\documentclass[10pt]{article} \usepackage[a4paper, hmargin={2.8cm, 2.8cm}, vmargin={2.5cm, 2.5cm}]{geometry} % Geometri-pakke: Styrer bl.a. maginer % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage[babel, lille, nat, farve, en]{ku-forside} % KU-forside \usepackage{amsmath,amssymb,amsfonts,amsthm,natbib} % Better mathematics \usepackage{nicefrac} % Include external graphics files \usepackage{enumerate,graphicx,nicefrac,verbatim} \usepackage{paralist} \usepackage{fancyhdr} \usepackage[all]{xy} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{problem}[theorem]{Problem} \newtheorem{definition}[theorem]{Definition} \newtheorem{eksempel}[theorem]{Eksempel} \newtheorem{bemaerkning}[theorem]{Bem�rkning} \renewcommand{\thesection}{\large Problem \arabic{section}} \renewcommand{\thesubsection}{\normalsize (\alph{subsection})} \renewcommand{\thesubsubsection}{\alph{subsubsection}.} \newcommand{\m}{\mathbb{#1}} \newcommand{\mf}{\mathfrak{#1}} %\newcommand{\mH}{{\mf{H}} \newcommand{\NN}{{\m{N}}} \newcommand{\KK}{{\m{K}}} \newcommand{\NZ}{{\m{N}_0}} \newcommand{\ZZ}{{\m{Z}}} \newcommand{\QQ}{{\m{Q}}} \newcommand{\CC}{{\m{C}}} \newcommand{\RR}{{\m{R}}} \newcommand{\eps}{\varepsilon} \newcommand{\q}{{\sigma}} \newcommand{\s}{\subseteq} \newcommand{\DD}{{\m{D}}} \newcommand{\OO}{{\m{O}}} \newcommand{\II}{{\m{I}}} \newcommand{\EE}{{\m{E}}} \newcommand{\BB}{{\m{B}}} \newcommand{\mX}{\mathfrak{X}} \newcommand{\mM}{\mf{M}} \newcommand{\mA}{\mathcal{A}} \newcommand{\mY}{\mathfrak{Y}} \newcommand{\mZ}{\mathfrak{Z}} \newcommand{\mL}{\mathcal{L}} \newcommand{\mF}{\mathcal{F}} \newcommand{\mE}{\mathcal{E}} \newcommand{\mC}{\mathcal{C}} \newcommand{\mD}{\mathcal{D}} \newcommand{\mB}{\mathcal{B}} \newcommand{\mS}{\mathcal{S}} \newcommand{\mU}{\mathcal{U}} \newcommand{\sD}{\q(\m{D})} \newcommand{\dd}{\mathrm{d}} \newcommand{\Syl}{\mathrm{Syl}} \newcommand{\op}{\mathrm{op}} \newcommand{\Ext}{\mathrm{Ext}} \newcommand{\Yext}{\mathrm{Yext}} \newcommand{\pd}{\mathrm{pd}} \newcommand{\GF}{\mathrm{GF}} \newcommand{\card}{\mathrm{card}} \newcommand{\Tor}{\mathrm{Tor}} \newcommand{\gldim}{\mathrm{gldim}} \newcommand{\Char}{\ \mathrm{char}\ } \newcommand{\sO}{\q(\m{O}_2)} \newcommand{\sI}{\q(\m{I}^2)} \newcommand{\sH}{\q(\m{H})} \newcommand{\rif}{\rightarrow\infty} \newcommand{\rw}{\rightarrow} \newcommand{\nl}{\vspace{12pt}\noindent} \newcommand{\mH}{\mathcal{H}} \newcommand{\co}{\mathrm{co}} \newcommand{\clin}{\mathrm{clin}} \newcommand{\Span}{\mathrm{span}} \newcommand{\Hom}{\mathrm{Hom}} \newcommand{\idd}{\mathrm{id}} \newcommand{\Rez}{\mathrm{Re}} \newcommand{\Imz}{\mathrm{Im}} \newcommand{\lin}{\mathrm{lin}} \newcommand{\imm}{\mathrm{im}} \newcommand{\Ra}{\Rightarrow} \newcommand{\Lra}{\Leftrightarrow} \newcommand{\lra}{\leftrightarrow} \newcommand{\NI}{\noindent} \newcommand{\ov}{\overline} \newcommand{\sm}{\setminus} \newcommand{\oo}{\emptyset} \newcommand{\dia}{\stackrel{#1}{\longrightarrow}} \titel{Homological Algebra} % \undertitel{Assignment 3} % \opgave{Overspringshandling} % Findes kun under 'titelside' \forfatter{\small Rasmus Sylvester Bryder}% \dato{\small January 24, 2012}% \vejleder{Doktoren} % Findes kun under 'titelside' %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% Her begynder dokumentet %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \maketitle\\\vspace{3pt} \section*{(1)} Let $\Gamma$ be a ring, let $x\in\Gamma$ be a central non-zero divisor and let $\Lambda=\Gamma/x\Gamma$. \subsection*{(a)} \textit{Suppose that $A$ and $B$ are $\Gamma$-modules on which multiplication by $x$ is zero respectively injective. Prove that there is an isomorphism} $$\Ext_\Lambda^n(A,B/xB)\simeq\Ext_\Gamma^{n+1}(A,B).$$ First of all, $A$ is easily made into a $\Lambda$-module by defining $[\omega]a:=\omega a$ for all $\omega\in\Gamma$ and $a\in A$, $[\omega]$ denoting the equivalence class of $\omega$; the multiplication is well-defined, since if $[\omega]=[\omega']$, then $\omega-\omega'=x\gamma$ for some $\gamma\in\Gamma$. Since $x$ is contained in the center of $\Gamma$, we obtain $\omega a-\omega' a=(\omega-\omega')a=(x\gamma)a=\gamma(xa)=0$ for any $a\in A$. $B/xB$ is made into a $\Lambda$-module as well by defining $[\omega][b]:=[\omega b]$ for all $\omega\in\Gamma$ and $b\in B$; this multiplication is also well-defined, since if $\omega-\omega'=x\gamma$ for some $\gamma\in\Gamma$ and $b-b'=x\beta$ for some $\beta\in B$, then $\omega b-\omega' b'=(\omega-\omega')b+\omega'(b-b')=x(\gamma b)+\omega'(x\beta)=x(\gamma b+\omega'\beta)\in xB$, so $[\omega b]=[\omega' b']$. \nl \begin{comment} Consider the short exact sequence of $\Gamma$-modules $\xymatrix{0\ar[r]&\Gamma\ar[r]^{\cdot x}&\Gamma\ar[r]^\pi&\Lambda\ar[r]&0,}$$\pi$ denoting the projection. Take a free resolution of $A$ as a $\Lambda$-module $\xymatrix{\cdots\ar[r]&\Lambda^{n_2}\ar[r]^{\partial_2}&\Lambda^{n_1}\ar[r]^{\partial_1}&\Lambda^{n_0}\ar[r]^{\eps}&A\ar[r]&0.}$ With the above short sequence, this yields a commutative diagram $\xymatrix{&0&0&0&&\\ \cdots\ar[r]&\Lambda^{n_2}\ar[r]^{\partial_2}\ar[u]&\Lambda^{n_1}\ar[r]^{\partial_1}\ar[u]&\Lambda^{n_0}\ar[r]^{\eps}\ar[u]&A\ar[r]&0\\ \ar@{.>}[r]^{\alpha_3}&\Gamma^{n_2}\ar[u]^{\pi^{n_2}}\ar@{.>}[r]^{\alpha_2}&\Gamma^{n_1}\ar[u]^{\pi^{n_1}}\ar@{.>}[r]^{\alpha_1}&\Gamma^{n_0}\ar[u]^{\pi^{n_0}}\ar@{.>}[ur]^{\alpha_0}&&\\ &\Gamma^{n_2}\ar[u]^{(\cdot x)^{n_2}}&\Gamma^{n_1}\ar[u]^{(\cdot x)^{n_1}}&\Gamma^{n_0}\ar[u]^{(\cdot x)^{n_0}},&& }$$n_i$ being the generating sets of the respective free modules. The homomorphisms $\alpha_i: \Gamma^{n_i}\to\Gamma^{n_{i-1}}$ for $i\in\NN$ are defined by for each $\gamma\in\Gamma^{n-i}$ taking a pre-image of $\partial_i\pi^{n_i}(\gamma)$ under $\pi^{n_{i-1}}$; it is well-defined, since if $\partial_i\pi^{n_i}(\gamma)-\partial_i\pi^{n_i}(\gamma')\in x\Gamma^{n_{i-1}}$ \end{comment} We can make any $\Lambda$-module $\mX$ into a $\Gamma$-module by defining $\gamma x:=[\gamma]x$ for all $\gamma\in\Gamma$ and $x\in\mX$. This technique becomes helpful in this next lemma.%If $\mX$ is projective as a $\Lambda$-module, it is projective as a $\Gamma$-module as well. Indeed, if $\eps: \mY\to\mZ$ is a surjective homomorphism of $\Gamma$-modules and $\rho: \mX\to\mZ$ is a homomorphism, $\eps$ and $\rho$ can be considered as homomorphisms of $\Lambda$-modules by just defining new'' homomorphisms $\tilde\eps: \mY\to\mZ$ and $\tilde\rho: \mX\to\mZ$ to be $\eps$ and $\rho$ respectively: indeed, $\tilde\eps([\gamma] y)=\eps(\gamma y)=\gamma\eps(y)=[\gamma]\tilde\eps(y)$ for $\gamma\in\Gamma$ and $y\in Y$ and the same goes for $\rho$. $\mX$ being projective as a $\Lambda$-module, there exists $\tilde\beta:\mX\to\mY$ (still over $\Lambda$-modules) such that $\tilde\eps\tilde\beta=\tilde\rho$. Making $\tilde\beta$ into a $\Gamma$-module homomorphism $\beta: \mX\to\mY$ by just making it equal to $\tilde\beta$, we obtain $\eps\beta=\rho$. \begin{lemma} If there is a surjective ring homomorphism $\xi: \Lambda\to\Lambda'$ and $\Lambda'$-modules $P$ and $Q$, it holds that $\Hom_\Lambda(P,Q)=\Hom_{\Lambda'}(P,Q)$.\end{lemma} \begin{proof}$P$ is made into a $\Lambda$-module by defining $\lambda p:=\xi(\lambda)p$ for $\lambda\in\Lambda$, $p\in P$; likewise for $Q$. If $\varphi'\in\Hom_{\Lambda'}(P,Q)$, then $\varphi'(\lambda p)=\varphi'(\xi(\lambda)p)=\xi(\lambda)\varphi'(p)=\lambda\varphi'(p)$ for all $p\in P$ and $\lambda\in\Lambda$, so $\varphi'\in\Hom_{\Lambda'}(P,Q)$. If $\varphi\in\Hom_\Lambda(P,Q)$, then for any $\lambda'\in\Lambda'$, there exists $\lambda\in\Lambda$ such that $\xi(\lambda)=\lambda'$ by surjectivity, so $\varphi(\lambda'p)=\varphi(\lambda p)=\lambda\varphi(p)=\lambda'\varphi(p)$ for all $p\in P$.\end{proof} %Let $P_\bullet\to A$ be a projective resolution of $A$ as a $\Lambda$-module. \noindent We run through the ordeal by induction, and will prove that for all $\Gamma$-modules $A$ and $B$ as described in the problem, there is an isomorphism $\Ext_\Lambda^n(A,B/xB)\simeq\Ext_\Gamma^{n+1}(A,B)$ for all $n\geq -1$ (the statement holds trivially for all $n<-1$). \nl \textit{Base cases ($n=-1,0,1$).} If $\varphi: A\to B$ is a $\Gamma$-module homomorphism, then $x\varphi(a)=\varphi(xa)=\varphi(0)=0$ for any $a\in A$, so $\varphi=0$ by injectivity of multiplication by $x$ in $B$ and $\Hom_\Gamma(A,B)=0$. \nl \textbf{Step 1: $n=-1$.