New answers tagged

1

You have \begin{align*} p(h.x)&=\sum_gg^{-1}.L(g.(h.x))\\&=\sum_g(hh^{-1}g^{-1}).L((gh).x)\\ &=\sum_gh.[(gh)^{-1}.L((gh).x)]\\ &=h.\left(\sum_g(gh)^{-1}.L((gh).x)\right)\\ &=h.p(x) \end{align*} where the last equality follows from the fact that $$\sum_g(gh)^{-1}.L((gh).x=\sum_{gh^{-1}}g^{-1}.L(g.x)=\sum_{g}g^{-1}.L(g.x)=p(x).$$


-2

Suppose $F$ is a free module of rank $n$ over a PID, with $N$ a submodule. Is $N$ always a direct summand of $F$? Yes this is true. First, $N$ is a free module, as it is a submodule of a free module over a PID. Now let $\{x_1, x_2, \ldots, x_n\}$ be a basis for $F$, and let $\{y_1, y_2, \ldots, y_k\}$ be a basis $N$. Since we are working over a PID, ...


0

Let $J$ be the annihilator of $N$, but suppose $JM\ne\{0\}$ (in particular $N\ne\{0\}$). Suppose also that $JM\cap N\ne\{0\}$ and take $x\in JM\cap N$, $x\ne0$. Then, of course, $x+JM=0+JM$, but $x+JN\ne 0+JN$. Now the task is to find such a situation. Let $R=\mathbb{Z}$, $p$ a prime and set $M=\mathbb{Z}/p^2\mathbb{Z}$. Let ...


1

In general, this is wrong. For instance, take $R=\Bbb Z = J$, $M=2\Bbb Z$, $N=4\Bbb Z$. Then:


3

It's not entirely clear to me what you are looking for exactly, but here is a short proof of your statement: Consider the $R$-module $N= \bigoplus_{x \in X} R_x$, where each $R_x$ is just a copy of $R_R$. Denote the unit of $R_x$ by $1_x$. Then the map of sets $f: X \to N, x \mapsto 1_x$ gives a unique module homomorphism $M \to N$ extending $f$ (and which ...


1

This never holds if $R$ is a nonzero ring and $B$ has more than one element: just take any two distinct elements $a,b\in B$, and let $N$ be the submodule generated by $a+b$.


4

This is not even true for vector spaces, which is just about the nicest possible case. Take $k$ a field, $M = k^2$ with basis $e_1, e_2$, and $N = \text{span}(e_1 + e_2)$.


0

I am aware of the fact, that passing to the quotient field by localization or tensoring gives us an easy proof, but the OP asked for something more elementary. First of all some notations: Denote the map $M \to M_2$ by $f$ and identify $M_1$ as the kernel of $f$, i.e. view $M_1$ as a subset of $M$. Here is an elementary proof for the inequality ...


2

It's convenient but unnecessary to pass to the fraction field. Your hunch that you should apply Cayley-Hamilton works. By Cayley-Hamilton, we have $$f^n = a_{n-1} f^{n-1} + \dots + a_i f^i$$ where $a_k \in R$ and $a_i \neq 0$ (of course if all the $a_i$ are zero then we are done). Multiplying by $f^{m-i-1}$ gives $$a_i f^{m-1} = 0$$ and hence, since $R$ ...


1

View the matix as lying in a matrix ring of the same size, but having entries in the algebraic closure of the field of factions for $R$. It therefore has a Jordan Normal Form, and since it is nilpotent, it must have all zeros on the diagonal. The normal form is then a strictly upper triangular. Finally, it is easy to prove that $M^n=0$ for any strictly ...


5

Let $Q$ be the field of fractions of $R$, and let $F:Q^n\to Q^n$ be the morphism of $Q$-modules which has the same matrix as $f$ (with respect to the standard bases of $R^n$ in the case of $f$ and of $Q^n$ in the case of $F$) Show that $F$ is nilpotent, and that therefore $F^n=0$. If $i:R^n\to Q^n$ is the obvious inclusion, then we have $F\circ i=i\circ ...


