New answers tagged

0

The chain in the theorem is an existence theorem and there is no uniqueness. As an example, let $R=M=\mathbb{Z}$. Then $Ass (M)=\{0\}$. But, we can write the chain $p\mathbb{Z}\subset \mathbb{Z}$ for any prime $p$ and we have $\mathcal{P}=\{0,p\mathbb{Z}\}$. In particular, the primes occurring in $\mathcal{P}$ can be fairly arbitrary.


4

There are plenty of cases where this happens. It's really only interesting when the module action interacts in an interesting way with the multiplication The most straightforward example, which works when $R$ is commutative, is the concept of an $R$-algebra, which I'll let you look up yourself. In this case it's easy to see what the action does. If you ...


1

What's the motivation for the question? If you pick $R = M = k$ a field and $f = 0$, then $C := C(f) \cong k[\varepsilon]/(\varepsilon^2)$ with the ${\mathbb Z}/2{\mathbb Z}$-grading coming from the natural $\mathbb Z$-grading. Then the residue field of $C$ is an irreducible graded $C$-module for which $\psi$ is not an isomorphism.


0

Hint: The polynomials $p_1,\dots,p_n$ must each annihilate a nontrivial submodule of $M$, so to identify these polynomials, try seeing if you can find elements of $\mathbb R[t]$ that annihilate some non-zero element of $M$.


0

I think $f$ is even an isomorphism of complexes. I take it that the complexes are complexes of abelian groups. Since $C$ and $D$ are isomorphic, we may assume that $f$ is a morphism of complexes from $C$ into itself. There are two cases: $f$ has a right inverse, or $f$ has a left inverse. In the former case, each $f_i\colon C_i\rightarrow C_i$ is a ...


1

For question 1, write the extended matrix $$\begin{pmatrix} 1&2&0\\ 1&0&2\\ \end{pmatrix}$$ and then bring it to Hermite Normal Form (HNF). The answer will always be of the form $$\begin{pmatrix} c | 0 \end{pmatrix}$$ ie a square lattice c that gives the basis you are asking for, followed by zero columns. You might want to search for "Hermite ...


0

Since $(n/1) = \iota (n) \in N'$, you have $$n/s = (1/s)(n/1) \in (S^{-1}A)N' = N'$$ because $N'$ is a $S^{-1}A$-module.


-1

Problem 2: It's not quite as straightforward as for a field but you can still get a very concrete answer. A square matrix over a commutative ring with unit $R$ defines an injective map $R^n \rightarrow R^n$ if and only if its determinant is not a zero divisor; and it defines a surjective map $R^n \rightarrow R^n$ if and only if its determinant is invertible. ...


0

Do this in another direction: $$f:N\rightarrow (P+N)/P,n\mapsto n+P$$ Check this is a well-defined module homomorphism. Check this is a surjection. Check the kernel is $N\cap P$. Then apply your epimorphism theorem (I think you are referring to the first isomorphism theorem. The result you are trying to prove is also called the second isomorphism theorem.) ...


2

First you have that $x=5k+1$ for $k\in \mathbb{N}$ by your first equation. Then, $5k+1\equiv 0\ (\mathrm{mod}\ 66)$, so, as you said in the second part, $k\equiv 13+66q \ (\mathrm{mod}\ 66)$. Now $x=5k+1=5(13+66q)+1=66+330q$. Finally you use the third equation ($66\equiv 3$ and $330\equiv 1$ both mod $7$): $x\equiv -1 \ (\mathrm{mod}\ 7)$, so ...


2

Let $U$ and $V$ be free modules over a nontrivial commutative ring $R$ and let $\alpha:U\to R$ and $\beta:V\to R$ be two $R$-linear maps which have $1$ in their image; such things are easily seen to exist using freeness. Then using the properties of tensor products you can show that there is a morphism of abeelian groups $f:U\otimes_RV\to R$ such that for ...


1

Let $R$ be an $k$-algebra and $M$ a simple $R$-module of dimension $n$, such that $M$ is also a finite vector space over $k$. Then the space $$ Hom_{k}(M,M)\cong Hom(k^{n},k^{n}) $$ has dimension $n^2$ over $k$ as a $k$-vector space, thus the dimension of the result above. To see that $Hom_{k}(M,M)\cong M^{n}$ as an $R$-module, it suffice to construction ...


