Tag Info

Hot answers tagged

4

That's an odd statement of a more general statement that may be easier to grasp. In any ring, the sum of all minimal left ideals is a two-sided ideal. (If there are no minimal left ideals, the sum is the empty sum $\{0\}$. This ideal is called the left socle of the ring and is denoted $Soc(_RR)$. There's no doubt at the start that $Soc(_RR)$ is a left ...


3

Suppose $a \in R$ is a non-invertible element. Suppose $f: K \longrightarrow R$ is an $R$-linear map such that $f(1) \neq 0$. Then $$ af(1)f \left( \frac{1}{af(1)} \right) = f \left( \frac{af(1)}{af(1)} \right) = f(1) $$ hence $1= a f \left( \frac{1}{af(1)} \right)$ and this contradicts that $a$ is not invertible. So it must be $f(1)=0$. And now for all $x, ...


2

The answer http://mathoverflow.net/a/10249 to Is it true that, as $\mathbb{Z}$-modules, the polynomial ring and the power series ring over integers are dual to each other? on MathOverflow shows that $$ \operatorname{Hom}(\mathbb{Z}^{\mathbb{N}},\mathbb{Z}) $$ is isomorphic to $\mathbb{Z}^{(\mathbb{N})}$. If $\mathbb{Z}^{\mathbb{N}}$ were free it would be ...


2

Start with listing generators $x_1, x_2, \ldots$. Now by induction we will find submodules $M_1, M_2, \ldots, M_n$ such that each $M_i$ is a member of $S$, the sum $M(n) = M_1 + \cdots + M_n$ is direct in $M$, the module $M(n)$ is a direct summand of $M$, and we have $x_1, \ldots, x_n \in M(n)$. To start apply the given to $x_1 \in K = M$ as suggested by ...


2

When I try to find (counter)examples, I don't consider random examples first, but rather try to simplify the general case so that, in the end, examples pop out automatically without any effort. This also works here: By the universal properties of $R$ and quotient modules, a homomorphism of left $R$-modules $R/I \to R/J$ corresponds to some element $[x] \in ...


2

Yes, this is very old work. In 1935, Koethe proved that the modules of Artinian principal ideal rings are all direct sums of cyclic submodules. Of course, your example falls into this category. In fact, all of the proper quotients of a principal ideal domain are Artinian principal ideal rings (in fact they are also self-injective, hence quasi-Frobenius.) If ...


2

Yes. Let $R$ be ring of endomorphisms of $\mathbb{R}^2$, $I$ and $J$ be annihilators of subspaces spanned by $(1,0)$ and $(0,1)$, respectively. Let $\theta\in R$ be given by $\theta(x,y)=(y,x)$, and $\phi(f)=f\circ \theta$ for $f\in R$. Then $\phi$ is an automorphism of $R$ as a module over itself, and a bijection between $I$ and $J$. Thus, it induces ...


2

I believe a slightly more elementary version of the example of Marcin Łoś consists in taking the ring $R$ of $2 \times 2$ matrices over $\mathbb{R}$, say, and the left ideals $$ I = \left\{ \begin{bmatrix}a & 0\\ b & 0\end{bmatrix} : a, b \in \mathbb{R} \right\} , \qquad J = \left\{ \begin{bmatrix} 0 & a\\0 & b\\\end{bmatrix} : a, b \in ...


1

This is a remark concerning points (e-g): your result is mentioned in the article R. B. WARFIELD, COUNTABLY GENERATED MODULES OVER COMMUTATIVE ARTINIAN RINGS, PACIFIC JOURNAL OF MATHEMATICS Vol. 60, No 2, 1975 right at the beginning, but without proof. However some pointers are given to articles about certain non-commutative rings in which the problem ...


1

The answer is yes. All of the proofs would go through with "left" replaced by "right" and with the arguments carried out with right modules instead of left modules. That goes for all theorems of this sort, not just the one you mention.


1

If not, then $R$ has a non-trivial ideal. Then $R/I$ has torsion as an $R$-module.


1

Proof in short: If we can construct a divisible cyclic submodule, it must be isomorphic to $R$, and therefore $R$ is a field. We will do this by showing that any element of a minimal set of generators of a divisible $R$-module generates a cyclic divisible submodule by showing that otherwise we obtain a smaller set of generators. Proof: Suppose $x_1,\ldots, ...


1

Let $M$ be a non-zero divisible $R$ module, finitely generated by $m_1,\ldots,m_n$. Say $r\in R$ is non-zero. We need to show $r$ has an inverse in $R$. Since $M$ is divisible, there exist elements $m’_1,\ldots,m’_n\in M$ such that $rm’_i=m_i$ for $i=1,2,\ldots,n$. We can write each $m’_i$ as an $R$-linear combination of the $m_j$, say $$ m’_i=\sum_{j=1}^n ...


1

Note that $$\text{End}_D(D^n)=\text{Mat}_n(D)=A$$ Taking the center on both sides yields $$Z(\text{End}_D(D^n))\cong D$$ But, the center of $\text{End}_D(D^n)$ is just the matrices which are $\text{Mat}_n(D)$-linear. So, by tracking this through the isomorphism $A\cong \text{Mat}_n(D)$, you see this turns into the statement $Z(\text{End}_D(D^n))\cong ...



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