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7

No, your argument isn't correct: if $1\otimes x=1\otimes y$ then not necessarily $x=y$. I'd do this as follows: if $M$ is a $\mathbb Z$-module, then $\mathbb Q\otimes_{\mathbb Z}M\simeq S^{-1}M$, where $S=\mathbb Z-\{0\}$. The isomorphism is given by $\dfrac ab\otimes x\mapsto\dfrac{ax}{b}$. This way $1\otimes(1,1,\dots)$ corresponds to the fraction ...


7

$\mathbb{Z}$-modules are precisely abelian groups. As every ring is an abelian group, it is a $\mathbb{Z}$-module. It is entirely possible to be a module over more than one ring. For example, if $M$ is an $R$-module then it is also an $S$-module for any subring $S$ of $R$ (you seem to be interested in the case where $M = R$). Another example is given by ...


6

For every nonzero $x, y\in\mathbb{Q}$, there are nonzero integers $m, n$ such that $$mx=ny$$ (where $mx, ny$ are interpreted in the obvious way). Now, any homomorphism $f$ between $(\mathbb{Q}, +)$ and another structure $(G, *)$ must preserve multiplication by integers: $$f(mx)=mf(x).$$ So if $(\mathbb{Q}, +)\cong(\mathbb{Q}^n, +)$, then $(\mathbb{Q}^n, +)$ ...


5

Since $M=\mathbb Z/n$ as a group is cyclic generated by $1$, an endomorphism $\psi:M\rightarrow M$ is determined by $\psi(1)$. If $\psi^2=-I_M$, then $a^2=-1$ in $M$. Thus, you're looking for all $n$ such that there is $a\in \mathbb Z$ such that $a^2 \equiv -1 \bmod n$. For instance, when $n$ is prime, this means that $n \equiv 1 \bmod 4$. But there are ...


5

It is enough to prove it preserves short exact sequences: $\;0\to M\to N\to P\to 0$. As the tensor product is right-exact, and $S^{-1}M\simeq M\otimes_A S^{-1}A$, it is even enough to prove it preserves injectivity. So consider an injective morphism $\varphi\colon M\to N$ and suppose $\;S^{-1}\varphi\Bigl(\dfrac ms\Bigr)=0$ in $S^{-1}N$. This means there ...


4

They aren't isomorphic. If $n>1$ then $\mathbb{Q}^n$ has a free finitely generated subgroup of rank at least $2$ (namely $\mathbb{Z}^n$). Now, every finitely generated subgroup of $\mathbb{Q}$ is cyclic, so they can't be isomorphic.


3

$\text{Ext}^1(-, -)$ sends direct sums in the first variable to direct products, and as mentioned in the comments, as an abstract abelian group $$S^1 \cong \left( \bigoplus_X \mathbb{Q} \right) \oplus \mathbb{Q}/\mathbb{Z}$$ where $X$ is uncountable. So it suffices to compute $\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$ and $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, ...


3

You actually don't want $y \in \ker h_2$. You have $f_2(y - f_1(w)) = f_2(y) = x$, and $$h_2(y-f_1(w)) = h_2(y) - h_2(f_1(w)) = h_2(y) - g_1(h_1(w)) = h_2(y) -g_1(z) = 0.$$ Exactly what we need.


2

Suppose first that $A$ is local with maximal ideal $\mathfrak m$. Then you want to show that if $\hat f:M/\mathfrak mM\to N/\mathfrak mN$ is injective $f$ is an isomorphism. Since this is an injective map of $k=A/\mathfrak m$ vector spaces of equal dimension it is also onto, and this means that $f$ itself is onto (use Nakayama). It suffices you show that an ...


2

For (1), I think it is easy to show that, if $v\in V$ is nonzero and $w\in V$, there exists $a\in E$ such that $va=w$. If $w\in \text{span}_k(v)$, then $a$ is just a scalar multiplication. If $w\notin\text{span}_k(v)$, then you can extend the $k$-linearly independent set $\{v,w\}$ to a basis of $V$, and then the rest should be trivial. Hence, $V$ is a ...


