). Then \begin{align*} F(B \circ A) = (x \mapsto BAx) = F(B) \circ (x \mapsto Ax) = F(B) \circ F(A) \end{align*} \itemFfully faithful as, by definition: \begin{align*} \Hom(m,n) = \Mat(n \times m,\C) = \Hom_{\textsf{C}}(\C^{m},\C^{n}) \end{align*} It is also essentially surjective, because ifV$is a$\C$-vector space of dimension$n < \infty, then \begin{align*} V \iso \C^{n} = F(n) \end{align*} \item First let's defineG$. 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$Gf \in \Hom_{\textsf{C}}(m,n)as follows: \begin{align*} Gf(f) = A = (a_{ij})_{i,j} \quad \text{where} \quad f(v_i) = \sum_{j=1}^{m}a_{ij} w_j \end{align*} Then define \begin{align*} \Phi: F \circ G \to \id_{\textsf{fd-Vec}} \quad \text{and} \quad \Psi: G \circ F \to \id_{\textsf{C}} \end{align*} given by \begin{align*} \Phi_V : \C^{n} \to V, \quad e_i \mapsto v_i\\ \Psi_n = \id_n: n \to n \end{align*} Check for yourself that the following diagrams commute: \begin{center} \begin{tikzcd}[ ] \C^{n} \arrow[]{d}{\Phi_V} \arrow[]{r}{(F \circ G) f} & \C^{m} \arrow[]{d}{} \\ V \arrow[]{r}{f} & W \end{tikzcd} \quad \begin{tikzcd}[ ] n \arrow[]{r}{A} \arrow[]{d}{\Psi_n} & m \arrow[]{d}{\Psi_m} \\ n \arrow[]{r}{A} & m \end{tikzcd} \end{center} \end{enumerate} \end{proof}  ... ... @@ -2,9 +2,39 @@ Letp$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$\pithe projection mapping \begin{align*} \iota: \Z \hookrightarrow R, \quad a \mapsto \frac{a}{1}\\ \pi: \Z \mapsto \Z/p\Z, \quad a \mapsto a \mod p \end{align*} First, we find the maximal idealm_R$. 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$m_R = \iota(p\Z). It's easy to see that this is a maximal ideal. The isomorphism is given by \begin{align*} \Phi: \Z/p\Z \to \Z_{p\Z}/\iota(p\Z), \quad a + p\Z \mapsto \tfrac{a}{1} + \iota(p\Z) \end{align*} Now we check that it is well-defined. Ifa-b \in p\Z$, 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$a,b \in \Z. Then \begin{align*} (\Phi \circ \pi)(a) - (\Phi \circ \pi)(b) = 0 \iff \tfrac{a - b}{1} \in \iota(p\Z) \iff \exists c \in p\Z:\ \tfrac{a - b}{1} = \tfrac{c}{1} \\ \iff \exists t \in \Z \setminus p\Z: \quad t(a - b - c) \in p\Z \iff a - b \in p\Z \end{align*} For surjectivity, letx + \iota(p\Z) \in k$. Then$xcan be written as \begin{align*} x = \frac{a}{b} \quad \text{for some} \quad a \in \Z, p\not| a, b \in (\Z \setminus p\Z) \end{align*} With the prime number decomposition ofa = p_1^{s_1} p_2^{s_2} \cdots p_m^{s_m}$, 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$T. Indeed one finds \begin{align*} &1 - \iota(s) X_S = 1 - s X_s = Y_s \in I \implies (\iota(s) X_s) I = 1 I \\ \implies &\psi(s) \pi(X_s) = 1 \in B \implies (f \circ \psi)(s) \cdot (f \circ \pi)(X_s) = 1 \in T &1 - \iota(s) X_S = 1 - s X_s = Y_s \in I \implies (\iota(s) X_s) + I = 1 + I \\ \implies &\psi(s) \pi(X_s) = \pi(\iota(s) X_s) = 1 \in B \implies (f \circ \psi)(s) \cdot (f \circ \pi)(X_s) = 1 \in T \end{align*} \item Letg: R \to T$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$g(S) \subseteq T^{\times}. By the characteristic property of polynomial rings (Proposition \ref{prop:polynomial-ring}), there exists a natural bijection \begin{align*} \Hom_{\Set}(S,T) \iso \Hom_{R\Alg}(A,T) \end{align*} sog$uniquely determines a morphism of$R-algebras \begin{align*} \tilde{g}: A \to T, \quad \text{given by} \quad \tilde{g}(X_{s_1}^{n_1} \dots X_{s_k}^{n_k}) = g(s_1)^{n_1} \dots g(s_k)^{n_k} \in T \end{align*} Now, we want to use the characteristic property of quotient rings (Proposition \ref{prop:quotient-ring}), but for this we first need to show that\Ker \tilde{g} \subseteq I. Let \item By (a), this map is well-defined. By the characteristic property of quotient rings (Proposition \ref{prop:quotient-ring}), there also exists a natural bijection \begin{align*} \left\{ \text{ring morphism }f: A \to T \text{ with } I \subseteq \Ker f \right\} \iso \Hom_{A\Alg}(A/I,T) \end{align*} so if we can show thatI \subseteq \Ker \tilde{g}$, 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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