1

Let $R$ be a ring with unit 1 as you say. By $R$ is "free as a bimodule" it often means $R$ is free as a module over $R\otimes R^{op}$, where $(a\otimes b)*r=arb$ (conventions differ in literature). Recall the definition of a free module is a direct sum of $R$. Thus in general $R$ is not free as an $R$ bimodule. In practice, often one work with the case $R$ ...


0

If $M$ is indecomposable, you're done. Otherwise, $M=X\oplus Y$, with $X$ and $Y$ both nonzero. Now prove that the lengths of $X$ and $Y$ are less than the length of $M$. Hint: build composition series for $X$ and $Y$ and paste them up.


4

$\mathbb{R}[x]$-submodules of $\mathbb{R}^2$ are vector subspaces $V$ such that $T(V)\subset V$. In particular, $V$ other than $0$ or $\mathbb{R}^2$ would be a line. Now can you find a line fixed by a rotation?


3

How do you use the splitness of your exact sequence? Once done, it is easy. So, let me call the middle term with basis $e_1,e_2$ and the map on the left by $i$. Then $i(1)=ae_1+be_2$. This splits says we have a map $j:Ae_1\oplus Ae_2\to A$ with $j\circ i$ identity. If $j(e_i)=c_i$, we see that $ac_1+bc_2=1$. Let $v_1=i(1), v_2=-c_2e_1+c_1e_2$. Then using the ...


4

It only holds for the trivial group. As it was mentioned in the comments, we consider $\mathbb{Z}$ as a trivial $\mathbb{Z}G$-module and in this way the augmentation map $\epsilon \colon \mathbb{Z}G \to \mathbb{Z}$ is a morphism of $\mathbb{Z}G$-modules. Let $ f \colon \mathbb{Z} \to \mathbb{Z}G$ be a morphism of $\mathbb{Z}G$-modules and say $ f(1) = ...


0

Hint: The size of a decomposition as a sum of nontrivial modules is bounded by the length of the module. Do you see why, and can you proceed from here? The crucial thing is that the length of any decomposition series bounds the size of any other nontrivial filtration - have you already proved this?


3

One reason why one needs the bilinear map (or multilinear in general), instead of just going ahead to define a map on the tensor product, is that one needs to show that the latter is well-defined. So it may be easy to define some $A \otimes B \rightarrow C$, but it actually may be quite hard to prove that it is well-defined. On the other hand, once a ...


2

For every x in M, there is a unique left R-module homomorphism from R to M that sends 1 to x. If x is nonzero, this homomorphism is surjective (why?). The kernel is precisely the annihilator of x. Now use the so-called correspondence theorem to see this is a maximal left ideal. Can you finish the argument from here?


2

Here's a counterexample. Take $A=k[x]/(x^2)$ and $M=N=A/(x)$. Then $\operatorname{Hom}(M,A)\cong M$, generated by the map $f:1\mapsto x$, and $\operatorname{Hom}(A,M)\cong M$, generated by the map $g:1\mapsto 1$. The tensor product $\operatorname{Hom}(M,A)\otimes \operatorname{Hom}(A,M)$ is then also isomorphic to $M$, generated by $f\otimes g$. But ...


1

Not really an answer, but I wanted to post this here in case anyone else ends up thinking about this thing. It might help set you on the right track. Anyway, after browsing through this collection of slides (see "Ingredients of Construction" slide) I now know that the morphisms in this category can be described using walled Brauer diagrams (see e.g. page ...


1

The group $\mathbb{Z}^\mathbb{N}$ satisfies your hypotheses but is not free. For a much smaller counterexample, let $\hat{\mathbb{Z}}$ be the profinite completion of $\mathbb{Z}$ and let $\alpha\in\hat{\mathbb{Z}}$ be an element such that $a\alpha+b$ is divisible by only finitely many integers whenever $a,b\in\mathbb{Z}$ with $a\neq 0$ (such an $\alpha$ can ...


5

$\prod_{i=1}^\infty \mathbb{Z}$ is a counterexample. It is torsion free, not free (Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?), and every nonzero element is only divisible by finitely many integers, so your extra hypotheses hold.


1

No, and the axiom of choice is irrelevant. For example, take $R = \mathbb{Z}, M = \mathbb{Q}/\mathbb{Z}$. $M$ is a divisible $R$-module but cannot be made into a module over the fraction field $\mathbb{Q}$: modules over $\mathbb{Q}$ correspond to uniquely divisible abelian groups.


