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5

Given a finitely generated module $M$ over a noetherian ring $A$, there exists a filtration of $M$ by submodules $M=M_0\supset M_1\cdots \supset M_n=0$ such that $M_i/M_{i+1}\cong A/\mathfrak p_i$ for some prime ideals $\mathfrak p_i\subset A$ (Bourbaki, Commutative Algebra, Chapter IV, ยง1, Theorem 1, page 261) Now $M$ has finite length iff all the ...


5

$\DeclareMathOperator{\Hom}{\operatorname{Hom}}$By tensor-Hom adjointness, $$\Hom_R(M, R/M) \cong \Hom_R(M, \Hom_{R/M}(R/M, R/M)) \cong \Hom_{R/M}(M \otimes_R R/M, R/M) \cong \Hom_{R/M}(M/M^2, R/M) = (M/M^2)^*$$ where $*$ denotes vector space dual over the field $R/M$. Thus $(M/M^2)^* = 0 \iff M/M^2 = 0 \iff M = 0$ by Nakayama's Lemma (since $M$ is finitely ...


4

Of course i find zcn's answer very instructive, however i think there is a more low-tech proof: Suppose $R$ is not a field. Then $M \neq 0$ or equivalently by Nakayama $M \neq M^2$ (since $M$ is finitely generated). Hence there exists an element $x \in M$ whose class $\bar{x}$ in $M/M^2$ is non-zero. Since $M/M^2$ is an $R/M$ vector space, we can define a ...


3

You're right to be skeptical. As you've already noted at your other question, Artinian rings are ruled out from consideration since the Krull-Schmidt theorem can be used to prove they have the aforementioned property. Happily, I can demonstrate a ring without the property that is even von Neumann regular. I was inspired by a theorem in Goodearl's book von ...


3

$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $\operatorname{pd}_RM<\infty$. Now we can apply the Auslander-Buchsbaum formula and get $\operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.


3

If $R$ is any ring and $M$ is a left $R$-module, then there is always an isomorphism of abelian groups $R \otimes_R M \cong M$ given by $r \otimes m \mapsto rm$ and $m \mapsto m \otimes 1$ in the other direction. In fact, $(r,m) \mapsto rm$ is $R$-balanced, hence induces a homomorphism $R \otimes_R M \to M$ of abelian groups. Clearly, $m \mapsto m \otimes ...


3

Clearly, the only candidate for a basis is $\{1\}$. However, is $\{1\}$ a linearly independent set? That is, is it true that for all $n \in \Bbb Z$, $n\cdot 1 = 0 \iff n = 0$?


3

We have $\operatorname{length}_{\mathcal{O}_P} \mathcal{O}_P/(f,g) = \operatorname{length}_{\mathcal{O}_P/(f,g)} \mathcal{O}_P/(f,g)$. So it is enough to show that the latter equals $\operatorname{length}_k \mathcal{O}_P/(f,g)$. Let $A = \mathcal{O}_P / (f,g)$, $\mathfrak{m}$ its maximal ideal, and assume that the two curves have no components in common, ...


3

The definition of the tensor product is its universal property, which is quite simple. What you are struggling with is the construction of the tensor product - this is something different. If you want to see a construction of the tensor product which avoids free modules at all, see here. $K$ is by definition a submodule, since it is defined as the submodule ...


3

Consider the multiplicative set $S= \mathbb Z\setminus \{0\}$. Then $$\mathbb Q\simeq S^{-1}\mathbb Z.$$ Now, in general you have that, if $M$ is an $A$-module and $S$ is a multiplicative set of $A$, then $$S^{-1}A\otimes_AM \simeq S^{-1}M, $$ that is naturally a $S^{-1}A$-module. This shows that this definition of rank of an $A$-module holds whenever $A$ ...


3

Take a Noetherian ring $R$. Then $R^n$ is a Noetherian module and all its submodules are Noetherian. For instance, PIDs are Noetherian, because all their ideals are principal, hence finitely generated. You could take $R = \mathbb Z$ or $R = k[x]$ (where $k$ is a field) and consider $M = R^n$. Looking at submodules of $M$, you get more Noetherian modules ...


3

I assume that, for you, an associated prime $P$ of $R$ is a prime ideal that is the annihilator of some nonzero $m\in M$, and that you want to show that this implies that there is some submodule of $M$ isomorphic to $R/P$. Here is my hint: As part of the condition, we are given a nonzero $m\in M$. Using this, construct a homomorphism of modules $R\to M$, ...


3

Let me show how a family of isomorphisms $\mathcal{C}(X,Z)\cong\mathcal{C}(Y,Z)$ natural in $Z$ gives us an isomorphism $Y\cong X$. I will assume that you are familiar with functors, in particular the $\operatorname{Hom}$-funcotrs in question and natural transformations, because otherwise discussing implications of Yoneda's Lemma won't be very fruitful. For ...


3

1) A one-dimensional space over $k$ always looks like $k$. Think about what copies of $k$ there are, and which ones are invariant. 2) Why do you say that $bxbx\notin W$? $bxbx = b^2 x^2 = 0 \in W$.