} Since $\Ext_\Lambda^{-1}(A,B/xB)=0$ by definition and $\Ext_\Gamma^0(A,B)\simeq\Hom_\Gamma(A,B)=0$, the result follows. \nl \textbf{Step 2: $n=0$.} Consider the short exact sequence of $\Gamma$-modules $$0\to B\stackrel{\delta}{\to} B\to B/xB\to 0,$$ where $\delta: B\to B$ is given by $\delta(b)=xb$. It induces a long exact $\Ext$-sequence in the second variable by (\cite{hilton}, Proposition IV.7.5):$\xymatrix{ \Hom_\Gamma(A,B)\ar[r]&\Hom_\Gamma(A,B/xB)\ar[r]^{\hspace{13pt}\omega_0}&\Ext^1_\Gamma(A,B)\ar[r]^{\delta'_*}&\Ext^1_\Gamma(A,B)\ar[r]&\cdots }$ We claim that the homomorphism $\delta'_*$ induced by $\delta$ is actually the zero map. Once that is proved, then it follows from the equality $\Hom_\Gamma(A,B)=0$ that the connecting homomorphism $\omega_0$ is an isomorphism. %Let $P_\bullet$ be a projective resolution of $A$ with homomorphisms $\partial_i$, $i\in\NN_0$. Then we have a commutative diagram $\xymatrix{ %\Hom_\Gamma(P_0,B)\ar[r]^{\partial_1^*}\ar[d]^{\delta_*}&\Hom_\Gamma(P_1,B)\ar[r]^{\partial_2^*}\ar[d]^{\delta_*}&\Hom_\Gamma(P_2,B)\ar[r]\ar[d]^{\delta_*}&\cdots\\ %\Hom_\Gamma(P_0,B)\ar[r]^{\partial_1^*}&\Hom_\Gamma(P_1,B)\ar[r]^{\partial_2^*}&\Hom_\Gamma(P_2,B)\ar[r]&\cdots %}$ since $\partial_i^*(\delta_*(\varphi))(p)=x\varphi(\partial_i(p))=\varphi(\partial_i(xp))=\delta_*(\partial_i^*(\varphi))(p)$ for all $\varphi\in\Hom_\Gamma(P_{i-1},B)$ and $p\in P_i$. $\delta'_*:\Ext^1_\Gamma(A,B)\to\Ext^1_\Gamma(A,B)$ is then defined by $\delta'_*([\varphi])=[\delta_*(\varphi)]=[\delta\varphi]$ for all $\varphi\in\ker\partial_2^*$. By the balancing of $\Ext$ we can consider an injective resolution $I^\bullet$ of $B$ with homomorphisms $\partial^i$, $i\in\NN_0$, yielding a commutative diagram $\xymatrix{ \cdots&\Hom_\Gamma(A,I^2)\ar[l]_{\hspace{-16pt}\partial^2_*}\ar[d]^{\delta^*}&\Hom_\Gamma(A,I^1)\ar[d]^{\delta^*}\ar[l]_{\hspace{4pt}\partial^1_*}&\Hom_\Gamma(A,I^0)\ar[d]^{\delta^*}\ar[l]_{\hspace{8pt}\partial^0_*}\\ \cdots&\Hom_\Gamma(A,I^2)\ar[l]_{\hspace{-16pt}\partial^2_*}&\Hom_\Gamma(A,I^1)\ar[l]_{\hspace{4pt}\partial^1_*}&\Hom_\Gamma(A,I^0)\ar[l]_{\hspace{8pt}\partial^0_*}. }$The homomorphism induced from $\delta$ under homology is the same as the homomorphism $\delta'_*$, but note that $\delta^*(\varphi)(a)=\varphi(xa)=0$ for all $\varphi\in\Hom_\Gamma(A,I^i)$ and $a\in A$. Thus $\delta'_*([\varphi])=[\delta^*(\varphi)]=$ for all $[\varphi]\in\Ext_\Gamma^1(A,B)$, implying $\delta'_*=0$. Thus $\Ext^1_\Gamma(A,B)\simeq\Hom_\Gamma(A,B/xB)$, but by Lemma 1, we then get $$\Ext^1_\Gamma(A,B)\simeq\Hom_\Gamma(A,B/xB)=\Hom_\Lambda(A,B/xB)\simeq\Ext_\Lambda^0(A,B/xB),$$ since the $\Gamma$-module structure induced by the quotient mapping $\Gamma\to\Lambda$ is in fact the original $\Gamma$-module structure of $A$. \nl \textbf{Step 3: $n=1$.} Choose a $\Lambda$-free presentation of $A$, i.e. a short exact sequence of $\Lambda$-modules $$0\hookrightarrow K\to P\to A\to 0$$ with $P\simeq\Lambda^{(S)}$ for some set $S$. The $\Lambda$-module structure induces a $\Gamma$-module structure on the above modules; in particular, multiplication on $K$ and $P$ by $x$ is zero, yielding two long exact $\Ext$-sequences in the first variable: $\xymatrix{ \Ext^0_\Lambda(_{\Lambda}P,B/xB)\ar[r]\ar[d]^\cong&\Ext^0_\Lambda(_{\Lambda}K,B/xB)\ar[r]\ar[d]^\cong&\Ext^1_\Lambda(_{\Lambda}A,B/xB)\ar[r]&\Ext^1_\Lambda(_{\Lambda}P,B/xB)\\ \Ext^1_\Gamma(_\Gamma P,B)\ar[r]&\Ext^1_\Gamma(_\Gamma K,B)\ar[r]&\Ext^2_\Gamma(_\Gamma A,B)\ar[r]&\Ext^2_\Gamma(_\Gamma P,B)}$ The last term in the upper row is zero by (\cite{hilton}, Proposition IV.7.2), and the two downward isomorphisms in the diagram is given by the induction hypothesis. Since $P\simeq\Gamma^{(S)}/x\Gamma^{(S)}$ as $\Gamma$-modules (see 1(c) for a proof), we have an exact sequence $\xymatrix{0\ar[r]&\Gamma^{(S)}\ar[r]^{\cdot x}&\Gamma^{(S)}\ar[r]^\pi&_\Gamma P\ar[r]&0,}$ of $\Gamma$-modules, $\pi$ denoting the quotient mapping. This yields a long exact $\Ext$-sequence in the first variable $\xymatrix{ \cdots\ar[r]&\Ext^{s-1}_\Gamma(\Gamma^{(S)},B)\ar[r]&\Ext^s_\Gamma(_\Gamma P,B)\ar[r]&\Ext^s_\Gamma(\Gamma^{(S)},B)\ar[r]&\cdots,}$ and since the outer two terms are zero by (\cite{hilton}, Proposition IV.7.2) for $s\geq 2$, $\Ext^s_\Gamma(_\Gamma P,B)=0$ for all $s\geq 2$. We therefore obtain a commutative diagram $\xymatrix{ \Ext^0_\Lambda(_{\Lambda}P,B/xB)\ar[r]\ar[d]^\cong&\Ext^0_\Lambda(_{\Lambda}K,B/xB)\ar[r]\ar[d]^\cong&\Ext^1_\Lambda(_{\Lambda}A,B/xB)\ar[r]&0\ar[d]^\cong\\ \Ext^1_\Gamma(_\Gamma P,B)\ar[r]&\Ext^1_\Gamma(_\Gamma K,B)\ar[r]&\Ext^2_\Gamma(_\Gamma A,B)\ar[r]&0}$ Indeed, the diagram is commutative if and only if the first square is, but since $\Ext^0_\Lambda(_{\Lambda}P,B/xB)\simeq\Ext^0_\Gamma(_{\Gamma}P,B/xB)$ and $\Ext^0_\Lambda(_{\Lambda}K,B/xB)\simeq \Ext^0_\Gamma(_{\Gamma}K,B/xB)$ by Lemma 1, and the diagram $\xymatrix{ \Ext^0_\Gamma(_{\Gamma}P,B/xB)\ar[r]\ar[d]^\cong&\Ext^0_\Gamma(_{\Gamma}K,B/xB)\ar[d]^\cong\\ \Ext^1_\Gamma(_\Gamma P,B)\ar[r]&\Ext^1_\Gamma(_\Gamma K,B)}$ is commutative by (\cite{hilton}, Proposition IV.7.6, diagram (7.5) with $\alpha$ being the inclusion $K\to P$), then the desired commutativity follows. In other to prove that there is an arrow between the third terms that is an isomorphism, we prove the following lemma: \begin{lemma}For any commutative diagram of $\Lambda$-modules with exact rows $\xymatrix{ A\ar[r]^{\mu_1}\ar[d]_\cong^\alpha&B\ar[r]^{\mu_2}\ar[d]_\cong^\beta&C\ar[r]&0\\ A'\ar[r]^{\mu_1'}&B'\ar[r]^{\mu_2'}&C'\ar[r]&0,}$ there is a induced arrow $\gamma: C\to C'$ that is an isomorphism.\end{lemma} \begin{proof}For any $c\in C$, take $b\in B$ such that $\mu_2(b)=c$ and consider $\mu_2'\beta(b)$. If another $b_0\in B$ satisfies $\mu_2(b_0)=c$, then $b-b_0\in\ker\mu_2=\imm\mu_1$, i.e. $\mu_1(a)=b-b_0$ for some $a\in A$. We now have $$\mu_2'\beta(b)-\mu_2'\beta(b_0)=\mu_2'\beta(b-b_0)=\mu_2'\beta\mu_1(a)=\mu_2'\mu_1'\alpha(a)=0,$$ so by defining $\gamma(c)=\mu_2'\beta(b)$, we obtain a well-defined mapping. It is a homomorphism, since if $c,c_0\in C$ and $\lambda\in\Lambda$, then by picking $b,b_0\in C$ such that $\mu_2(b)=c$ and $\mu_2(b_0)=c_0$, then $\mu_2(b+b_0)=c+c_0$ and $\gamma(c+c_0)=\mu_2'\beta(b+b_0)=\gamma(c)+\gamma(c_0)$, and additionally $\gamma(\lambda c)=\mu_2'\beta(\lambda b)=\lambda \gamma(c)$. By adding an extra column of zeros at the end, we obtain a commutative diagram with isomorphisms in the last two columns, the Five Lemma yielding that $\gamma$ is an isomorphism.\end{proof} \noindent Alas $\Ext_\Lambda^1(A,B/xB)\simeq\Ext_\Gamma^2(A,B)$. \nl \textit{Inductive step.} Assume that the theorem holds for $n=m$. Choose a $\Lambda$-free presentation of $A$, i.e. a short exact sequence of $\Lambda$-modules $$0\to K\hookrightarrow P\to A\to 0$$ with $P\simeq\Lambda^{(S)}$ for some set $S$. We induce the same $\Gamma$-module structure as in Step 3, so multiplication on $K$ and $P$ by $x$ is zero, and we obtain a diagram $\xymatrix{ \Ext^m_\Lambda(_{\Lambda}P,B/xB)\ar[r]\ar[d]^\cong&\Ext^m_\Lambda(_{\Lambda}K,B/xB)\ar[r]\ar[d]^\cong&\Ext^{m+1}_\Lambda(_{\Lambda}A,B/xB)\ar[r]&\Ext^{m+1}_\Lambda(_{\Lambda}P,B/xB)\\ \Ext^{m+1}_\Gamma(_\Gamma P,B)\ar[r]&\Ext^{m+1}_\Gamma(_\Gamma K,B)\ar[r]&\Ext^{m+2}_\Gamma(_\Gamma A,B)\ar[r]&\Ext^{m+2}_\Gamma(_\Gamma P,B)}$ The outer terms in the upper row is zero by (\cite{hilton}, Proposition IV.7.2), and the two downward isomorphisms in the diagram are given by the induction hypothesis. Since $\Ext^s_\Gamma(_\Gamma P,B)=0$ for all $s\geq 2$ as proved in Step 3 of the induction start, then the two outer terms in the lower row are zero, and we obtain the desired isomorphism by composing $\xymatrix{ \Ext^m_\Lambda(_{\Lambda}K,B/xB)\ar[r]^\cong\ar[d]^\cong&\Ext^{m+1}_\Lambda(_{\Lambda}A,B/xB)\\ \Ext^{m+1}_\Gamma(_\Gamma K,B)\ar[r]^\cong&\Ext^{m+2}_\Gamma(_\Gamma A,B)}.$ \begin{comment} Let $P_\bullet$ be a $\Lambda$-projective resolution of $A$ with homomorphisms $\partial_i$ and let $K_0$ denote the kernel of $\ker\partial_0$. It's clear that we obtain a short exact sequence $$0\to K_0\hookrightarrow P_0\stackrel{\partial_0}{\to} A\to 0.