2

The localisation of a projective module is projective since a projective module is a direct summand of a free module, and localisation preserves direct summands and freeness. A counter-example for injective modules was built by E. Dade in 1981, and you can find it in his paper Localization of injective modules, in which he also gives a sufficient condition ...


2

We're assuming that both $A$ and $B$ are symmetric $R$-algebras, which means that $A\cong\operatorname{Hom}_R(A,R)$ as $A$-bimodules, and similarly for $B$. And we're assuming that $_AM_B$ is an $(A,B)$-bimodule finitely generated and projective over $A$ and over $B$. For left $A$-modules $_AX$ and $_AY$, there is a natural $R$-module homomorphism ...


2

Since $N$ is finitely generated, we have an exact sequence $$M\stackrel{f}{\rightarrow} N\rightarrow \mathbb{Z}^k\times T\rightarrow 1$$ where $T$ is finite and $k \ge 0$. If $f$ is not surjective then either $k\ge 1$ or $T$ is nontrivial. Since tensoring is right-exact, tensoring this sequence with $\mathbb{Z}/p$ gives: ...


2

It's an endomorphism of $\mathbb{C}G$ as a vector space, not as a $\mathbb{C}G$-module. I'm guessing that this comes from a development of inner products of characters, where the trace of this endomorphism is calculated, which doesn't require it to be a $\mathbb{C}G$-module endomorphism.


2

Theorem: For any ring $R$ and any right $R$-module $M$, the functor $M\otimes_ R - : R\textsf{-Mod}\to\textsf{Ab}$ commutes with products if and only if $M$ is finitely presented. This is Proposition 4.44 in Lam, Lectures on Modules and Rings. In your case, $M$ is finite-dimensional over $k$, so in particular finitely generated over $R$, but not ...


2

The intersection of any family of submodules is a submodule. For the kernels, it is true because the different $\ker u^k$ are linearly ordered (actually, it is a direct limit), hence if you take $x\in\ker u^k$, $y\in\ker u^l$ for some $k,l$, one of them in contained in the other, say $\ker u^k\subseteq\ker u^l$, hence $x+y$ exists in $\ker u^l$.


2

Yes. Use the following facts to prove it: kernels and images are submodules intersection of submodules are submodules. So $\operatorname{Im}(u^{\infty})$ is a submodule If $C$ is a chain of submodules, then $\bigcup C$ is a submodule. Since $\ker(u^n) \subset \ker (u^{n+1})$ for all $n$, $\ker (u^{\infty})$ is a submodule.


2

The answer to your Question 1 is that matrices don't induce homomorphisms in general. In the notation of your Question 2, the mapping $w_i \mapsto \sum a_{ij}w_j$ can only be expected to extend to a module homomorphism for arbitrary $(a_{ij})$, if the $w_i$ generate $M$ freely. For a counter-example take $A = M = \mathbb{Z}$, $w_1 = 1$ and $w_2 = 2$, and try ...


2

Say $A/MA=R\widehat{a}_1+\cdots+R\widehat{a}_n$, where $\widehat{a}_i$ is the residue class of $a_i\in A$ modulo $MA$. Then for $a\in A$ we have $\widehat{a}=r_1\widehat{a}_1+\cdots+r_n\widehat{a}_n$, so $a-(r_1a_1+\cdots+r_na_n)\in MA$. This shows that $A=(Ra_1+\cdots+Ra_n)+MA$, and by NAK we get $A=Ra_1+\cdots+Ra_n$.


2

Let $A = \mathbb Z$ and $M = \mathbb Z/2\mathbb Z$.


