New answers tagged

0

Consider the simpler case where $n=2$. If $M_1$ and $M_2$ have finite length, then also $M_1\oplus M_2$ has finite length. Let $0=X_0<X_1<X_2<\dots<X_h=M_1$ and $0=Y_0<Y_1<Y_2<\dots<Y_k=M_2$ be composition series. Then $$ 0=X_0\oplus0<X_1\oplus0<\dots<X_h\oplus 0 <X_n\oplus Y_1<X_n\oplus Y_2<\dots<X_h\oplus ...


1

Since $R$ is a ring $(R,+)$ is an abelian group. Hence $(M_{m\times n}(R),+)$ is an abelian group where "$+$" the usual matrix addition. Now, let $a,b\in R$ and $(a_{ij}),(b_{ij})\in M_{m\times n}(R)$. Let's check the three module axioms and we are done (i) \begin{align} a((a_{ij})+(b_{ij}))&=a((a_{ij}+b_{ij})) =(a(a_{ij}+b_{ij}))\\ ...


2

It isn't true that if every module of the form $R/aR$ is projective, then $R$ is semisimple. For instance, let $R=\mathbb{F}_2^X$ for some infinite set $X$. Then for any $a\in R$, the ideal $aR$ is a direct summand of $R$, with complement $(1-a)R$ (this follows from the fact that $a^2=a$). Thus $R/aR\cong (1-a)R$ is projective. But not every ideal in $R$ ...


0

Not necessarily. For instance, take $R=M=k[x,y]/(x^2,xy,y^2)$ for some field $k$. Then $M$ has a unique submodule $N$ such that $M/N$ is simple (namely $N=(x,y)$), so any composition series for $M$ must have $M_2=N$. But any simple submodule of $N$ is a direct summand, since $x$ and $y$ both act trivially on $N$ (so a submodule of $N$ is just a ...


1

The ring $\mathbb{C}\{t\}$ is local and its maximal ideal is generated by $t$, see Ring of Convergent Power Series in R and C is a Local Ring In other words, power series $f(t)$ with $f(0)\ne0$ are invertible. So every nonzero convergent power series can be written as $$ t^k g(t) $$ for some integer $k\ge0$ and $g(0)\ne0$. Therefore $t^kg(t)m=0$ if and only ...


2

Check that for any $$\;m\Bbb Z\le\Bbb Z\;\;,\;\;2\Bbb Z\cap m\Bbb Z\neq0$$ and thus $\;2\Bbb Z\;$ , or for that matter any non-trivial subgroup of the integers, cannot be a non-trivial direct summand.


0

If you start from $\phi: B\to End_F(M)$ then $M$ is a left $B$-module, by $b\cdot m = \phi(b)(m)$. Then by definition $M$ is a right $B^o$-module, because in general a structure of right $A$-module on $X$ is a morphism $A^o\to End(X)$, so here $\phi: B=B^{oo}\to End_F(M)$. The definition of the right action is then $m\cdot b^o = b\cdot m = \phi(b)(m)$ ...


1

Here is a very low dimensional example: consider a two dimensional vector space $V$ with basis $\left\{v_1,v_2\right\}$. Then $\left\{v_1\otimes v_1, v_2\otimes v_1,v_1\otimes v_2, v_2\otimes v_2\right\}$ is a basis of $V\otimes V$. You can easily show that $$v_1\otimes v_2+v_2\otimes v_1\neq u\otimes w$$ for all $u,w\in V$. Edit: Be sure to work out the ...


2

Consider $G=\Bbb Z[x], H= \Bbb R$ treated as $\Bbb Z$-modules. Then for everything to be an elementary tensor would mean that every polynomial in $\Bbb R[x]\cong\Bbb Z[x]\otimes_{\Bbb Z}\Bbb R$ is of the form $r\cdot p(x)$ for some $r\in\Bbb R$ and $p(x)\in\Bbb Z[x]$.


3

The condition is equivalent to being a torsion $\mathbb{Z}$-module. If $X$ is a torsion module, then any finitely generated submodule is a finitely generated torsion $\mathbb{Z}$-module : by the structure theorem for finitely generated modules over PID, it's finite. If $X$ is not a torsion module, then any non-torsion element generates an infinite ...


0

More generally, by the Chinese remainder theorem, $$\mathbf Z/m\mathbf Z\times\mathbf Z/n\mathbf Z\simeq\mathbf Z/mn\mathbf Z $$ if $m\wedge n=1$, hence it is a torsion cyclic group.It is not even cyclic if $m\wedge n>1$, be cause it has no element of order $mn$.