$$ The $\Lambda$-module structure induces a $\Gamma$-module structure on the above modules; in particular, multiplication on $K_0$ and $P_0$ by $x$ is zero, and the above sequence, now considered as a sequence of $\Gamma$-modules, is exact. This induces the long exact $\Ext$-sequence in the second variable, specifically the exact sequence$\xymatrix{ \Ext^{m+1}_\Gamma(P_0,B)\ar[r]&\Ext^{m+1}_\Gamma(K_0,B)\ar[r]&\Ext^{m+2}_\Gamma(A,B)\ar[r]&\Ext^{m+2}_\Gamma(P_0,B),}$ yielding an isomorphism $\Ext^{m+1}_\Gamma(K_0,B)\simeq\Ext^{m+2}_\Gamma(A,B)$, since the outer terms of the sequence are zero (\cite{hilton}, Proposition IV.7.2). Since $K_0$ is a $\Gamma$-module on which multiplication by $x$ is zero, the induction hypothesis yields $$\Ext^{m+2}_\Gamma(A,B)\simeq\Ext^{m+1}_\Gamma(K_0,B)\simeq\Ext^m_\Lambda(K_0,B/xB).$$ \end{comment} \subsection*{(b)} \textit{Let $A$ be a $\Lambda$-module such that $\pd_\Lambda A=n<\infty$. Prove that there exists a free $\Lambda$-module $F$ such that $\Ext_\Lambda^n(A,F)\neq 0$.} \nl Let $P_\bullet\to A$ be a minimal $\Lambda$-projective resolution of $A$ with homomorphisms $\partial^i: P_i\to P_{i-1}$, $i\in\NN$. For any $\Lambda$-module $B$ we obtain the induced sequence $\xymatrix{ \cdots\ar[r]&\Hom_\Lambda(P_{n-1},B)\ar[r]^{\partial^n_*}&\Hom_\Lambda(P_n,B)\ar[r]^{\hspace{24pt}\partial^{n+1}_*}&0. }$ Since $P_n$ is projective, there exists a $\Lambda$-module $Q$ such that $F:=P_n\oplus Q$ is free (\cite{hilton}, Theorem I.4.7). We now want to prove that $\partial_n^*$ is not surjective in the case of this $F$, so that $$\Ext_\Lambda^n(A,F)=H^n\Hom_\Lambda(P_\bullet,F)=\frac{\ker\partial^{n+1}_*}{\imm\partial^n_*}=\frac{\Hom_\Lambda(P_n,F)}{\imm\partial^n_*}\neq0.$$ Assume that $\partial_n^*$ is surjective and consider the inclusion $\iota: P_n\to F$ given by $\iota(p)=(p,0)$. By the assumption of surjectivity, there exists $\varphi: P_{n-1}\to F$ such that $\varphi\partial_n=\partial_n^*\varphi=\iota$. Let $\pi: F\to P_n$ denote the standard projection; then $(\pi\varphi)\partial^n=\pi(\varphi\partial^n)=\pi\iota=1_{P_n}$, so $\partial_n$ has a left inverse. Consider now the short exact sequence $\xymatrix{ 0\ar[r]&P_n\ar[r]^{\partial^n}&P_{n-1}\ar[r]^{\partial^{n-1}}&\imm\partial^{n-1}\ar[r]&0; }$ It splits, since $\partial^n$ has a left inverse. Thus by the Splitting Lemma (\cite{hilton}, Exercise I.3.7), we obtain an isomorphism $P_n\oplus\imm\partial^{n-1}\simeq P_{n-1}$, so $\imm\partial^{n-1}$ is a projective $\Lambda$-module by (\cite{hilton}, Proposition I.4.5). Consider now an altered version $P'_\bullet$ of the chain complex $P_\bullet$:$\xymatrix{ P_\bullet':\quad\cdots\ar[r]&0\ar[r]&\imm\partial^{n-1}\ar[r]^{i}&P_{n-2}\ar[r]^{\partial^{n-2}}&\cdots\ar[r]^{\partial^1}&P_0\ar[r]&0, }$ $i$ denoting the inclusion. Then the sequence $\xymatrix{ P_\bullet'\to A: \quad 0\ar[r]&\imm\partial^{n-1}\ar[r]^{i}&P_{n-2}\ar[r]^{\partial^{n-2}}&\cdots\ar[r]^{\partial^1}&P_0\ar[r]&A\ar[r]&0, }$is clearly exact at $\imm\partial^{n-1}$ and at $P_{n-2}$ since $\ker\partial^{n-2}=\imm\partial^{n-1}=i(\imm\partial^{n-1})$. It inherits exactness at all other $P_i$ and at $A$ from exactness of $P_\bullet\to A$. Thus $P'_\bullet$ is a $\Lambda$-projective resolution of $A$, but this contradicts the assumption that $P_\bullet$ was a minimal $\Lambda$-projective resolution of $A$, since this new projective resolution has length $n-1$. Alas $\partial^n_*$ is not surjective, so we have found a free module $F$ such that $\Ext_\Lambda^n(A,F)\neq 0$. \subsection*{(c)} \textit{Prove that $\gldim\Gamma\geq\gldim\Lambda+1$ if $\Lambda$ has finite global dimension.} \nl Assume $\gldim\Lambda=n<\infty$. Then there exists a $\Lambda$-module $A$ such that $\pd_\Lambda A=n$. By 1(b), there exists a free $\Lambda$-module $F$ such that $\Ext_\Lambda^n(A,F)\neq 0$. Define a $\Gamma$-module multiplication on $A$ by $\gamma a:=[\gamma]a$ for all $\gamma\in\Gamma$, $a\in A$. Thus $A$ becomes a $\Gamma$-module such that multiplication by $x$ is 0, since $x\in x\Gamma$. \nl Since $F$ is free, $F\simeq\Lambda^{(S)}$ for some set $S$; $\Lambda^{(S)}$ obtains a $\Gamma$-module structure by defining $\gamma([\lambda_s])_{s\in S}=([\gamma\lambda_s])_{s\in S}$ for all $\gamma,\lambda_s\in\Gamma$. Define a $\Gamma$-module $B=\Gamma^{(S)}$; then it is clear that the map $B/xB\to\Lambda^{(S)}$ given by $[(\gamma_s)_{s\in S}]\mapsto([\gamma_s])_{s\in S}$ is a well-defined $\Gamma$-module homomorphism, since if $(\gamma_s)_s\in xB$, then each $\gamma_s\in x\Gamma$, so $([\gamma_s])_{s\in S}=0$. It's also clearly bijective; surjectivitity is clear from the outset, and if $([\gamma_s])_{s\in S}=0$, then $\gamma_s\in x\Gamma$ for all $s\in S$, so $(\gamma_s)_{s\in S}\in xB$, so it is injective as well. If $x(\gamma_s)_{s\in S}=(x\gamma_s)_{s\in S}=0$, then because $x$ is a non-zero divisor, we obtain $(\gamma_s)_{s\in S}=0$, so multiplication by $x$ on $B$ is injective. \nl By 1(a), we now obtain $\Ext^{n+1}_\Gamma(A,B)\simeq\Ext^n_\Lambda(A,B/xB)\simeq\Ext^n_\Lambda(A,F)\neq0$. By the Projective Dimension Theorem, we obtain $\pd_\Gamma A>n$, or $\pd_\Gamma A\geq n+1$, but then $$\gldim\Gamma\geq\pd_\Gamma A\geq n+1=\gldim\Lambda+1.$$ \subsection*{(2)} Let $\Lambda$ be a ring and let $\Lambda[x]$ denote the polynomial ring over $\Lambda$. If $A$ is a $\Lambda$-module then let $A[x]=\Lambda[x]\otimes_\Lambda A$. Clearly, $A[x]$ is a $\Lambda[x]$-module by defining the left multiplication as $Q\cdot(P\otimes a):=(QP\otimes a)$. \subsection*{(a)} \textit{Prove that $\pd_{\Lambda[x]}A[x]=\pd_\Lambda A$ for any $\Lambda$-module $A$.} \nl First a lemma: \begin{lemma}If the right $\Lambda$-module $B$ is flat, the functor $B\otimes_\Lambda-:\mM^\ell_\Lambda\to\mathbf{Ab}$ preserves long exact sequences.\end{lemma} \begin{proof}Let $\xymatrix{ A_\bullet:\quad\cdots\ar[r]&A_{n+1}\ar[r]^{\partial_{n+1}}& A_n\ar[r]^{\partial_n}&A_{n-1}\ar[r]&\cdots }$ be a long exact sequence. For $n\in\ZZ$, this gives rise to the short exact sequence $\xymatrix{ 0\ar[r]&\imm\partial_{n+1}\ar[r]^i& A_n\ar[r]^{\partial_n}&\imm\partial_n\ar[r]&0 }$ $i$ denoting the inclusion. For any $\Lambda$-module homomorphism $\varphi: C_1\to C_2$ and a right $\Lambda$-module $R$, it holds that the map $R\times\imm\varphi\to \imm(R\otimes_\Lambda\varphi)$ given by $(r,\varphi(c))\mapsto r\otimes\varphi(c)$ is bilinear and thus allows for a homomorphism $R\otimes_\Lambda\imm\varphi\to \imm(R\otimes_\Lambda\varphi)$ given by $r\otimes\varphi(c)\mapsto r\otimes\varphi(c)$. It has an inverse given in the same way, so it is an isomorphism. With these considerations, we obtain a commutative diagram $\xymatrix{ 0\ar[r]&B\otimes_\Lambda\imm\partial_{n+1}\ar[rr]^{B\otimes_\Lambda i}\ar[d]^\cong&& B\otimes_\Lambda A_n\ar[rr]^{B\otimes_\Lambda\partial_n}\ar[d]^=&&B \otimes_\Lambda\imm\partial_n\ar[r]\ar[d]^\cong&0\\0\ar[r]&\imm (B\otimes_\Lambda\partial_{n+1})\ar[rr]^\iota&& B\otimes_\Lambda A_n\ar[rr]^{B\otimes_\Lambda\partial_n}&&\imm (B \otimes_\Lambda\partial_n)\ar[r]&0, }$$\iota$ denoting an inclusion. Since $B$ is flat, the upper and thus lower row is exact. Therefore $$H_n(B\otimes_\Lambda A_\bullet)=\frac{\ker(B\otimes_\Lambda\partial_n)}{\imm(B\otimes_\Lambda\partial_{n+1})}=\frac{\imm(B\otimes_\Lambda\partial_{n+1})}{\imm(B\otimes_\Lambda\partial_{n+1})}=0,$$ so exactness is preserved.\end{proof} \noindent Let $P_\bullet\to A$ be a minimal $\Lambda$-projective resolution of $A$ with homomorphisms $\partial_i: P_i\to P_{i-1}$, $i\in\NN$. $\Lambda[x]$ is a free $\Lambda$-module, as it has generators $\{1,x,x^2,x^3,\ldots\}$, and therefore it is flat. Thus the sequence \begin{eqnarray}\label{cigar} \xymatrix{ \quad\cdots\ar[r]&P_n[x]\ar[rr]^{\hspace{7pt}\Lambda[x]\otimes_\Lambda\partial_n }&&\cdots\ar[rr]^{\hspace{-5pt}\Lambda[x]\otimes_\Lambda\partial_1}&&P_0[x]\ar[r]&A[x]\ar[r]&0 }\end{eqnarray} is exact by the above lemma. Each $P_i[x]$ is also a $\Lambda[x]$-module, and the homomorphisms $\Lambda[x]\otimes_\Lambda\partial_i$ are $\Lambda[x]$-module homomorphisms since for each $Q,P\in\Lambda[x]$ and $p\in P_i$, we have $$(\Lambda[x]\otimes_\Lambda\partial_i)(Q(P\otimes p))=QP\otimes\partial_i(p)=Q(P\otimes\partial_i(p))=Q(\Lambda[x]\otimes_\Lambda\partial_i)(P\otimes p).