### com alg w03

parent 9c06bb54
 ... ... @@ -189,6 +189,7 @@ Packages for sectioning and positioning. \DeclareMathOperator{\ggT}{ggT} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Alg}{-Alg} % Cat \newcommand*{\Top}{\text{\sffamily Top}} ... ... @@ -317,6 +318,7 @@ Setup numbering system and define a function that creates environment. \defboxenv{edfn}{(Definition)}{orange!20} \defboxenv{thm}{Theorem}{blue!40} \defboxenv{lem}{Lemma}{blue!20} \defboxenv{prop}{Proposition}{orange!30} \defboxenv{cor}{Corollary}{blue!30} \defboxenv{xmp}{Example}{green!50!blue!30!white} \defboxenv{exr}{Exercise}{black!30} ... ...
 ... ... @@ -178,7 +178,7 @@ The Yoneda Lemma is a first non-trivial result about categories. The Hom-functor $h_X = \Hom_{\textsf{C}}(-,X)$ \emph{characterizes} an object $X$ up to a canonical'' isomorphism. \begin{dfn}[] \begin{dfn}[]\label{dfn:natural-transformation} Let $F$ and $G$ be two functors $\textsf{C} \to \textsf{D}$. A \textbf{natural transformation} $\Phi: F \to G$ is a rule that to every object $X$ of $\textsf{C}$ assigns a morphism $\Phi(X) : FX \to GX$ such that for any $X \stackrel{f}{\to}Y$, the following diagram (in $\textsf{D}$) commutes: ... ... @@ -198,39 +198,72 @@ A \textbf{natural transformation} $\Phi: F \to G$ is a rule that to every object \end{tikzcd} \end{center} $\Phi$ is called a \textbf{(natural) isomorphism}, if $\Phi(X)$ is an isormophism for all $X$. If that is the case, then $\Psi(X) = \Phi(X)^{-1}$ defines a natural transformation $\Psi: G \to F$. We write $\Nat(F,G)$ for the set\footnote{Since Categories can be large, it is not clear why this would be a set. This however follows from Yoneda's Lemma.} of natural transformations from $F$ to $G$. We write $\Nat(F,G)$ for the set of natural transformations from $F$ to $G$. \end{dfn} The morphism $\Phi(X)$ is called the \textbf{component} of $\Phi$ at $X$ and we will sometimes write $\Phi_X$ instead. \begin{xmp}[] Fix a morphism $X \stackrel{f}{\to}Y$ and consider the (contravariant) Hom-functors $h_X = \Hom(-,X),h_Y = \Hom(-,Y): \textsf{C} \to \Set$. For all $A$ in $\textsf{C}$, we can define a mapping Take $\textsf{C} = \textsf{D} = \Vec_{K}$ for some field $K$. Let $F = \id_{\textsf{C}}$ be the identity functor and $G$ the bidual functor \begin{align*} \Phi_f(A) = (f \circ -) = (a \mapsto f \circ a): h_X(A) \to h_Y(A) GV = (V')' = \Hom(\Hom(V,K),K), Gf = (f')' \end{align*} \begin{center} \begin{tikzcd}[ ] A \arrow[]{r}{a} We know from linear algebra that for any vector space $V$, there exists a canonical map \begin{align*} V \to (V')', \quad v \mapsto \left( \alpha \mapsto \alpha(v) \right) \quad \text{for} \quad \alpha \in V', v \in V \end{align*} This defines a natural transformation $\Phi: \id_{\textsf{C}} \to G$. Check for yourself that the diagram commutes. \end{xmp} Recall the (contravariant) Hom-functors $h_X = \Hom(-,X),h_Y = \Hom(-,Y): \textsf{C} \to \Set$. \begin{lem}[The Yoneda Lemma] Let $X,Y$ be objects of $\textsf{C}$. There is a bijection \begin{align*} \Hom_{\textsf{C}}(X,Y) \stackrel{\sim}{\to} \Nat(h_X,h_Y) \end{align*} \end{lem} \begin{proof} \begin{itemize} \item Let $X \stackrel{f}{\to}Y \in \Hom(X,Y)$. We will use this to define a natural transformation. For all $A$ in $\textsf{C}$, we can define a mapping \begin{align*} \Phi_f(A) = (f \circ -) = (a \mapsto f \circ a): h_X(A) \to h_Y(A) \end{align*} \begin{center} \begin{tikzcd}[ ] A \arrow[]{r}{a} & X \arrow[]{r}{f} & Y \end{tikzcd} \end{center} we can check that $\Phi_f$ is a natural transformation $\Phi_f: h_X \to h_Y$: \end{tikzcd} \end{center} we