We take cohomology with coefficients in a commutative ring $R$ and we write $\otimes$ for $\otimes_R$.  Let $X = \Sigma Y = Y \wedge S^1$. How do I see that the cup product$$\tilde{H}^p(X) \otimes \tilde{H}^q(X) \to \tilde{H}^{p+q}(X)$$is the zero homomorphism?
120714
Let $X$ be a CW-complex. Let $\Sigma$ be suspension. Let $R$ be a commutative ring. Is the cup product of  $$ H^*(\Sigma X;R)$$ trivial? How to prove? Where can I find the result?
49979
What is the cohomology ring  $$ H^*(\mathbb{R}P^\infty;\mathbb{Z})?$$ $$ H^*(\mathbb{R}P^n;\mathbb{Z})?$$  for mod 2 coefficient, the answer is on Hatcher's book and . For the graded module structure, it is obtained from the homology struture.
53094
How to show that $\mathbb Z_{mn}$ is isomorphic to $\mathbb Z_m \times\mathbb Z_n$ when $m$ and $n$ are coprime? It is easy to show that the natural map from $\mathbb Z_{mn}$ to $\mathbb Z_m \times\mathbb Z_n$ is a ring homomorphism. How to show that it is bijective?
98474
Question. Do you know a specific example which demonstrates that the tensor product of monoids (as defined below) is not associative?    Let $C$ be the category of algebraic structures of a fixed type, and let us denote by $|~|$ the underlying functor $C \to \mathsf{Set}$. For $M,N \in C$ we have a functor $\mathrm{BiHom}(M,N;-) : C \to \mathsf{Set}$ which sends an object $K \in C$ to the set of bihomomorphisms $M \times N \to K$, i.e. maps $|M| \times |N| \to |K|$ which are homomorphisms in each variable when the other one is fixed. Then one can show as usual that $\mathrm{BiHom}(M,N;-)$ is representable and call the universal bihomomorphism $M \times N \to M \otimes N$ the tensor product of $M,N$. This is a straight forward generalization of the well-known case $C=\mathsf{Mod}(R)$ for a commutative ring $R$.  Actually, this is a special case of a more general tensor product in concrete categories, studied in the paper "Tensor products and bimorphisms", Canad. Math. Bull. 19 (1976) 385-401, by B. Banaschewski and E. Nelson.   Here are some examples: For $C=\mathsf{Set}$, the tensor product equals the usual cartesian product. This is also true for $C=\mathsf{Set}_*$. For $C=\mathsf{Grp}$, we get $G \otimes H \cong G^{\mathsf{ab}} \otimes_{\mathbb{Z}} H^{\mathsf{ab}}$, using the . (This differs from the "tensor product of groups" studied in the literature). The case $C=\mathsf{CMon}$ is very similar to the well-known case $C=\mathsf{Ab}$ and is spelled out ; namely, we have internal homs and therefore a hom-tensor-adjunction. The same is true for $C=\mathsf{Mod}(\Lambda)$ for a commutative algebraic monad $\Lambda$, see , Section 5.3.  Note that the tensor product is commutative, and that it commutes with filtered colimits in each variable. However, the case $C=\mathsf{Grp}$ shows that it does not have to commute with coproducts. In particular, tensoring with some object is no left adjoint. Also, the free object on one generator is not a unit in general:  Let us consider $C=\mathsf{Mon}$. Then, we have  $\mathbb{N} \otimes M = M / \{ (mn)^p = m^p n^p \}_{m,n \in M, p \in \mathbb{N}}$  The usual proof of the associativity of the tensor product breaks down: There is a map $\beta : M \times (N \otimes K) \to (M \otimes N) \otimes K$ mapping $(m, n \otimes k) \mapsto (m \otimes n) \otimes k$, which is a homomorphism in the second variable. But what about the first variable? The equation $\beta(mm',t) = \beta(m,t) \beta(m',t)$ is clear if $t \in N \otimes K$ is a pure tensor. But for $t=(n \otimes k) (n' \otimes k')$ we end up with the unlikely equation  $((m \otimes n) \otimes k) ((m' \otimes n) \otimes k) ((m \otimes n') \otimes k') ((m' \otimes n') \otimes k')$ $=((m \otimes n) \otimes k)  ((m \otimes n') \otimes k') ((m' \otimes n) \otimes k) ((m' \otimes n') \otimes k')$
107939