3

your morphism is correct. For the surjectivity, if you take any $(m,s)\in M\oplus S$, take $f(m)+s'-f\circ\psi(s')$ where $s'$ is any preimage of $s$ under $g$. For injectivity, if $(\psi(n),g(n))=(0,0)$, then $n$ is in the image of $f$ because the sequence is exact, i.e. $n=f(m)$, but then $0=\psi (n)=\psi\circ f (m)=m$ and so $n=0$


2

For a field $k$, $k[x]$ is a principal ideal ring. By the correspondence theorem, the only ideals of $k[x]/(x^2)$ are those generated by divisors of $x^2$. Thus $k[x]/(x^2)$ has exactly three submodules: $(x^2)/(x^2),(x)/(x^2)$ and $k[x]/(x^2)$. So the existence of $(x)/(x^2)$ immediately tells you why $k[x]/(x^2)$ isn't irreducible. And how could it be ...


2

I claim that $I=J=0$ already yields a family of counterexamples to $R\otimes_R R \cong R$. We must have $R\cdot R \neq R$ for this to work (I think this condition might also be sufficient for the natural map to not be an isomorphism, but I haven't worked out all the details). $R=n\mathbb{Z}$ is such a rng: $n\mathbb{Z} \otimes_{n\mathbb{Z}} n\mathbb{Z} ...


2

The Krull-Schmidt theorem literally says: If you decompose a module with finite composition length into a direct sum of indecomposables in two ways, then the two decompositions are the same length, and there's some permutation that pairs up factors from each decomposition into isomorphic pairs. So the idea is that you take the two decompositions ...


2

If $\mathfrak{p}$ is any prime ideal of $A$, we have $$ A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p} \cong \text{Frac}(A/\mathfrak{p}).$$ Let's assume this for now. In general, we have $$M_\mathfrak{p}/\mathfrak{p}M_\mathfrak{p} = (A_\mathfrak{p} \otimes_A M) / (\mathfrak{p} A_\mathfrak{p} \otimes_A M) = (A_\mathfrak{p} / \mathfrak{p}A_\mathfrak{p})\otimes_A ...


2

Yes, the trivial homomorphism. But apart from that, there is no natural homomorphism. Proof. Suppose that $\eta_{A,B} : A \otimes B \to A \times B$ is a natural homomorphism, meaning that $\eta_{A,B}$ defined for all $R$-modules $A,B$ and $\eta = (\eta_{A,B})_{A,B}$ is a natural transformation. Let $(r_1,r_2) := \eta_{R,R}(1 \otimes 1) \in R \times R$. If ...


2

No need any assumption on $I$ like being invertible. $I/mI$ is an $R/m$-vector space. (If $ax=0$ in a $K$-vector space, then $a=0$ or $x=0$.)


2

You can always start with a ring $R$ and an $R$-module $M$. Considering the localization $M_{\mathfrak p}$ of the module $M$ at the prime ideal $\mathfrak p$ (which can be realized as $R_{\mathfrak p} \otimes_R M$), you can assume the ring local. Then you can consider the tensor product $R_{\mathfrak p}/\mathfrak p R_{\mathfrak p} \otimes_{R_{\mathfrak p}} ...


2

If $V$ is finite dimensional then $V$ is a finitely generated torsion $K[x]$-module. Now apply the structure theorem and find that $V$ is a direct sum of cyclic $K[x]$-modules. Since $\chi_T$ is the product of all invariant factors and $m_T$ is the last of them, we can conclude that $V$ has only one invariant factor equal to $\chi_T=m_T$, so $V\simeq ...


2

Here's a basic argument without using any structure theorem (though I shall use that $K[x]$ is a PID). First off, $K[x]$ does not have any simple modules that are infinite dimensional over $K$. Any element$~v\neq0$ of a simple module generates the whole module, and if $P.v=0$ for some nonzero $P\in K[x]$, the generated module would be of dimension $\deg ...


2

She could probably define singular (co)homology and do a couple of applications in two hours, if she just stated the main technical theorems. Possibly axiomatic homology theory would be a beginning point suited to the abstract and algebraic pitch of the course. Something more interesting would involve spectra and triangulated categories, but that's surely ...


2

Given that you know that the right ideals are precisely the powers of the Jacobson radical, I'm guessing you also know that they coincide with the left ideals, and that $J^k=p^kR=Rp^k$ for any $p\in J\setminus J^2$? If $\alpha:p^kR\to R$ is a right $R$-module homomorphism then $\operatorname{im}(\alpha)$ is a right ideal of $R$ and can't have length greater ...


2

This is true if $M$ is finitely generated. By the structure theorem, there exists a presentation $M = \mathbb{Z} \langle e_i \rangle$ where the only relations are of the form $n_ie_i = 0$, for some $n_i \in \mathbb{Z}$. Then $M \otimes M = \mathbb{Z}\langle e_i \otimes e_j \mid (n_i, n_j) \ne 1 \rangle$, and to define a group homomorphism on $M \otimes M$ it ...


2

Say $p$ is a continuous antiperiodic function that is cyclic. Note any multiple of $p$ must be zero wherever $p$ is zero. Since there exist anti periodic functions with disjoint zero sets (e.g. sine and cosine), what must be true about the zero set of $p$? Does there exist such a $p$?


2

Since then $D^i$ would be a finite dimensional $\Bbb R$ division algebra, the Frobenius theorem says $D^i$ has to be $\Bbb R$, $\Bbb C$ or $\Bbb H$. But $\Bbb R$ and $\Bbb H$ are not $\Bbb C$ algebras because their centers are both $\Bbb R$, and so neither center can contain a copy of $\Bbb C$. So $D^i=\Bbb C$ for every $i$. That means each matrix ring has ...



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