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6

Yes, there is such a finite subset. Because $X$ is a generating set for the module, given any $m \in M$ there exist $x_1,\dots, x_k \in X$ and scalars $r_1,\dots,r_k \in R$ such that $m = r_1 x_1 + \cdots + r_k x_k$. Now suppose $m_1,\dots, m_n \in M$ is a finite generating set. Then for each $m_i$ there exist $x^{(i)}_1, \dots, x^{(i)}_{k(i)} \in X$ such ...


3

If $A$ is a finite dimensional algebra, then every simple $A$-module is finite dimensional, since it is a quotient of the regular module (if $S$ is a simple left module, then for any $0\neq s\in S$, $a\mapsto as$ is a non-zero homomorphism $A\to S$, which is surjective since $S$ is simple).


3

It is enough to show that the homomorphism $\operatorname{Hom}_{R_S}(R_S,I_S) \rightarrow \operatorname{Hom}_{R_S}(J_S,I_S)$ is surjective for every ideal $J$ of $R$. We have an exact sequence $0 \rightarrow J \rightarrow R \rightarrow R/J \rightarrow 0$. Since $I$ is injective, the functor $\operatorname{Hom}_R(-,I)$ is exact. Hence we have an exact ...


3

For any commutative group $G$, you have an isomorphism $\,\,\varphi\colon \operatorname{Hom}(\mathbf Z, G) \simeq G$, defined by $\varphi(f)=f(1)$.


3

First of all, since $M$ is Artinian and Noetherian, its Krull dimension is zero. Hence all associated primes of $M$ are maximal ideals of $R$. Moreover, these finitely many associated primes - let's call them $m_1,\dots,m_n$ - constitute the entire support of $M$. The above remarks imply that $S := R/Ann(M)$ is Artinian ring and ...


3

Sure; one does this all the time, in fact. Let me use $B$ for an $A$-algebra instead of $N$. It is, in particular, an $A$-module so we can at least form the $A$-module $M \otimes_A B$. The additional feature is that $M \otimes_A B$ is also a $B$-module in a natural way — to see this, you can use the $A$-trilinear map \begin{align*} B \times M \times B ...


3

The answer is yes, as $M'$ and $M$ are both free the inclusion map $M' \to M$ is given by a matrix and this matrix can be put into Smith Normal Form which is a matrix of the form $$\begin{bmatrix}a_1 \\ & a_2 \\ && \ddots \\ &&& a_n \\ &&& \mbox{} \\ &&& \mbox{} \\ &&& \mbox{} \end{bmatrix}$$ ...


2

Since any finitely generated module $M$ over a PID $R$ can be decomposed as $F\oplus T$, where $F$ is free and $T$ is torsion, it follows that $$ \operatorname{rank}M=\dim M\otimes_R Q $$ where $Q$ is the field of fractions of $R$ and $M\otimes_RQ$ is considered as $Q$-vector space in the obvious way. Note that $T\otimes_RQ=0$, if $T$ is torsion. Since $Q$ ...


2

This comes from $M\otimes_{A}N\simeq M\otimes_{A/I}(A/I\otimes_{A}N)$, and $A/I\otimes_{A}N\simeq N/IN=N$.


2

Consider $\mathbb{Z} \rightarrow \mathbb{Z}$ multiplication by 2.


2

Let $A=\mathbb Z$, $B=\mathbb Q$, and $M=\oplus_{n\ge1}\mathbb Z/2^n\mathbb Z$. Then $\operatorname{Ann}_A(M)=(0)$, while $\operatorname{Ann}_B(M_B)=B$ since $M\otimes_AB=0$. Remark. In this example $A\to B$ is flat, but $M$ is not finitely generated.


2

I'm not an expert on $\mathcal{D}$-modules, but hopefully I can help a bit. For 1, you're probably right. If $M = N$ is also equal to the base ring $A$, then yes, $\operatorname{Diff}(M,M)$ will be an almost commutative ring, sometimes denoted $\mathcal{D}(X)$ if $X$ is the affine scheme $X = \operatorname{Spec}A$ (of course using the sheaf definition in ...


2

It means that there is a polynomial $p(n)=a_{r-1}n^r+...+a_0$, and a number $N$ such that $H_M(s)=p(s)$ for $s>N$. Maybe more important to understand than that statement is the following equivalent one. Define $\Delta f(n):=f(n+1)-f(n)$ and $\Delta^{m}f(n)=\Delta\Delta^{m-1}f(n)$. Then the statement above is the same as $\Delta^{r}H_M(s)=0$ for all ...


2

Yes, that's true: You have the decomposition $R = eR \oplus (1-e) R$ as right $R$-modules, hence $$\text{Hom}_R(eR,M)\cong\text{ker}\left(\text{Hom}_R(R,M)\xrightarrow{\text{res}}\text{Hom}_R((1-e)R,M)\right)\stackrel{(\dagger)}{\cong}\text{ker}(M\xrightarrow{\cdot (1-e)} M) = Me,$$ Explicitly, this map is given by $\varphi\mapsto\varphi(e)$ and hence right ...


2

a) One solution, not depending on the construction of the tensor product but on its universal property, is to realize that the map $f: M\times N\to M\otimes_B N:(m,n)\mapsto m\otimes n$ is $A$-bilinear and thus factors through an $A$-linear morphism $f: M\otimes_A N\to M\otimes_B N:m\otimes n \mapsto m\otimes n$, which is obviously surjective. Very nice, ...


