# Tag Info

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 ...

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 $p(x)\in\Bbb Z[x]$.

2

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

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 $R$ ...

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 $6$.

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

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 $\;\sum r_i\otimes m_i=\sum 1\otimes r_im_i=1\otimes0=0$.

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.

1

By Schur's Lemma, the two endomorphism algebras on the right are iso to $\mathbb{C}$ (so long as $\dim M$ and $\dim N$ are finite, which I will assume). So in that case you really only need that $M\otimes N$ is a simple $R\otimes S$-module. Let $\sum m_i \otimes n_i$ be a non-zero element of a submodule $U$ of $M\otimes N$ with the $n_i$ linearly ...

1

Assume $I+J \neq A$. Let $\mathfrak m$ be maximal ideal containing $I+J$. We get a surjection $A/I \oplus A/J \twoheadrightarrow A/\mathfrak m \oplus A/\mathfrak m$. If $A/I \oplus A/J$ was cyclic, we can compose this surjection with $A \twoheadrightarrow A/I \oplus A/J$ to get a surjection $$A \twoheadrightarrow A/\mathfrak m \oplus A/\mathfrak m$$ ...

1

The submodule $\mathfrak{m}^n$ is exactly the same as $$\mathfrak{m}R^n=\{a_1x_1+\dots+a_kx_k: a_i\in\mathfrak{m}, x_i\in R^n\}$$ Indeed, if $x\in\mathfrak{m}R^n$, then $x\in\mathfrak{m}^n$, by looking at the components. If $x=(a_1,\dots,a_n)\in\mathfrak{m}^n$, then $$x=a_1(1,0,\dots,0)+\dots+a_n(0,\dots,0,1)\in\mathfrak{m}R^n$$ Since $f$ is an ...

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 ...

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 $I\cdot A/I=0$.

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 ...

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}))\\ ...

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 ...

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