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6

The set $\{1\}$ isn't independent, because $2(1)=1+1=0$. (The $2$ here lives in $\Bbb{Z}$, so the "multiplication" is the module action; all $1$s and $0$s live in $\Bbb{Z}/2\Bbb{Z}$.)


5

$\mathbb R $ over $\mathbb Q$ is a vector space with dimension $2^{\mathbb N}$. and also $\mathbb R^n $ over $\mathbb Q$ is a vector space with dimension $2^{\mathbb N}$. So the additive group of $\mathbb R$ is isomorphic with additive group of $\mathbb R^n$. for example consider $f:\mathbb R\rightarrow \mathbb R^n$ be such isomorphism , now define new ...


4

That's an odd statement of a more general statement that may be easier to grasp. In any ring, the sum of all minimal left ideals is a two-sided ideal. (If there are no minimal left ideals, the sum is the empty sum $\{0\}$. This ideal is called the left socle of the ring and is denoted $Soc(_RR)$. There's no doubt at the start that $Soc(_RR)$ is a left ...


3

(2) is incomplete. Recall that not every element in the tensor product is a pure tensor. In order to construct the algebra structure, I suggest the following (well-known) reformulation: If $A$ is some $R$-module, then an $R$-algebra structure on $A$ is the same as $R$-linear maps $\eta : R \to A$ and $\mu : A \otimes_R A \to A$ such that certain diagrams ...


3

Suppose $a \in R$ is a non-invertible element. Suppose $f: K \longrightarrow R$ is an $R$-linear map such that $f(1) \neq 0$. Then $$ af(1)f \left( \frac{1}{af(1)} \right) = f \left( \frac{af(1)}{af(1)} \right) = f(1) $$ hence $1= a f \left( \frac{1}{af(1)} \right)$ and this contradicts that $a$ is not invertible. So it must be $f(1)=0$. And now for all $x, ...


2

This is not true. For example, $\{1\}$ and $\{2,3\}$ are minimal generating sets of the $\mathbb{Z}$-module $\mathbb{Z}$.


2

The inclusion $E\to E+F$ induces a map $\phi: E\to (E+F)/F$ given by $x\mapsto x+F$. This defines a homomorphism with kernel $E\cap F$. That is, $$E/E\cap F \cong Im \ \phi.$$ To show $\phi$ is surjective, we let $x+y+F\in (E+F)/F$ where $x\in E$ and $y\in F$. Observe that $x+y+F=x+F$ in $(E+F)/F$ since $y\in F$ and notice that $$\phi(x)=x+F=x+y+F.$$ Thus, ...


2

The answer http://mathoverflow.net/a/10249 to Is it true that, as $\mathbb{Z}$-modules, the polynomial ring and the power series ring over integers are dual to each other? on MathOverflow shows that $$ \operatorname{Hom}(\mathbb{Z}^{\mathbb{N}},\mathbb{Z}) $$ is isomorphic to $\mathbb{Z}^{(\mathbb{N})}$. If $\mathbb{Z}^{\mathbb{N}}$ were free it would be ...


2

Năstăsescu and van Oystaeyen wrote two books, "Graded and Filtered Rings and Modules" and the more recent "Methods of Graded Rings". Even if you don't want to read the more specialised material in them, the first chapter of the former and the first two of the latter contain a fairly comprehensive introduction for most purposes.


2

Hint: every element is associate to a power of $7$.


2

The direct limit of $\Bbb Z$-modules: $$\Bbb Z/p\to \Bbb Z/p^2\to \Bbb Z/p^3\to\cdots$$ is not finitely generated as a $\Bbb Z$ module but every proper submodule is isomorphic to $\Bbb Z/p^k$ for some $k$. Edit: For more information, please see this wiki page.


2

Just as a heads-up, take any non-Noetherian ring $R$. Then $R$ is itself a finitely-generated $R$-module (generated by $1$), and so is $R/\mathfrak a$ for any ideal $\mathfrak a \trianglelefteq R$ (since it is again generated by $1$), but your ideal $\mathfrak a$ can be picked non-finitely generated if $R$ is picked not Noetherian. About your example, ...


2

Start with listing generators $x_1, x_2, \ldots$. Now by induction we will find submodules $M_1, M_2, \ldots, M_n$ such that each $M_i$ is a member of $S$, the sum $M(n) = M_1 + \cdots + M_n$ is direct in $M$, the module $M(n)$ is a direct summand of $M$, and we have $x_1, \ldots, x_n \in M(n)$. To start apply the given to $x_1 \in K = M$ as suggested by ...


2

Hint: for all matrices $A \in M_2(R)$ and all elements of $x \in M^2$, we have $$ \Phi(Ax) = A\Phi(x) $$ Assuming $R$ is a ring with unity, consider $x = (m,0)^T \in M^2$ and $A = \pmatrix{0&1\\1&0}$.


2

If you want an example where the modules are free over $\mathbb{Z}$, then let $G=\langle g\rangle$ be a cyclic group of order $2$, let $M=\mathbb{Z}G$ be the regular $\mathbb{Z}G$-module, and let $N=U\oplus V$ be a direct sum of two copies of $\mathbb{Z}$, where $g$ acts trivially on $U$ and by multiplication by $-1$ on $V$.


