@@ -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 $\{a^{n}\big\vert n \geq0\}$ 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 $r \neq0$, 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 $R_p =(R - p)^{-1}R$.

@@ -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 $f(0)\neq0$, 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 $I$.

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 $I = m_R$.

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

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

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}\\

\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 $(m,n)$.

@@ -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 $\dim(R)\geq1$.

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 $p$ 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

\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.