2

For $s\in U$ and $x\in M$ we have $sx=0\implies sx\in Q$, where $Q$ is a primary submodule of $M$ which appears in a primary decomposition of $(0)$ and $r_M(Q)\cap U=\emptyset$. Conclude that $x\in Q$. For the converse, $x\in\cap Q_j$ with $r_M(Q_j)\cap U=\emptyset$. On the other side, for some $Q_i$ such that $r_M(Q_i)\cap U\ne\emptyset$ we get an $s_i\in ...


1

The notation here is a little awkward, so I'm going to adjust it a bit. You have that $V$ is a $\mathbb{C}G$-module, so for each $v\in V$ and $g\in G$, you should know what $g.v$ means (and similarly for $g.w$, $w\in W$). Now, starting from a linear transformation $L:V\to W$ and $x\in V$ we can compute $g^{-1}.L(g.x)$ (that is, act on $x$ with $g$, map it ...


0

For $g\in G$ and $x\in V$, $g^{-1}Lg(x)$ is the element of $W$ obtained by first multiplying $x$ by $g$ using the $\mathbb{C}G$-module structure of $V$, then applying the linear map $L$ to get an element of $W$, and then multiplying that element of $W$ by $g^{-1}$ using the $\mathbb{C}G$-module structure of $W$. Then $p(x)$ is just the sum of the elements ...


8

Sometimes it is better to look at a more general situation. In fact, this makes it easier to see what is really going on. 1) Let $V$ be a $K$-vector space and let $L/K$ be a field extension. Then $L \otimes_K V$ carries the structure of a vector space over $L$ via (linear extension of) $\alpha(\beta \otimes x) = \alpha\beta \otimes x$. 2) If ...


5

Let $\mathbb{F}$ be a field and let $V$ be a module over $\mathbb{F}$ (i.e. a vector space over $\mathbb{F}$), then $V^*$ is also a module over $\mathbb{F}$. The tensor product $V^*\otimes_{\mathbb{F}}V$ is canonically isomorphic to $\operatorname{End}_{\mathbb{F}}V$ via the map induced by the bilinear map $V^*\times V \to ...


4

Take $R$ to be a field $k$ and take $A = B = k^2$, with basis $e_1, e_2$. The tensor product $A \otimes_k B$ is $k^4$ with basis $e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2$, and most elements of it are indecomposable. For example, $e_1 \otimes e_1 + e_2 \otimes e_2$ is indecomposable. There's no more reason to expect all tensors to ...


0

For a commutative ring this is the case, and you should be able to see the axioms for an ideal say exactly that. To make the statement apply to noncommutative rings, you just have to be a bit more specific and say that the submodules of the ring module $R_R$ are exactly the right ideals.


2

The "finitely generated" isn't really important here. You have to understand what are free modules. A $R$-module $M$ is free if it has a basis, i.e., there exists $\{x_i\}_I \subset M$ (where $I$ can be chosen the be finite in the finitely generated case) such that the $x_i$'s form a $R$-basis of $M$. This is a very special property, and every free module ...


0

For a module to be a free module, it not only has to have a basis, but that basis has to be linearly independent. If you look at $Z/2Z$ as an $R$-module for some $R$, it is certainly generated by $1+2Z$ as you say, but since $Z/2Z$ has so few elements, I'll bet that there is some nonzero $r\in R$ such that $(a+2Z)r=0$.


0

I think the conditions should be for the ideals chain $\;M_1\subset M_2\subset\ldots\;$ that $$\begin{cases}M_1\cap N\subset M_2\cap N\subset\ldots\;\text{stabilizes}\\{}\\M_1N/N\subset M_2N/N\subset\ldots\;\text{stabilizes}\end{cases}\;\;\implies\;\;M_1\subset M_2\subset\ldots\;\text{stabilizes}$$ Let us take $\;k\;$ to be the minimal index for which both ...