can check that $\Phi_f$ is a natural transformation $\Phi_f: h_X \to h_Y$: Indeed, let $A \stackrel{g}{\to}B$ be a morphism in $\textsf{C}$, and consider the diagram \begin{center} \begin{tikzcd}[column sep=4em] h_X(A) \arrow[swap]{d}{\Phi_f(A) = (f \circ -)} Indeed, let $A \stackrel{g}{\to}B$ be a morphism in $\textsf{C}$, and consider the diagram \begin{center} \begin{tikzcd}[column sep=4em] h_X(A) \arrow[swap]{d}{\Phi_f(A) = (f \circ -)} & h_X(B) \arrow[swap]{l}{h_X(g) = (- \circ g)} ... ... @@ -240,24 +273,22 @@ The morphism $\Phi(X)$ is called the \textbf{component} of $\Phi$ at $X$ and we & h_Y(B) \arrow[]{l}{h_Y(g) = (- \circ g)} \end{tikzcd} \end{center} Starting from the top-right with a morphism $B \stackrel{h}{\to} X \in h_X(B)$, the left-down path gives us \begin{align*} (B \stackrel{h}{\to}X) \mapsto (A \stackrel{h \circ g}{\to} X) \mapsto (A \stackrel{f \circ (h \circ g)}{\to} Y) \end{align*} and the down-left path gives us \begin{align*} (B \stackrel{h}{\to}X) \mapsto (B \stackrel{f \circ h}{\to} Y) \mapsto (A \stackrel{(f \circ h) \circ g}{\to} Y) \end{align*} which checks out. \end{tikzcd} \end{center} Starting from the top-right with a morphism $B \stackrel{h}{\to} X \in h_X(B)$, the left-down path gives us \begin{align*} (B \stackrel{h}{\to}X) \mapsto (A \stackrel{h \circ g}{\to} X) \mapsto (A \stackrel{f \circ (h \circ g)}{\to} Y) \end{align*} and the down-left path gives us \begin{align*} (B \stackrel{h}{\to}X) \mapsto (B \stackrel{f \circ h}{\to} Y) \mapsto (A \stackrel{(f \circ h) \circ g}{\to} Y) \end{align*} which checks out. What is more is that every element $(X \stackrel{f}{\to}Y) \in h_Y(X)$ uniquely determines a natural transformation. \end{xmp} \item For the other way, let $\Phi: h_X \to h_Y$ be a natural transformation. We show that it is of the form $\Phi_f$ for some $f \in \Hom(X,Y)$. \begin{proof} Let $\Phi: h_X \to h_Y$ be a natural transformation. So for every $A \stackrel{g}{\to}B$, we get a commutative diagram Since $\Phi$ is given, for every $A \stackrel{g}{\to}B$, we get a commutative diagram \begin{center} \begin{tikzcd}[ ] h_X(A) ... ... @@ -283,21 +314,28 @@ The morphism $\Phi(X)$ is called the \textbf{component} of $\Phi$ at $X$ and we \id_X \mapsto \Phi_X(\id_X) \mapsto (\Phi_X(\id_X) \circ g) \end{align*} If we chose any $f \in h_Y(X)$ and set $\Phi_X(\id_X)$ equal to it, then commutativity of the diagram reads But if we chose our $f \in h_Y(X)$ to be $f := \Phi_X(\id_X)$, then commutativity just reads \begin{align*} \Phi_A(g) \stackrel{!}{=} (f \circ g) \Phi_A(g) \stackrel{!}{=} (\underbrace{\Phi_X(\id_X)}_{=: f} \circ g) = (f \circ g) = \Phi_f(A)(g) \end{align*} so the component $\Phi_A$ is determined by the choice of $f$, \emph{for all objects} $A$. for all objects $A$. Another way to view this is that the component $\Phi_A$ is determined by the element $\Phi_X(\id_X) \in h_Y(X)$, \emph{for all objects} $A$. \end{itemize} \end{proof} %The Yoneda Lemma generalizes this fact, where instead of talking about Hom-functors $h_Y$, this generalizes to arbitrary Set-valued functors. %In particular, we will see that there exists a bijection between the sets %\begin{align*} %h_Y(X) \iso \Nat(h_X,h_Y) %\quad \text{where} \quad \Nat(h_X,h_Y) = \{\Phi: h_Y \to h_Y \text{ is a natural transformation}\} %\end{align*} In the proof, we didn't need any special properties of $h_Y$, so the proof generalizes for any other set-valued functors $F$, i.e. \begin{align*} \Hom(X,Y) = h_Y(X) \iso \Nat(h_X,h_Y) \text{ generalises to } F(X) \iso \Nat(h_X,F) \end{align*} \begin{cor}[] \begin{cor}[] \label{cor:iso-with-hom} The functors $h_X$ and $h_Y$ are isomorphic if and only $X$ and $Y$ are isomorphic. \end{cor} What this means is that to define an object $X$, it is enough to describe its hom-functor. \begin{exr}[] The Yoneda Lemma is also true for the covariant functor $h^{X}$. \end{exr}