2

Consider $\prod \limits_{i \in I} M_i$. The sum is $$( a_i)_{i \in I } + ( b_i)_{i \in I } = ( a_i + b_i)_{i \in I }$$ The action of R : $$ r \cdot ( a_i)_{i \in I } = ( r \cdot a_i)_{i \in I }$$


2

We have the short exact sequence $$ 0 \to N \xrightarrow[]{i} M \xrightarrow[]{\pi} M/N \to 0$$ Tensor on the right by $A$ and get a right exact sequence $$N \otimes A \xrightarrow[]{i\otimes 1_A} M \otimes A \xrightarrow[]{\pi\otimes 1_A} (M/N)\otimes A \to 0$$ Therefore, the kernel is the submodule of $M\otimes A$ generated by $i(n)\otimes a$ with ...


2

$\DeclareMathOperator{\Tor}{Tor}$ Let $N\subseteq M$ and $A$ be as in the question. In general, there is not a simple description of the kernel of the map $M\otimes A\to M/N\otimes A$. Given modules $X$ and $Y$ (on the right and on the left, respectively) there is a standard way to construct an abelian group $\Tor(X,Y)$; this can be done in several ways, ...


2

Use the third isomorphism theorem. If $G$ is a group (or ring, or module) and $H$ and $K$ are normal subgroups (or ideals, or submodules, respectively) of $G$, with $H\subseteq K$, then there is a natural isomorphism $$(G/H)/(K/H)\cong G/K.$$


2

If $\mathcal{C}$ is closed symmentric monoidal category, and $\mathbb{T}$ is symmetric monoidal monad on $\mathcal{C}$, then the category of $\mathbb{T}$-algebras caries a structure of closed symmetric monoidal category. this result is well known, and was proved be Anders Kock in the '70. For the definition of symmetric monoidal monad, you can look at ...


2

For a finitely generated module over a local ring being flat is the same as being free. Can the fraction field of a domain have two elements linearly independent over the domain? Try this for $\mathbf{Z}$, for example.


2

Here is an example: $R = \mathbb{Z}/4$. Take a non-zero $R$-module $M$ which is simply an abelian group annihilated by $4$. If $2M$ is non-zero, it will be a non-zero $R/2R \simeq \mathbb{Z}/2$ module, but $\mathbb{Z}/2$ is a field, done. If $2M$ is $0$ then $M$ is a $R/2R$ module, ... , again done. Obs: Works similarly for $\mathbb{Z}/p^2$, for ...


2

I would like to post an alternative proof: By Proposition 2.12 of the ops answer (in other words by the universal property of tensor product) the following bilinear map \begin{eqnarray*} \psi:A/\mathfrak{a}\times M & \to & M/\mathfrak{a}M \\ (x+\mathfrak{a},M) & \mapsto & xm+\mathfrak{a}M \end{eqnarray*} induces the unique ...


2

Yes, there is a surjective homomorphism of $A$-modules $L\to N$ and denote its kernel by $K$. Then, by the fundamental isomorphism theorem we have $N\simeq L/K$. The diagram chasing isn't that hard: let $x\in M\otimes N$ which is sent to $0$; since $\beta$ is surjective there is $y\in M'\otimes L$ such that $x=\beta(y)$. Now let $f=(M\otimes L\to M\otimes ...


2

Let $R$ be a commutative ring, let $M$ be an $R$-module, and let $I \subseteq R$ be an ideal. Then $M/IM$ is naturally an $R/I$-module, with scalar multiplication given by $(r + I)(m + IM) = rm + IM$ for all $r \in R$ and $m \in M$. (Easy exercise: check that this is well-defined and makes $M/IM$ into an $R/I$-module.) Similarly, let $U$ be a ...


2

Hint. If $C$ is $A$-flat, then $(sA\cap tA)C=sC\cap tC$. Does this equality hold?


1

Consider the diagram of the proof: $$\require{AMScd} \begin{CD} {} @. M'\otimes K @>f>> M\otimes K @>g>> M''\otimes K \\ @. @VVhV @VViV @VV{\alpha}V \\ 0 @>>> M'\otimes L @>j>> M\otimes L @>k>> M''\otimes L \\ @. @VV{\beta}V @VVlV \\ {} @. M'\otimes N @>m>> M\otimes N \\ @. @VVV @VVV \\ {} @. 0 @. 0 ...


1

This is mostly similar to your argument, but much easier. From the exact sequence $0\to\mathfrak{a}\to A\to A/\mathfrak{a}\to 0$ we get the commutative diagram with exact rows $$\require{AMScd}\def\ma{\mathfrak{a}} \begin{CD} {} @. \ma\otimes_AM @>>> A\otimes_AM @>>> A/\ma\otimes_AM @>>> 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> ...


1

Reposting from my comments since other proofs have already been posted. By tensoring the exact sequence $0 \to \mathfrak a \to A \to A / \mathfrak a \to 0$ with $M$, we get the exact sequence $$ \mathfrak a \otimes_A M \to A \otimes_A M \to A / \mathfrak a \otimes_A M \to 0. $$ Consider the canonical isomorphism $A \otimes_A M \cong M$. By composing the ...


1

$$0=Tor^1(M^{\prime\prime},N)\to M^\prime \otimes N\to M\otimes N\to M^{\prime\prime} \otimes N\to 0$$ $0=Tor^1(M^{\prime\prime},N)$, since $M^{\prime\prime}$ is flat.



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