0

A chain $(N_\alpha)_{\alpha\in A}$ of submodules always has the property that, given a finite subset $\{x_1,x_2,\dots,x_n\}$ of $Q$, then there exists $\alpha$ such that $\{x_1,x_2,\dots,x_n\}\subseteq N_\alpha$. This is clear for $n=1$; suppose it is for subsets having $n$ elements. Suppose $\{x_1,x_2,\dots,x_n,x_{n+1}\}\subseteq Q$ and take $\alpha$ and ...


1

A direct sum of $FH$-modules is also a direct sum of vector spaces, so it is enough to consider a single term $M \otimes t_i$ in the decomposition. Now $M \otimes t_i = \{ m \otimes t_i : m \in M \}$. So, if $a_1,\ldots,a_n$ is a basis of $M$ as a vector space over $F$, then every element of $M$ can be written uniquely as $\sum_{i=1}^n f_i a_i$ with $f_i ...


3

when are ideals also rings with unity? Proposition: An ideal $I\lhd R$ will be a ring with identity iff there exists a central idempotent $e$ such that $eR=I$. Proof: ($\implies$) The identity of $I$, call it $e$, is an idempotent element of $R$ and satisfies $I=eI$. Then $I=eI\subseteq eR\subseteq I$, so $I=eR$. Since $e\in I$, we have ...


1

If $R$ is a ring then clearly is generated as a $R$ Module by $1\in R$. Considering $R^n$ the direct sum of n copies of $R$ we know that there are canonical injections of $i_i: R \hookrightarrow R^n$ sending $r \mapsto (0,0,..r,..)$. We denote $e_i=i_i(1)$ with this is clear that the set $\{e_i\}_{i=1}^n$ generates $R^{n}$ For your question, since you have ...


2

You're absolutely right that (ii) does not obviously follow from (i) as stated. What (i) should say is not just that $I_i$ is ring-isomorphic to $M(n_i,F)$ but that it is isomorphic to $M(n_i,F)$ as an $F$-algebra. Concretely, this means that if you take an element $a\in F$, consider it as an element of $FG$, project it to $I_i$, and then map it to ...


2

Direct limit is generalisation of the notion of union of a family of sets. Two other examples: Lazard's theorem in commutative algebra asserts that a flat $R$-module is a direct limit of free $R$-modules. Germs of continuous function at a point $a$ of a topological space is defined as the direct limit of the system of pairs $(U,f)$, where $U$ is an open ...


2

Given a prime $p$, the Prüfer $p$-group is the direct limit of the cyclic groups of order a power of $p$. If $\Omega$ is an infinite set, the group of finitary permutations on $\Omega$ is the direct limit of all groups of permutations on the finite subsets of $\Omega$.


1

The first step is to adapt the proof that $$End_R(V_1\oplus V_2)\cong End_R(V_1)\oplus End_R(V_2)$$ to the case of arbitrary $V_1$ and $V_2$ satisfying $Hom_R(V_1,V_2)=0$. Then, prove the following: For $\phi\in\mathrm{End}_R(S^n)$, define $$\phi_{ij}=\pi_j\circ\phi\circ \iota_i$$ where $\pi_j:S^n\to S$ is the projection onto the $j$th factor and ...


0

Show that the left ideal $F[G]N_G=FN_G$, and you will have shown that it is $1$-dimensional as an $F$-subspace, and that is necessarily a simple left module.


2

You are right: $P(A,t)=(1-t)^{-s}$, and this can be proved by induction on $s$. For the Hilbert polynomial just count the number of monomials of a fixed degree. For the characteristic polynomial note that the associated graded ring of $A$ with respect to $Q$ is nothing but $A$. Then the characteristic polynomial is $\binom{X+s}{s}$.