 \section{Constructions of rings} \subsection{Quotient rings [ACL 1.5]} Recall that if $R$ is a ring and $I \subseteq R$ an ideal, then the abelian group $R/I$ is a ring with multiplication \begin{align*} (r + I) \cdot (s + I) = rs + I \end{align*} Another way to characterize this is to say that the projection mapping \begin{align*} \pi_I : R \to R/I, \quad r \mapsto r +I \end{align*} 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 $R/I$. \begin{prop}[] \label{prop:quotient-ring} 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\} \\ g &\mapsto g \circ \pi_I \end{align*} \begin{center} \begin{tikzcd}[ ] \faktor{R}{I} \arrow[]{r}{g} & S \\ R \arrow[]{ur}{g \circ \pi_I} \arrow[]{u}{\pi_I} \end{tikzcd} \end{center} \end{prop} We can understand the ideal of the quotient ring with ideals from the original ring. \begin{prop}[(ACL 1.5.4, 1.5.5)] \label{prop:ideals-and-quotients} Thre are reciprocal bijections \begin{align*} \left\{\text{ideals of } R/I\right\} &\leftrightarrow \left\{ \text{ideals of $R$ containing $I$} \right\}\\ J &\mapsto \pi_I^{-1}(J)\\ \pi_I(J) &\mapsfrom J \end{align*} which preserve inclusion and the sets of prime ideals and maximal ideals. \end{prop} \begin{proof} If $J \subseteq R$ corresponds to $\tilde{J} = \pi_I(J)$, then we get a surjection \begin{align*} R \stackrel{\pi_I}{\to} \faktor{R}{I} \to \faktor{(R/I)}{\tilde{J}} \end{align*} with kernel $\pi_I^{-1}(\tilde{J}) = J$, so this induces an isomorphism \begin{align*} R/J \stackrel{\sim}{\to} \faktor{(R/I)}{\tilde{J}} \end{align*} \end{proof} If $S$ is an $R$-algebra and $H$ is an $S$-algebra, then $H$ is an $R$-algebra aswell. In particular, modules over $R/I$ are also $R$-modules by \begin{align*} r \cdot m = \underbrace{\pi_I(r)}_{\in R/I} m \end{align*} Conversely, an $R$-module $M$ can be viewed as an $R/I$ module if (and only if) $i \cdot m = 0_M$ for all $i \in I, m \in M$. Check that defining the multiplication as \begin{align*} (r + I) \cdot m := r \cdot m \end{align*} is well-defined. Of course, not every $R$-module satisfies $i \cdot m = 0$, but there is a way to construct a ring that does. For any $R$-module $M$, let $IM$ be the submodule generated by $im$, $i \in I, m \in M$. Then $M/IM$ is naturally an $R/I$ module. \subsection{Polynomial rings [ACL 1.3.5]} In constrast to last years lectures Algebra I \& II, we will construct polynomial rings in potentially infinite variables. Let $R$ be a base ring and let $I$ be a (possibly infinite) set with a family of variables $(X_i)_{i \in I}$. Then there is a ring $A = R[(X_i)_{i \in I}]$ which has a basis over $R$ given by the \textbf{monomials} of the form \begin{align*} Q = X_{i_1}^{n_1} \dots X_{i_k}^{n_k} \quad \text{for} \quad k \geq 0, (i_{1}, \ldots, i_{k}) \text{ distinct elements of }I \text{ and } n_i \geq 1 \end{align*} up to re-ordering of the product. This ring $A$ is in a natural way an $R$-algebra with the obvious structure morphism of embedding $R$ into the constant polynomials of $A$. Again, we can describe the Hom-functor easily. \begin{prop}[] \label{prop:polynomial-ring} Given any $R$-algebra $S$, there is a natural bijection \begin{align*} \Hom_{R\text{-alg}}(A,S) &\to \Hom_{\Set}(I,S) \\ f &\mapsto (i \mapsto f(X_i)) \end{align*} and in the opposite