Commit 11e366a1 by Han-Miru Kim

### comalg fix errors, exercises, tensor beginning

parent 8873a238
 ... ... @@ -172,6 +172,7 @@ Packages for sectioning and positioning. \DeclareMathOperator{\ev}{ev} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\kernel}{Ker} \DeclareMathOperator{\Ann}{Ann} \DeclareMathOperator{\Image}{Im} \DeclareMathOperator{\spn}{span} ... ...
 ... ... @@ -9,7 +9,7 @@ Another way to characterize this is to say that the projection mapping is a ring morphism. As remarked in the introduction, this means that $R/I$ is an $R$-algebra, where $\pi_I$ is the structure morphism. We will look at the category of $R$-modules and describe the hom-functor for $R/I$. We will look at the category of $R$-algebras and describe the hom-functor for $R/I$. \begin{prop}[] \label{prop:quotient-ring} For any $R$-algebra $S$, there is a natural'' bijection \begin{align*} ... ...
 ... ... @@ -9,8 +9,8 @@ Let $R$ be a ring. A subset $S \subseteq R$ is called \textbf{multiplicative}, i \begin{xmp}[] \begin{enumerate} \item For $R$ an integral domain, $S = R \setminus \{0\}$ is multiplicative. \item For a ring $R$ with $P \subseteq R$ a prime ideal, then $S = R \setminus P = P^{c}$ is multiplicative. \item For $R$ an integral domain, $S = R - \{0\}$ is multiplicative. \item For a ring $R$ with $P \subseteq R$ a prime ideal, then $S = R - P = P^{c}$ is multiplicative. \item Let $a \in R$, then $\{a^{n} \big\vert n \geq 0\}$ is multiplicative. \item Let $R_1 \stackrel{f}{\to}R_2$ to be a morphism of rings. \begin{itemize} ... ... @@ -50,7 +50,7 @@ such that for any $R$-algebra $A$, there exists a bijection \end{thm} \begin{xmp}[] For $R$ an integral domain, $S = R \setminus \{0\}$, the localisation $S^{-1}R$ is isomorphic to the fraction field $\text{Quot}(R)$. For $R$ an integral domain, $S = R - \{0\}$, the localisation $S^{-1}R$ is isomorphic to the fraction field $\text{Quot}(R)$. Given a morphism $f: R \to A$ with $f(r) \in A^{\times}$ if $r \neq 0$, we can extend $f$ to $\text{Quot}(R)$ by defining $f(\frac{s}{r}) = \frac{f(s)}{f(r)}$. ... ... @@ -118,7 +118,7 @@ $\phi_S$ is not always injective. \begin{xmp}[] Set $R = \Z$ \begin{itemize} \item For $S = \Z \setminus \{0\}$, $S^{-1}R = \Q$ \item For $S = \Z - \{0\}$, $S^{-1}R = \Q$ \item If $S = \Z - p\Z$, then $S^{-1}R = \{\frac{a}{b} \in \Q \big\vert p \not| b\} ... ... @@ -130,11 +130,11 @@$\phi_S$is not always injective. \textbf{Notation}: For$R$a ring \begin{enumerate} \item If$a \in R$, write$R_a = \{a^{n}\}^{-1}R$\item For$p \subseteq R$a prime ideal, write$R_p = (R \setminus p)^{-1}R$. \item For$p \subseteq R$a prime ideal, write$R_p = (R - p)^{-1}R$. \end{enumerate} \begin{prop}[ACL 2.2.7] \begin{prop}[ACL 2.2.7]\label{prop:prime-ideals-localisation} Let$S \subseteq R$multiplicative. The map$q \mapsto \phi_S^{-1}(q)$gives a bijection ... ... @@ -238,15 +238,15 @@ This set is called the \textbf{nilradical} of$R$. \end{proof} \begin{cor}[] Let$q \subseteq R$be a prime ideal and let$S = R \setminus q$. Let$q \subseteq R$be a prime ideal and let$S = R - q. Then there exists a bijection \begin{align*} \{ \text{prime ideals in }R_q\} \stackrel{\sim}{\to} \{\text{prime ideals }p \subseteq R \text{ with }p \cap S = \emptyset \} \end{align*} Note thatp \cap S = p \cap (R \setminus q)$so$p \cap S = \emptyset \iff p \subseteq q$Note that$p \cap S = p \cap (R - q)$so$p \cap S = \emptyset \iff p \subseteq q$\end{cor} So when we want to find out things about subsets of prime ideals$q$, we can just look at all prime ideals of$(R \setminus q)^{-1}R$So when we want to find out things about subsets of prime ideals$q$, we can just look at all prime ideals of$(R - q)^{-1}R$In particular, the maximal ideals in$R_q$correspond to the prime ideals$p \subseteq R$that are maximal under those that are contained in$q$. But the only such ideal is$q$itself, so$R_qhas the property of having a unique maximal ideal. ... ... @@ -296,7 +296,7 @@ You might wonder is local'' local rings. \end{align*} Since this mapping is surjective, we have that\faktor{R}{m} \iso \C$is a field$\impliesm$is maximal. For uniqueness, it is enough to show that if$a = \left(a_{n}\right)_{n \in \N} \in R \setminus m$, then$a$is invertible, because from this follows that any ideal not contained in$m$contains units and thus is equal to$R$. For uniqueness, it is enough to show that if$a = \left(a_{n}\right)_{n \in \N} \in R - m$, then$a$is invertible, because from this follows that any ideal not contained in$m$contains units and thus is equal to$R$. To see this, we can use the fact that if$f(0) \neq 0$, then$\frac{1}{f}$is holomorphic in some small region around$0$. ... ... @@ -306,7 +306,7 @@ You might wonder is local'' local rings. If we pick a point$a \in \R$and add an inverse to the element$(X -a) \in \R[X]$, then the localisation consists of rational functions that are defined everywhere, \emph{except possibly} at the point$a$. This is called \textbf{localisation away from}$a$, or localisation away from the ideal$I$generated by$(X-a)$. If we put an inverse to every element of$\R[X]$that is \emph{not} in$I$, i.e. set$S = \R[X] \setminus I$, we obtain the ring of rational functions that are defined \emph{at least} at the point$a$. If we put an inverse to every element of$\R[X]$that is \emph{not} in$I$, i.e. set$S = \R[X] - I$, we obtain the ring of rational functions that are defined \emph{at least} at the point$a$. We call this \textbf{localisation at$a$}, or localisation at the ideal$I$. \end{itemize} \end{rem} ... ...  ... ... @@ -34,18 +34,18 @@ Let$R$be a ring. Then \begin{rem}[] Let$R$be a ring and$I \subseteq Ran ideal. Then \begin{align*} R^{\times} = R \setminus I \iff R \text{ is local and } I = m_R R^{\times} = R - I \iff R \text{ is local and } I = m_R \end{align*} \end{rem} \begin{proof} \begin{itemize} \item[ \impliedby$''] If$R$is local, then$m\cap R^{\times} = \emptyset$, so$R^{\times} \subseteq R \setminus m$. Conversely, if$x \in R \setminus m$, then$Rx \subseteq R$is an ideal not contained in any maxmial ideal or$R$, so$Rx = R \implies x \in R^{\times}$. If$R$is local, then$m\cap R^{\times} = \emptyset$, so$R^{\times} \subseteq R - m$. Conversely, if$x \in R - m$, then$Rx \subseteq R$is an ideal not contained in any maxmial ideal or$R$, so$Rx = R \implies x \in R^{\times}$. \item[ $\implies$''] Since$R^{\times} = R \setminus I$, it's clear that$I$is maximal, because any ideal$J \nsupseteq I$contains an$x \in R \setminus I = R^{\times}$, so$J = R$. Moreover, any ideal$J \neq R$is contained in$I$, because$J \cap R^{\times} = \emptyset \implies J \subseteq R \setminus R^{\times} = I$, which shows that$R$is local and$I = m_R$. Since$R^{\times} = R - I$, it's clear that$I$is maximal, because any ideal$J \nsupseteq I$contains an$x \in R - I = R^{\times}$, so$J = R$. Moreover, any ideal$J \neq R$is contained in$I$, because$J \cap R^{\times} = \emptyset \implies J \subseteq R - R^{\times} = I$, which shows that$R$is local and$I = m_R$. \end{itemize} \end{proof} ... ...  \section{Noetherian Rings and Modules} \section{Noetherian Rings and Modules}\label{sec:noetherian} The goal of this chapter is to obtain a class of rings with some good finiteness properties that are motivated by vector spaces. In linear algebra, if$V$is some finite-dimensional vector space and$W \subseteq V$is a subspace then$W$is also finite-dimensional. ... ...  The tensor product is a way to linearize the concept of \emph{bilinear maps} of$R$-models, where we can view them as a \emph{linear} map from some other module. \begin{dfn}[] Let$M,N,P$be$R$-modules. An$R-\textbf{bilinear map} \begin{align*} b: M \times N \to P \end{align*} is a function such that it is linear in both arguments, i.e. \begin{itemize} \itemb(rm,n) = r b(m,n)$\item$b(m,rn) = r b(m,n)$\item$b(m + m',n) = b(m,n) + b(m',n)$\item$b(m,n+n') = b(m,n) + b(m,n')$. \end{itemize} for$r \in R, m,m' \in M, n,n' \in N$. \end{dfn} \begin{lem}[] The set$\Bil_R(M,N;P)$of bilinear maps$M \times N \to P$is an$R-module with \begin{align*} (rb + r'b')(m,n) := rb(m,n) + r'b'(m,n) \end{align*} given anR$-module$Q$and an$R$-linear map$P \stackrel{f}{\to}Q$, there is an$R-linear map \begin{align*} \Bil_R(M,N;P) \to \Bil_R(M,N,Q), \quad b \mapsto f \circ b \end{align*} In other words, there is a functor\Bil_R(M,N;-)$on the category of$R$-modules. \end{lem} \begin{xmp}[] \begin{itemize} \item For$M = N = P = R$, the multiplication map$\cdot: R \times R \to R$is$R$-bilinear. \item For$M = R$,$P = N$, the scalar multipliation$R \times N \to N$is$R$-bilinear. \item (Uncurrying) Let$f: M \to \Hom_R(N,P)$be an$R$-linear map. Since$f$returns'' a linear map, we can think of$f$being a higher-order'' map. By \emph{uncurrying}, we get a first-order''$R-bilinear map \begin{align*} b_f : M \times N \to P \quad b_f(m,n) = (f m)(n) \end{align*} In fact, any bilinear mapb: M \times N \to P$is of this form, as given such a$b, we can set \begin{align*} f : M \to \Hom_R(N,P), \quad m \mapsto b(m,-) \end{align*} and similarly by switching the roles ofM,N$. So there exist$R-linear isomorphisms \begin{align*} \Hom_R(M,\Hom_R(N,P)) \iso \Bil_R(M,N;P) \iso \Hom_R(N,\Hom_R(M,P)) \end{align*} \end{itemize} \end{xmp} Our goal is to view\Bil_R(M,N;P)$as$\Hom_R(T,P)$for some other$R$-module$T$in a natural way. The tensor product let us find out the what$T$should look like. Before we give the definition, we make a small detour to the category$\Set$\footnote{That is not part of the lecture, so you can skip until the definition, but I think it is very insightful.