$$ We now prove that each $P_i[x]$ is a projective $\Lambda[x]$-module. Let $i\in\NN_0$ and $\eps: B\to C$ and $\beta: P_i[x]\to C$ be $\Lambda[x]$-module homomorphisms with $\eps$ surjective. They are clearly also $\Lambda$-module homomorphisms, and defining $\hat\beta: P_i\to C$ by $\hat\beta(p)=\beta(1\otimes p)$, it is clearly a $\Lambda$-module homomorphism, since $$\hat\beta(\lambda p)=\beta(1\otimes\lambda p)=\beta(\lambda\otimes p)=\lambda\beta(1\otimes p)=\lambda\hat\beta(p),\quad \lambda\in\Lambda,\ p\in P_i.$$ By $P_i$ being projective, there now exists $\hat\rho: P_i\to B$ such that $\eps\hat\rho=\hat\beta$. Define $\tilde\rho: \Lambda[x]\times P_i\to B$ by $\tilde\rho(Q,p)=Q\hat\rho(p)$, it is clearly bilinear and so allows for a homomorphism $\rho: P_i[x]\to B$ defined by $\rho(Q\otimes p)=Q\hat\rho(p)$. It is a $\Lambda[x]$-module homomorphism, and since $$\eps\rho(Q\otimes p)=\eps(Q\hat\rho(p))=Q\eps(\hat\rho(p))=Q\hat\beta(p)=Q\beta(1\otimes p)=\beta(Q\otimes p),\quad Q\in\Lambda[x],\ p\in P_i,$$ $P_i[x]$ is projective. The sequence (\ref{cigar}) therefore yields $\pd_{\Lambda[x]}A[x]\leq\pd_\Lambda A$. In particular, if $\pd_{\Lambda[x]}A[x]=\infty$, then $\pd_{\Lambda[x]}A=\infty$, so it only remains to show the equality for $\pd_{\Lambda[x]}A[x]$ finite. \nl Assume now that $\pd_{\Lambda[x]}A[x]=n<\infty$, and let $\xymatrix{ P_\bullet\to A:\quad\cdots\ar[r]&P_{n+1}\ar[r]^{\partial_{n+1}}& P_n\ar[r]^{\partial_n}&\cdots\ar[r]&P_0\ar[r]&A\ar[r]&0 }$ be a projective resolution of $A$. Letting $K_i=\ker\partial_i$ for all $i\in\NN_0$, we obtain an exact sequence $\xymatrix{ 0\ar[r]& K_{n-1}\ar@{^{(}->}[r]& P_{n-1}\ar[r]^{\partial_{n-1}}&\cdots\ar[r]&P_0\ar[r]&A\ar[r]&0. }$ Now, by the Projective Dimension Theorem, $\pd_\Lambda A\leq n$ if and only if $K_{n-1}$ is projective. Construct a projective resolution of $A[x]$ as before, i.e. $\xymatrix{ \quad\cdots\ar[r]&P_n[x]\ar[rr]^{\hspace{7pt}\Lambda[x]\otimes_\Lambda\partial_n }&&\cdots\ar[rr]^{\hspace{-5pt}\Lambda[x]\otimes_\Lambda\partial_1}&&P_0[x]\ar[r]&A[x]\ar[r]&0 .}$ Since $\pd_{\Lambda[x]}A[x]=n$ by assumption, then by the Projective Dimension Theorem, $L=\ker(\Lambda[x]\otimes_\Lambda\partial_{n-1})$ is projective. We will now prove that $K_{n-1}$ is projective using this fact. Let $\beta: K_{n-1}\to C$ and $\eps: B\to C$ be $\Lambda$-module homomorphisms with $\eps$ surjective. $K_{n-1}$, $B$ and $C$ can be given a $\Lambda[x]$-module multiplication by defining $$(\lambda_0+\lambda_1x+\cdots+\lambda_nx^n)b=\lambda_0b,\quad\lambda_i\in\Lambda,\ b\in B;$$ likewise for the two others. Define the homomorphism $\hat\beta: L\to C$ by $\hat\beta(P\otimes a)=\beta(Pa)$ (possible by clear bilinearity). Letting $P_0$ and $Q_0$ denote the constant terms of the $P,Q\in\Lambda[x]$ respectively, we obtain $$\hat\beta(Q(P\otimes a))=\hat\beta(QP\otimes A)=\beta((QP)a)=Q_0\beta(P_0a)=Q\beta(Pa)=\hat\beta(Q(P\otimes a)),$$ for all $a\in A$, so $\beta$ is a $\Lambda[x]$-module homomorphism. $\eps$ is clearly a $\Lambda[x]$-module homomorphism as well, seen in the same way. Since $L$ is projective, there exists $\hat\rho: L\to B$ such that $\eps\hat\rho=\hat\beta$. If $a\in K_{n-1}$, then $(\Lambda[x]\otimes_\Lambda\partial_{n-1})(1\otimes a)=1\otimes\partial_{n-1}(a)=0$, and therefore it is possible to define a map $\rho: K_{n-1}\to B$ by $\rho(a)=\hat\rho(1\otimes a)$. It is clearly additive and $$\rho(\lambda a)=\hat\rho(1\otimes(\lambda a))=\hat\rho(\lambda\otimes a)=\lambda\hat\rho(1\otimes a)=\lambda\rho(a)\quad a\in K_{n-1},\ \lambda\in\Lambda,$$ so it is a homomorphism and $\eps\rho(a)=\eps\hat\rho(1\otimes a)=\hat\beta(1\otimes a)=\beta(1a)=\beta(a)$ for all $a\in K_{n-1}$. Therefore $K_{n-1}$ is projective, so $\pd_\Lambda A\leq n=\pd_{\Lambda[x]}A[x]$. \subsection*{(b)} \textit{Let $A$ be a $\Lambda[x]$-module. Prove that there is a short exact sequence of $\Lambda[x]$-modules $$0\to A[x]\to A[x]\stackrel{\eps}{\to} A\to 0,$$ where $\eps(P\otimes a)=Pa$.} \nl Note first that for any $P=\lambda_0+\lambda_1x+\cdots+\lambda_nx^n\in\Lambda[x]$ and $a\in A$, an elementary tensor of $A[x]$ can be written $$P\otimes a=\sum_{i=0}^n\lambda_i x^i\otimes a=\sum_{i=0}^nx^i\otimes\lambda_i a.$$ Thus any element $\Psi\in A[x]$ is of the form $\Psi=\sum_{i=0}^nx^i\otimes a_i$ for some $n\in\NN$ and $a_i\in A$, and we define $\deg(\Psi)=n$ in this case. It is clear that $\deg(\Psi)\geq 0$ for all $\Psi\in A[x]$. \nl Clearly, $\eps(1\otimes a)=a$ for all $a\in A$, so $\eps$ is surjective. Let $\tilde\mu: \Lambda[x]\times A\to A[x]$ be given by $\tilde\mu(P,a)=Px\otimes a-P\otimes xa.$ $\tilde\mu$ is bilinear, since it is linear in both variables and $$\tilde\mu(P,\lambda a)=Px\otimes \lambda a-P\otimes x(\lambda a)=(P\lambda)x\otimes a-(P\lambda)\otimes(xa)=\tilde\mu(P\lambda,a)$$ for all $P\in\Lambda[x]$, $a\in A$ and $\lambda\in\Lambda$. It thus allows for a homomorphism $\mu: A[x]\to A[x]$ given by $\mu(P\otimes a)=Px\otimes a-P\otimes xa$ that is clearly a $\Lambda[x]$-module homomorphism as well. Clearly $\imm\mu\s\ker\eps$, since $\eps(\mu(x^i\otimes a))=x^{i+1}a-x^{i+1}a=0$ for all $i\in\NN$, $a\in A$. We will show that it is injective and onto $\ker\eps$, so that we obtain a short exact sequence $$0\to A[x]\stackrel{\mu}{\to} A[x]\stackrel{\eps}{\to} A\to 0.$$ \noindent Assume that $\mu$ doesn't map onto $\ker\eps$; then $\ker\eps\sm\imm\mu$ is non-empty, and we can define $$n=\min\{\deg(\Psi)\,|\,\Psi\in\ker\eps\sm\imm\mu\}.$$ Assume that $n\geq 1$. Then take $\Psi\in\ker\eps\sm\imm\mu$ with $\deg(\Psi)=n$ and write $\Psi=\sum_{i=0}^nx^i\otimes a_i$. Now, define $$\hat{\Psi}=\sum_{i=0}^{n-1}x^i\otimes a_i+x^{n-1}\otimes xa_n.$$ Since $$\eps(\hat{\Psi})=\sum_{i=0}^{n-1}x^ia_i+x^{n-1}xa_n=\sum_{i=0}^nx^ia_i=\eps(\Psi)=0,$$ we have $\hat{\Psi}\in\ker\eps$. Now, since $\Psi=\hat{\Psi}+x^n\otimes a_n-x^{n-1}\otimes xa_n=\hat{\Psi}+\mu(x^{n-1}\otimes a)$, then assuming $\hat{\Psi}\in\imm\mu$ leads to an immediate contradiction, since $\Psi\notin\imm\mu$ by assumption. Therefore $\hat{\Psi}\notin\imm\mu$, but $\deg(\hat{\Psi})=n-1$, contradicting minimality of $n$. Therefore $n=0$, so all elements $\Psi\in\ker\eps\sm\imm\mu$ have degree 0, i.e. $\Psi=1\otimes a_0$ for some $a_0\in A$, but then $0=\eps(\Psi)=\eps(1\otimes a_0)=1a_0=a_0$, so $\Psi=1\otimes 0=0\in\imm\mu$, another contradiction. Therefore $n<0$ which is absurd, so we conclude that $\mu$ maps onto $\ker\eps$. \nl Assuming that $\mu(\Psi)=0$ for $\Psi\in A[x]$, then by writing $\Psi=\sum_{i=0}^nx^i\otimes a_i$, we obtain \begin{eqnarray*}0=\mu(\Psi)&=&\sum_{i=0}^n\mu(x^i\otimes a_i)\\&=&\sum_{i=0}^nx^{i+1}\otimes a_i-\sum_{i=0}^nx^i\otimes xa_i\\&=&\sum_{i=1}^{n+1}x^i\otimes a_{i-1}-\sum_{i=0}^nx^i\otimes xa_i\\&=&x^{n+1}\otimes a_n+\sum_{i=1}^nx^i\otimes (a_{i-1}-xa_i)-1\otimes xa_0.\end{eqnarray*}Note that there is an isomorphism $A[x]\to\bigoplus_{i=0}^\infty A$ given by $\sum_{i} x^i\otimes a_i\mapsto (a_i)_{i=0}^\infty$ with the obvious inverse $(a_i)_{i=0}^\infty\mapsto\sum_{i} x^i\otimes a_i$; thus the above sum being equal to zero yield the equalities $$a_n=0,\quad a_{n-1}-xa_n=0,\quad a_{n-2}-xa_{n-1}=0,\quad\ldots,\quad a_0-xa_1=0,\quad -xa_0=0,$$ where $a_n=0$ implies $a_{n-1}=0$ which in turn implies $a_{n-2}=0$ and so on. Thus $\Psi=\sum_{i=0}^nx^i\otimes a_i=0$, so $\mu$ is injective as well. \subsection*{(c)} \textit{Prove that $\gldim\Lambda[x]\leq\gldim\Lambda +1$.} \nl Let $A$ be a $\Lambda[x]$-module and $n:=\gldim\Lambda$. $A$ is clearly also a $\Lambda$-module, and thus from 2(a) we obtain that $\pd_{\Lambda[x]}A[x]=\pd_\Lambda A\leq n$. The short exact sequence from 2(b) yields a long exact $\Ext$-sequence in the first variable for all $\Lambda[x]$-modules $B$ (\cite{hilton}, p. 139):$\xymatrix{ \cdots\ar[r]&\Ext^{n+1}_{\Lambda[x]}(A[x],B)\ar[r]&\Ext^{n+2}_{\Lambda[x]}(A,B)\ar[r]&\Ext^{n+2}_{\Lambda[x]}(A[x],B)\ar[r]&\cdots }$ By the Projective Dimension Theorem, we obtain for all $\Lambda$-modules $B$ that $$\Ext^{n+1}_{\Lambda[x]}(A[x],B)=\Ext^{n+2}_{\Lambda[x]}(A[x],B)=0$$ since $\pd_{\Lambda[x]}A[x]\leq n$, and therefore $\Ext^{n+2}_{\Lambda[x]}(A,B)=0$. By the Projective Dimension Theorem, we obtain $\pd_{\Lambda[x]} A\leq n+1$. Since $A$ was arbitrarily chosen, we obtain $$\gldim\Lambda[x]\leq n+1=\gldim\Lambda+1.