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A finitely generated module over a noetherian ring is of finite length iff its support is contained into the maximal spectrum. (See here, Proposition 1.6.9.) Now let $\mathfrak p\in\operatorname{Supp}(\operatorname{Coker}f)$. If $\mathfrak p$ is not maximal, then it is minimal since $\dim R\le 1$. But $(\operatorname{Coker}f)_{\mathfrak ... 3 You got something wrong there.$\mathbb{Q}$is not artinian as a$\mathbb{Z}$-module because$\mathbb{Z}\supseteq 2\mathbb{Z}\supseteq 4\mathbb{Z}\supseteq 8\mathbb{Z} \supseteq \cdots$is a strictly descending sequence of submodules. And similarly$\mathbb{Z}_{(2)}\supseteq 2\mathbb{Z}_{(2)}\supseteq 4\mathbb{Z}_{(2)}\supseteq 8\mathbb{Z}_{(2)} \supseteq ...

2

This isn't true. For instance, if $R=\mathbb{Z}$, then since every simple module is finite, any finite length module must be finite. But there are finitely generated modules that are not finite (e.g., $\mathbb{Z}$ itself). For your followup question, you can take $M=\mathbb{Z}^2$ and $N=\mathbb{Z}$ to get a counterexample.

2

You just apply the distributive property. The element $\alpha=(0,a,b,0,0,\dotsc)$ is just $a+b$ and $\xi=(0,x_1,x_2,0,0,\dotsc)$ is $x_1+x_2$; then $$\alpha\xi=(a+b)(x_1+x_2)=ax_1+ax_2+bx_1+bx_2= (0,0,ax_1,ax_2+bx_1,bx_2,0,0,\dotsc)$$ because $ax_1$ has degree $2$, $bx_2$ has degree $4$ and both $ax_2$ and $bx_1$ have degree $3$. It's no different at all ...

2

It indeed follows from looking at how the ideal looks. Consider what happens when you multiply $H_n$ by an element of $H$ from the left. This clearly (after a substitution) leaves us with $H_n$ again (it is invariant). Now consider an element $$\sum_{h \in H}{f_h h } \in F[H]$$ Then using the above remark, we find that the left multiplication of this element ...

2

A ring over which this is not possible is called a stably finite ring (to relate your definition to the one on Wikipedia, take $A$ and $B$ to be the matrices of the projection $M\to N$ and the inclusion $N\to M$, respectively). Here's an example of a ring with invariant basis number that is not stably finite. Let $k$ be a field and let $R=k\langle ... 2 Endomorphisms of finite direct sums are most conveniently described with matrices. Suppose$M$and$N$are modules; write the elements of$M\oplus N$as columns; then an endomorphism of$M\oplus N$is described by a matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix},$$ where$a\colon M\to M$,$b\colon N\to M$,$c\colon M\to N$and$d\colon N\to ...

1

You are right: $P(A,t)=(1-t)^{-s}$, and this can be proved by induction on $s$. For the Hilbert polynomial just count the number of monomials of a fixed degree. For the characteristic polynomial note that the associated graded ring of $A$ with respect to $Q$ is nothing but $A$. Then the characteristic polynomial is $\binom{X+s}{s}$.

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I assume that $D$ is a division ring as in one of your previous questions and that $\{e_i\}_{1≤i≤n}$ is the canonical basis. For the first part you know that $\phi$ is determined on $e_1,\dots, e_n$. Then you can use elementary matrices to show that $\phi$ is actually only determined on $e_1$. For the second part, you know that you can write $$\phi(e_1) ... 1 Let R=k[X,Y]. We have H_{R/I}(t)=1+t, and H_{R/J}(t)=1+2t+t^2. On the other side, their Hilbert polynomial is 0. 1 1) The functor - \otimes_A M is right exact. Therefore, the exact sequence I \to A \to A/I \to 0 yields the exact sequence I \otimes_A M \to A \otimes_A M \to A/I \otimes_A M \to 0. Now use that A \otimes_A M \cong M and that the image of I \otimes_A M \to M is exactly IM. This shows A/I \otimes_A M \cong M/IM. You can also show directly that ... 1 ‘\overline\phi is surjective’ means N=\phi(M)+IN. From this you deduce$$N=\phi(M)+I(\phi(M)+IN)=\phi(M)+I\phi(M)+I^2N=\phi(M)+I^2N,$$and by a silly induction:$$N=\phi(M)+I^kN\quad\text{for all}\enspace k\ge 1. Now choose for $k$ the nilpotency index of $I$, and you get $\;N=\phi(M)$, i.e. you get the surjectivity of $\phi$.

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I'm surprised no one has mentioned the following very simple example. Let $A=\mathbb{Z}$ and let $M=\mathbb{Q}$. Then $M$ is free at the generic point of $\operatorname{Spec} A$, but is not free in any open neighborhood. Or, if you want $x$ to be a closed point, you can instead take $M=\mathbb{Z}_{(p)}$ for some prime $p\in\mathbb{Z}$. More generally, if ...

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For $n\in\mathbb{N}$, let $M_n\subseteq\mathbb{Q}$ be the submodule generated by $2^n$. Then the $M_n$ form an infinite descending chain of submodules. (This is essentially the same example as in the linked question, just instead of using different primes, you use higher and higher powers of the only prime you have.)

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Here's a counterexample. Let $k$ be a field and take $A=B=k[x]/(x^2)$ and $M=C=k$ (considered as an $A$-algebra via $x\mapsto 0$). Then $\mathrm{Hom}_B(M,B)\cong k$ (spanned by $1\mapsto x$), so $\mathrm{Hom}_B(M,B) \otimes_A C\cong k$ as well. But $1\mapsto x$ maps to $0$ in $\mathrm{Hom}_B(M,B \otimes_A C)$, since $x=0$ in $B\otimes_A C=C$. On the ...

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