} \begin{rem}[] Let$X,Y,Zbe \emph{sets}. Given a higher-order function \begin{align*} f: X \to \Hom_{\Set}(Y,Z), \quad x \mapsto f_x \end{align*} we can uncurryfto obtain a first-order function \begin{align*} f: X \times Y \to Z \end{align*} which (since we are in\Set), induces an isomorphism \begin{align*} \Hom_{\Set}(X,\Hom_{\Set}(Y,Z)) \iso \Hom_{\Set}(X \times Y,Z) \end{align*} so the opposite'' of uncurrying a function is to curry a function from the productX \times Y$. If we write$Y^{X}$for$\Hom_{\Set}(X,Y), this isomorphism can be written in the suggestive form \begin{align*} (Z^{Y})^{X} \iso Z^{X \times Y} \end{align*} hinting at the power rule for real numbers ((a^{b})^{c} = a^{bc}$for$a > 0$. Now, the set$X \times Y$isn't the only set that let us uncurry higher-order functions$X \to (Y \to Z)$. We can, for example take the set$X \times Y \times \{0,1\}$, and define the curried function by ignoring the value of the point in$\{0,1\}$. Clearly, this solution is not optimal as the added degree of freedom removes the uniqueness of the uncurried function$X \times Y \to \{0,1\} \to Z$. If we return to the category of modules, we can no longer just take$X \times Y$, as the uncurried function is no longer linear, but bilinear. \end{rem} \begin{thm}[Whitney] Let$M,N$be$R$-modules. Then there exists an$R$-module$M \otimes_R N$($M$tensor$N$over$R) and a bilinear map \begin{align*} \beta: M \times N \to M \otimes_R N, \quad \beta \in \Bil_R(M,N;M \otimes_R N) \end{align*} which is \emph{universal} in the sense that for anyR$-module$P$and any$R$-bilinear map$b: M \times N \to P$, there exists a \emph{unique}$R$-linear map$f: M \otimes_R N \to P$such that the following diagram commutes: \begin{center} \begin{tikzcd}[ ] M \times N \arrow[]{r}{b} \arrow[]{d}{\beta} & P\\ M \otimes_R N \arrow[dashed]{ur}{!\exists f} \end{tikzcd} \end{center} The pair$(M \otimes_R N,\beta)$is unique in the sense that if$(T,\beta')$satisfies the same condition, there exists a unique$R$-linear morphism$T \stackrel{f}{\to} M \otimes_R Nsuch that the folllowing diagram commutes \begin{center} \begin{tikzcd}[column sep=0.8em] & M \times N \arrow[swap]{dl}{\beta'} \arrow[]{dr}{\beta}\\ T \arrow[dashed]{rr}{!\exists f} && M \otimes_R N \end{tikzcd} \end{center} \end{thm} In other words, there is a natural equivalence \begin{align*} \Phi: \Hom_R(M \otimes_R N,-) \to \Bil_R(M,N;-), \Phi_P = (f \mapsto f \circ \beta) \end{align*} \textbf{Notation:} We will write \begin{align*} \beta(m,n) = m \otimes n \quad \text{for} \quad (m,n) \in M \times N \end{align*} \begin{proof}[Proof Sketch] \textbf{Uniqueness:} This actually follows from the Yoneda Lemma (more precisely, Corollary \ref{cor:iso-with-hom}) since we have just described the Hom-functor\Hom_R(M \otimes_R N,-)$. We obtain a less abstract proof by just applying the trick'' we used in the proof of Yoneda's Lemma (Taking$P = T$and$\id_T \in h^{T}(T)$). Assume we have two modules with bilinear maps$(T,\beta), (T',\beta')$satisfying the universal properties. From this, we obtain$R$linear maps$a,b$such that the following diagram commutes: \begin{center} \begin{tikzcd}[ ] M \times N \arrow[swap]{d}{\beta} \arrow[]{dr}{\beta'} \arrow[]{r}{\beta} & T\\ T \arrow[]{r}{g} & T' \arrow[]{u}{f} \end{tikzcd} \end{center} Note that the morphism$(f \circ g)satisfies the universal property \begin{align*} (f \circ g) \circ \beta = f \circ (g \circ \beta) = f \circ \beta' = \beta \end{align*} But the identity\id_T$also satisfies this universal property. Since this map must be unique, it follows$(f \circ g) = \id_T$. Similarly, one can show$(g \circ f) = \id_{T'}$which proves$T \iso T'$. \textbf{Existence}: (The proof pattern we use