2

Hint: If we let $D=N$, there is a special homomorphism in $\mathrm{Hom}_R(N,N)$, namely the identity morphism on $N$. Since the sequence of Hom sets is exact, there must be some $R$-module map $N\to M$ which maps to the identity morphism in $\mathrm{Hom}_R(N,N)$. What does this mean with respect to the map $M\to N$ in the original sequence?


2

Yes. Let $R$ be ring of endomorphisms of $\mathbb{R}^2$, $I$ and $J$ be annihilators of subspaces spanned by $(1,0)$ and $(0,1)$, respectively. Let $\theta\in R$ be given by $\theta(x,y)=(y,x)$, and $\phi(f)=f\circ \theta$ for $f\in R$. Then $\phi$ is an automorphism of $R$ as a module over itself, and a bijection between $I$ and $J$. Thus, it induces ...


2

I believe a slightly more elementary version of the example of Marcin Łoś consists in taking the ring $R$ of $2 \times 2$ matrices over $\mathbb{R}$, say, and the left ideals $$ I = \left\{ \begin{bmatrix}a & 0\\ b & 0\end{bmatrix} : a, b \in \mathbb{R} \right\} , \qquad J = \left\{ \begin{bmatrix} 0 & a\\0 & b\\\end{bmatrix} : a, b \in ...


2

Well this is certainly wrong as long as you don't assume M to be finitely generated (just take an infinite dimensional vector space). If M is finitely generated this should be true, even without the assumption that supp(M) is finite (which will rather be a consequence). First note that this is obviously true if R is artinian,since M is a quotient of some ...


2

Yes, this is very old work. In 1935, Koethe proved that the modules of Artinian principal ideal rings are all direct sums of cyclic submodules. Of course, your example falls into this category. In fact, all of the proper quotients of a principal ideal domain are Artinian principal ideal rings (in fact they are also self-injective, hence quasi-Frobenius.) If ...


2

No. If $V$ is any vector space with underlying set $S$, then it is a quotient of $K^{\oplus S}$, which has $S$ as a basis. But the existence of bases in arbitrary vector spaces is equivalent to AC (by a result by Andreas Blass).


2

In Definition 2 the action of $k$ on $M$ induced by the action of $k$ on $A$ can be different from the given $k$-action on $M$.


2

The claim is not true. If $A$ is an integral domain which has a non-trivial Picard group, there is some invertible $A$-module $M$ which is not free. But then $M$ has no quotient isomorphic to $A$, since any epimorphism $M \to A$ is an isomorphism (using that $M$ is locally free of rank $1$). However, we can prove the following: Let $M$ be a finitely ...


1

$\DeclareMathOperator{\Hom}{\operatorname{Hom}}$There is a well-known criterion for $\Hom$ to vanish: Proposition: Let $A$ be a Noetherian ring, $M, N$ f.g. $A$-modules. Then $\Hom_A(M, N) = 0$ iff $\text{ann}_A(M)$ contains a nonzerodivisor on $N$. Taking $N = A/p$ for $p \in V(\text{ann}_A(M))$ gives that $\Hom_A(M, A/p) \ne 0$, i.e. some nonzero ...


1

As you say in your reasoning, the $A$-module structure on $M$ induces a $k$-module structure. So does the $B$-module structure. But without added conditions these two $k$-module structures could be different. For example, take $A=B=M=\mathbb{C}$ with the bimodule structure $$a\cdot m\cdot b=am\overline{b},$$ where $\overline{b}$ is the complex conjugate of ...


1

The first definition is what you might call a "ring" bimodule. It only asks for $A$ to be a ring, and it doesn't mind if $A$ has any $k$ algebra structure. I would call your second definition an algebra-bimodule, but I think you're missing an axiom. Not only does $M$ have a $k$-module structure, but this structure is compatible with $A$'s $k$-structure! You ...


1

If the isomorphism $M/L\cong M/N$ is canonical it is easy to show that $M=N$. But in general is not true see the example: $M=\Bbb R[X]$ and $N=\Bbb R_n[X]$ and $L=\Bbb R_{n+1}[X]$ (as $\Bbb R$ module $=$ $\Bbb R$ vector space), it is clear that $L$ and $N$ are not isomorph, but $M/N\cong (x^k, k\geq n+1)\cong (x^k,k\geq n+2)\cong M/L$. $(x^k, k\geq n+1)$ ...


1

The paper that I am reading assumes rank of a module always exists. Hence, it may be using a different definition of rank, which may (or may not) coincide with the definition provided in the previous answer, when $M\otimes_RQ$ is free over $Q$. After I posted my question I found the following definition of rank in the book Syzygies (By E. Graham Evans, ...


1

the way you write it it is not correct, because you could choose $x=y=p$, with $p$ irreducible and not a unit. then you would have counterexample As far as i know there is a definition for "prime" in integral domains: $p \mbox{ is prime } :\Leftrightarrow \forall x,y \in R : p|xy \Rightarrow p|x \mbox{ or } p|y$. "prime" and "irreducible" are equivalent ind ...


1

Note that $$\text{End}_D(D^n)=\text{Mat}_n(D)=A$$ Taking the center on both sides yields $$Z(\text{End}_D(D^n))\cong D$$ But, the center of $\text{End}_D(D^n)$ is just the matrices which are $\text{Mat}_n(D)$-linear. So, by tracking this through the isomorphism $A\cong \text{Mat}_n(D)$, you see this turns into the statement $Z(\text{End}_D(D^n))\cong ...



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