2

In general this is not true: for instance that $\mathbb{Z}_p \otimes_\mathbb{Z} \mathbb{Z}_q \cong \{ 0\}$ if $\gcd(p,q)=1$.


0

There's a classification theorem for finitely-generated modules over a Dedekind domain. See for instance these notes.


2

With conclusion $\,a_1\equiv a_2,\,$ it is true iff $\,b\,$ is invertible mod $n\,$ (iff $\,\gcd(b,n)=1)$ Else $b,n$ share a divisor $c>1$ so $\,\color{#c00}{(n/c)}b = n(b/c)\equiv 0\equiv \color{#c00} 0\cdot b,\,$ but $\,\color{#c00}{n/c\not\equiv 0}\pmod n$ But your hypothesis does not imply that $b$ is invertible. Indeed it is true for all ...


2

This is definitely not true in general unless $\gcd(n, b) = 1$. For instance $2 \cdot 3 = 4 \cdot 3 \pmod 6$ but $2 \not \equiv 4 \pmod 6$. The problem is that here $3$ is not invertible since it is not relatively prime to $6$.


1

For question 1, let $A = k[x, y]/(x^2, xy, y^2)$ and $I = (x, y)$. The $A$-module endomorphisms $\text{End}_A(I)$ can be identified with $2 \times 2$ matrices, and of those, only the scalar matrices show up as multiplication by an element of $A$. For question 2, this is true for every ideal $I$ if and only if $M$ is flat. This comes from thinking of $IM$ as ...


2

No. To show it's injective, you have to show that, if $\;\sum r_i\otimes m_i\mapsto\sum r_im_i=0$, then $\;\sum r_i\otimes m_i=0$. But that is because $\;\sum r_i\otimes m_i=\sum 1\otimes r_im_i=1\otimes0=0$.


1

If $\;r\otimes m\to rm=0\;$ then $$r\times m=r(1\otimes m)=1\otimes rm=1\otimes0=0$$ and we've finished. I am assuming, of course, the tensor product is over $\;R\;$ without any further conditions.


1

Here is an example for question 2: Let $J$ be an ideal, and $M=A/J$. The $M\otimes I\simeq I/IJ$. $IM=I\cdot A/J=(I+J)/J\simeq I/(I\cap J)$ Now usually, $IJ\neq I\cap J$, unless $I$ and $J$ are coprime, i.e. $I+J=A$. In particular, if $J=I$, $A/I\otimes I\simeq I/I^2$, while $I\cdot A/I=0$.


2

If you look at the dimensions : $\dim(C) = |S|$, $\dim(D) = |T|$, and $\dim(A\otimes B) = |S|+|T|$. But then $\dim(C\otimes D) = |S|\cdot |T|\neq \dim(A\otimes B)$ (if the dimesnions are finite).


1

$\ker(B)/\operatorname{im}(A)$ is just the kernel of the map $$\Phi:\operatorname{cok}(A)=R^n/\operatorname{im}(A)\to R^o$$ induced by $B$, from the cokernel of $A$ to $R^o$. But $\operatorname{cok}(A)$ is isomorphic to $$R/a_1R\oplus\dots\oplus R/a_r R \oplus R^{n-r},$$ and $\ker(\Phi)$ must contain the torsion submodule $R/a_1 R\oplus\dots\oplus R/a_r ...


1

Since $M=N+IN'\subseteq N+IM$ you get $M=N+IM$. Can you continue from here? Since $M=N+IN'\subseteq N+N'$ you get $M=N+N'$. Can you see now why $M/N$ is finitely generated?


1

Hint: if $I$ is nilpotent, it is contained in the Jacobson radical, you deduce that $N+J(A)N'=N+IN'=M$. You have $I\subset J(A)$, thus $IN'\subset J(A)N'$ and $M=N+IN'\subset N+J(A)N'\subset M$. You can apply the Nakayama lemma. statement 3 https://en.wikipedia.org/wiki/Nakayama_lemma#Statement


3

Let $R$ be a ring such that every left $R$-module is free, and let $I \subset R$ be a maximal left ideal. Then $R/I$ is a simple nonzero $R$-module, and is free by hypothesis, so $R/I$ has a basis. Take any basis element $x$, and let $\varphi \colon R \to R/I$ be the $R$-module homomorphism given by $\varphi(r) = rx$. Since $x$ is nonzero and $R/I$ is ...


