# Tag Info

9

No: For a local ring $(R,{\mathfrak m})$ with ${\mathfrak m}^2=0$ you have $\text{Hom}_R(R/{\mathfrak m},R)\cong\{x\in R\ |\ {\mathfrak m}x=0\}={\mathfrak m}$, so it suffices to choose $R$ such that ${\mathfrak m}$ is not finitely generated, e.g. $R := {\mathbb k}[x_1,x_2,\ldots]/(x_i^2, x_i x_j)$.

8

Nakayama Lemma. Let $N$ be a finitely generated $R$-module, and $J\subseteq R$. Suppose that $J$ is closed under addition and multiplication and $JN=N$. Then there is $a\in J$ such that $(1+a)N=0$. (Here by $JN$ we denote the subset of linear combinations of $N$ with coefficients in $J$.) Cayley-Hamilton Theorem. Let $A$ be a commutative ring, $I$ an ...

5

It is still true because: Any $A$-module is the direct limit of its finitely generated submodules (for any ring $A$). Tensor products commute with direct limits. Direct limit is an exact functor. Btw, a submodule of an $A$-module with this property (the morphism $M'\hookrightarrow M$ is universally exact) is called a pure submodule of $M$, and the ...

4

I'm not sure if the statement is obvious to you but in general $+$ and $\cap$ don't distribute over each other. I wouldn't expect them to either since $\cap$ is a set theoretic operation and $+$ is only defined for modules/ideals/abelian groups. However, when $A\subset C$ they do distribute over one another. But the proof shouldn't go something like this: ...

4

Consider $R=\mathbb Z$, $M=\mathbb Z$ and $N=\mathbb Z/2\mathbb Z$.

4

If $m = n$, the $0 \to \mathbb{Z}/n\mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z}/n\mathbb{Z} \to 0$ is a free resolution. If $m < n$, it will be useful to write $n = km$, where $k > 1$. The first step of the resolution looks like $F_0 \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_0$. ...

4

If I'm not mistaken this is already true for $I={\mathbb N}$. For any abelian group $A$, the subgroup $D(A)$ of divisible elements (those $a\in A$ for which for any $n\in {\mathbb N}$ there exists some $b\in A$ such that $nb=a$) is the unique maximal injective subgroup of $A$. In your case of $A := {\mathbb Z}^{\mathbb N}/{\mathbb Z}^{({\mathbb N})}$ these ...

3

Take $R=\mathbb{Z}$, $M=\mathbb{Q}\oplus \mathbb{Z}/2\mathbb{Z}$ and $N= \mathbb{Z}/2\mathbb{Z}$, with inclusion $R\to M:n\mapsto (n,\overline{0})$. Then, in $M\otimes_\mathbb{Z} N$, we have that $$(1,\overline{0})\otimes \overline{1} = (\frac{1}{2},\overline{0})\otimes 2\cdot\overline{1} = (\frac{1}{2},\overline{0})\otimes \overline{0} = 0.$$ Therefore ...

3

if the $R$-module $M$ is simple then for $a \in R$ we have $aM=0$ or $aM=M$. suppose $aM \ne 0$ then $aM=M$. if $a \ne 0$ is nilpotent, then $\exists n \gt 1$ such that $a^n=0$. then $$a^nM=0$$ this requires $$a(a^{n-1}M) =0$$ but now $a^{n-1}M=M \Rightarrow aM=0 \Rightarrow a^{n-1}M=0$, a contradiction. so $a^{n-1}M=0$. if $n-1 \gt 1$ we may repeat ...

3

An elementary proof that $\mathbf Q$ is not projective over $\mathbf Z$: if $\mathbf Q$ were projective, it would be a direct summand of a free $\mathbf Z$-module $L$, hence there would be an injective homomorphism from $\mathbf Q$ into $L$. However the only homorphism from $\mathbf Q$ into a free module is the null homomorphism: indeed, for any $n$, and ...

3

For the second question, the finite rings are precisely those for which every finitely generated module has finitely many submodules. For a ring $R$ and $r\in R$, let $M_r$ be the submodule of $R\oplus R$ generated by $(1,r)$. Then $(1,r)$ is the only element of $M_r$ whose first coordinate is $1$, and so $M_r\neq M_s$ for $r\neq s$, and so if $R$ is ...

3

If $x$ is a non-unit, then so is $ux$ for any $u$. This is true in any commutative ring. Otherwise we'd have $$(ux)^{-1}ux=((ux)^{-1}u)x=1$$ This contradicts the assumption that $x$ is not a unit. If the ring is noncommutative we can still obtain the result that $ux$ is not a unit from the fact that the nonunits form an ideal in a local ring.

3

I think $1$ is false, even when $R$ is a PID. Let $R=\mathbb{Z}$, and $M=\mathbb{Z}$, $N=\mathbb{Z}\oplus \mathbb{Z}$, and $\mathbb{P}=\oplus_{n=1}^{\infty} \mathbb{Z}$. Then $M \oplus P \simeq N \oplus P$ as they are both isomorphic to $\oplus_{n=1}^{\infty} \mathbb{Z}$, but $M\not\simeq N$ as $\mathbb{Z}$-modules (they have different rank). For part (2), ...

2

You are almost there. There is a section $s:I\to K\oplus I$ such that $p_2s=id_I$. All you need to do is multiply the equality $vf=p_2$ on the right by $s$, to obtain $$v(fs)=id_I.$$ Thus $fs$ is a section whose image is in $\ker g$, and the claim is proved.