here comes up quite regularly for other algebraic constructions, so it is good to have it in mind.) We construct$M \otimes N$as a quotient of a huge'' free module with a submodule that imposes the bilinearity by applying the$\Hom$properties of free modules aswell as those of quotient modules. Let$\tilde{T}$be the free$R$-module with basis induced by the elements of$M \times N$and write$[(m,n)]$for the basis vector corresponding to$(m,n)$. \end{proof}  \begin{thm}[Krull]\label{thm:krull-prime-ideal} Let$R$be a noetherian ring,$a \in R \setminus R^{\times}$. Let$R$be a noetherian ring,$a \in R - R^{\times}$. Then \begin{enumerate} \item there exists a minimal prime ideal$p \subseteq R$such that$a \in p$. \item For any such$p$, we have$ht(p) \leq 1$with equality if$a$is not a zero-divisor in$R$. \end{enumerate} \end{thm} For example, if$R = \Z$and$p \in \Z \setminus \{\pm 1\}$,$p = p\Z$for any prime divisor$p$of$a$. For example, if$R = \Z$and$p \in \Z - \{\pm 1\}$,$p = p\Z$for any prime divisor$p$of$a$. For (a), we just have to apply the following proposition: ... ... @@ -30,7 +30,7 @@ For example, if$I = \{0\}$, then there are minimal prime ideals in$R$. There i \end{proof} \begin{proof}[Proof of (b) in Theorem \ref{thm:krull-prime-ideal}] Let$a \in R \setminus R^{\times}$and$p$minimal with$a \in p$. Let$a \in R - R^{\times}$and$p$minimal with$a \in p$. Assume that there exists a$q \subseteq R$prime such that$q \subsetneq q$. (Otherwise,$ht(p) = 0$), so$ht(p) \geq 1$. Let$q_1 \subseteq q$prime. We need to show$q_1 = q$. ... ...  ... ... @@ -14,7 +14,7 @@ Let$(R,v)$be a valuation ring with$\Gamma_v \neq \{0\}$.$y = \underbrace{x}_{\in I} \cdot \underbrace{\frac{y}{x}}_{\in R} \in I$. So if$J \not\subseteq I$, then there exists a$y_0 \in J$such that$v(y_0) < v(x)$for every$x \in I$which means$I \subseteq J$. \item We have a chain of length$1$given by$\{0\} \subsetneq m_v$, so$\dim(R) \geq 1$. Let$p \neq \{0\} \subseteq R$be a prime ideal and let$x \in p \setminus \{0\}$and$\alpha = v(x) > 0$. Let$p \neq \{0\} \subseteq R$be a prime ideal and let$x \in p - \{0\}$and$\alpha = v(x) > 0$. Let$y \in m_v$. Then for all$n \geq 1we have \begin{align*} y^{n} = \underbrace{x}_{\in p} \cdot \underbrace{\frac{y^{n}}{x}}_{\in R?} ... ... @@ -49,7 +49,7 @@ The proof of this is given in the next section \ref{thm:krull-intersection}. \item[(a)\implies$(b):] Let$\pi$be a uniformizer and$\alpha = v(\pi)$. Let$I \subseteq R$be an ideal with$I \neq \{0\}, and \begin{align*} k = \min \{v(x) \big\vert x \in I \setminus \{0\}\} \subseteq \alpha \Z k = \min \{v(x) \big\vert x \in I - \{0\}\} \subseteq \alpha \Z \end{align*} We claim thatI = \pi^{k}R$(and in particular a principal ideal). ... ... @@ -73,12 +73,12 @@ The proof of this is given in the next section \ref{thm:krull-intersection}. By induction,$I$is principal. \item[(b)$\implies$(a):] Since$R$is a PID, the maximal ideal$m$is principal. Let$\pi$be a generator of$m$. Let$x \in R \setminus \{0\}$. Let$x \in R - \{0\}. By Krull's Intersection Theorem \begin{align*} \bigcap_{k \geq 1}\pi^{k}R = \{0\} \end{align*} So for eachx \in R \setminus \{0\}$, there exists a unique$k \geq 0$such that$x \in \pi^{k}R$but not$x \in \pi^{k+1}R$. So for each$x \in R - \{0\}$, there exists a unique$k \geq 0$such that$x \in \pi^{k}R$but not$x \in \pi^{k+1}R$. For each$x$, let$\tau(x)$denote this number (with$\tau(0) = \infty$) Then \begin{enumerate}[{(}i{)}] ... ...  \begin{exr}[] \begin{exr}[]\label{exr:2-1} 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$\Z_{p\Z} = (Z - p\Z)^{-1}\Z$and$k = R/m_R$. Denote by$\iota$be inclusion mapping and by$\pithe projection mapping \begin{align*} \iota: \Z \hookrightarrow R, \quad a \mapsto \frac{a}{1}\\ ... ... @@ -10,7 +10,7 @@ Denote by\iota$be inclusion mapping and by$\pithe projection mapping \end{align*} First, we find the maximal idealm_R$. 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$m_R = \iota(p\Z)$. Since any ideal containing a unit must equal the whole ring and every element$a \in \Z - p\Z$is invertible (in$\Z_{p\Z}$when identified with$\iota(a) = \tfrac{a}{1}$), we claim that$m_R = \iota(p\Z). It's easy to see that this is a maximal ideal. The isomorphism is given by \begin{align*} ... ... @@ -27,11 +27,11 @@ To show injectvivity, leta,b \in \Z. Then (\Phi \circ \pi)(a) - (\Phi \circ \pi)(b) = 0 \iff \tfrac{a - b}{1} \in \iota(p\Z) \iff \exists c \in p\Z:\ \tfrac{a - b}{1} = \tfrac{c}{1} \\ \iff \exists t \in \Z \setminus p\Z: \quad t(a - b - c) \in p\Z \iff a - b \in p\Z \iff \exists t \in \Z - p\Z: \quad t(a - b - c) \in p\Z \iff a - b \in p\Z \end{align*} For surjectivity, letx + \iota(p\Z) \in k$. Then$xcan be written as \begin{align*} x = \frac{a}{b} \quad \text{for some} \quad a \in \Z, p\not| a, b \in (\Z \setminus p\Z) x = \frac{a}{b} \quad \text{for some} \quad a \in \Z, p\not| a, b \in (\Z - p\Z) \end{align*} With the prime number decomposition ofa = p_1^{s_1} p_2^{s_2} \cdots p_m^{s_m}$, we see that ... ... @@ -131,15 +131,16 @@ Let$\psi: R \to B$be the canonical morphism (composition of$R \to A \to A/I) \end{align*} \item Letg: R \to T$with$g(S) \subseteq T^{\times}$. By the characteristic property of polynomial rings (Proposition \ref{prop:polynomial-ring}), there exists a natural bijection By the characteristic property of polynomial rings (Proposition \ref{prop:polynomial-ring}), there exists a bijection\footnote{There also exists a \emph{natural} bijection, but we are \emph{not} using that one, as we define$\tilde{g}differently.} \begin{align*} \Hom_{\Set}(S,T) \iso \Hom_{R\Alg}(A,T) \end{align*} sog$uniquely determines a morphism of$R$-algebras so we can extend$g$to a morphism of$R-algebras \begin{align*} \tilde{g}: A \to T, \quad \text{given by} \quad \tilde{g}(X_{s_1}^{n_1} \dots X_{s_k}^{n_k}) = g(s_1)^{n_1} \dots g(s_k)^{n_k} \in T \tilde{g}: A \to T, \quad \text{given by} \quad \tilde{g}(X_{s_1}^{n_1} \dots X_{s_k}^{n_k}) = g(s_1)^{-n_1} \dots g(s_k)^{-n_k} \in T \end{align*} Note that\tilde{g}(Y_s) = 1 - g(s) g(s)^{-1} = 0$, so$I \subseteq \Ker \tilde{g}. By the characteristic property of quotient rings (Proposition \ref{prop:quotient-ring}), there also exists a natural bijection \begin{align*} \left\{ ... ... @@ -148,7 +149,7 @@ Let\psi: R \to B$be the canonical morphism (composition of$R \to A \to A/I) \iso \Hom_{A\Alg}(A/I,T) \end{align*} so if we can show thatI \subseteq \Ker \tilde{g}$, there exists a ring morphism$f: A/I \to T$such that the following diagram commutes: therefore there exists an$A$-algebra morhpism i.e.