2

Note that since $A$ is not only semi-simple but also simple, there is a unique simple $A$-module. Call it $I$. Then any finitely generated $A$-module is isomorphic to $I^n$ for some $n$ (in particular they are all projective). Then $\operatorname{End}_A(P) \simeq \operatorname{End}_A(I^n) = M_n(\operatorname{End}_A(I))$ (the last equality is valid for any ...


1

$a \bmod p$ is the only integer $r$ such that $a \equiv r \bmod p$ and $0 \le r \le p-1$. Therefore, if $a \equiv -1 \bmod p$, then $a \bmod p = p - 1$, because $p-1 \equiv -1 \equiv a \bmod p$ and $0 \le p-1 \le p-1$.


1

if $ a \equiv -1\ (\textrm{mod}\ p)$, then $a \equiv (- 1) + n * p, \ \forall n \in Z$. Let $n = 1$, then $a \equiv -1 + 1 * p$, Hence $a \equiv p - 1$.


3

Since $a \text{ mod } p = -1$, you already know that $p| a+1$. Since $p|-p$, you can infer $p| ( a+1)+(-p)$ and $p| a-(p-1)$, that is to say $a \text{ mod } p = p-1$.


2

Since $x^p$ and $y^p$ are in the center, they act as scalars on your simple module $V$. This means that in fact $V$ is a module over the algebra $k\langle x,y\mid yx-xy-1,x^p-\alpha,y^q-\beta\rangle$ forsome scalars $\alpha$, $\beta$ in the field. Show that this algebra is central and simple (imitating the proofs for the Weyl algbra in characteristc zero, ...


0

Is the following reasoning correct? Define $V_{\lambda,n}:=\left\{v\in V\mid (y-\lambda)^nv=0\right\}$. Claim: $\dim(V_{\lambda,n})=n$ for all $1\leq n \leq p-1$. We have that $V_{\lambda,k}\subset V_{\lambda,k+1}$ for all $k$ and $x^{n-1}v\in V_{\lambda,n}$. Now suppose that $w,w'\in V_{\lambda,n}$ are independent vectors for some $n$, and $w,w'\notin ...


0

Let $H$ be a subgroup of the finite group $G$, and let $M$ be a $G\times H^{\mathrm{opp}}$-bimodule. Then there is a functor from left $H$-modules to left $G$-modules given by $$V\mapsto M\otimes_{\mathbb{C}[H]}V.$$ If you take $M=\mathbb{C}[G]$ (with the left $G$-translation and the right $H$-translation), then this is the usual induction functor, which ...


0

We have $R^n \cong R \oplus \ldots \oplus R$, and $\otimes$ distributes over $\oplus$ so in fact $M \otimes R^n \cong (M \otimes R) \oplus \ldots \oplus (M \otimes R)$. Each $M \otimes R$ is just an $M$, and clearly in $M \oplus \ldots \oplus M$ we have each element having a unique expression as claimed. We can also see this more directly: the construction ...


0

Given a trilinear map $\mu:M_1\times M_2\times M_3\to T$, for each $m_3\in M_3$ the map $(m_1,m_2)\mapsto \mu(m_1,m_2,m_3)$ is an $(A_0,A_1,k)$-bilinear map $M_1\times M_2\to T$. There is thus a unique map $\nu:(M_1\otimes_{A_1} M_2)\times M_3\to T$ such that $\nu(m_1\otimes m_2,m_3)=\mu(m_1,m_2,m_3)$ and $\nu$ is $A_0$-linear in the first variable. It is ...


3

For $A$-modules and homomorphisms $0\to M'\stackrel{u}{\to}M\stackrel{v}{\to}M''\to 0$ is exact, if $M'$ and $M''$ are fintely generated then $M$ is finitely generated. There is exact sequence: $0 \to M_1 \cap M_2 \to M_1 \oplus M_2 \to M_1 + M_2 \to 0 $ .


1

You want to find two vectors which, together with $(m,n,k)$ make a matrix with determinant $\pm 1$. Then the three vectors will form a free basis of ${\mathbb Z}^3$. One way of doing that is to perform some elementary unimodular column operations on $(m,n,k)$ to transform it to $(1,0,0)$. This you can easily extend to a free basis of ${\mathbb Z}^3$. Then, ...


2

The main point is to realise that linear combinations can by definition only have finitely many nonzero coefficients. They must be defined this way, because in pure linear algebra there is no way to take the sum of infinitely many nonzero vectors (this cannot be defined by repeated addition: one never reaches the goal). In analysis some (convergent) infinite ...