direction, given a $g: I \to S$, define \begin{align*} f(X_{i_1}^{n_1} \dots X_{i_k}^{n_k}) := g(i_1)^{n_1} \dots g(i_k)^{n_k} \in S \end{align*} If we write $A = R[(X_i)_{i \in I}]$ as the image of the functor $G: \Set \to R\text{-alg}$ that adjoints variables indexed by a set to $R$, then this gives an adjunction to the forgetful functor. \end{prop} \begin{exr}[] Show that \begin{align*} \Hom_{R\text{-alg}}(R[X],S) \stackrel{\sim}{\to} S \end{align*} \end{exr} \begin{proof}[Solution] This is trivial as we can identify $\Hom_{\Set}(\{\ast\},S)$ with $S$. \end{proof} Similarly to the previous section, in order to give an $R$-module $M$ the structure of an $R[(X_i)_{i \in I}]$-module, it is sufficient to give a family of \textbf{commuting} $R$-linear maps \begin{align*} u_i: M \to M \quad \text{with} \quad X_i \cdot m := u_i(m) \end{align*} \begin{exr}[] Find a bijection \begin{align*} \left\{ M \stackrel{u}{\to}M \text{ $R$-linear}\right\} \stackrel{\sim}{\to} \left\{ R[X]\text{-module structures on $M$} \right\} \end{align*} \end{exr} A consequence of this is the Jordan normal form for $R = \C$.
 Given an $R$-algebra $S$ and an arbitrary subset $G \subseteq S$, there are two ways to define the subalgebra of $S$ \emph{generated by } $G$. One is to first note that the intersection of subalgebras is again a subalgebra. So we can look at the intersection of all subalgebras that contain $G$ to get the subalgebra generated by $G$. The above gives a nice description of what the elements are, but a more constructive definition uses the polynomial algebra, where the set of indeterminates is indexed by the set $G$. \begin{dfn}[] Let $S$ be an $R$-algebra and $G \subseteq S$ a subset. For the polynomial algebra $A = R\left[(X_g)_{g \in G}\right]$, there is a canonical'' morphism\footnote{ In fact, this morphism is the one induced by the universal property of polynomial rings for the inclusion mapping $G \hookrightarrow S$. } \begin{align*} \phi_G: \left\{\begin{array}{lll} A \to S\\ X_g \mapsto g \end{array}\right. \end{align*} the \textbf{subalgebra generated by $G$} is the image of $\phi_G$. If $\phi_G$ is surjective, we say that $S$ is generated by $G$.. If there is a finite set $G$ generating $S$, we say that $S$ is finitely generated. \end{dfn} \begin{cor}[] Any commutative ring with unit is isomorphic to a quotient of a polynomial ring. \end{cor} \begin{proof} Since every ring is an algebra over $\Z$, we take the set $G = S$ and use the above construction. This gives a morphism \begin{align*} \phi_S: \Z[(X_s)_{s \in S}] \to S \end{align*} which is surjective and thus induces an isomorphism \begin{align*} \faktor{\Z[(X_s)_{s \in S}}{\Ker \phi_S} \stackrel{\sim}{\to} S \end{align*} \end{proof} \subsection{Localisation [ACL 1.6]} From Algebra I, we know that if $R$ is an integral domain, we can form a field of fractions with elements of the form $\frac{a}{b}$ with $a,b \in R, b \neq 0$. We can generalize this definition by restricting what sort of elements can appear in the denominator. However, if we want to multiply fractions $\frac{a}{b} \cdot \frac{c}{d}$, we need the element $bd$ to be eligible as a denominator. \begin{dfn}[] Let $R$ be a ring. A subset $S \subseteq R$ is called \textbf{multiplicative}, if $1 \in S$ and $a,b \in S \implies ab \in S$. \end{dfn} \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 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} \item If $S_1 \subseteq R_1$ is multiplicative, then $f(S_1) \subseteq R_2$ is multiplicative. \item If $S_2 \subseteq R_2$ is multiplicative, then $f^{-1}(S_2) \subseteq R_1$ is multiplicative. \end{itemize} \item $R^{\times} \subseteq R$ is multiplicative. \end{enumerate} Note, the constructions (b) and (c) are very different. Consider $R = \Z$ and $p = 2\Z$ and $a = 2$. Construction (b) gives the odd integers, whereas (c) gives $S = \{1,2,4,8,\ldots\}$. \end{xmp} Our goal is for $S \subseteq R$ multiplicative, we want to construct a ring $S^{-1}R$, where the elements of $S$ become invertible, which we will calll \textbf{$R$ localized at $S$}. \begin{thm}[] \label{thm:localisation} Let $S \subseteq R$ be multiplicative. There exists an $R$-algebra \begin{align*} \phi_S: R \to S^{-1}R \end{align*} such that for any $R$-algebra $A$, there exists a bijection \begin{align*} \Hom_{R\Alg}(S^{-1}R,A) &\stackrel{\sim}{\to} \{g \in \Hom_{R\Alg}(R,A) \big\vert g(S) \subseteq A^{\times}\}\\ f &\mapsto f \circ \phi_S \end{align*} \begin{center} \begin{tikzcd}[] R \arrow[]{r}{\phi_S} \arrow[swap]{dr}{f \circ \phi_S} & S^{-1}R \arrow[]{d}{f} \\ & A \end{tikzcd} \end{center} \end{thm} \begin{xmp}[] For $R$ an integral domain, $S = R \setminus \{0\}$, the localisation $S^{-1}R$ is isomorphic to the fraction field $\text{Quot}(R)$. 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 $f(\frac{s}{r}) = \frac{f(s)}{f(r)}$. \end{xmp} \begin{rem}[] Since we have identified $\Hom_{R\Alg}(S^{-1}R,-)$, the ring $S^{-1}R$ is unique up to isomorphism by the Yoneda lemma (see \ref{cor:iso-with-hom}). A less abstract nonsense'' proof of uniqueness can be obtained by using the same trick in the Yoneda Lemma. For any other $\psi: R \to T$ with the same property, we plug in $A = S^{-1}R$ and $f = \id_{S^{-1}R}$ and in the end, we obtain an isomorphism $T \iso S^{-1}R$. \end{rem} \begin{proof}[Proof Theorem] We construct the algebra $S^{-1}R$ similar to how we construct $\text{Quot}(R)$, but by only allowing elements of $S$ to be in the denominator. However this restriction introduces a subtlety on when two fractions are considered equivalent. Let $X = R \times S$. We define \begin{align*} \tilde{+}: \quad &\left\{\begin{array}{lll} X \times X & \to & X\\ ((r_1,s_1),(r_2,s_2)) & \mapsto & (r_1s_2 + r_2s_1, s_1s_2) \end{array} \right. \\ \tilde{\cdot}: \quad &\left\{\begin{array}{lll} X \times X & \to & X\\ ((r_1,s_1),(r_2,s_2)) & \mapsto & (r_1r_2, s_1s_2) \end{array} \right. \\ \tilde{\phi_S} &: R \to X, \quad r \mapsto (r,1_R) \end{align*} Next, we define the quivalence relation as follows \begin{align*} (r_1,s_1) \sim (r_2,s_2) \iff \exists t \in S: \quad t (r_1s_2 - r_2s_1) = 0 \end{align*} The $\exists t$ clause is only necessary if $S$ has zero divisors in order to ensure that it does indeed form an equivalence relation. If $S$ is an integral domain, this definition is equivalent to $r_1s_2 - r_2s_1 = 0$. \begin{enumerate} \item $\sim$ forms an equivalence relation and we set $S^{-1}R = X/\sim$. \item We can inherit addition and multiplication from $X$. Since $X$ is not a ring, we have to check that it is well defined on $S^{-1}R$. \item The mapping $R \stackrel{\tilde{\phi_S}}{\to}X \stackrel{\iota}{\to} S^{-1}R$ defines a ring morphism $\phi_S: R \to S^{-1}R$. \end{enumerate} \begin{center} Missing rest of proof. Just check the above claims. Then check that the Hom-functor does indeed follow the characterisation \end{center} \end{proof} \begin{rem}[] $\phi_S$ is not always injective. \begin{align*} \Ker \phi_S = \left\{r \in R \big\vert \frac{r}{1} = \frac{0}{1} \text{ in }S^{-1}R\right\} = \left\{r \in R \big\vert \exists t\in S \big\vert t(r-0) = rt = 0\right\} \end{align*} \begin{itemize} \item If $S$ has no zero-divisors, then $\phi_S$ is injective. \item If $0 \in S$ or if $S$ contains a nilpotent element, then $S^{-1}R$ is the zero ring $\{0\}$. \end{itemize} \end{rem}