\ 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} ... ... @@ -157,8 +158,18 @@ Let$\psi: R \to B$be the canonical morphism (composition of$R \to A \to A/I$) 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 construction of$\tilde{g}$we have$\tilde{g} \circ \iota = g$, so$fsatisfies \begin{align*} f \circ \psi = f \circ \pi \circ \iota = \tilde{g} \circ \iota = g \end{align*} \item Part (a) shows that the mapf \mapsto f \circ \psi$is well-defined. Part (b) shows surjectivity. For injectivity, let$f,f' \in \Hom_{R\Alg}(B,T)$with$f \circ \psi = f' \circ \psi$. This means that for all$a \in R\begin{align*} f([a]) = f'([a]) \implies f = f' \end{align*} \item Comparing the Hom-functor\Hom_{R\Alg}(B,-)$in the result of (c) with the description of$\Hom_{R\Alg}(S^{-1}R,-)$in Theorem \ref{thm:localisation}, we see that they are isomorphic (in the sense of Definition \ref{dfn:natural-transformation}). By the Yoneda Lemma, (in particular Corollary \ref{cor:iso-with-hom}), the proof follows. ... ... @@ -193,26 +204,45 @@ Let$\psi: R \to B$be the canonical morphism (composition of$R \to A \to A/I$) To show that$m_A$is the unique maximal ideal, we show that any ideal not contained in$m_A$equals the whole ring$A$. We do this by letting$f = \sum_{n = 0}^{\infty}a_n X^{n}\in A \setminus m_A$and showing that$f$is invertible. Let$f = \sum_{n = 0}^{\infty}a_n X^{n}\in A - m_A$. We show that$f$is invertible. Since$a_0 \neq 0, we can define its inverse by \begin{align*} g = \sum_{n = 0}^{\infty} b_n X^{n} \quad \text{where} \quad b_0 = a_0^{-1}, \quad \text{and} \quad b_n = \sum_{i=0}^{n-1}b_i a_{i-k} \end{align*} Which shows(f) = A$. \item \item We claim that the inverse is$(1 + X)^{-1} = \sum_{n=0}^{\infty}X^{n}$. By the definition of the Cauchy-product, one finds \item Let$I \subseteq A$be a non-zero ideal. For each$f = \sum_{n=0}^{\infty}a_n X^{n} \in I, set \begin{align*} k_f := \min \left\{ n \geq 0 \big\vert a_n \neq 0 \right\} \end{align*} Then pickp \in I$such that$k_p = \min \{k_f \big\vert f \in I\} =: k_{\min}$. (The minimum exists, as the set in question is bounded from below). So we can write$p = X^{k_{\min}}q$for some$q \in A - m_A$. We claim that$I = (X^{k_{\min}})$. Since$q \in A^{\times}$, we immediately have$I \supseteq (X^{k})$. For $\subseteq$'', let$f = \sum_{n=0}^{\infty}a_nX^{n} \in I$. Clearly$k_f \geq k$and$a_n = 0$for all$n \leq k_{\min}$. So we can factor out$X^{k_{\min}}$and write$f = X^{k_{\min}}t$for some$t \in A$. This shows$I \subseteq (X^{k_{\min}}). \item By the definition of the Cauchy-product, one finds \begin{align*} (1 + X)(1 - X + X^{2} - \ldots + (-1)^{n}X^{n}) = 1 + (-1)^{n}X^{n} \end{align*} It's easy to see that \begin{align*} (1 + X) \cdot \sum_{n=0}^{\infty}X^{n} = \sum_{n=0}^{\infty}c_n X^{n} \quad \text{for} \quad c_n = (1 + X)^{-1} = \sum_{n=0}^{\infty}(-1)^{n}X^{n} \end{align*} \end{enumerate} \end{proof} \begin{exr}[] \begin{exr}[]\label{exr:nilpotent-polynomials} LetA$be a ring and consider the polynomial ring$A[X]$. Let$f = \sum_{i=0}^{n}a_i X^{i} \in A[X]$be a polynomial. Prove that: ... ... @@ -230,11 +260,9 @@ Let$\psi: R \to B$be the canonical morphism (composition of$R \to A \to A/I$) \begin{proof}[Solution] \begin{enumerate} \item Let$f$as above. \begin{itemize} \item[$\impliedby$] \item[$\implies$] \end{itemize} \item This is fairly standard. \item Since the set of nilpotent