$$ \noindent\textbf{Observe that} (1) and (2) imply that $\gldim\Lambda[x]=\gldim\Lambda+1$ for any ring $\Lambda$ with finite global dimension. Indeed, since $\Lambda\simeq\Lambda[x]/x\Lambda[x]$ and $x$ is a central non-zero divisor in $\Lambda[x]$, 1(c) yields $\gldim\Lambda[x]\leq\gldim\Lambda+1$, the other inequality following from 2(c). As a corollary, we obtain Hilbert's syzygy theorem: $\gldim k[x_1,\ldots,x_n]=n$ if $k$ is a field. This follows from the fact that all modules over fields are free and thus projective, so $\gldim k=0$, and therefore $$\gldim k[x_1,\ldots,x_n]=\gldim(k[x_1,\ldots,x_{n-1}])[x_n]=1+\gldim k[x_1,\ldots,x_{n-1}]=\cdots=n+\gldim k=n.$$ \section*{(3)} For a group $G$, let $B_\bullet G$ denote the chain complex $(E_\bullet G)_G$ where $E_\bullet G$ is some choice of a projective resolution of the trivial $G$-module $\ZZ$ (i.e. $\ZZ$ being viewed as a $\ZZ G$-module via augmentation: $(\sum_{g\in G}z_gg)z:=(\sum_{g\in G}z_g)z$, $z\in\ZZ$). \subsection*{(a)} \textit{Let $G$ and $H$ be groups. Prove that there is a homotopy equivalence of chain complexes of abelian groups $B_\bullet(G\times H)\simeq B_\bullet G\otimes_\ZZ B_\bullet H$.} \nl Let $D=G\times H$ and choose free resolutions $E_\bullet G$, $E_\bullet H$ and $E_\bullet D$ of $\ZZ$ as a trivial $G$-, $H$- and $D$-module respectively; alas \begin{eqnarray*}E_\bullet G:&&\cdots\to(\ZZ G)^{G_n}\to\cdots\to(\ZZ G)^{G_1}\to(\ZZ G)^{G_0}\to0,\\ E_\bullet H:&&\cdots\to(\ZZ H)^{H_n}\to\cdots\to(\ZZ H)^{H_1}\to(\ZZ H)^{H_0}\to0,\end{eqnarray*} with generating sets $G_i,H_i$; recall that $H_0(E_\bullet G)\simeq\ZZ\simeq H_0(E_\bullet H)$ and $H_n(E_\bullet G)=H_n(E_\bullet H)=0$ for all $n\geq 1$. Consider the chain complex of abelian groups $E_\bullet G\otimes_\ZZ E_\bullet H$. The above chain complexes $E_\bullet G$ and $E_\bullet H$ consist of free abelian groups, so by the K�nneth theorem (\cite{hilton}, Theorem V.2.1) it follows that \begin{eqnarray*}H_n(E_\bullet G\otimes_\ZZ E_\bullet H)&\simeq&\left(\bigoplus_{p+q=n}H_p(E_\bullet G)\otimes_\ZZ H_q(E_\bullet H)\right)\oplus\left(\bigoplus_{p+q=n-1}\Tor_1^\ZZ(H_p(E_\bullet G),H_q(E_\bullet H))\right)\\&\simeq&\left\{\begin{array}{ll}0&n\geq 2\\\Tor_1^\ZZ(\ZZ,\ZZ)&n=1\\\ZZ\otimes_\ZZ\ZZ&n=0\end{array}\right.\\&\simeq&\left\{\begin{array}{ll}0&n\geq1\\\ZZ&n=0\end{array}\right.\end{eqnarray*} by $\ZZ$ being $\ZZ$ projective (\cite{hilton}, p. 161) and using properties of the tensor product. The abelian groups of the chain complex are given by $$(E_\bullet G\otimes_\ZZ E_\bullet H)_n=\bigoplus_{p+q=n}(\ZZ G)^{G_p}\otimes_\ZZ(\ZZ H)^{H_q}\simeq\bigoplus_{p+q=n,\,g\in G_p\,h\in H_q}\ZZ G\otimes_\ZZ\ZZ H$$ for $n\geq 0$, using (\cite{hilton}, Proposition III.7.3). We can give $\ZZ G\otimes_\ZZ\ZZ H$ a $D$-module structure: for all $(g,h)\in D$, define maps $\bar\Psi_{g,h}: \ZZ G\times\ZZ H\to\ZZ G\otimes_\ZZ \ZZ H$ by $\bar\Psi_{g,h}(a_1,a_2)=(ga_1)\otimes(ha_2)$ for $a_1\in\ZZ G,$ $a_2\in\ZZ H.$ They are clearly $\ZZ$-bilinear and induce homomorphisms $\Psi_{g,h}: \ZZ G\otimes_\ZZ\ZZ H\to\ZZ G\otimes_\ZZ \ZZ H$. The desired $D$-module structure is then given by $$(g,h)(a_1\otimes a_2):=\Psi_{g,h}(a_1\otimes a_2)=(ga_1)\otimes(ha_2),\quad g\in G,\ h\in H,\ a_1\in\ZZ G,\ a_2\in\ZZ H,$$ All abelian groups in $E_\bullet G\otimes_\ZZ E_\bullet H$ thus become $D$-modules and it is clear that the differentials on $E_\bullet G\otimes_\ZZ E_\bullet H$ respect this structure. Also, $\ZZ G\otimes_\ZZ\ZZ H$ is a free $D$-module: define the mapping over $\ZZ$-modules $\bar\xi: \ZZ G\times\ZZ H\to\ZZ D$ by $$\bar\xi\left(\sum_{g\in G}z_gg,\sum_{h\in H}z_h h\right)=\sum_{(g,h)\in D}z_gz_h(g,h);$$ it is clearly bilinear, and so induces a homomorphism $\xi: \ZZ G\otimes_\ZZ\ZZ H\to\ZZ D$. It is a $D$-module homomorphism, and a composition of isomorphisms of direct sums and tensor products (again using \cite{hilton}, Proposition III.7.3) by considering the commutative diagram $\xymatrix{ \ZZ G\otimes_\ZZ\ZZ H\ar[d]^\simeq\ar[rr]^\xi&&\ZZ D&&\bigoplus_{(g,h)\in D}\ZZ\ar[ll]^\simeq\\ \bigoplus_{g\in G}\ZZ\otimes_\ZZ\bigoplus_{h\in H}\ZZ\ar[rr]^\simeq&&\bigoplus_{h\in H}\left(\left(\bigoplus_{g\in G}\ZZ\right)\otimes_\ZZ\ZZ\right)\ar[rr]^\simeq&&\bigoplus_{(g,h)\in D}\ZZ\otimes_\ZZ\ZZ\ar[u]^\simeq }$ Thus it follows that $(E_\bullet G\otimes_\ZZ E_\bullet H)_n$ is $D$-free as well, being a direct sum of free $D$-modules, so that $E_\bullet G\otimes_\ZZ E_\bullet H$ becomes a free $D$-resolution of $\ZZ$, using the above discussion. Since any two projective resolutions of a module are homotopy equivalent, it follows now that the chain complexes $E_\bullet G\otimes_\ZZ E_\bullet H$ and $E_\bullet D$ are homotopy equivalent. Since the functor $-_D$ is an additive covariant functor $\mM^\ell_D\to\mathbf{Ab}$, it then follows from (\cite{hilton}, Lemma IV.3.4) that the chain complexes of abelian groups $B_\bullet D$ and $(E_\bullet G\otimes_\ZZ E_\bullet H)_D$ are homotopy equivalent. As \begin{eqnarray*}\left((E_\bullet G\otimes_\ZZ E_\bullet H)_n\right)_D&\simeq&\ZZ\otimes_{\ZZ D}\left(\bigoplus_{p+q=n,\,g\in G_p,\,h\in H_q}\ZZ G\otimes_\ZZ\ZZ H\right)\\&\simeq&\ZZ\otimes_{\ZZ D}\left(\bigoplus_{p+q=n,\,g\in G_p,\,h\in H_q}\ZZ D\right)\\&\simeq&\bigoplus_{p+q=n,\,g\in G_p,\,h\in H_q}\ZZ\otimes_{\ZZ D}\ZZ D\\&\simeq&\bigoplus_{p+q=n,\,g\in G_p,\,h\in H_q}\ZZ\\&\simeq&\bigoplus_{p+q=n,\,g\in G_p,\,h\in H_q}\ZZ\otimes_\ZZ \ZZ\\&\simeq&\bigoplus_{p+q=n,\,g\in G_p,\,h\in H_q}\left(\ZZ\otimes_{\ZZ G}\ZZ G\right)\otimes_\ZZ\left(\ZZ\otimes_{\ZZ H}\ZZ H\right)\\&\simeq&\bigoplus_{p+q=n,\,g\in G_p,\,h\in H_q}\left(\ZZ G\right)_G\otimes_\ZZ\left(\ZZ H\right)_H\\&\simeq&\bigoplus_{p+q=n}\left((\ZZ G)_G\right)^{G_p}\otimes_\ZZ\left((\ZZ H)_H\right)^{H_q}\\&\simeq&\bigoplus_{p+q=n}\left((\ZZ G)^{G_p}\right)_G\otimes_\ZZ\left((\ZZ H)^{H_q}\right)_H\\&=&(B_\bullet G\otimes_\ZZ B_\bullet H)_n,\end{eqnarray*} by using \textit{a lot} of tensor product isomorphisms of abelian groups and the fact that $-_A$ for all groups $A$ is additive. One can verify that the isomorphism is also a chain map, and so we obtain the desired homotopy equivalence $B_\bullet D\simeq B_\bullet G\otimes_\ZZ B_\bullet H$ by composing this (complicated) isomorphism of chain complexes with the homotopy equivalences $B_\bullet D\simeq(E_\bullet G\otimes_\ZZ E_\bullet H)_D$. \subsection*{(b)} \textit{Calculate the integral homology of the group $C_2\times C_2$ in all dimensions.} \nl We first calculate the integral homology of $C_2$ in all dimensions, as it will become useful in the following. Since $\ZZ C_2=\ZZ[x]/(x^2-1)$, a free resolution of the trivial $C_2$-module $\ZZ$ is given by $\xymatrix{ E_\bullet C_2\to\ZZ:\quad\cdots\ar[r]&\ZZ C_2\ar[r]^{\cdot(x-1)}&\ZZ C_2\ar[r]^{\cdot(x+1)}&\cdots\ar[r]^{\cdot(x+1)}&\ZZ C_2\ar[r]^{\cdot(x-1)}&\ZZ C_2\ar[r]^{\eps}&\ZZ\ar[r]&0, }$$\eps$ denoting the augmentation map. By the isomorphism of functors $-_{C_2}\cong\ZZ\otimes_{\ZZ C_2}-$, we then obtain$\xymatrix{ B_\bullet C_2:\quad\cdots\ar[r]&\ZZ\ar[r]^{0}&\ZZ\ar[r]^{2}&\ZZ\ar[r]^{0}&\cdots\ar[r]^{2}&\ZZ\ar[r]^{0}&\ZZ}$ as we regard $\ZZ$ as a $C_2$-module via augmentation (so multiplication by $x\pm 1$ becomes multiplication by $1\pm 1$). Thus the integral homology of $C_2$ is $$H_n(C_2;\ZZ)=H_n((E_\bullet C_2)_{C_2})=H_n(B_\bullet C_2)=\left\{\begin{array}{ll}\ZZ&n=0\\\ZZ_2&n>0,\,n\ \text{odd}\\0&n>0,\,n\ \text{even and all }n<0.\end{array}\right.$$%; it is exact at degrees greater than 0, since \textit{(1)} $P(x+1)(x-1)=P(x-1)(x+1)=P(x^2-1)$ for all $P\in\ZZ[x]$, \textit{(2)} if $P=\sum_{i=0}^nz_ix^i\in\ZZ[x]$, then $\eps([P(x-1)])=\eps([\sum_{i=0}^nz_ix^{i+1}-\sum_{i=0}^nz_ix^i])=0$ and \textit{(3)} if $$\eps([\sum_{i=0}^{2n-1}z_ix^i])=\eps([\sum_{i=0}^{n-1}(-1)^i(z_{2i}+z_{2i+1}x)]=\sum_{i=0}^{n-1}(-1)^i(z_{2i}+z_{2i+1})=0,$$ then $[\sum_{i=0}^{2n-1}z_ix^i]=[\sum_{i=0}^{n-1}(-1)^i(z_{2i}+z_{2i+1}x)] \noindent Let$D=C_2\times C_2$; we now calculate$H_n(D;\ZZ)$for all$n\in\ZZ$. Take a$D$-projective resolution$E_\bullet D$of$\ZZ$and a$D$-projective resolution$E_\bullet D$of$\ZZ$; then$H_n(D;\ZZ)=H_n((E_\bullet D)_D)$. Using the homotopy equivalence obtained in 3(a), we obtain $$H_n(D;\ZZ)=H_n((E_\bullet D)_D)=H_n(B_\bullet D)\simeq H_n(B_\bullet C_2\otimes_\ZZ B_\bullet C_2),$$$B_\bullet C_2$denoting the chain complex found earlier. Since$\ZZ$is a PID and$B_\bullet C_2$consists entirely of free and hence flat$\ZZ$-modules, by the K�nneth theorem (\cite{hilton}, Theorem V.2.1), we obtain $$H_n(D;\ZZ)\simeq\left(\bigoplus_{p+q=n}H_p(B_\bullet C_2)\otimes_\ZZ H_q(B_\bullet C_2)\right)\oplus\left(\bigoplus_{p+q=n-1}\Tor_1^\ZZ(H_p(B_\bullet C_2),H_q(B_\bullet C_2)\right).