 \begin{xmp}[] Set $R = \Z$ \begin{itemize} \item For $S = \Z \setminus \{0\}$, $S^{-1}R = \Q$ \item If $S = \Z - p\Z$, then $S^{-1}R = \{\frac{a}{b} \in \Q \big\vert p \not| b\}$ \item $S = \{1,a,a^{2},\ldots\} \implies S^{-1}R = \{\frac{x}{y} \in \Q \big\vert y = a^{n}\}$. \end{itemize} \end{xmp} \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 $R_p = (R \setminus p)^{-1}R$. \end{enumerate} \begin{prop}[ACL 2.2.7] Let $S \subseteq R$ multiplicative. The map $q \mapsto \phi_S^{-1}(q)$ gives a bijection \begin{align*} \{\text{prime ideals } q \subseteq S^{-1}R\} \to \{\text{prime ideals }p \subseteq R \text{ with } p \cap S = \emptyset\} \end{align*} \end{prop} \begin{proof} In fact, we will provide the inverse map \begin{align*} p \mapsto \text{ideal generated by } \phi_S(p) \end{align*} First, we check that the first map is well defined, so if $q \subseteq S^{-1}R$ is a prime ideal, then $\phi_S^{-1}(q)$ should be a prime ideal and doesn't intersect $S$. It's easy to check that $\phi_S^{-1}(q)$ is prime and if $x \in \phi_S^{-1}(q) \cap S$, then \begin{align*} \phi_S(x) \in q \cap (S^{-1}R)^{\times} \implies q \text{ contains a unit } \implies q = R \lightning \end{align*} Now let $p \subseteq R$ prime with $p \cap S = \emptyset$. The ideal generated by $\phi_S(p)$ is \begin{align*} q = \left\{\frac{r}{s} \big\vert r \in p, s \in S\right\} \end{align*} this is indeed a prime ideal because \begin{align*} \frac{r_1}{s_1} + \frac{r_2}{s_2} = \frac{r_1s_2 + r_2s_1}{s_1s_2} \end{align*} where the numerator is clearly in $p$ and the denominator is an element of $S$. Now we check that $\phi_S^{-1}(q) = p$. The inclusion $\subseteq$ is trivial by definition. Conversely, if $r \in \phi_S^{-1}(q)$, then $\frac{r}{1} = \frac{r_1}{s_1}$ for some $r_1 \in p$. So there exists a $t \in S$ such that \begin{align*} t(r s_1 - r_1) = 0 \implies \underbrace{ts_1}_{\notin p} \implies r \in p \end{align*} where we used that $p \cap S = \emptyset$ and $p$ is a prime ideal. Now we check that the inverse map is well-defined, i.e. that $q$ is prime in $S^{-1}R$. Suppose $\frac{r_1}{s_1} \frac{r_2}{s_2} \in q$. Then \begin{align*} \exists r \in p, \exists s \in S: \quad \frac{r_1r-2}{s_1s_2} \in \frac{r}{s} \end{align*} so there exists a $t \in S$ such that \begin{align*} t (s r_1r_2 r s_1s_2) 0 \implies \underbrace{ts}_{\notin p}r_1r_2 \in p \implies r_1r_2 \in p \end{align*} so either $r_1 \in p$ or $r_2 \in p$. By this either $\frac{r_1}{s_1} \in q$ or $\frac{r_2}{s_2} \in q$. Now we check that the two maps are inverses: \begin{enumerate} \item Start with $p \subseteq R$, then construct $q$. By showing both sides of the inclusion, we already showed $p = \phi_S^{-1}(q)$. \item Starting with $q$, let $p = \phi_S^{-1}(q) \subseteq R$ and let $q_1$ be the ideal generated by $\phi_S(p)$. Then $\phi_S(p) \subseteq q \implies q_1 \subseteq q$. Conversely, let $\frac{r}{s} \in q$. Then \begin{align*} \phi_S(r) = \frac{r}{s} = \frac{s}{1} \frac{r}{s} \in Q \implies r \in p \end{align*} which shows that \begin{align*} \frac{r}{s} = \frac{r}{1} \cdot \frac{1}{s} = \phi_S(p) \frac{1}{S} \in q_1 \end{align*} \end{enumerate} \end{proof} \begin{rem}[] The bijections respect inclusion, but are not maximal ideals in general. A counterexample would be the inclusion \begin{align*} \Z &\hookrightarrow \Q\\ \{0\} &\mapsto \{0\} \end{align*} \end{rem} \begin{xmp}[]