# Tag Info

0

"Torsion" is more module theoretic. Most often zero divisors are talked about in the context of products of ring elements, but you can also talk about a ring element "being a zero divisor on" module elements. The term "torsion" is much more overloaded than zero-divisor is. For example: one definition of a torsion element of an $R$ module is that its ...

0

Throughout, I assume you are looking at left $R$-modules. Since $R$ is a finite-dimensional $\mathbb{Z}_p$-module, then using the duality $$D={\rm Hom}_{\mathbb{Z}_p}(-, \mathbb{Z}_p): {\rm mod}\ R^{op} \to {\rm mod}\ R,$$ where $R^{op}$ is the opposite algebra, is a good way to answer your question. More precisely, projective $R^{op}$-modules will be ...

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 $F_0$. ...

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.

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.

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.)

1

Note that an $R$-module is the same as an abelian group $M$ together with a ring homomorphism $R\longrightarrow\operatorname{End}_\mathbb{Z}(M)$. The kernel of this homomorphism is precisely $\operatorname{Ann}_R(M)$ (actually this is how one should define the annihilator, so one does not have to check it is an ideal). Now the homomorphism theorem yields a ...

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 ...

1

I want to add a categorical phrasing of this problem. Consider the quotient category $R\text{-Mod}/\langle N\rangle_\oplus$ of the category $R\text{-Mod}$ at the additive closure $\langle N\rangle_\oplus$ of $N$: Its objects are again the $R$-modules, but for two $R$-modules $A,B$ we define $\text{Hom}_{R\text{-Mod}/\langle N\rangle_\oplus}(A,B)$ to be the ...

1

It seems like your best bet for both questions will be to consider finite rings and their finitely generated modules. These at least will be closed under products.

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. $r_1m=r_2m$.

1

When a projection map $R\to S$ is in play, the notation $\bar{r}$ often denotes the image of $r$ in $S$ under the projection map. E.g. the elements of $\Bbb Z/n\Bbb Z$ are the residues $\overline{0},\overline{1},\overline{2},\cdots,\overline{n-1}$. A priori one would have to wonder if $\overline{r}m:=rm$ is well-defined, since $\overline{r}=\overline{s}$ is ...

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 ...

1

Since $I$ is a principal ideal, it is isomorphic to $R$ as an $R$-module. Hence $$I \otimes_R I \cong R \otimes_R R \cong R$$ which is obviously torsionfree.

1

1. "Yes" to the second question. More precisely, he is regarding $B$ as a $\left(B,A\right)$-bimodule, where the left $B$-module structure is obvious (i.e., given by multiplication) and where the right $A$-module structure is the one you define. "No" to the first question. He is tensoring a $\left(B,A\right)$-bimodule with a left $A$-module. This yields a ...

1

Since $k$ is algebraically closed and $A$ is basic, $A$ is isomorphic to a quotient $kQ/I$ of a path algebra on a quiver $Q$ (which is, in fact, the Gabriel quiver of $A$) by an admissible ideal $I$ of $kQ$. It follows easily from this the enveloping algebra $A\otimes A^{op}$ is the quotient of the path algebra on a certain quiver $Q\times Q$, which we can ...

0

After some works, I found that the problem lies in the fact that finitely generated maximal ideal might not exist in every ring, as demonstrated by Jeremy. However, finitely generated proper ideal does exist in any nonzero ring (like the zero ideal). Then if we replace every "maximal" by "proper" in the statement (and specifying that the rings are nonzero, ...

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.

0

If you have an algebraic closed field $K$ then $A/rad(A)$ is isomorphic to full matrix algebras over $K$. Therefore the radical factor structure is separable and your claim follows by W-M-T.

1

If $R$ is a local commutative integral domain (not a field) with infinitely generated maximal ideal, and $S$ is its field of fractions, then $S$ is flat but not faithfully flat over $R$, but since $R$ has no finitely generated maximal ideals the "finitely generated" version of $S\neq SI$ is vacuously satisfied.

1

The relationship between the three numbers can be summed up in two inequalities: $$rk_R(M)\leq mass(M) \leq length(M) .$$ There isn't much more that can be said, see below for an idea why. First, let us prove the inequalities. Assume that $mass(M)=m<\infty$, and let $\{x_1, \ldots, x_m\}$ be a minimal generating set for $M$. Then we have a strictly ...

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. $R := {\mathbb k}[x_1,x_2,\ldots]/(x_i^2, x_i x_j)$.

0

One can show that $M$ is locally free. The begin with, suppose that $R$ is local. Let $n$ be the minimal number of generators of $M$, and let $\varphi:R^n\to M$ be a surjective homomorphism. Moreover, it is an isomorphism after tensoring by $Q(R/\mathfrak p)$, the field of fraction of $R/\mathfrak p$, for any minimal prime $\mathfrak p$. Now conclude that ...

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 ...

1

The nilideal is contained in the nilradical, since $1-ax$ is a unit whenever $x$ is nilpotent. The nilradical is the intersection of all the kernels of irreducible representations.

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

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.

0

That is because for any ring $A$, the centre of $M_n(A)$ is made up of scalar matrices with coefficients in the centre of $A$. Cf. Bourbaki, Algebra, Ch. 8, Modules and semi-simple rings, § 5 Commutation, no 3, cor. 2 of theorem 2.