$$If any of the above$p$or$q$are negative, then the corresponding terms in the sum are 0 by the integral homology of$C_2$; thus we need only consider$p$and$q$that are simultaneously non-negative. We consider the various cases: \begin{itemize} \item If$n<0$, it's clear that$H_n(D;\ZZ)=0$. \item If$n=0$, the second term is 0; the first term boils down to$H_0(B_\bullet C_2)\otimes_\ZZ H_0(B_\bullet C_2)$in the same way. As$H_0(B_\bullet C_2)\otimes_\ZZ H_0(B_\bullet C_2)=\ZZ\otimes_\ZZ\ZZ\simeq\ZZ$, we obtain$H_0(D;\ZZ)\simeq\ZZ$. \item If$n=1$, the second term boils down to$\Tor_1^\ZZ(H_0(B_\bullet C_2),H_0(B_\bullet C_2))=\Tor_1^\ZZ(\ZZ,\ZZ)=0$, since$\ZZ$is free; the first term boils down to $$(H_1(B_\bullet C_2)\otimes_\ZZ H_0(B_\bullet C_2))\oplus(H_0(B_\bullet C_2)\otimes_\ZZ H_1(B_\bullet C_2))\simeq(\ZZ_2\otimes_\ZZ\ZZ)\otimes(\ZZ\otimes_\ZZ\ZZ_2)\simeq(\ZZ_2)^2.$$ Alas$H_1(D;\ZZ)\simeq(\ZZ_2)^2$. \item If$n>1$and$n$is odd, then$n-1$is even. The second term has terms equal to 0 if and only if$q$is even and$\geq 0$, as$\ZZ$is free, and thus the second term boils down to$\nicefrac{n-1}{2}$copies of$\Tor_1^\ZZ(H_p(B_\bullet C_2),H_q(B_\bullet C_2))$for$p,q$odd, i.e. $$\Tor_1^\ZZ(H_p(B_\bullet C_2),H_q(B_\bullet C_2))=\Tor_1^\ZZ(\ZZ_2,\ZZ_2).$$ A$\ZZ$-free resolution of$\ZZ_2$is$\ZZ\stackrel{2}{\to}\ZZ$whose first homology group when tensored with$\ZZ_2$is isomorphic to$\ZZ_2$; alas$\Tor_1^\ZZ(\ZZ_2,\ZZ_2)\simeq\ZZ_2$. The first term has terms equal to 0 if and only if$01$and$n$is even, then$n-1$is odd. The second term then has terms equal to 0 if$q=0$or$q$is even (as$\ZZ$is free), so we need only check the cases where$q$is odd. Thus$p$must be even, but for the term to be non-zero, then$p$must be 0. Thus the second term boils down to calculating$\Tor_1^\ZZ(H_0(B_\bullet C_2),H_{n-1}(B_\bullet C_2))=\Tor_1^\ZZ(\ZZ,\ZZ_2)$. Using the balancing of$\Tor$and that$\ZZ$is free, we then obtain$\Tor_1^\ZZ(\ZZ,\ZZ_2)=0$, so there are no contributions from the second term. The first term has terms equal to 0 if and only if$p$and$q$are both even; summing over all$p$and$q$odd, we then obtain$\nicefrac{n}{2}$copies of$(H_p(B_\bullet C_2)\otimes_\ZZ H_q(B_\bullet C_2))=\ZZ_2\otimes_\ZZ\ZZ_2.$Since$\ZZ_2\otimes_\ZZ\ZZ_2$is an abelian group of order 2 (its only non-zero element being$1\oplus 1$, out of the four possible'' elements), we obtain$H_n(D;\ZZ)\simeq(\ZZ_2)^{\nicefrac{n}{2}}$.%Therefore$H_n(D;\ZZ)\simeq(\ZZ_2\otimes_\ZZ\ZZ_2)^{\nicefrac{n}{2}}$. \end{itemize} To summarize: $$H_n(C_2\times C_2;\ZZ)\simeq\left\{\begin{array}{ll}0&n<0\\\ZZ&n=0\\(\ZZ_2)^{\nicefrac{n+3}{2}}&n>0,\,n\ \text{odd}\\(\ZZ_2)^{\nicefrac{n}{2}}&n>0,\,n\ \text{even}.\end{array}\right.$$ \subsection*{(4)} \subsection*{(a) \cite{hilton}, Exercise IV.9.4} \textit{Define addition in$\Yext_\Lambda^m(A,B)$, independently of the equivalence$\Ext_\Lambda^m\simeq\Yext_\Lambda^m$. Describe a representative of$0\in\Yext_\Lambda^m(A,B)$,$m\geq 2$and show that$\xi+0=\xi$,$\xi\in\Yext_\Lambda^m(A,B)$.} \nl First follows an extensive explanation of$\Yext_\Lambda^m(A,B)$and the tools needed to define addition. For$\Lambda$-modules$A$and$B$,$\Yext_\Lambda^m(A,B)$consists of all equivalence classes (under a specific relation to be mentioned later) of$m$-extensions of$A$by$B$; these are exact sequences of$\Lambda$-modules \begin{eqnarray}\label{cigar2}\xymatrix{ \mathfrak{E}:\quad 0\ar[r]&B\ar[r]^\mu&E_m\ar[r]^{\eps_m}&\cdots\ar[r]&E_1\ar[r]^{\eps}&A\ar[r]&0. }\end{eqnarray} A morphism$f: \mathfrak{E}\to\mathfrak{E'}$of$m$-extensions of$A$by$B$is a sequence of homomorphisms$f_i: E_i\to E'_i$,$i=1,\ldots,m$such that the diagram $\xymatrix{ \mathfrak{E}:\quad 0\ar[r]&B\ar[r]\ar[d]^=&E_m\ar[r]\ar[d]^{f_m}&\cdots\ar[r]&E_1\ar[r]\ar[d]^{f_1}&A\ar[r]\ar[d]^=&0\\ \mathfrak{E}':\quad 0\ar[r]&B\ar[r]&E'_m\ar[r]&\cdots\ar[r]&E'_1\ar[r]&A\ar[r]&0 }$ commutes. Two$m$-extensions$\mathfrak{E}$and$\mathfrak{E}'$of$A$of$B$are equivalent (we write$\mathfrak{E}\sim\mathfrak{E}'$) if there exists a sequence of morphisms between$\mathfrak{E}$and$\mathfrak{E}'$, in the sense that there are a finite number of$m$-extensions$\mX_1,\ldots,\mX_n$and morphisms such that we obtain a diagram $$\mathfrak{E}\lra\mX_1\lra\cdots\lra\mX_n\lra\mathfrak{E}',$$$\lra$meaning that there is a morphism from one of the$m$-extensions to the other; this is clearly an equivalence relation, and$\Yext_\Lambda^n(A,B)$is the set of its equivalence classes. \nl \textbf{1. Induced homomorphisms in the first variable.} If$\alpha: A'\to A$is a homomorphism of$\Lambda$-modules, it induces a map$\alpha^*: \Yext_\Lambda^m(A,B)\to\Yext_\Lambda^m(A',B)$as follows: given$[\mathfrak{E}]\in\Yext^m_\Lambda(A,B)$, we obtain a commutative diagram $\xymatrix{ E_3\ar[r]^{\eps_3}&E_2\ar[r]^{\eps_2}&E_1\ar[r]^\eps&A\ar[r]&0\\ && E_1^\alpha\ar@{.>}[r]^{\eps'}\ar@{.>}[u]^{\beta}&A'\ar[r]\ar[u]_\alpha&0, }$ where$E_1^\alpha$denotes the$\Lambda$-module$\{(e,a')\in E_1\oplus A'\,|\,\eps(e)=\alpha(a')\}$and$\eps'$and$\beta$are defined by$\eps'(e,a')=a'$and$\beta(e,a')=e$for all$e\in E_1$,$a\in A'$. It is a pullback of the diagram since if$\gamma: Z\to A'$and$\rho: Z\to E_1$are morphisms such that$\alpha\gamma(z)=\eps\rho(z)$for all$z\in Z$, then$(\rho(z),\gamma(z))\in E_1^\alpha$for all$z\in Z$, so defining the map$(\rho,\gamma): Z\to E_1^\alpha$by$z\mapsto(\rho(z),\gamma(z))$, we obtain a morphism making the new triangles commute, and it is unique, since if a morphism$\zeta: Z\to E_1^\alpha$satisfies$\eps'\zeta=\gamma$and$\beta\zeta=\rho$, then$\zeta=(\rho,\gamma)$; therefore$(E^1_\alpha;\beta,\eps')$is unique with this property (which is indeed necessary for the induced map to be well-defined). Note what follows by exactness of the upper row: \begin{itemize}\itemsep-2pt \item$\eps'$is surjective by surjectivity of$\eps$, since if$a'\in A$, then$\eps(e)=\alpha(a')$for some$e\in E_1$, so$\eps'(e,a')=a'$. \item Define a homomorphism$d_2: E_2\to E_1^\alpha$by$d_2(e)=(\eps_2(e),0)$. Then$\beta d_2=\eps_2$and it's clear that$\imm d_2\s\ker\eps'$; to prove the other inclusion, let$(e,a')\in\ker\eps'$. Then$a'=0$and$\eps(e)=\alpha(0)=0$, so$e\in\ker\eps=\imm\eps_2$by exactness of the upper row; choosing$e'\in E_2$such that$\eps_2(e')=e$, then$d_2(e')=(e,a')$. Thus$\imm d_2=\ker\eps'$. Note that for any$e''\in E_3$,$\eps_2(\eps_3(e''))=0$, so$\eps_3(e'')\in\ker d_2$. On the other hand, then if$d_2(e)=(\eps_2(e),0)=0$, then$\eps_2(e)=0$, so$e\in\imm\eps_3$. \end{itemize} Thus, we obtain a commutative diagram with exact rows $\xymatrix{ \mathfrak{E}:\quad 0\ar[r]&B\ar[r]&E_m\ar[r]&\cdots\ar[r]^{\eps_3}&E_2\ar[r]^{\eps_2}&E_1\ar[r]^\eps&A\ar[r]&0\\ \mathfrak{E}^\alpha:\quad 0\ar[r]&B\ar[r]\ar[u]_=&E_m\ar[r]\ar[u]_=&\cdots\ar[r]^{\eps_3}&E_2\ar[r]^{d_2}\ar[u]^=&E_1^\alpha\ar[r]^{\eps'}\ar[u]^{\beta}&A'\ar[r]\ar[u]_\alpha&0, }$ that is unique by the property of the last square that is a pullback (note that this holds no matter what$m$is). If$\mf{E}'\sim\mf{E}'$, then there exist morphisms of extensions connecting them; i.e. $$\mathfrak{E}\lra\mX_1\lra\cdots\lra\mX_n\lra\mathfrak{E}',$$ we obtain$\mathfrak{E}^\alpha\lra\mX_1^\alpha\lra\cdots\lra\mX_n^\alpha\lra\mathfrak{E}'^\alpha$by composing the morphisms with the homomorphisms obtained by the ones induced by$\alpha$in the right way, depending on which ways the arrows go. Thus we obtain a well-defined homomorphism$\alpha^*([\mf{E}])=[\mf{E}^\alpha]$. \nl \textbf{2. Induced homomorphisms in the second variable.