1

If you consider the Hopf algebra $k^G$ of $k$-valued functions on a finite group $G$, then $k^G\text{-coMod}$ is equivalent to $G\text{-Rep}_k$, so your question reduces to finding irreducible finite group representations whose endomorphism division algebras are strictly bigger than the scalar field. For this, e.g. see About the Schur's lemma


0

$0\rightarrow Hom_A(P,M')\rightarrow Hom_A(P,M)\rightarrow Hom_A(P,M'') \rightarrow 0$ is exact So $\phi :Hom_A(P,M)\rightarrow Hom_A(P,M'')$ is surjective. Hence there is an $h\in Hom (P,M)$ such that $\phi (h)=f$


0

You're on the right track. There is really no need for $L$ and $L'$ since you have obvious candidates to complete the idempotent to an orthogonal set. What I mean is this: from $R=Re\oplus R(1-e)=Re'\oplus R(1-e')$ and $Re\cong Re'$, you can apply the Krull-Schmidt theorem to conclude $R(1-e)\cong R(1-e')$. Then using your previous exercise, there exists ...


0

$0 \rightarrow \mathrm{Im}(A_{i+1}) \rightarrow A_i \rightarrow A_{i-1} \rightarrow 0$ must be exact. Because if $A_{i-2}=0$, then $\mathrm{Im}(A_{i-1})=0$, then $\mathrm{Im}(A_{i})=\mathrm{Ker}(A_{i-1})=A_{i-1}$. This is just a special case of the short exact sequence you get.


1

Let $V$ have basis $e_1, \ldots, e_n$. There is a basis $\delta_, \ldots, \delta_n$ of $V^\vee$ called the dual basis characterized by the property $\delta_i(e_j) = \begin{cases}1 & \text{if }i=j \\ 0 & \text{otherwise}\end{cases}$. The element $"\mathrm{id}" \in T \otimes T^\vee$ corresponding to the identity $V \to V$ is then $\sum_{i=1}^n e_i ...


0

I think your example is wrong: $$ 18 \mod{4}=2 \not= 1= 9 \mod{}4$$


1

Error: $9\not \equiv 18 \pmod{4}$


5

For a very simple counterexample to both statements, take $R=F=\mathbb{Z}$. Then $\{2,3\}$ is a generating set which contains no basis, and $\{2\}$ is a linearly independent set which cannot be extended to a basis.


2

As a counterexample for the second consider $\mathbb{Z}^2$. This is a free $\mathbb{Z}$-module of rank $2$, but the set $\{(0,1),(2,0)\}$ does not extend to a basis


0

Consider the map $\Phi: Hom_R(R/I,R/I)\rightarrow R/I$ defined by $\Phi(f)= f([1])$ where $[1]$ is the class of $1$ in $R/I$. $\Phi(f)=\Phi(g)$ implies that $f([1])=g([1])$. This implies that for every $r\in R$, $rf([1])=f([r])=rg([1])=g([r])$. Thus $\Phi$ is injective. For every $[u]$ in $R/I$, define $f:R/I\rightarrow R/I$ by $f([r])=[ru]$, $\Phi(f)=[u]$, ...


1

Yes, because $\mathbb{Z}$ is the initial ring, so using that tensor products of rings agree with coproducts, we have: $$\mathbb{Z} \otimes R \cong \mathbb{Z} \sqcup R \cong R$$ for an arbitrary ring $R$.


1

Consider the following maps: $$\phi:x\mapsto x\otimes 1:\mathbb Z/m\mathbb Z\to\mathbb Z/m\mathbb Z\otimes\mathbb Z$$ and $$\psi:(a+m\mathbb Z,b)\mapsto ab+m\mathbb Z:\mathbb Z/m\mathbb Z\times\mathbb Z\to\mathbb Z/m\mathbb Z$$ Then $\psi$ is $\mathbb Z$-bilinear, hence induces a abelian group homomorphism $\bar\psi:\mathbb Z/m\mathbb Z\otimes\mathbb ...


2

HINT: show that every element of $\mathbb{Z}/m\mathbb{Z}\otimes\mathbb{Z}$ is equal to one of the form $[k]\otimes 1$ for some $k\in\{0, . . . , m-1\}$. It will be enough to show this for elements of the form $[x]\otimes y$.


