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Here is another more complicated example that shows Julian's ideas aren't just hiding the problem in a sneaky way, but that my dichotomy/theorem of the alternative is just completely wrong for simple modules: Consider the (circular) quiver on 4 vertices with edges $1\to 2 \to 3 \to 4 \to 1$ and add the zero relations $1\to2\to3\to4$, $2\to3\to4\to1\to2$, ...

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If I understood your question correctly then the following provides a counterexample: $$1\to 2\to 3\to 4\to 2$$ where the two $2$'s should be identified. Take $A$ to be the path algebra of this quiver modulo $\operatorname{rad}^2 A$. If you take the projective dimension of $S_1$, then $P_1$ just appears in the beginning as $P_0$ and then $P_2, P_3$ and $P_4$ ...

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Following Jack's gentle nudging, I'm posting the part of my comments not covered by rschwieb's nice post more prominently as an answer: In general, this type of question can be answered quickly by consulting Lam's Lectures on Modules and Rings, Springer Graduate Texts, Vol. 189, whose numerous exercises are solved and put into further context by Lam himself ...

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Over a von Neumann regular ring, every right module (and every left module) is flat. Let $V$ be an countable dimensional $F$ vector space, and let $R$ be the ring of endomorphisms of that vector space. It's known that $R$ is a von Neumann regular ring with exactly three ideals. The nontrivial ideal $I$ consists of the endomorphisms with finite dimensional ...

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Rings over which every module is flat are called absolutely flat, or Von Neumann regular. If $I$ is any nontrivial ideal of an absolutely flat ring, then $R/I$ is finitely generated flat over $R$, but not free (since $R/I \neq 0$ and $\mathrm{Ann}(R/I)=I \neq 0$). Products of fields are absolutely flat.

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Here is an example that one can come up with from algebraic number theory. Consider $K = \Bbb{Q}(\sqrt{-6})$ and the ideal $I = (2,\sqrt{-6}) \subseteq \mathcal{O}_K$. This ideal is clearly finitely generated as an $\mathcal{O}_K$ module and furthermore not only is it flat but it is a projective module: we have $$I \oplus ( 3,\sqrt{-6}) \cong (\sqrt{-6}) ... 6 Let R be a semisimple ring which isn't a division ring, and take an idempotent e\notin \{0,1\}. Then we have that R=eR\oplus(1-e)R is a nontrivial decomposition of R, and both pieces are cyclic and projective (since they are summands of R) hence flat. Actually neither piece is a free module, but we'll argue here that at least one of them isn't ... 1 You are overcomplicating the problem too much. First recall that an R - submodule of R/(p^k) is just an ideal of this ring. By the correspondence theorem, the ideals of R/(p^k) are in one to one correspondence with the ideals of R that contain (p)^k. You should check that the only ideals that contain this are$$(p)^i, \hspace{5mm} 0 \leq i \leq k$$... 1 Since Jack Schmidt has mentioned direct limits here is how we can prove the first part of (b) using direct limits. Let I be any ideal of R; we can write I = \operatorname{colim} I_\alpha where the colimit is over all finitely generated ideals contained in I. Now consider the ses$$0 \to I\to R \to R/I \to 0$$from which we get a long exact ... 1 Thanks to Gerry Myersons comment, I was able to find the general outline of the proof: Let us denote \tilde{\sigma} as the divisor counting function in \mathbb{Z}[i] (\sigma is the divisor counting function in \mathbb{Z}). We use the identity$$a=\prod\limits_{k=1}^n p_k^{e_k}\sigma(a)=\prod\limits_{k=1}^n (e_k+1)$$and that (according to Gerry ... 2 First note that that the condition A \otimes_R K \to A \otimes_R F is injective for all free R-modules F and submodules K is equivalent to A being flat. So you will definitely need to assume that A \otimes_R I \to A \otimes_R R is injective for all ideals I \subseteq R. We will prove that \operatorname{Tor}_1^R(A, F/K) = 0 for any such F ... 2 When you do the operation 4 R_2 - 5 R_1, you are altering the determinant by a factor of 4, since you are multiplying your matrix on the left by$$ \begin{bmatrix}1 & 0 & 0\\-5&4&0\\0&0&1\end{bmatrix}. $$Ditto for 3 R_1 + R_2, which alters the determinant by a factor of 3. These two operations have introduced the spurious ... 0 Using Fermat's Little Theorem, a^{p-1}\equiv1\pmod p where p is prime and a is any integer relatively prime to p If modulo/Multiplicative order of a is ord_pa, "as a consequence of Lagrange's theorem, ord_pa always divides \phi(p)=p-1", we need to test for those powers of a which divides p-1 Now,$$13^1\equiv1\pmod 3\implies ...