elements forms an ideal, if$a_{0}, \ldots, a_{n}$are nilpotent, then so is$f$. For $\implies$'': Missing \end{enumerate} \end{proof} ... ... @@ -256,8 +284,4 @@ Prove that$S$is a multiplicative subset of$A[X], and construct an isomorphis \end{align*} show well defined-ness blablabla. \end{proof}  \begin{exr}[] LetR$be a ring and$I \subseteq R$the nilradical of$R$. \begin{enumerate} \item Show that if$R$is noetherian, there exists an integer$k \geq 0$such that$x^{k} =0$for all$x \in I$. \item Show that this is not true for all rings. \end{enumerate} \end{exr} \begin{proof}[Solution] \begin{enumerate} \item Since$R$is noetherian,$I$is finitely generated. Let$x_{1}, \ldots, x_{n}$be generators of$I$. As$x_i \in I$, there exist$k_i \in \N$such that$x_i^{k_i} = 0$. Set$\ell := \max_i k_i$. Let$x = \sum_{i=1}^{n}a_ix_i \in I$for$a_i \in R$. Then for$k = m \ellwe have \begin{align*} x^{k} \in (x_1^{l},\ldots,x_n^{l}) = 0 \end{align*} \item The counter example will have to be a non-noetherian ring. The example we saw in chapter \ref{sec:noetherian} wasR = k[(X_n)_{n \in \N}]$, as the ideal$(X_{1}, X_2,\ldots)is not finitely generated. But this ideal is not the nil-radical so it won't work. If we want the indeterminates to be nilpotent, we can set them to be equal to zero for some power by taking the quotient \begin{align*} R = \faktor{k[(X_n)_{n \geq 1}]}{(X_1,X_2^{2},X_3^{3},\ldots)} \end{align*} soI = (X_1,X_2,\ldots)$is the nilradical and it is not true that$I^{n} = 0$for any$n \geq 1$. \end{enumerate} \end{proof} \begin{exr}[] Let$R$be a noetherian ring and$S \subseteq R$a multiplicative set. Show that$S^{-1}R$is noetherian. Give a non-trivial example ($S^{-1}R \neq \{0\})$) where the converse does not hold. \end{exr} \begin{proof} Let$R$be noetherian and$S \subseteq Rmultiplicative and let \begin{align*} \iota: R \hookrightarrow S^{-1}R, \quad r \mapsto \tfrac{r}{1} \end{align*} denote the canonical inclusion. Assume we have an ascending chain of ideals inS^{-1}R\begin{align*} I_1 \subseteq I_2 \subseteq \ldots \end{align*} and consider the sets \begin{align*} J_i = \left\{ r \in R \big\vert \iota(r) \in I_i \right\} \end{align*} We claim that this induces an ascending chain of ideals inR\begin{align*} J_1 \subseteq J_2 \subseteq \ldots \end{align*} Indeed, leta,b \in J_i. Then \begin{align*} \iota(a + b) = \tfrac{a+b}{1} = \tfrac{a}{1} + \tfrac{b}{1} = \iota(a + b) \in I_i \implies a + b \in J_i \end{align*} and for allx \in Rwe have \begin{align*} \iota(xa) = \tfrac{xa}{1} = \tfrac{x}{1} \tfrac{a}{1} \in I_i \implies xa \in J_i \end{align*} Which proves the claim. AsR$is noetherian, the sequence$J_1 \subseteq J_2 \subseteq \ldots$must stabilize. This means that for some$n_0 \in \N$, we have that for all$n \geq n_0: \begin{align*} \forall a \in R: \iota(a) \in I_{n} \implies \iota(a) \in I_{n_0} \end{align*} AsI_{n_0}$is an ideal, for all$\tfrac{a}{b} \in I_{n}it holds \begin{align*} \tfrac{a}{b} = \underbrace{\tfrac{1}{b}}_{\in S^{-1}R} \cdot \underbrace{\iota(a)}_{\in I_{n_0}} \in I_{n_0} \end{align*} which showsI_n = I_{n_0}$, so the sequence also stabilizes. Therefore$S^{-1}R$is noetherian. For the converse example, we start with a non-noetherian ring$R$and try to chose$S$, such that$S^{-1}R$becomes noetherian. We take$R = k[X_1,X_2,\ldots]\$ and set