Commit 11e366a1 authored by Han-Miru Kim's avatar Han-Miru Kim
Browse files

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 that $p \cap S = p \cap (R \setminus q)$ 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 $R_q$ has 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 $\implies$ $m$ 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 $I \subseteq R$ an 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}
\item $b(rm,n) = r b(m,n)$
\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 an $R$-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 map $b: M \times N \to P$ 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 of $M,N$.
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 $X,Y,Z$ be \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 uncurry $f$ to 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 product $X \times Y$.
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 any $R$-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 $T \stackrel{f}{\to} M \otimes_R N$ such 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 $n \geq 1$ we 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 that $I = \pi^{k}R$ (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 each $x \in R \setminus \{0\}$, 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 $\pi$ the 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 $\pi$ the projection mapping
\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 $m_R = \iota(p\Z)$.
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, let $a,b \in \Z$. 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, 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)
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 of $a = p_1^{s_1} p_2^{s_2} \cdots p_m^{s_m}$, 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 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
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*}
so $g$ 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 that $I \subseteq \Ker \tilde{g}$, 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 $f$ satisfies
\begin{align*}
f \circ \psi = f \circ \pi \circ \iota = \tilde{g} \circ \iota = g
\end{align*}
\item Part (a) shows that the map $f \mapsto f \circ \psi$ 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 pick $p \in I$ 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}
Let $A$ 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}[]
Let $R$ 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 $k = m \ell$ we 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} was $R = k[(X_n)_{n \in \N}]$, 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*}
so $I = (X_1,X_2,\ldots)$ 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 $S \subseteq R$ multiplicative 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 in $S^{-1}R$
\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 in $R$
\begin{align*}
J_1 \subseteq J_2 \subseteq \ldots
\end{align*}
Indeed, let $a,b \in J_i$. 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 all $x \in R$ we 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.
As $R$ 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*}
As $I_{n_0}$ 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 shows $I_n = I_{n_0}$, 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