2

Take the regular bimodule $_RR_R$: it is free (hence projective) as a left/right $R$-module, but projective as an $R$-$R$-bimodule only if $\text{Hom}_{R-R}(_RR_R,-)\cong \text{HH}^0(R;-): M\mapsto \{m\in M\ |\ \forall r\in R: rm=mr\}$ is exact. Example where this is not the case: Consider $R = {\mathbb Z}[X]$. Identifying $R$-$R$-bimodules with ${\mathbb ...


1

You've done most of the work already. Just observe that $\ker(\phi) = \langle p^{n-1}\rangle \times \{0\} \subset R\times R$. So indeed $$(R\times R) / \ker(\phi) \,\,\,\,\simeq \,\,\,\,(R/\langle p^{n-1}\rangle) \times R\,\,\,\, \simeq \,\,\,\,\Bbb Z/p^{n-1} \Bbb Z \times \Bbb Z / p^n \Bbb Z$$


1

I think the following should work, but take it with a grain of salt. Let $\mathbb Z(p^\infty)$ denote the $p$-Prüfer group, that is, the injective envelope of $\mathbb Z(p)$. The module $I=\prod_p \mathbb Z(p^\infty)$ is an injective module containing your module $M=\prod_p \mathbb Z(p)$ as a submodule. Now consider the module $$ E = M + \bigoplus_p \mathbb ...


1

Consider first the case $M=R/Rp^k$ and the map $m\mapsto p^{k-1}m$.


1

Given a coalgebra, there is naturally a dual algebra. Namely, the coalgebra map $\Delta: A \rightarrow A \otimes A$ can be dualized to get a map $\Delta^*: (A \otimes A)^* \rightarrow A^*$. Composing with the natural map $\eta: A^* \otimes A^* \rightarrow (A \otimes A)^*$ gives the multiplication map. Explicitly, this natural map is given by $$\eta(\sum ...


1

The second hypothesis says $$R+IS/IS=S/IS\iff S=R+IS$$ As $S$ is a finite $R$-module, Nakayama's lemma says there exists an element $a\in I$ such that $(1+a)S\subset R $. As $I$ is contained in the radical of $R$, $1+a$ is a unit in $R$, so really $S\subset R$. As the reverse inclusion is in the hypotheses, this proves $S=R$.


1

I'll use the terminology of Hovey's book on Model Categories. If $X$ is an $R$-module, and $\iota:X\to I$ a monomorphism from $X$ to an injective module, then $X\oplus I$ is a cylinder object for $X$ with the maps $$X\oplus X\stackrel{\begin{pmatrix}1&1\\\iota&0\end{pmatrix}}{\to}X\oplus I\stackrel{\begin{pmatrix}1&0\end{pmatrix}}{\to}X.$$ All ...


1

Given any coalgebra $C$ over a field ${\mathbb k}$, the category of $C$-comodules fully embeds into the category of $C^{\ast}$-modules by sending a $C$-comodule $(M, \nabla: M\to C\otimes M$) to the $C^{\ast}$-module having $M$ as the underlying ${\mathbb k}$-vector space, and with $C^{\ast}$-action given by $C^{\ast}\otimes M\to C^{\ast}\otimes C\otimes ...


1

Yes and yes. You switch notation from $m$ to $m_0$, so I'm going to stick with $m\in M$ non-zero. You can deduce the maximality of the left ideal $A(m)$ from the second claim as follows. Define $\varphi:R\to M$ by $\varphi(r)=rm$. This is an $R$-module homomorphism whose image is a submodule of $M$ containing the non-zero element $m$, and hence must be all ...


1

Let $M$ be a coherent $A$-module and $S\subset A$ a multiplicative set. Then $S^{-1}M$ is a coherent $S^{-1}A$-module. If $N'$ is a finitely generated $S^{-1}A$-submodule of $S^{-1}M$, then there is a finitely generated $A$-submodule $N$ of $M$ such that $N'=S^{-1}N$. (Suppose that $N'$ is generated by $x_1/s_1,\dots,x_n/s_n$. Set $N=Ax_1+\cdots+Ax_n$. ...



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