2

Hint. Let $R=k[X,Y]$. We have $H_{R/I}(t)=1+t$, and $H_{R/J}(t)=1+2t+t^2$. On the other side, their Hilbert polynomial is $0$. Edit. It seems the OP looked for $H_I(t)$, $H_J(t)$ and the corresponding Hilbert polynomial. Then he could have figured this out by himself using the exact sequence $0\to I\to R\to R/I\to 0$ (and the corresponding one for $J$) ...


0

Your construction is correct and you also identified the remaining problem: to prove that your construction is well defined. Proceed as follows: (You should use the left side of the diagram somehow...) Suppose that also $f_2(y^{\prime})=x$. Then $f_2(y^{\prime}-y)=0$ and therefore $y^{\prime}-y\in \text{Ker} f_2=\text{Im} f_1$. So there is an $z\in ...


1

Here's a counterexample. Let $k$ be a field and take $A=B=k[x]/(x^2)$ and $M=C=k$ (considered as an $A$-algebra via $x\mapsto 0$). Then $\mathrm{Hom}_B(M,B)\cong k$ (spanned by $1\mapsto x$), so $\mathrm{Hom}_B(M,B) \otimes_A C\cong k$ as well. But $1\mapsto x$ maps to $0$ in $\mathrm{Hom}_B(M,B \otimes_A C)$, since $x=0$ in $B\otimes_A C=C$. On the ...


2

A ring over which this is not possible is called a stably finite ring (to relate your definition to the one on Wikipedia, take $A$ and $B$ to be the matrices of the projection $M\to N$ and the inclusion $N\to M$, respectively). Here's an example of a ring with invariant basis number that is not stably finite. Let $k$ be a field and let $R=k\langle ...


2

You just apply the distributive property. The element $\alpha=(0,a,b,0,0,\dotsc)$ is just $a+b$ and $\xi=(0,x_1,x_2,0,0,\dotsc)$ is $x_1+x_2$; then $$ \alpha\xi=(a+b)(x_1+x_2)=ax_1+ax_2+bx_1+bx_2= (0,0,ax_1,ax_2+bx_1,bx_2,0,0,\dotsc) $$ because $ax_1$ has degree $2$, $bx_2$ has degree $4$ and both $ax_2$ and $bx_1$ have degree $3$. It's no different at all ...


0

Maybe this is another description of Bernard's answer; so if it dont help you I will delete: Let $I^n=0$. As above: $N=Im(\phi)+IN$. So $I(\frac{N}{Im(\phi)})=\frac{IN+Im(\phi)}{Im(\phi)}=\frac{N}{Im(\phi)}.$ Hence $$0=I^n(\frac{N}{Im(\phi)})=\frac{N}{Im(\phi)}.$$ So $N=Im(\phi).$


1

I'm surprised no one has mentioned the following very simple example. Let $A=\mathbb{Z}$ and let $M=\mathbb{Q}$. Then $M$ is free at the generic point of $\operatorname{Spec} A$, but is not free in any open neighborhood. Or, if you want $x$ to be a closed point, you can instead take $M=\mathbb{Z}_{(p)}$ for some prime $p\in\mathbb{Z}$. More generally, if ...


1

‘$\overline\phi$ is surjective’ means $N=\phi(M)+IN$. From this you deduce $$N=\phi(M)+I(\phi(M)+IN)=\phi(M)+I\phi(M)+I^2N=\phi(M)+I^2N,$$ and by a silly induction: $$N=\phi(M)+I^kN\quad\text{for all}\enspace k\ge 1.$$ Now choose for $k$ the nilpotency index of $I$, and you get $\;N=\phi(M)$, i.e. you get the surjectivity of $\phi$.


0

Your condition is not even sufficient. Consider a module $A$ such that $A\cong A\oplus A$; let $g\colon A\to A\oplus A$ be an isomorphism. Now take $L=A$, $L'=A$, $M=L\oplus L'$, $P=A\oplus A$, $P'=0$, $N=P\oplus P'$. Define $f\colon L\to P$ as $g$. Then you have an isomorphism $f\colon L\to P$ that obviously cannot be extended to an isomorphism $M\to N$. ...