} If$\beta: B\to B'$is a homomorphism of$\Lambda$-modules, it induces a map$\beta_*: \Yext_\Lambda^m(A,B)\to\Yext_\Lambda^m(A,B')$as follows: given$[\mathfrak{E}]\in\Yext^m_\Lambda(A,B)$, we obtain a commutative diagram $\xymatrix{ 0\ar[r]&B\ar[r]^\mu\ar[d]_\beta&E_m\ar[r]^{\eps_m}\ar@{.>}[d]_\omega&E_{m-1}\ar[r]^{\eps_{m-1}}&E_{m-2}\\ 0\ar[r]&B'\ar@{.>}[r]^{\mu'}&E_{m\beta},&& }$where$E_{m\beta}$denotes the$\Lambda$-module$E_m\oplus B'/\{(\mu(b),-\beta(b))\,|\,b\in B\}$and$\mu'$and$\omega$are defined by$\mu'(b')=[(0,b')]$and$\omega(e)=[(e,0)]$for all$e\in E_m$,$b'\in B'$. It is a pushout of the diagram since if$\gamma: E_m\to Z$and$\rho: B'\to Z$are morphisms such that$\gamma\mu(b)=\rho\beta(b)$for all$b\in B$, then consider the homomorphism$E_m\oplus B'\to Z$given by$(e,b')\mapsto\gamma(e)+\rho(b')$. If$(e_1-e_2,b_1'-b_2')=(\mu(b),-\beta(b))$for$(e_1,b_1'),(e_2,b_2')\in E_m\oplus B'$and some$b\in B$, then $$\gamma(e_1)+\rho(b'_1)-(\gamma(e_2)+\rho(b'_2))=\gamma(e_1-e_2)-\rho(b'_1-b'_2)=\gamma\mu(b)-\rho\beta(b)=0$$ so defining the map$\psi: E_{m\beta}\to Z$by$[(e,b')]\mapsto\gamma(e)+\rho(b')$, we obtain a well-defined homomorphism making the new triangles commute, and it is unique, since if a morphism$\zeta: E_{m\beta}\to Z$satisfies$\zeta\mu'=\rho$and$\zeta\omega=\gamma$, then $$\zeta([(e,b')])=\zeta([(e,0)])+\zeta([(0,b')])=\rho(e)+\gamma(b)=\psi([(e,b')]);$$ therefore$(E_{m\beta};\mu',\omega)$is unique with this property (which is indeed necessary for the induced map to be well-defined). Note what follows by exactness of the upper row: \begin{itemize}\itemsep-2pt \item$\mu'$is injective, since$\mu'(b')=[(0,b')]=0$, then$(0,b')=(\mu(b),-\beta(b))$for some$b\in B$; since$\mu$is injective,$b=0$, so$b'=-\beta(b)=0$. \item Define a homomorphism$d_m: E_{m\beta}\to E_{m-1}$by$d_m([(e,b')])=\eps_m(e)$. It is well-defined, since if$(e_1-e_2,b_1'-b_2')=(\mu(b),-\beta(b))$for$(e_1,b_1'),(e_2,b_2')\in E_m\oplus B'$and some$b\in B$, then$\eps_m(e_1)-\eps_m(e_2)=\eps_m\mu(b)=0$. It's clear that$d_m\omega=\eps_m$and that$\imm\mu'\s\ker d_m$. On the other hand, if$[(e,b')]\in\ker d_m$, then$\eps_m(e)=0$, so$e=\mu(b)$for some$b\in B$; therefore$[(e,b')]=[(0,b'+\beta(b))]$and$\mu'(b'+\beta(b))=[(e,b')]$. It's clear that$\imm d_m([(e,b')])\s\ker\eps_{m-1}$; if$\eps_{m-1}(e')=0$, then$\eps_{m_2}(e)=e'$for some$e\in E_m$and$d_m([e,0])=e'$. \end{itemize} Thus, we obtain a commutative diagram with exact rows $\xymatrix{ \mf{E}:\quad 0\ar[r]&B\ar[r]^\mu\ar[d]_\beta&E_m\ar[r]^{\eps_m}\ar[d]_\omega&E_{m-1}\ar[r]^{\eps_{m-1}}\ar[d]_=&\cdots\ar[r]&E_1\ar[r]\ar[d]_=&A\ar[r]\ar[d]_=&0\\ \mf{E}_\beta:\quad 0\ar[r]&B'\ar[r]^{\mu'}&E_{m\beta}\ar[r]^{d_m}&E_{m-1}\ar[r]^{\eps_{m-1}}&\cdots\ar[r]&E_1\ar[r]&A\ar[r]&0, }$that is unique. In the same manner as before, we obtain a well-defined homomorphism$\beta_*([\mf{E}])=[\mf{E}_\beta]$. \nl \textbf{3. The addition.} Note that for two$m$-extensions$\mathfrak{E}=(E_i)_{i=1}^m$,$\mathfrak{E}'=(E'_i)_{i=1}^m$of$A$by$B$, by taking the direct sum of the sequences, we obtain an$m$-extension of$A\oplus A$by$B\oplus B$: $\xymatrix{ \mathfrak{E}\oplus\mathfrak{E}':\quad 0\ar[r]&B\oplus B\ar[r]&E_m\oplus E'_m\ar[r]&\cdots\ar[r]&E_1\oplus E'_1\ar[r]&A\oplus A\ar[r]&0. }$ By defining homomorphisms$\Delta: A\to A\oplus A$,$\Delta(a)=(a,a)$and$\nabla: B\oplus B\to B$,$\nabla(b,b')=b+b'$, we can now define $$[\mathfrak{E}]+[\mathfrak{E'}]:=\Delta^*\nabla_*([\mathfrak{E}\oplus\mathfrak{E}']),\quad [\mathfrak{E}],[\mathfrak{E'}]\in\Yext_\Lambda^m(A,B),$$ which is well-defined by well-definedness of the induced homomorphisms. This composition in fact allows for an abelian group structure on$\Yext$as we shall see in the next problem. \nl We will now find a representative of a neutral element under this composition. Consider the$m$-extension of$A$by$B$given by $\xymatrix{ 0:\quad 0\ar[r]&B\ar[r]^1&B\ar[r]&0\ar[r]&\cdots\ar[r]&0\ar[r]&A\ar[r]^1&A\ar[r]&0. }$ It is clearly exact, and there is reason behind its name as we shall now see. Let$\mf{E}$be the extension (\ref{cigar2}). With$\Delta$and$\nabla$as defined above, recall how we defined the induced homomorphisms and consider the following diagram with exact rows:$\xymatrix{ 0\ar[r]&B\oplus B\ar[r]^{(\mu,1)}\ar[d]_\nabla&E_m\oplus B\ar[r]^{\hspace{6pt}\binom{\eps_m}{0 }}\ar[d]_\omega&E_{m-1}\ar[r]^{\eps_{m-1}}\ar[d]_=&\cdots\ar[r]^{\hspace{-6pt}(\eps_2,0)}&E_1\oplus A\ar[r]^{(\eps,1)}\ar[d]_=&A\oplus A\ar[r]\ar[d]_=&0\\ 0\ar[r]&B\ar[r]^{\hspace{-16pt}\mu'}&(E_m\oplus B)_\nabla\ar[r]^{\hspace{8pt}d_m}&E_{m-1}\ar[r]^{\eps_{m-1}}&\cdots\ar[r]^{\hspace{-6pt}(\eps_2,0)}&E_1\oplus A\ar[r]^{(\eps,1)}&A\oplus A\ar[r]&0, }$the upper row being$\mf{E}\oplus 0$and the lower row being$(\mf{E}\oplus 0)_\nabla$. Consider now$\xymatrix{(\mf{E}\oplus 0)_\nabla:\quad 0\ar[r]&B\ar[r]^{\hspace{-16pt}\mu'}\ar[d]_=&(E_m\oplus B)_\nabla\ar[r]^{\hspace{8pt}d_m}\ar[d]_\xi&E_{m-1}\ar[d]_=\ar[r]^{\eps_{m-1}}&\cdots\ar[r]^{\hspace{-6pt}(\eps_2,0)}&E_1\oplus A\ar[r]^{(\eps,1)}\ar[d]_=&A\oplus A\ar[r]\ar[d]_=&0\\\mf{E}':\quad 0\ar[r]&B\ar[r]^{-\mu}&E_m\ar[r]^{\eps_m}&E_{m-1}\ar[r]^{\eps_{m-1}}&\cdots\ar[r]^{\hspace{-6pt}(\eps_2,0)}&E_1\oplus A\ar[r]^{(\eps,1)}&A\oplus A\ar[r]&0, }$ with$\xi:(E_m\oplus B)_\nabla\to E_m$defined by$[((e,b),b')]\mapsto e+\mu(b-b')$. We will show that it makes the diagram commute. It is well-defined, since if$((e_1,b_1),b'_1)-((e_2,b_2),b_2')=((\mu(b),b'),b+b')$for$e_1,e_2\in E_m$and$b,b',b_1,b_1',b_2,b_2'\in B$(using the definition of$(E_m\oplus B)_\nabla$), then$e_1-e_2=\mu(b)$,$b_1-b_2=b'$and$b_1'-b_2'=b+b'$. Thus $$e_1+\mu(b_1-b'_1)-(e_2+\mu(b_2-b'_2))=\mu(b)+\mu(b_1-b_2)-\mu(b'_1-b'_2)=\mu(b)+\mu(b')-\mu(b+b')=0,$$ so$\xi$is well-defined. For$[((e,b),b')]\in (E_m\oplus B)_\nabla$, then $$d_m([((e,b),b')])=\binom{\eps_m}{0}(e,b)=\eps_m(e)=\eps_m(e+\mu(b-b'))=\eps_m\xi([((e,b),b')])$$ and$\xi\mu'(b)=\xi([((0,0),b)])=-\mu(b)$for all$b\in B$. The lower row is exact, and thus$(\mf{E}\oplus 0)_\nabla\sim\mf{E}'$, so$\nabla_*([\mathfrak{E}\oplus0])=[\mf{E}']$. Now consider the diagram $\xymatrix{\mf{E}':\quad 0\ar[r]&B\ar[r]^{-\mu}&E_m\ar[r]^{\eps_m}&\cdots\ar[r]&E_2\ar[r]^{\hspace{-15pt}(\eps_2,0)}&E_1\oplus A\ar[r]^{(\eps,1)}&A\oplus A\ar[r]&0\\ (\mf{E}')^\Delta:\quad 0\ar[r]&B\ar[r]^{-\mu}\ar[u]^=&E_m\ar[u]^=\ar[r]^{\eps_m}&\cdots\ar[r]&E_2\ar[u]^=\ar[r]^{\hspace{-15pt}d_2}&(E_1\oplus A)^\Delta\ar[r]^{\hspace{18pt}\eps'}\ar[u]&A\ar[u]^\Delta\ar[r]&0, }$ and then the diagram$\xymatrix{\mf{E}:\quad 0\ar[r]&B\ar[r]^{\mu}&E_m\ar[r]^{\eps_m}&\cdots\ar[r]&E_2\ar[r]^{\eps_2}&E_1\ar[r]^{\eps}&A\ar[r]&0\\ (\mf{E}')^\Delta:\quad 0\ar[r]&B\ar[r]^{-\mu}\ar[u]^{-1}_\cong&E_m\ar[u]^=\ar[r]^{\eps_m}&\cdots\ar[r]&E_2\ar[u]^=\ar[r]^{\hspace{-15pt}d_2}&(E_1\oplus A)^\Delta\ar[r]^{\hspace{18pt}\eps'}\ar[u]^\Psi&A\ar[u]^=\ar[r]&0, }$ with$\Psi: (E_1\oplus A)^\Delta\to E_1$given by$\Psi((e,a),a')=e$for all$((e,a),a')\in (E_1\oplus A)\oplus A$such that$(\eps(e),a)=(a',a')$. The last diagram commutes, since$\eps\Psi(((e,a),a')=\eps(e)=a'=\eps'((e,a),a')$for such$((e,a),a')$and$\Psi d_2(e)=\Psi((\eps_2(e),0),0)=\eps_2(e)$for all$e\in E_2$. Thus$(\mf{E}')^\Delta\sim\mf{E}$since the upper row is exact, so we obtain $$[\mf{E}]+=\Delta^*\nabla_*([\mathfrak{E}\oplus0])=\Delta^*[\mf{E}']=[\mf{E}].$$ Since$\mathfrak{E}\oplus\mathfrak{E}'\sim\mathfrak{E}'\oplus\mathfrak{E}$by using a coordinate-swapping isomorphism, we obtain$+[\mf{E}]=[\mf{E}]$as well, so$0$is a representative of the neutral element of the addition, and$\xi+0=\xi$,$\xi\in\Yext_\Lambda^m(A,B)$. \subsection*{(b) \cite{hilton}, Exercise IV.9.6} \textit{Prove that the addition given above is compatible with the equivalence$\Ext_\Lambda^m\simeq\Yext_\Lambda^m$.} \nl Let$A$and$B$be$\Lambda$-modules; then the natural isomorphism$\theta: \Yext_\Lambda^m(A,B)\to\Ext_\Lambda^m(A,B)$is defined as follows. Let \begin{eqnarray*}\label{cigar3}\xymatrix{ \mathfrak{E}:\quad 0\ar[r]&B\ar[r]^\mu&E_m\ar[r]^{\eps_m}&\cdots\ar[r]&E_1\ar[r]^{\eps}&A\ar[r]&0 }\end{eqnarray*} be an$m$-extension of$A$by$B$and take a projective resolution$P_\bullet$of$A$. Then as$\Ext_\Lambda^m(A,B)$is isomorphic to the homotopy classes of morphisms of complexes$P_\bullet\to\Sigma_m B$, we only need to find a representative morphism for$\mf{E}$. Let$D_\bullet$be the chain complex given by $\xymatrix{ D_\bullet:\quad \cdots\ar[r]&0\ar[r]&B\ar[r]^\mu&E_m\ar[r]^{\eps_m}&\cdots\ar[r]&E_2\ar[r]^{\eps_2}&E_1\ar[r]&0\ar[r]&\cdots }$ with$B$in degree$m$and$E_1$in degree 0.