0

Take $L_{ij}=e^ie_j$ and suppose that $M$ is in the center of $\mathfrak{M}(K)$. Then $ML_{ij}e_i=0$ and $L_{ij}Me_i=M_{il}e_l=M_{ij}$ if $i\neq j$, so that $M_{ij}=0$. Thus $M$ is a diagonal matrix. Now $L_{ij}+L_{ji}$ shows that all the diagonal elements coincide, so that the matrix is a multiple of the identity. Now clearly the identity matrix commutes ...

1

Why would the centre of $\mathfrak{M}_n(K)$ be a field? To commute with all of the unit matrices, an element of the center must be contained in the set $D=\{kI_n\mid k\in K\}$. This is already isomorphic to $K$. Then in order for an element to commute with all the elements of $D$, the scalar multiplier must be in the center of $K$ (which is sometimes ...

0

The statement is false even for irreducible modules. Take $k=\mathbb{R}$, and $A=B=\mathbb{C}$ viewed as $\mathbb{R}$-algebras. Then the tensor product algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$ is isomorphic to the algebra $\mathbb{C}\times \mathbb{C}$ (see, for instance, this question). Now, take $M=N=\mathbb{C}$. Then $M$ and $N$ are ...

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.

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 ...

1

Judging from your diagram, it looks like (?) you are interpreting $+$ as set union. Certainly, if you are working with sets using union and intersection, your picture is accurate. But using using the $+$ to join submodules of a module does not behave the same way. If $A$ and $B$ are submodules of a module, $A+B$ contains many more elements than just $A\cup ... 4 I'm not sure if the statement is obvious to you but in general$+$and$\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$they do distribute over one another. But the proof shouldn't go something like this: ... 0 I'll show$\mathbb Z_n$is not projective for every$n>1$(hence it can't be free for every free module is projective). Let$\jmath:n\mathbb Z\longrightarrow \mathbb Z$be the inclusion and let$p:\mathbb Z\longrightarrow \mathbb Z_n$be the canonical projection . Then one gets a short exact sequence of$\mathbb Z$-modules: $$0\longrightarrow n\mathbb ... 3 An elementary proof that \mathbf Q is not projective over \mathbf Z: if \mathbf Q were projective, it would be a direct summand of a free \mathbf Z-module L, hence there would be an injective homomorphism from \mathbf Q into L. However the only homorphism from \mathbf Q into a free module is the null homomorphism: indeed, for any n, and ... 1 The rationals as a module over the integers is flat but not projective. 1 Let 0 = X_0 \subseteq X_1 \subseteq X_2 \subseteq \cdots \subseteq X_a = N be a composition series for N. Let 0 = Y_0 \subseteq Y_1 \subseteq Y_2 \subseteq \cdots \subseteq Y_b = M/N be a composition series for M/N. This gives rise to a sequence N = Z_0 \subseteq Z_1 \subseteq Z_2 \subseteq \cdots \subseteq Z_b = M of submodules of M defined by ... 2 Aside from some problems with the question (lengths not necessarily being finite, making it problematic to take differences in general), the answer is still no. For instance, consider$$ 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}))$$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), ... 1 Suppose that there are$n+1$linearly independent elements$x_1, x_2, …, x_{n+1}$in$M=R^{n}$. Consider the submodule$F$generated by these$n+1$elements. Then clearly$F\cong R^{n+1}$. Since$F$is a submodule of$M=R^{n}$, it follows that there is an injective$R$-linear map$R^{n+1}\to R^{n}$. But this is impossible! That there is no injective ... 1 First of all we are looking for an ideal$\mathfrak a$which isn't flat. (A simple example is$\mathfrak a=(X,Y)$in$A=K[X,Y]$.) It would be nice if$\mathfrak a$contains an$A$-sequence$a,b$, that is,$a$is a non-zero divisor on$A$and$b$is a non-zero divisor on$A/(a)$. (For instance, in the previous example$a=X$and$b=Y$.) Then take$M=N=A/(a)$, ... 1 I believe this works. Consider$A=k[x,y]$,$I=(x,y)$(note that the ideal correspond to a codimension 2 subset). Consider the map$\varphi:k[x,y]/(x)\rightarrow k[x,y]/(x)$which is the multiplication by$y$. It is obviously injective. Now consider$(x,y)\otimes\varphi:(x,y)\otimes_A k[x,y]/(x)\rightarrow (x,y)\otimes_A k[x,y]/(x)$. The element$x\otimes ...

0

If I understand well you have an infinite PID which has a finite non-zero module, say $M$. In particular, $M$ is finitely generated and torsion, so it is a (finite) direct sum of indecomposable torsion modules, that is, of modules of the form $R/(p^m)$ with $p\in R$ prime, $p\ne0$. This shows that you have to impose on $R$ the following condition: ...

1

Yes, this is true. Let $B$ be a basis for $F$; then there is a countable subset $B_0\subseteq B$ such that $F_0$ is contained in the subgroup $G$ generated by $B_0$. Let $H$ be the subgroup generated by $B\setminus B_0$; we then have $F=G\oplus H$ and $F/F_0=G/F_0\oplus H$. If $I\subseteq F/F_0$ is injective, then its projection onto $H$ must be trivial, ...

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.

0

The isomorphism theorem states that $im(f) \cong M/ ker( f )$, but this isomorphism does not need to be the injection of the submodule. In the example above, $2\mathbb{Z} \cong \mathbb{Z}$, but not by the injection.

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