0

It is not necessary: take $R=\mathbb{Z}$, $M=N=\mathbb{Q}$, with the usual structure; and let $L=\mathbb{Z}$, $P=2\mathbb{Z}$ with $\psi:x\mapsto 2x$ the obvious module isomorphism. Then $\psi$ extends to an appropriate automorphism of $\mathbb{Q}$, while $\mathbb{Q}$ does not decompose as a direct sum at all (exercise).


0

Take $R = M = N = L = \mathbb{Z}, P = 2 \mathbb{Z}$ for a fairly simple example where no such isomorphism exists. I don't see a useful sufficient condition. An equivalent question can be stated using fewer modules: suppose $M$ is a module. When does $\text{Aut}(M)$ act transitively on the set of all submodules of $M$ abstractly isomorphic to a given ...


4

A finitely generated module over a noetherian ring is of finite length iff its support is contained into the maximal spectrum. (See here, Proposition 1.6.9.) Now let $\mathfrak p\in\operatorname{Supp}(\operatorname{Coker}f)$. If $\mathfrak p$ is not maximal, then it is minimal since $\dim R\le 1$. But $(\operatorname{Coker}f)_{\mathfrak ...


3

You got something wrong there. $\mathbb{Q}$ is not artinian as a $\mathbb{Z}$-module because $\mathbb{Z}\supseteq 2\mathbb{Z}\supseteq 4\mathbb{Z}\supseteq 8\mathbb{Z} \supseteq \cdots$ is a strictly descending sequence of submodules. And similarly $\mathbb{Z}_{(2)}\supseteq 2\mathbb{Z}_{(2)}\supseteq 4\mathbb{Z}_{(2)}\supseteq 8\mathbb{Z}_{(2)} \supseteq ...


1

For $n\in\mathbb{N}$, let $M_n\subseteq\mathbb{Q}$ be the submodule generated by $2^n$. Then the $M_n$ form an infinite descending chain of submodules. (This is essentially the same example as in the linked question, just instead of using different primes, you use higher and higher powers of the only prime you have.)


2

This isn't true. For instance, if $R=\mathbb{Z}$, then since every simple module is finite, any finite length module must be finite. But there are finitely generated modules that are not finite (e.g., $\mathbb{Z}$ itself). For your followup question, you can take $M=\mathbb{Z}^2$ and $N=\mathbb{Z}$ to get a counterexample.


2

It indeed follows from looking at how the ideal looks. Consider what happens when you multiply $H_n$ by an element of $H$ from the left. This clearly (after a substitution) leaves us with $H_n$ again (it is invariant). Now consider an element $$\sum_{h \in H}{f_h h } \in F[H]$$ Then using the above remark, we find that the left multiplication of this element ...


1

I assume that $D$ is a division ring as in one of your previous questions and that $\{e_i\}_{1≤i≤n}$ is the canonical basis. For the first part you know that $\phi$ is determined on $e_1,\dots, e_n$. Then you can use elementary matrices to show that $\phi$ is actually only determined on $e_1$. For the second part, you know that you can write $$\phi(e_1) ...


2

Endomorphisms of finite direct sums are most conveniently described with matrices. Suppose $M$ and $N$ are modules; write the elements of $M\oplus N$ as columns; then an endomorphism of $M\oplus N$ is described by a matrix $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix}, $$ where $a\colon M\to M$, $b\colon N\to M$, $c\colon M\to N$ and $d\colon N\to ...


1

1) The functor $- \otimes_A M$ is right exact. Therefore, the exact sequence $I \to A \to A/I \to 0$ yields the exact sequence $I \otimes_A M \to A \otimes_A M \to A/I \otimes_A M \to 0$. Now use that $A \otimes_A M \cong M$ and that the image of $I \otimes_A M \to M$ is exactly $IM$. This shows $A/I \otimes_A M \cong M/IM$. You can also show directly that ...



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