$\mf{E}$can be considered as a surjective quasi-isomorphism of chain complexes$\xi: D_\bullet\to A$, using exactness of$\mf{E}$in the following way: $\xymatrix{ D_\bullet:\quad \cdots\ar[r]&0\ar[d]\ar[r]&B\ar[d]\ar[r]^\mu&E_m\ar[d]\ar[r]^{\eps_m}&\cdots\ar[r]&E_2\ar[d]\ar[r]^{\eps_2}&E_1\ar[d]^{\eps}\ar[r]&0\ar[r]&\cdots\\ A:\quad \cdots\ar[r]&0\ar[r]&0\ar[r]&0\ar[r]&\cdots\ar[r]&0\ar[r]&A\ar[r]^{\eps}&0\ar[r]&\cdots. }$ In the same way, we obtain a morphism of complexes$\omega: P_\bullet\to A$. There now exists a morphism$\lambda: P_\bullet\to D_\bullet$unique up to homotopy equivalence such that$\omega=\xi\lambda$since the chain complexes are bounded below and$P_\bullet$is projective. Since we also have a morphism of chain complexes$\psi: D_\bullet\to\Sigma_m B$given by $\xymatrix{ D_\bullet:\quad \cdots\ar[r]&0\ar[d]\ar[r]&B\ar[d]^1\ar[r]^\mu&E_m\ar[d]\ar[r]^{\eps_m}&\cdots\\ \Sigma_m B:\quad \cdots\ar[r]&0\ar[r]&B\ar[r]&0\ar[r]&\cdots, }$ we obtain a morphism of complexes$\psi\lambda: P_\bullet\to\Sigma_m B$. This is our representative, and we define$\theta([\mf{E}])=[\psi\lambda]$; it is well-defined and natural. The compatibility of the addition will follow once we prove for$[\mf{E}],[\mf{E}']\in\Yext_\Lambda^m(A,B)$that$\theta([\mf{E}]+[\mf{E}'])=\theta([\mf{E}])+\theta([\mf{E}'])$. As$\theta$is natural, we obtain $$\theta([\mf{E}]+[\mf{E}'])=\theta([\Delta^*\nabla_*([\mathfrak{E}\oplus\mathfrak{E}'])])=\Delta^*\nabla_*\theta([\mathfrak{E}\oplus\mathfrak{E}']),$$ using the same notation$\Delta^*$and$\nabla^*$for the homomorphisms induced from$\Delta: A\to A\oplus A$and$\nabla: B\to B\oplus B$over$\Ext$. \nl Let the extension$\mf{E}$, the complex$D_\bullet$and the morphisms$\xi$,$\omega$,$\lambda$and$\psi$given in the way above. In the case of another$m$-extension$\mathfrak{E}'$of$A$by$B$(with$E_i'$,$i=1,\ldots,m$being the modules in between$B$and$A$in the required exact sequence), we construct a chain complex$D'_\bullet$analoguously along with a surjective quasi-isomorphism$\xi': D'_\bullet\to A$, a morphism$\lambda': P_\bullet\to D'_\bullet$unique up to homotopy equivalence such that$\omega=\xi'\lambda'$and a morphism$\psi': D'_m\to\Sigma_m B$such that the diagrams $\xymatrix{&D_\bullet\ar[d]^\xi\ar[r]^\psi&\Sigma_mB&&D'_\bullet\ar[d]^{\xi'}\ar[r]^{\psi'}&\Sigma_mB\\ P_\bullet\ar[ur]^\lambda\ar[r]^\omega&A&&P_\bullet\ar[ur]^{\lambda'}\ar[r]^\omega&A }$ commute. With this, we obtain$\theta([\mf{E}])=[\psi\lambda]$and$\theta([\mf{E}'])=[\psi'\lambda']$. To consider$\mf{E}\oplus\mf{E}'$, note that we only need take direct sums of the complexes and the morphisms, since$P_\bullet\oplus P_\bullet$is a projective resolution of$A\oplus A$; since the diagram $\xymatrix{&D_\bullet\oplus D'_\bullet\ar[d]^{\xi\oplus\xi'}\ar[r]^{\psi\oplus\psi'}&\Sigma_m(B\oplus B)\\ P_\bullet\oplus P_\bullet\ar[ur]^{\lambda\oplus\lambda'}\ar[r]^{\omega\oplus\omega}&A\oplus A& }$ then commutes,$\lambda\oplus\lambda'$given by$(p,p')\mapsto(\lambda_n(p),\lambda'_n(p'))$for all$n\in\ZZ$and$p,p'\in P_n$etc., we obtain$\theta([\mf{E}\oplus\mf{E}'])=[(\psi\oplus\psi')(\lambda\oplus\lambda')]$; note here that all morphisms making the diagram commute are homotopy-equivalent to$\lambda\oplus\lambda'$. We then get $$\theta([\mf{E}]+[\mf{E}'])=\theta([\Delta^*\nabla_*([\mathfrak{E}\oplus\mathfrak{E}'])])=\Delta^*\nabla_*[(\psi\oplus\psi')(\lambda\oplus\lambda')]=[\nabla(\psi\oplus\psi')(\lambda\oplus\lambda')\Delta].$$ The representative of the homotopy class is a morphism of complexes$P_\bullet\to\Sigma_m B$. For$n\in\ZZ$, note \begin{eqnarray*}\nabla(\psi_n\oplus\psi'_n)(\lambda_n\oplus\lambda'_n)\Delta(p)&=&\nabla(\psi_n\oplus\psi'_n)(\lambda_n\oplus\lambda'_n)(p,p)\\ &=&\nabla(\psi_n\oplus\psi'_n)(\lambda_n(p),\lambda_n'(p))\\ &=&\nabla(\psi_n\lambda_n(p),\psi_n'\lambda_n'(p))\\ &=&\psi_n\lambda_n(p)+\psi_n'\lambda_n'(p)\end{eqnarray*} for all$p\in P_n$, so that$\theta([\mf{E}]+[\mf{E}'])=[\psi\lambda+\psi'\lambda']=[\psi\lambda]+[\psi'\lambda']=\theta([\mf{E}])+\theta([\mf{E}']),$as wanted. %Consider$\theta([\mathfrak{E}\oplus\mathfrak{E}'])$with$\mf{E}$being the extension (\ref{cigar3}) and$\mf{E'}=(E_i')_{i=1}^m$given analogously. Let$P_\bullet$be a projective resolution of$A$. In the same way as above,$\mathfrak{E}$and$\mathfrak{E}'$can be considered as surjective quasi-isomorphisms$\xi: D_\bullet\to A$and$\xi': D'_\bullet\to A$with$B$in degree$m$, and we obtain a morphism of complexes$\omega: P_\bullet\to A$from the projective resolution. Then there exists morphisms$\lambda: P_\bullet\to D_\bullet$and$\lambda': P_\bullet\to D'_\bullet$unique up to homotopy equivalence such that$\xi\lambda=\omega=\xi'\lambda'$(by the chain complexes being bounded below). The morphisms of chain complexes$\psi: D_\bullet\to\Sigma_mB$and$\psi': D'_\bullet\to\Sigma_mB$yield$\theta([\mathfrak{E}])=[\psi\lambda]$and$\theta([\mathfrak{E}'])=[\psi'\lambda']$. Note now that$P_\bullet\oplus P_\bullet$is a projective resolution of$A\oplus A$, yielding the morphism of complexes$\omega\oplus\omega: P_\bullet\oplus P_\bullet\to A\oplus A$given by$\omega\oplus\omega(c,d)=(\omega(c),\omega(d))$.$\mathfrak{E}\oplus\mathfrak{E}'$is a surjective quasi-isomorphism$\xi\oplus\xi': D_\bullet\oplus D'_\bullet\to A\oplus A$in disguise with$B\oplus B$in degree$m$. Then there exists a morphism$\lambda^\oplus: P_\bullet\oplus P'_\bullet\to D_\bullet\oplus D'_\bullet$unique up to homotopy equivalence such that$(\xi\oplus\xi')\lambda^\oplus=\omega\oplus\omega$, but since$\omega\oplus\omega=\xi\oplus\xi'(\lambda,\lambda')$, we obtain$\lambda^\oplus\simeq(\lambda,\lambda')$.%. Let %Let$R_\bullet$be a projective resolution of$A\oplus A$. In the same way as above,$\mathfrak{E}\oplus\mathfrak{E}'$can be considered as a surjective quasi-isomorphism$\xi^\oplus: D^\oplus_\bullet\to A\oplus A$with$B\oplus B$in degree$m$, and we obtain a morphism of complexes$\omega^\oplus: R_\bullet\to A\oplus A$from the projective resolution. Then there exists a morphism$\lambda^\oplus: R_\bullet\to D^\oplus_\bullet$unique up to homotopy equivalence such that$\xi^\oplus\lambda^\oplus=\omega^\oplus$. The morphism of chain complexes$\psi^\oplus: D^\oplus_\bullet\to\Sigma_m(B\oplus B)$yields$\theta([\mathfrak{E}\oplus\mathfrak{E}'])=[\psi^\oplus \lambda^\oplus]$. Now,$R_\bullet$can be split into two projective resolutions$P_\bullet$and$P'_\bullet$of$A$, using the projections$\pi: R_\bullet\to P_\bullet$and$\pi': R_\bullet\to P'_\bullet$; this leads to an abundance of new chain complexes and morphisms: %\begin{itemize}\itemsep-2pt % \item$\mathfrak{E}$and$\mathfrak{E}'$are once again surjective quasi-isomorphisms$\xi: D_\bullet\to A$and$\xi': D'_\bullet\to A$in disguise (note that$\xi^\oplus(c,d)=(\xi(c),\xi'(d))$by how$\xi^\oplus$arises), with$B$in degree$m$and$D_\bullet\oplus D'_\bullet=D^\oplus_\bullet$; % \item we possess morphisms of complexes$\omega: P_\bullet\to A$and$\omega': P'_\bullet\to A$defined by$\omega(p)=\pi_A^1\omega^\oplus(p,0)$and$\omega'(p)=\pi_A^2\omega^\oplus(0,p)$,$\pi_A^1$and$\pi_A^2$denoting the projections$A\oplus A\to A$and % \item morphisms of chain complexes$\psi: D_\bullet\to\Sigma_mB$and$\psi': D'_\bullet\to\Sigma_mB$with$\psi^\oplus=(\psi,\psi')$. %\end{itemize} %As there exist morphisms$\lambda: P_\bullet\to D_\bullet$and$\lambda': P'_\bullet\to D'_\bullet$such that$\xi\lambda=\omega$and$\xi'\lambda'=\omega'$, we can define$\alpha=\psi\lambda$and$\beta=\psi'\lambda'$. %Consider now the morphism of complexes$\gamma=\alpha\oplus\beta=(\psi\lambda,\psi'\lambda'): R_\bullet\to\Sigma_m(B\oplus B)$defined by$\gamma(c,d)=(\psi\lambda(c),\psi'\lambda'(d))$. Since$\omega^\oplus(c,d)=(\xi\lambda^\oplus_1(c,d),\xi'\lambda^\oplus_2(c,d))$, we obtain$\omega(p)=\xi\lambda^\oplus_1(p,0)$and$\omega'(p)=\xi'\lambda^\oplus_2(0,p)$, so$\lambda^\oplus_1\simeq\lambda$and$\lambda^\oplus_2\simeq\lambda'$. There thus exist homotopies$\q_1,\q_2$such that$\lambda^\oplus_1-\lambda=\q_1\partial^{P}+\partial^{P}\q_1$and$\lambda^\oplus_2-\lambda'=\q_2\partial^{P'}+\partial^{P'}\q_2\$ \bibliographystyle{phjcp} \bibliography{bib} \end{document}