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There is a bit of theory to go through to explain how the invariant factors of the matrix you wrote down is actually related to $M,$ you will have to go through your notes to see the precise connection. As for the calculation, perhaps recheck your operations, or post them here so we might we able to say where you went wrong. It turns out that for this ...

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$\operatorname{im}(\varphi)=\{\varphi(x_1,x_2,x_3):(x_1,x_2,x_3)\in\mathbb Z^3\}=\{x_1\varphi(1,0,0)+x_2\varphi(0,1,0)+x_3\varphi(0,0,1):$ $(x_1,x_2,x_3)\in\mathbb Z^3\}=\{x_1(6,4)+x_2(4,8)+x_3(4,0):(x_1,x_2,x_3)\in\mathbb Z^3\}$ $=$ $\langle(6,4),(4,8),(4,0)\rangle=N$. The form of the canonical diagonal matrix $D$ shows that there exists a $\mathbb ... 1 Some ideas: The homomorphism$\,\phi\,$described there could be thought of as an embedding one: note that$\,N\cong R^s\le R^k\,$, so one can embed the former in the latter. Of course, this embedding is usually far from being unique. From here$\,\ker\phi=0\,$is immediate. The notation$\,\langle x\rangle\,$usually denotes the cyclic (module, group, ... 3$\mathrm{Hom}_R(R/I,N)\simeq (0:_NI)$, where$(0:_NI)=\{x\in N:Ix=0\}$. Since$R/J\otimes M\simeq M/JM$one can take$N=M/JM$and get$\mathrm{Hom}_R(R/I, R/J\otimes M)\simeq(JM:_MI)/JM$. 3 A possibly instructive, concrete example is$\Bbb{Q}$as an abelian group (i.e. as a$\Bbb{Z}$-module). Of course, this is just a special case of Dedalus's answer (and YACP's comment), but let's still prove the result directly. Clearly$\Bbb{Q}$is not a faithfully flat abelian group, e.g. the zero morphism $$\Bbb{Z}/2\Bbb{Z} \rightarrow 0$$ is not ... 3$0$is always flat, but not faithfully flat (unless the base is$0$). 4 Take$f: A \rightarrow B$to be a ring homomorphism such that the corresponding morphism of affine schemes$Spec B \rightarrow Spec A$is not surjective, but only flat. There is an easy way to do this: Remember that localizing a ring R in a multiplicative subset$S$gives a flat ring homomorphism$R \rightarrow S^{-1}R$. However, this ring homomorphism is ... 3 As far as I know, field should not matter, unless your algebra has "weird relations", and in most case, you are pretty safe as long as characteristic of$k$is not 2. I hope somebody can give some supplement answer here... Back to your specific example, (say if we work over characteristic 0..) the (minimal) projective resolution of$S(1)$is actually: $$... 1 For your first question, f is given by \epsilon_1\mapsto \beta, or equivalently p_1:\epsilon_1 x \mapsto \beta x for any x\in kQ. You just need to stare at your diagram long enough to realise this...Note that the squares have to commute. I usually prefer computing in (sort of) Loewy diagram form, as the algebra is basic and the field (so all ... 0 I like to prove things like this as algorithmically as possible, using matrices and row and column operations that are as elementary as possible. I construct an invertible matrix A such that A \matrix(a_1,\ldots,a_n)^T = (1,0,\ldots,0)^T. Let B be the linear map corresponding to A relative to the basis \{y_1,\ldots,y_n\}. Then the desired basis ... 7 If I is an ideal of R, the dual of R/I is isomorphic to \mathrm{Ann}(I) = \{r \in R : rI = 0\}, and this doesn't have to be finitely generated. Take for instance R = k[y,x_1,x_2,\dotsc]/(y x_i : i \geq 1) and I=(y). A more natural question would be: How can we characterize commutative rings with the property that duals of f.g. modules over that ... 2 Let R be a PID M a free R-module of finite rank n. Let \{x_1,\dots,x_k\} be k\leq n elements in M, and denote N their R-span in M. Then the claim is that the k-ple \{x_1,\dots,x_k\} can be completed to a basis of M over R if and only if$$ rk(N)=k\qquad\text{and}\qquad\text{$M/N$is torsion-free.} $$Indeed, if ... 1 The second isomorphism theorem says we have an exact sequence$$0 \to K \cap N \to K \to (K+ N)/N \to 0$$and so l(K) = l(K\cap N) + l\left( K+N/N\right) = l(K \cap N)+ l(K+N) - l(N) . Thus$$l(K) + l(N) = l(K \cap N) + l(K+N).$$1 Hint 1: Consider (K+N)/K. If you have a composition series for this module, then it can "stack on top of" a composition series of K to make a composition series for K+N. Hint 2: By an isomorphism theorem, (N+K)/K\cong N/(N\cap K). Hint 3: Convince yourself that \ell(M/N)=\ell(M)-\ell(N), (at least when all the lengths are finite.) 2 If you ask the question if M is an R-module of projective dimension n, is \operatorname{Ext}^n(M,R)\neq0? then the answer is surpring: it depends. When R=\mathbb Z, this is essentially known as Whitehead's problem and Shelah proved that its answer depends on the specific set theory that you choose. Indeed, he showed that depending on the ... 1 The two theorems are not the same. One is about graded modules and the other is about modules. The two results are related, though. As you say, every projective graded module (that is every projective object of the category of graded modules and homogeneous maps of degree zero) is free as a module, and the conection between the two theorems stems from ... 0 When you use the modulo operator, multiplication is conserved, e.g:$$20 \equiv 3 \pmod {17}40 \equiv 6 \pmod {17}$$So that:$$40*20 \equiv 6*3 \pmod {17}\equiv 1 \pmod {17}$$The same holds for powers (since this is just multiplication many times), so:$$20^{15} \equiv 3^{15} \pmod {17}16^{18} \equiv (-1)^{18} \pmod {17}$$Now use:$$3^{15} = ... 6 The key thing to remember about the operation "mod" is that it behaves "well" with respect to product (hence powers), and of course addition. This means that if you can simplify your life a lot distributing the calculation into many steps and taking "mod" at each stage. To compute$20^{15}$, you can first notice that$20 \pmod{17} = 3$. Then$20^{15} ...