Hypothesis: Let $G$ and $H$ be defined in terms of the following presentations:  $$ G \cong \left\langle a,b \mid abab^{-1}\right\rangle $$  $$ H \cong \left\langle c,d \mid c^2 d^2 \right\rangle $$  Goal: Show that $G \cong H$.  Hint: Show that $G \cong \pi_1(K) \cong H$ where $\pi_1(K)$ denotes the fundamental group of the Klein Bottle.  Attempt:  (1) First I tried ignoring the hint and directly showing that $abab^{-1} = e$ if and only if $c^2d^2 = e$.  Here all I could obtain is that $abab^{-1} = e \implies ba^{-1} b^{-1}a^{-1} = e$ as well as that $dc^{-1}d^{-1}cd^2 = e$  (2) Using the hint, all I know at this point in my studies is that the Klein bottle $K$ is formally defined as the quotient space of $[0,1] \times [0,1]$ arising from the equivalence relations  $$ (s,0) \equiv (s,1) $$  and  $$ (0,t) \equiv (1,1-t) $$  From here I'm not sure how to proceed.
79217
Possible Duplicate:          In  the following claim was mentioned:  $R/I \otimes_R M \cong M / IM$  So I tried to prove it. Can you help me finish my proof? Thanks!    We recall that $M \otimes_R -$ is a covariant right-exact functor that it is exact if $M$ is flat and we observe that the following is an exact sequence: $$ 0 \to IM \xrightarrow{i} M \xrightarrow{\pi} M / IM \to 0$$  $R/I$ is an $R$ module since $R/I$ is a subring of $R$ closed with respect to multiplication from $R$ but it is not necessarily flat hence we only get exactness on one side:  $$  (R/I) \otimes_R IM \xrightarrow{id \otimes i} (R/I) \otimes_R M \xrightarrow{id \otimes \pi} (R/I) \otimes_R M / IM \to 0$$  Then $$ \mathrm{Im(id \otimes \pi)} = (R/I) \otimes_R M / IM  \cong (R/I) \otimes_R M / \mathrm{Ker} (id \otimes i) $$  so we want to show two things:   (i) that $\mathrm{Ker} (id \otimes i) = \{0\}$. To this end let $r + I \otimes im \in (R/I) \otimes_R  IM $ and assume $ id \otimes i(r + I \otimes im ) = r + I \otimes im = 0 + I \otimes 0$. I'm not sure how to proceed from here.   (ii) and that $M /IM \cong (R/I) \otimes_R M / IM $  Can you help me? Thanks.
114339
Let $(A_i)_{i\in I}$ be a family of right $R$-modules and $M$ be a left $R$-module, where $I$ is an index set. The natural homomorphism $$\varphi:(\prod_{i\in I}A_i)\otimes_RM\to \prod_{i\in I}(A_i\otimes_RM)$$ given by $(a_i)\otimes m\mapsto(a_i\otimes m)$ is not always a bijection. It is  easy to obtain $\varphi$ is a surjection provided $M$ is finitely generated. If we assume that $M$ is finitely presented, can we prove $\varphi$ is a bijection?
54656
I have a finite abelian group $A$ and I have shown that $A\otimes\mathbb{Q} \cong \{0\}$, the trivial group.  Now I want to show that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$, i.e we only have the trivial zero group homomorphism from $A$ to $\mathbb{Q}$.  The question I'm trying to solve is phrased in such a way that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$ should fall out almost automagically once you have that the tensor product collapses but I'm really stuck on this even though some people I talked to said it was really easy.  Some thoughts:   I've tried to assume the existence of a nontrivial homorphism $A \to    \mathbb{Q}$ to generate a contradiction but with no success. We have the Tensor-hom adjunction theorem stating that for abelian groups $A,B,C$ we have a natural bijection  $$\text{Hom}_{\textbf{Ab}}(A\otimes B,C) \cong \text{Hom}_{\textbf{Ab}}(B ,\text{Hom}_{Ab}(A,C))$$ and by plugging in our particular groups we get that:   $$\text{Hom}_{\textbf{Ab}}(A\otimes \mathbb{Q},\mathbb{Q})\cong\text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q}))$$  Since $A\otimes\mathbb{Q} \cong \{0\} \implies \text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q})) = \{0\}$ so we only have the trivial morphism from $\mathbb{Q} \to \text{Hom}_{Ab}(A,\mathbb{Q})$ but I don't think this implies that $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ and I'm not sure if this reasoning helps me at all.
