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1

Yes, unless $V$ is trivial. I'll help you unpack the definitions, but I'll leave the meat of the problem to you. Observation. Let $V$ denote a $k$-module. Then $V$ is a simple $\text{End}_k V$-module iff: $V$ has at least two distinct $k$-submodules, Every non-trivial $k$-submodule of $V$ that is closed under the action of $\text{End}_k V$ ...


1

Hint: The answer is yes (as is often the case for finite dimensional vector spaces). Note that for any $v \in V \setminus \{0\}$, we may select maps $T_1,\dots,T_n \in A$ so that $\{T_j(v)\}$ forms a basis (or a spanning set, if you prefer).


0

Here is the important part. Let $v $ be an eigenvector for $h $. Then $$ hev = (2e+eh)v = (\lambda +2)ev, $$ so $ev $ is also an eigenvector for $h $ with eigenvalue $(\lambda+2) $. But $M $ is finite dimensional. Can you complete the argument?


2

Assuming we are additionally given that $0\le m<N$, then, yes, it is possible to reveal $m$: Just try the candidates one by one and check if $(m+r)^e\bmod B$ turns out to be $C$. Mathematically, this solves the problem. Also, we have found an explicit algorithm, albeit with exponential running time ($O(N)$, which is linear in $N$, but exponential in the ...


1

Yes, the inclusion of a factor into a direct product $M \to M \oplus N$ defined by $m \mapsto (m, 0)$ is always an injective homomorphism no matter what modules you choose for $M$ and $N$. In particular, you can inject $M \to M \oplus R[G]^n$ not just for some $n$, but in fact for any $n$. If $M$ and $N$ are $G$-modules then it would be a good exercise for ...


0

Easy: let $A$ be an integral domain (hence it is commutative), and $I$ a non-zero ideal. Then $$\operatorname{Hom}(A/I,A)=\operatorname{Ann}_A(I)=0.$$


2

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


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 ...


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 ...


1

Yes, they are. $A$ is a simple Artinian ring. Therefore it has, up to isomorphism, a unique simple right $A$-module $S$ and every $A$-module is isomorphic to a direct sum $S^{(I)}$ for some index set $I$. Moreover, the module $S$ has some finite dimension $n$ over $k$. (Explicitly, $A$ is isomorphic to $M_r(D)$ for a division ring $D$ which is ...


1

Yes, it can be confusing talking about modules one moment and then talking about representations the next. In your case there isn't much difference. In fact, the confusion shouldn't confuse you. A Lie algebra module is a vector space such that ... A Lie algebra module is irreducible is there are no invariant (non trivial) proper submodules. Since the ...


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}, ...


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 ...


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 'must be faulty' proof (or proofs, with the addendum) above looks (look) correct, and the original exercise, as it appeared in the text, was wrong. In fact - see http://www.math.rutgers.edu/~weibel/Hbook-corrections.html for corrections to Charles Weibel's Introduction to Homological Algebra. The page (currently) has links to the 1994 hard- and 1995 ...


1

I was getting to it, but the addendum convinces me totally. I don't think you will find this exact question in my book Acyclic Models, but you will find a lot of arguments of this sort.


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 $(\mathbb{Q}^n, +)$ ...


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.


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

If $M$ is a nontrivial projective module and $N=R^n$ for $n\geq 1$ and you want a nontrivial homomorphism $M\to N$, you could do the following. Since every projective module is (isomorphic to) a direct summand of a free module $F$, you would have an injective homomorphism $g:M\to F$. If $F$ has lower rank than $M$, then it is easy to come up with another ...


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 ...


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 ...


1

I believe the adjunction in question is likely the classic tensor hom adjunction $\text{Hom}_C(A \otimes_R B, X) \cong \text{Hom}_R(A, \text{Hom}_C(B, X))$. Where all the objects in question have the appropriate structures such that this makes sense (X is a module over C, A is a module over R, B is a bi-module over both R and C, and R, C are just rings). ...


