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4

Let $A=M_2(\Bbb Z)$ be $2\times 2$ matrices over $\Bbb Z$ and let $x=2I$ be twice the identity.


3

I googled the question verbatim and the first result had a solution to the problem. As it turns out, one should perhaps look at more than just StackExchange when hunting for solutions. Here's a slightly expanded version of the proof (under Problem 2 here near the start of the document): Let $\bar k$ be the algebraic closure of $k$. Then $\bar k$ is ...


2

By re-denoting $y$ as $\frac{d}{dx}$ you can see that your ring is nothing else but polynomial differential operators on the affine line $\mathbb{A}^1_{\mathbb{Q}}$, where $x$ act by multiplication by $x$, and $\frac{d}{dx}$ acts by differentiation. It is easy to see that any differential operator can be put into the form $p_n(x) ...


2

Let $f:R\to\operatorname{Hom}_K(R,K)$ be an $R$-module monomorphism. This is also a $K$-vector space monomorphism. Since $R$ is a finitely dimensional $K$-vector space, $\dim_KR=\dim_KR^*$, and thus every monomorphism of $K$-vector spaces $R\to R^*$ must be an isomorphism.


2

The neutral element (or, as I would call it, the multiplicative identity) is an element $1_A \in A$ such that for any $f \in A$, we have $f \circ 1_A = 1_A \circ f = f$. Verify that this is true for $1_A(x) = x$. It seems that you've correctly shown that $A$ satisfies the distributivity axiom for both right and left multiplication. In order to prove that ...


2

No, I don't believe that this version of Nakayama's lemma is valid for noncommutative rings in general. There is a notion of a right weakly regular ring (or right fully idempotent ring) where all right ideals are idempotent. Now, all von Neumann regular ring are right and left weakly regular, but there must be examples of right weakly regular rings that ...


2

Let $R$ be the subalgebra of $K^{\mathbb N}$ (a countable direct product of copies of a field $K$; for simplicity one can choose $K=\mathbb Z/2\mathbb Z$) generated by $1$ and $K^{(\mathbb N)}$ (a countable direct sum of copies of the field $K$), that is, $R$ consists of all sequences $(a_n)_{n\ge 0}$ such that $a_n\in K$ and which are constant from some $n$ ...


2

Notice that $A \cong \hom_R(R/J,M)$. Now we have the following fact: If $R \to S$ is a ring homomorphism and $M$ is an injective $R$-module, then $\hom_R(S,M)$ is an injective $S$-module. In fact, $\hom_S(-,\hom_R(S,M)) \cong \hom_R((-)|_R,M)$ is exact.


2

In Definition 2 the action of $k$ on $M$ induced by the action of $k$ on $A$ can be different from the given $k$-action on $M$.


1

As you say in your reasoning, the $A$-module structure on $M$ induces a $k$-module structure. So does the $B$-module structure. But without added conditions these two $k$-module structures could be different. For example, take $A=B=M=\mathbb{C}$ with the bimodule structure $$a\cdot m\cdot b=am\overline{b},$$ where $\overline{b}$ is the complex conjugate of ...


1

The first definition is what you might call a "ring" bimodule. It only asks for $A$ to be a ring, and it doesn't mind if $A$ has any $k$ algebra structure. I would call your second definition an algebra-bimodule, but I think you're missing an axiom. Not only does $M$ have a $k$-module structure, but this structure is compatible with $A$'s $k$-structure! You ...


1

Take $g_1=k_1\circ \sigma_1$ and $h_1=k_1\circ \sigma_2$ where $\sigma_1,\sigma_2:M\to M\oplus M$, $\sigma_1(x)=(x,0)$ and $\sigma_2(y)=(0,y)$.


1

Apologies in advance for this answer. The problem with it is that it is too advanced, and also relies on a lemma that is very similar to your question. I will continue to seek a more elementary answer. Lemma: Every left $R$ module over a left Artinian ring $R$ has a projective cover. Lemma: Every nonzero projective module has a maximal submodule. Lemma: ...


1

Given any ascending sequence of sets $X_1\subset X_2\subset X_3\cdots\subset X$ with $\bigcup_{n\ge1}X_n=X$ and any collection of functions $g_i:X_i\to Y$ such that $g_i|_{X_{i-1}}=g_{i-1}$, we can form the map $g:X\to Y$ defined by the relation $g(x)=g_i(x)$ whenever $x\in X_i$. Each function $g_i$ is a bigger and bigger "glimpse" of $g$. One can check ...


1

Let $N_n=\sum_{i=1}^{k}c_iR$; then $D$ is the union of the increasing family of submodules $N_n$ and as such it is its direct limit with inclusions as transition maps. If you consider the restriction $h_n$ of $g_n$ to $N_n$, the given condition translates into the fact that $h_n\colon N_n\to B$ is a family of morphisms compatible with the inclusion maps, so ...



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