42711
Let $R$ be a commutative ring with $1 \neq 0$.  It is known that $R^{n} \hookrightarrow R^{m}$ implies $n \leq m$ and $R^{n} \twoheadrightarrow R^{m}$ implies $n \geq m$, and I might use this without mentioning (though I don't think I did). Isomorphisms are $R$-module maps if not mentioned otherwise.  Let $m \geq n$ and consider $R^{n} = R^{n} \times 0 \times \cdots \times 0 \subseteq R^{m}$. We have $\text{Hom}_{R}(R^{n}, R^{m}) \simeq M_{m \times n}(R)$ via $\phi \mapsto M(\phi) = (c_{ij})$ where $\phi(e_{j}) = \sum_{i=1}^{m}c_{ij}e_{i}$. Take any $R$-linearly independent $y_{1}, \cdots, y_{n} \subseteq R^{m}$ and define an $m \times n$ matrix $A = (a_{ij})$ via $y_{j} = b_{1j}e_{1} + \cdots + b_{mj}e_{m}$.  If we understand $e_{j}$ with $1 \leq j \leq n$ as $n \times 1$ column matrix in $R^{n}$ when we multiply an $m \times n$ matrix to it on its left, then $Ae_{j} = y_{j}$ for $1 \leq j \leq n$. Thus, we can understand this matrix as an injection $A : R^{n} \hookrightarrow R^{m}$.  By definition $A$ is an $m \times n$ matrix, but if what I wrote is correct, then $A|_{R^{n}}$ must be an isomorphism $R^{n} \simeq \bigoplus_{j=1}^{n} Ry_{j}$. I am not trying to find an invertible matrix to $A$, but is it okay to think of it as an embedding like how I described?
109254
What I want to ask is the proof of the Proposition 10. in "Linear Representations of Finite Groups" by Jean-Pierre Serre.   Let $\rho_i : G \rightarrow GL(W_i)$ be the distinct irreducible representations of $G$.$(1 \le i \le h)$ We can extend $\rho_i$ to $\widetilde{\rho_i} : \mathbb{C}[G] \rightarrow \text{End}(W_i)$ by $\widetilde{\rho_i}(\sum_{g \in G} a_g g) = a_g \sum_{g \in G} \rho_i(g)$.   Then $\widetilde{\rho} = (\widetilde{\rho_i}) : \mathbb{C}[G] \rightarrow \prod_{i=1}^{h} \text{End}(W_i) \cong \prod_{i=1}^h M_{n_i}(\mathbb{C})$ is a homomorphism.  Then Proposition 10 states that it is an isomorphism.  Since dimensions of domain and codomain are same, it suffices to show that $\widetilde{\rho}$ is surjective.   In the proof of the book,   If $\widetilde{\rho}$ is not surjective, then there exists a nonzero linear form on $\prod M_{n_i}(\mathbb{C})$ vanishing on the image of $\widetilde{\rho}$. This gives a nontrivial relation on the coefficients of the representations $\rho_i$ which is impossible because of the orthogonality formulas of 2.2.  But I can not follow the above three lines in the proof. I will appreciate it if you give an explanation for the proof.   Thank you.