2

I) More generally, consider a short exact sequence $0\longrightarrow I\xrightarrow{\ \ \iota\ \ }R\xrightarrow{\ \ \pi\ \ }R/I\longrightarrow 0$. Tensoring with $M$ gives a right exact sequence $$I\otimes M\xrightarrow{\iota\otimes\operatorname{id}}R\otimes M\xrightarrow{\pi\otimes\operatorname{id}} (R/I)\otimes M\longrightarrow 0\,.$$ Now there is an ...

2

The symbol $\bar r$ denotes the equivalence class of $r$ in $\text{Ann}_R(M)$. The point here is that you want to define the action of the quotient on the module. This is well-defined because if $\bar{r_1}=\bar{r_2}$, this means that $r_1-r_2\in\text{Ann}_R(M)$, so for any $m$ we have $(r_1-r_2)m=0$, i.e. $r_1m=r_2m$.

2

This is true in a more general setting. Let $R$ be a principal ideal domain, $M$ a torsion $R$-module, and let $p$ be a prime element of $R$. Define $$M_1 = \{a\in M \ | \ p^n a = 0 \textrm{ for some } n\in \mathbb{N} \};$$ $$M_2 = \{a\in M \ | \ qa=0 \textrm{ for some } q\in R \textrm{ coprime with } p \}.$$ Now, let $a\in M$. Since $M$ is a torsion ...

2

The matrix $A$ corresponds to an endomorphism $\varphi_A$ of the $K$-vector space $K^n$. Then, by Fitting's Lemma, there is a decomposition $K^n=V_1\oplus V_2$ such that $\varphi_A$ restricted to $V_1$ is nilpotent, and $\varphi_A$ restricted to $V_2$ is invertible, that is, there is a $K$-basis of $K^n$ such that the matrix associated to $\varphi_A$ in this ...

2

$1-ux$ can not be in the maximal ideal as $x$ is and hence $ux$ is. And since we have a local ring, any element not in the maximal ideal must be a unit.

2

$\Phi$ is indeed an isomorphism. For injectivity, suppose $$\Phi(m)=0$$ Then $\psi(m)=0$ and $g(m)=0$. $g(m)=0$ means that $m$ is in the image of $f$. Thus there is an $n\in N$ such that $f(n)=m$. Then $\psi(f(n))=n=0$, so we must have had that $n=0$, hence $m$ was $0$ in the first place. For surjectivity, given $(n,l)$ we want to find $m$ such that ...

2

Is my answer to the first question on the right track? (Show that there is only one simple right R−module up to isomorphism) You made a comment about a chain and a minimal ideal, but this isn't going anywhere. The question has been asked a few times before, and you can find an explanation here: http://math.stackexchange.com/a/1011301/29335 How do I ...

2

If $r\in R$ and $A$ is a minimal right ideal, then $rA$ is either $0$ or a minimal right ideal, because the mapping $R\to R$ defined by $x\mapsto rx$ is a homomorphism of right modules. In any case $rA$ is contained in the sum of the minimal right ideals.

2

I would suggest the following proof which one can find it in Rotman's book (An Introduction to homological algebra p134). I also assume $_{\mathbb{Z}}N$ is f.g as you used it like that in your proof. Let $N$ be a f.g $\mathbb{Z}$-module, since $N$ is torsion free hence it is free, by using the fundamental theorem of finitely generated modules over PID. ...

2

A submodule that is generated by two elements. (This isn't exactly a standard or well-known term, but in that context there's nothing else it could sensibly mean.)

2

Aside from some problems with the question (lengths not necessarily being finite, making it problematic to take differences in general), the answer is still no. For instance, consider $$M \rightarrow M \rightarrow 0$$ where $M$ is some nonzero module of finite length, and with the map $M \rightarrow M$ the identity. Then $\ker(M\to 0)=M$, $\ker(M\to M)=0$, ...

2

Because of: $l(coker(f_i))=l(M_{i+1})-l(\ker(f_{i+1}))$ and $l(M_i)-l(\ker(f_i))=l(M_{i+1})-l(coker(f_i))$ We have: $$l(M_i)=l(\ker(f_i))+l(\ker(f_{i+1}))$$ And $$l(M_{i+1})=l(\ker(f_{i+1}))+l(\ker(f_{i+2}))$$ Hence: $$l(M_i)-l(M_{i+1})=l(\ker(f_i))-l(\ker(f_{i+2}))$$

2

I think you want $I$ to be a right ideal and $J$ a left ideal, not the other way round. You are almost done. Consider the canonical homomorphism $\mathbf{m} :A\otimes_{A}A\rightarrow A$ of $\mathbb{Z}$-modules which sends every $a\otimes b\in A\otimes_{A}A$ to $ab\in A$. It is well-known that $\mathbf{m}$ is an isomorphism. Thus, $\left( ... 2 If$R=\mathbb C[X]$, and$M=\mathbb C[X]/(X^2)$, then there is no such chain for the simple reason that$M$is a noetherian$R$-module, so any ascending chain of submodules must stop. 2 This result is known as Schanuel's lemma. A quick proof is given by introducing the pullback of$f$and$f'$. It is the submodule of$P\oplus P'$given by $$X = \{(p,p')\in P\oplus P'\mid f(p) = f(p')\}.$$ Then, the following sequences are exact : $$0 \longrightarrow \ker(f')\simeq K' \longrightarrow X \longrightarrow P \longrightarrow 0$$ and$$0 ... 2$Z_n\left(M\right)$is the$n$-th homogeneous component of the$0$-th cyclic homology of the tensor algebra$T\left(M\right)\$. There is a nice expository note about this by Clas Löfwall, including the relation to the logarithm: Clas Löfwall, Cyklisk homologi, 28 Aug 2012. Despite the title, it is in English.

Only top voted, non community-wiki answers of a minimum length are eligible