-3

The actual isomorphism is (in category theory): hom(AxB,C) = hom(A,hom(B,C)) It is normally known as "currying". (the hom's are in Set, not in C or B) I'm a little confused why they call it "adjunction", since that means slightly different thing, but the word doesnt seem to explain your isomorphism.


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

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 ...


1

Given any $R$-modules $M$ and $N,$ there is at least one $R$-module homomorphism $M\to N.$ In particular, there is always the trivial homomorphism $m\mapsto 0_N.$


1

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


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 $S=R$.


0

Since $M$ is simple we have $M=R/\mathfrak m$ where $\mathfrak m$ is a maximal ideal of $R$. But $R/\mathfrak m$ is a finitely generated $k$-algebra, and by Zariski's Lemma it is a finite field extension of $k$.


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

It is true because $K$ is torsion-free and divisible. So I'll prove that, on an integral domain, a torsion-free, divisible module $E$ is injective. Let $I$ be an ideal in $A$. We may suppose $I\ne 0$. Using Baer's criterion, we have to prove that any homomorphism $f\colon I\to E$ can be extended to $A$. As $I\ne 0$, we can pick a non-zero element $x\in ...


1

Fortunately this has little do to with bifunctors. First of all, note that limits are functorial: given two diagrams $D$, $D' : \mathscr I → \mathscr D$ and a natural transformation $α : D → D'$, we can get a morphism $\lim α : \lim D → \lim D'$. This is easy to construct and write out in some explicit examples of limits (eg. try to do it for products). In ...


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 ...


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.


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.


1

Let $R$ be a commutative unitary ring, $M$ an $R$-module, $N$ a finitely generated $R$-module, and $f:M\to N$ a homomorphism. Suppose that $\bar f:M/\mathfrak mM\to N/\mathfrak mN$ is surjective for every maximal ideal $\mathfrak m$ of $R$. Then $f$ is surjective. $f$ surjective $\Leftrightarrow$ $f_{\mathfrak m}:M_{\mathfrak m}\to N_{\mathfrak m}$ ...


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: ...


-1

If you have an $N\times N$ matrix $A=[a_{n,m}]$ over a field, then the cofactor expansion of the determinant of $A$ gives you $$ \mbox{adj}(A)A=\mbox{det}(A)I, $$ where $\mbox{adj}(A)$ is the adjunct or adjugate matrix consisting of the cofactors of $A$. Therefore, $$ \mbox{adj}(\lambda I-A)(\lambda I-A)=p(\lambda)I $$ where ...


2

$V$ is a $k[T]$-module. $V \otimes $ Something is a ($k[T] \otimes$ Something) - module.


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 $T\colon M\to M$ be an $R$-homomorphism such that $T(x)\ne0$ for some $x\in M$. Then $x\notin\ker T$, which forces $\ker T=\{0\}$, because the $R$-submodules of $M$ are $\{0\}$ and $M$. Similarly the image of $T$ is an $R$-submodule of $M$ and it is not $\{0\}$, so…


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.


-1

The reason why this is not a nonexample: $k(x_i)_{i \in \mathbb{N}}$ is noetherian as a $k(x_i)_{i \in \mathbb{N}}$-module but not as a $k[x_i]_{i \in \mathbb{N}}$-module, since ideals in an ascending chain of $k[x_i]_{i \in \mathbb{N}}$ are $k[x_i]_{i \in \mathbb{N}}$-submodules of $k(x_i)_{i \in \mathbb{N}}$.


1

You might want take a look at adjoint matrix to get the idea.


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 ...


0

I think I have it figured out. Define \begin{align*} P^{\alpha} := p^{\alpha}R = \{p^{\alpha}r | r \in R\} \end{align*} and it should follow that $P^{\alpha}$ is an $R$-submodule and is isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$


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 ...


0

I'm not sure that the following considerations will be helpful. I believe that $p$ is a prime ideal in $R$. Also I believe that $(S^{-1}M)_p$ means $C(p)^{-1}M$, where $C(p) = R\setminus p$ (as sets; $C(p)$ is a multiplicatively closed subset due to the fact that $p$ is a prime ideal). So if $(S^{-1}M)_p\neq 0$ then there is $x\in M$ such ...



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