24182
This question has been modified...  Problem  Given Hilbert spaces.  In general, their algebraic tensor product isn't complete: $$\mathcal{H}\hat{\otimes}\mathcal{K}=\mathcal{H}\otimes\mathcal{K}\iff\dim\mathcal{H}&lt;\infty\lor\dim\mathcal{K}&lt;\infty$$ How to prove this from scratch?  Attempt  Choose orthonormal bases: $$\mathcal{S}\otimes\mathcal{T}:=\{\sigma\otimes\tau:\sigma\in\mathcal{S},\tau\in\mathcal{T}\}$$  One obtains some candidates: $$\sigma_k\otimes\tau_l\in\mathcal{S}\otimes\mathcal{T}:\quad\sum_{kl=1}^\infty\frac{1}{kl}\sigma_k\otimes\tau_l\quad\sum_{k=1}^\infty\frac{1}{kl}\delta_{kl}\sigma_k\otimes\tau_l$$ However the former one drops out: $$\sum_{kl=1}^\infty\frac{1}{kl}\sigma_k\otimes\tau_l=\left(\sum_{k=1}^\infty\frac{1}{k}\sigma_k\right)\otimes\left(\sum_{l=1}^\infty\frac{1}{l}\tau_l\right)$$ So it is not obvious at all wether the latter one works out!  Reference  Build-up on:
58810
I was studying the following example from Atiyah &amp; MacDonald's Introduction to Commutative Algebra:   Let $x$ be the non-zero element in $N := \mathbf{Z}/ 2\mathbf{Z}$, $M := \mathbf{Z}$, and $M' := 2 \mathbf{Z}$. The element $2 \otimes x$ is zero in $M \otimes N$, but non-zero in $M' \otimes N$. I suppose this can be seen by the fact that $2 \otimes x$ generates $M' \otimes N$, and the tensor-product $M' \otimes N$ is non-zero.   However, I was wondering, what one would do if that element weren't a generator of $M' \otimes N$. So my question is:   What are the methods to prove that an element is non-zero in a tensor-product of modules?   Thanks a lot!
79013
Let $R = \mathbb{C}[t]$ be a ring of polynomials in variable $t$ with coefficients in the field of complex numbers $\mathbb{C}$ and let $$N = R[x]/(tx-t).$$  I claim that $N$ is not a flat $R$- module. If we consider the exact sequence $$0 \rightarrow(t) \rightarrow R$$ such that the ideal $(t)$ is viewed as an $R$-module.   We know that the above sequence exact iff the map $(t) \rightarrow R$ is injective. Let us assume that N is flat then this means that $N \otimes  (t) \rightarrow N \otimes R$ has to be injective. How do I go from here?   Thanks.
108020
Let $A$, $B$ be two reduced $k$-algebras. Then if an element of the form $$\sum a_{i}\otimes b_{j}$$ is nilpotent, we can compose it with any $k$-homomorphism $f$ from $A$ to $k$ to get a homomorphism from $A\otimes B$ to $B$. This map's image is $$\sum f(a_{i})\otimes b_{j}$$must be nilpotent as well. So if we assume $b_{j}$ is linearly independent in the first place, we may reason that $f(a_{i})=0$. This showed that $a_{i}$ are in the Jacobson radical of $A$. However, we only know $\mathrm{Nil}(A)=0$, and it is not clear to me how to pass from the Jacobson radical to the nilradical without using some commutative algebra machinery.   Here is what I think that might work. If $A$ is of finite type, then by Nullstellensatz we can assert the above is true. But if $A$ is not of finite type, then I can write $A=k[x_{1},...,x_{n}][X]$, where $n$ is the transcendence degree of $k(A)$ and $k[X]$ is finite over $k$. Then I should be able to pass from $k[X]$ to $A$ by inductive argument on the transcendental degree of $k(A)$, since prime ideals are maximal in a UFD (I am working with $k[x]$ by introducing $[x]$).   The proof I have for the case of finite type is not satisfying. Given $I\subset k[X]$, using Nullstellensatz one can show that $$I_{X}V_{X}[I]=\sqrt{I}$$ where $V_{X}[I]$ is the set of all points in $X$ on which $I$ vanishes, and $I_{X}(V_{X}[I])$ be the ideal in $k[X]$ which vanishes on $V_{X}[I]$. Granting this, let $I=(a_{i})$, then we want to show $I=0$, which is $V_{X}[I]=X$. This follows since if $U\subset W$, then $V_{X}[U]\supset V_{X}[W]$. Since $V_{X}[I]\supset V_{X}[m_{x}]$ for all maximal ideal corresponding to points, it must be $X$. This proof is unnecessarily winding by using the ideal-variety correspondence. Ideally I should be able to construct a pure algebraic proof.   