# Tag Info

6

If $A$ is any ring, then $A$ and $M_n(A)$ have equivalent categories of modules, and usually $A$ and $M_n(A)$ are not isomorphic. This is the simplest example of a Morita equivalence.

3

If $R$ is a commutative ring and $M$ is some $R$-module, then $R[x]$-module structures on $M$ extending the given $R$-module structure correspond 1:1 to $R$-linear endomorphisms of $M$. So, any $\mathbb{C}$-linear map $\mathbb{C}^2 \to \mathbb{C}^2$ (and surely, there are plenty of them!) corresponds to a $\mathbb{C}[x]$-module structure on $\mathbb{C}^2$.

2

There exist a lot of ways to show this. I think that the simplest is to say that we can define a $\Bbb{C}[x]$-module structure on $M$ by composing $$\Bbb{C}[x] \longrightarrow \Bbb{C} \longrightarrow \operatorname{End}(M)$$ where the map $\Bbb{C}[x] \longrightarrow \Bbb{C}$ is any ring morphism (for example $f \mapsto f(0)$). This can be generalized, saying ...

1

To make an abelian group $M$ into a module over $R$ we need a ring homomorphism from $R$ into the (non-commutative) ring of group endomorphisms of $R$. To make $\mathbf{C^2}$, a module we can construct a homomorphism that goes into the subring of vector space endomorphisms. Here we can easily specify a ring homomorphism $\mathbf{C}[X]\to ... 1 Suppose$M$were a finite-dimensional vector space over$\mathbb{C}$. Let$f : M \to M$be the action of$x$, ie$f(m) = x \cdot m$. Then by the Cayley–Hamilton theorem, there is a polynomial$p \in \mathbb{C}[x]$,$p \neq 0$(eg. the characteristic polynomial of$f$) such that$p(f) = 0$; in other words, for all$m \in M$,$(p(f))(m) = 0 = p(x) \cdot ...

1

No. Let $A = 0$. Then your question is whether, for every submodule $B \subseteq C$, we can find a submodule $L \subseteq C$ such that $C/L \cong B$. In other words, the question is whether every submodule of $C$ is isomorphic to a quotient module of $C$. This is true if $B$ is a direct summand, in which case we can take $L$ to be its complement, but not ...

1

Note that if either side is $\infty$ then the other is as well. So assume at least $\lambda(M)$ is finite. Let $$N=N_0\supset N_1 \supset \cdots \supset N_t=(0)$$ be a normal series series for $N$. By the correspondence theorem, any submodule of $M/N$ is of the form $L'=L/N$ for some submodule $L\subset M$ containing $N$. Then we have a normal series for ...

1

$aL=0$ and $bN=0$ implies $(ab)x=0$ for all $x\in M$:

1

Consider $R = \mathbb{Z}$ and $A=C= \mathbb{Z}/(p)$ and $B=\mathbb{Z}/(p^2)$ and the short exact sequence $$0 \to \mathbb{Z}/(p) \to \mathbb{Z}/(p^2) \to \mathbb{Z}/(p) \to 0$$ where the first morphism is given by $1\mapsto p$, the second is the projection mod$(p)$. Clearly for $p$ prime, this does not split, since $\mathbb{Z}/(p^2)$ is not isomorphic to ...

1

No, Let $A,B = \mathbb{Z}$ as $\mathbb{Z}$-modules. Let $M$ be the submodule of $A\oplus B$ generated by $(2,1)$ and let $N$ be the submodule generated by $(0,1).$ Then $M\oplus N \subset A\oplus B$ but $M$ is not a subset of $A$ or $B.$

1

No, in either cases really. Consider $V=\bigoplus_{p\in\Bbb P}\Bbb Z/p\Bbb Z$ (where $\Bbb P$ is all the prime integers). Now partition $\Bbb P$ into $A,B$ and $A',B'$ such that neither $A\subseteq A'$ nor $B\subseteq B'$, and consider the decompositions defined by $A,B$ and $A',B'$ as counterexamples.

1

$${}_pM=\bigoplus_{i=1}^t(p^{e_i-1})/(p^{e_i})\simeq\bigoplus_{i=1}^tR/(p).$$ Indeed, if $\bar x=x+(p^{e_i})$ is an element of $R/(p^{e_i})$, we have: $$p \bar x=\bar 0\iff px\in(p^{e_i})\iff x\in (p^{e_i-1})$$ since $p$ is a prime element.

1

Hint: look at $A=F[x]/(x^2)$ with $A=M$, and $L=I=(x)/(x^2)$

1

If I understood well your question, the answer is negative: in $\mathbb Z[i]$ the principal ideal $I=(2+i)$ is a free $\mathbb Z$-module of rank two. On the other side, a two-generated ideal of a quadratic number field is necessarily a free $\mathbb Z$-module of rank two (why?).

1

An ideal is a $\mathbb{Z}$-submodule. It is possible for two elements in $\mathbb{Z}^2$ to generate a proper submodule, but it is not possible for three elements to be linearly independent over $\mathbb{Z}$ (we can always find a nontrivial linear combination that is $0$). Furthermore, any submodule is free and therefore has a linearly independent generating ...

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