1

For one thing, there is a linear map on your vector space which takes the value 1 on each vector in your basis. Is this map a linear combination of the elements of the dual basis?


0

Hint: Relate the modules in question to the quotient field of the given integral domain and use linear algebra.


2

Hints. Let $p_N:F\to N$ be the projection on $N$. Use a basis of $F$ and find $g:F\to M$ such that $f\circ g=\phi\circ p_N$. Now set $\psi=g_{|N}$, the restriction of $g$ to $N$.


0

If $I\subset R$ is an $M$-primary ideal then $R/I$ is artinian. In particular, every ideal $J/I$ if $R/I$ is an $R/I$-module of finite length, hence an $R$-module of finite length.


1

Since $f$ is injective, just define $\phi(n)=f^{-1}(\psi(n))$. To know that this definition makes sense, you need to know that $\psi(n)$ is always in the image of $f$. But the image of $f$ is the kernel of $g$, and $g(\psi(n))=0$ by assumption.


1

I would say that the orientation sheaf of a topological manifold is a helpful example. It's defined in terms of singular homology, if you have some familiarity with that. It's one way to formulate rigorously what we mean by a consistent choice of orientation at each point (even when there is no smooth structure). Also look up the connection between analytic ...


1

Linearity is not satisfied: $$(rf)(ax)=r\Bigl(f(ax)\Bigr)=r\Bigl(af(x)\Bigr)=(ra)f(x)\neq a\Bigl((rf)(x)\Bigr)=(ar)f(x),$$ unless $ar=ra$ for all $a,r$.


2

A $\mathbb Z[i]$-module structure on $F$ is the same as a ring map $\mathbb Z[i] \to \operatorname{End}(F)$. We have $Z[i] = \mathbb Z[X]/(X^2+1)$, i.e. a ring map $\mathbb Z[i] \to \operatorname{End}(F)$ is the same as a ring map $\mathbb Z[X] \to \operatorname{End}(F)=F$, which maps $X$ to an element, which satisfies $x^2+1=0$. Hence $F=\mathbb ...


1

This is true in a much more general setting; let $R$ be a ring (commutative or not is irrelevant) and $M$ a (left) $R$-module, with a submodule $L$. Then, for each positive integer $n$, $$ (M/L)^n\cong M^n/L^n $$ To prove this, consider the canonical projection $\pi\colon M\to M/L$ and define $f\colon M^n\to (M/L)^n$ by $$ ...


3

First notice that $\phi$ must be supposed injective, else it is impossible to extend $\phi$ to the fraction fields. If this is the case then, yes, the extended morphism $\text{Frac }\phi:\text{Frac } A\to \text{Frac }B$ is a finite extension. The trick is to consider the multiplicative set $S=A\setminus \{0\}$ and the morphism $S^{-1}\phi: ...


1

This is false in general. For example if $B=\mathbb Z/p\mathbb Z$ and $A=\mathbb Z$, then certainly $B$ is a finitely generated $A$-module, but $\mathbb F_p$ is not even a field extension of $\mathbb Q$. Alternatively, if $A=\mathbb Z[X]$, $B=\mathbb Z$ and $\varphi: f(X)\mapsto f(0)$, then $F_1$ is isomorphic to an infinite extension of $F_2$. However, ...


1

There is a standard dualization trick : for any $R$-module $A$, write $A^* = Hom_\mathbb{Z}(A,\mathbb{Q}/\mathbb{Z})$, which is a $R$-module. This gives a contravariant functor from the category of $R$-modules to itself. Then write $A^*$ as quotient of a free module : $\bigoplus_{i\in I} R \to A^*\to 0$. Dualizing : $0\to A^{**}\to \prod_{i\in I} R^*$ (this ...


2

This is true, because your "third condition" always holds for sets $C$ satisfying your first two conditions. Indeed, if $C$ is nonempty (say $c\in C$) and satisfies your first two conditions, then it contains $c^0=1$. Now suppose $g^k\in C$ for some $k>0$. Note that $$g^k=(g^k)^1\cdot 1^{k-1},$$ so by the second condition on $C$ (with $m=2$, $k_1=1$, ...


1

By Euler's theorem we have $x^{\varphi(321)}\equiv 1 \bmod 321$ for every unit $\bmod 321$. This implies the order of every unit is a divisor of $\varphi(321)$. As you noted $\varphi(321)=\varphi(3\times 107)=2\times 106=2^253^1$. Therefore the orders of units must be divisors of $\varphi(321)$, having the prime factorization we rapidly find there are only ...



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