The thing not clear to me is whether the above strategy really works. Namely I am not sure if I can really induct on the transcendental degree of $k(A)$. To make matters simple assume we showed the statement holds for $A$ with transdental degree $&lt;n$. For $A$ with transdental degree $n$ we can thus write it as $k[A'][x]$. Then any maximal ideal $m'$ of $k[A'][x]$ must correspond to a maximal ideal $m=m'\cap k[A']$ in $k[A']$. But it seems not possible to show that any prime ideal is maximal. Indeed considering the simple example of $k[x,y]$, where $(x)$ is a prime ideal but not maximal since $k[x,y]/(x)=k[y]$ is not a field. The problem is in the inducting process the base ring is not neccessarily a field, and so this strategy cannot work.   I am thinking maybe I can show $m'=m[f(x)]$ with $f(x)$ some irreducible polynomial in $k[A'][x]$. Because we know $k[A'][x]$ is noetherian by Hilbert's basis theorem, so in particular $m'$ must have a finite generating set. And we can reduce the number of irreducible polynomial to 1 since $m'$ is a maximal ideal. But I still do not know how to use this to finish the proof.
70945
I'm supposed to find all the ring homomorphisms $\mathbb Z[x]\rightarrow \mathbb Q$.  Here's my attmept: $\mathbb Z[x]$ is generated by $\left\{ x^n \right\} _{n\geq 0}$, so the images of the generators determines the homomorphism. Since ring homomorphisms also respect multiplication, the images of $1,x$ determine everything. Homomorphisms respect units, so the image of $x$ determines our homomorphism. Hence, the only homomorphisms are the evaluations: for each $q\in \mathbb Q$ we have $\mathrm{eval}_q:\sum_na_nx^n\mapsto \sum _n a_n q^n$.  Is this correct? If so, isn't the codomain completely irrelevant? It seems we can say the same for any ring $R$ in place of $\mathbb Q$.
113148
I am trying to understand the proof of Theorem 3.4.3 from Weibel's book Introduction to homological algebra.  The statement is the following.     Let $R$ be a ring. Given $R$-modules $A$ and $B$, an extension $\xi$ of $A$ by $B$ is an exact sequence $0\to B\to X\to A\to0$. Then the connecting homomorphism $$\delta:\text{Hom}(A,A)\to\text{Ext}^1(A,B) \tag{1}$$ induces the 1-1 correspondence $\theta$ between equivalence classes of extensions of $A$ by $B$ and $\text{Ext}^1(A,B)$, which is given by $\theta:\xi\mapsto\delta(\text{id}_A)$.   In the beginning of the proof it is written the following     Fix an exact sequence $0\stackrel{j}\to M\to P\to A\to0$ with $P$ projective. Applying $\text{Hom}(-,B)$ yields an exact sequence $$\text{Hom}(P,B)\to\text{Hom}(M,B)\stackrel{\delta}\to\text{Ext}^1(A, B)\to0.\tag{2}$$   I don't understand why the connecting homomorphism in the sequence (2) is denoted by the same letter as in (1). Are they really the same?
119186
let $G$ be   a multiplicative group of non-zero complex analysis.consider the group homomorphism $\phi:G\rightarrow G$ defined by $\phi(z)=z^4$.  1.Identify kernel of $\phi=H$.  2.Identify $G/H$  My try:  let $z\in \ker \phi$  then $\phi(z)=1\implies z^4=1$ let $z=re^{i\theta}\implies r^4\cos 4\theta =1;r^4\sin 4\theta =0$   then $r=1$ and $\tan 4\theta=0\implies 4\theta=0\implies \theta=\frac{n\pi}{2}$  Is it correct?  I cant proceed in the 2nd problem.  Any hints in this regard.