1

Hint. Show that the ideal generated by $n\notin\mathfrak m$, when $\mathfrak m$ runs over the maximal ideals of $A$, is the whole ring.


3

Since $I^n=0$ the ideal $I^{n-1}$ is a finitely generated $R/I^{n-1}$-module: it is finitely generated (any power of a finitely generated ideal is finitely generated) and $I^{n-1}\cdot I^{n-1}=0$; see also here. Now use the exact sequence of $R$-modules $$0\to I^{n-1}\to R\to R/I^{n-1}\to 0.$$


0

So since $K=S^{-1}R$ where $S=R\backslash \{0\}$ and $U^{-1}R\otimes_RM\cong U^{-1}M$ for any multiplicatively closed subset $U\subset R$, one has $K\otimes M= S^{-1}R\otimes M\cong S^{-1}M$ with $1\otimes m\mapsto \frac{m}{1}$ and it now follows from $0=1\otimes m$ that there exists an $r\neq 0$ with $rm=0$ by definition of the localization $S^{-1}M$. $S$ ...


4

Schur's lemma tells you that an endomorphism of $M$ is either zero or invertible. That means precisely that $\text{End}_R(M)$ is a division ring. If $R$ is noncommutative this is the most that you can say. But in fact, because $R$ is commutative, the simple $R$-modules have the form $R/m$ where $m$ is a maximal ideal, and their endomorphism rings are again ...


1

Yes, your attempt is correct; you have nothing more to do. Here's an unabridged version of your proof. Let $x\in N$; then $$ x=fg(x)+(x-fg(x)) $$ Note that $fg(x)\in\operatorname{Im}(f)$; moreover $g(x-fg(x))=g(x)-gfg(x)=g(x)-g(x)=0$, therefore $x-fg(x)\in\ker g$. Hence $N=\operatorname{Im}(f)+\ker g$. Thus we only have to prove that ...


1

Suppose that $A$ and $B$ are $m \times n$ integer matrices with the same image. So, for any integer vector $x$, there exists an integer vector $y$ such that $$ Ax = By $$ Now, take $x \in \mathcal B = \{e_1,\dots,e_n\}$ (the standard basis), and let $y_j$ denote the vector $y$ corresponding to $e_j$ as above. Let $P$ be the matrix whose columns are ...


1

Hints: For you second question, note that if $M=\oplus_k M_k$, then (valid in any commutative ring) $$(0:I)_M=\bigoplus_k(0:I)_{M_k}$$ Furthermore, $$(0:r)_{\mathbf Z/n\mathbf Z}=\begin{cases}d\mathbf Z/n\mathbf Z&\text{if}\enspace r\mid n,\enspace n=rd,\\0&\text{if}\enspace r\wedge n=1.\end{cases} $$


1

If $I$ is any ideal of $R$, then the quotient $R/I$ is not only a ring, but also an $R$-module, with scalar multiplication given by the formula $r \cdot (x+I) = rx+I$. Let $M$ be a cyclic $R$-module. So there exists an $m \in M$ such that $M = \{rm : r \in R\}$. The function $\varphi: R \rightarrow M$ defined by $$\varphi(r) = rm$$ is a homomorphism of ...


3

Zorn's lemma If every chain in a partially ordered set has an upper bound, then the partially ordered set has a maximal element. The condition you are mentioning is about any (non empty) collection of submodules having a maximal element. Such collections need not satisfy the condition that every chain has an upper bound, so Zorn's lemma cannot apply. ...


0

For an example of how it is useful, let $R=F[X]$ for $F$ a field, and let $V$ be a finite-dimensional $F$-vector space and $T$ a linear operator on $V$. Then $V$ can be made into an $F[X]$-module by letting $X$ act via $T$, and the decomposition $$ V\simeq F[X]/(q_1(X))\oplus \dots\oplus F[X]/(q_s(X)) $$ gives you the rational canonical form of $T$. ...


3

Zorn's lemma says that if $X$ is a poset and every chain has an upper bound then $X$ has a maximal element. It does not say that every chain has a maximal element, which is what's relevant here.


7

Zorn's lemma tells you that if you consider the set of all submodules, then this set has a maximal element. It does not tell you whether every nonempty collection contains this maximal element, so you cannot conclude that the collection has one. Now, the collection indeed does have an upper bound (as you've remarked, $ M $ is one such submodule, as it is a ...



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