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Hints: $$20=3\pmod{17}\;,\;\;16=-1\pmod{17}\implies$$ $$20^{15}+16^{18}=3^{15}+(-1)^{18}=(**)$$ But for any integer $\,a\;,\;\;(a,17)=1\,$ , we have that $\,a^{16}=1\,$ , so $$(**)=3^{16}\cdot 3^{-1}+1=1\cdot 6+1=7\ldots$$ and so the claim is false: the remainder is $\,7\,$ .

0

consider this over $\mathbb{Z}/17\mathbb{Z}$. Addition and multiplication are well defined, so we can without harm say $20^{15} + 16^{18} \equiv 3^{15} + (-1)^{18}$ mod 17. You want to show what it is equivalent to mod 17. To do this you would just crunch it out using Fermats litle theorem and stuff like that.

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To understand $\Hom$, think about generators. Since any generator of $\mathbb{Z}_{p^\infty}$ is torsion, it has to map to $0$ in $\mathbb{Z}$. So, $$\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}) = 0.$$ In order to understand $\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}_{p^\infty})$, consider the definition of $\mathbb{Z}_{p^\infty}$ as a direct limit. $$... 1 \mathbb{Z}/p^{\infty} is divisible, this implies \hom(\mathbb{Z}/p^{\infty},\mathbb{Z})=0. We have \mathbb{Z}/p^{\infty} \cong \mathrm{colim}_n ~\mathbb{Z}/p^n, hence \hom(\mathbb{Z}/p^{\infty},\mathbb{Z}/p^{\infty}) \cong \lim_n \hom(\mathbb{Z}/p^n,\mathbb{Z}/p^{\infty}) Now \hom(\mathbb{Z}/p^n,\mathbb{Z}/p^{\infty}) is isomorphic to the ... 0 \def\Z{\mathbb{Z}}\def\Hom{\mathrm{Hom}}We have \Hom(\Z_{p^\infty},\Z)=0 because \Z_{p^\infty} is divisible and \Z is reduced. On the other hand \Hom(\Z_{p^\infty},\Z_{p^\infty}) is the ring of p-adic integers. You find its description in any Algebra book (Cohn, for instance, or Atiyah-McDonald), or http://en.wikipedia.org/wiki/P-adic_number 3 Let me expand on the comment I made (the other answer is great, by the way. But, I did have fun working it all out by hand). First, since M is cyclic it is of the form \mathbb{Z}/m\mathbb{Z} for some m \mid n and m > 1. I claim that M is projective (resp. flat, injective) if and only if (m,n/m) = 1. Suppose that m\mid n and m'\mid n. The ... 3 This question was answered in comments: This is called the Noether-Deuring theorem and a proof can be found at the MO link. 0 Unfortunately the answer is no, at least when 2 \in R^*. If C is a cocomplete R-linear symmetric monoidal category with an object T satisfying 1 \cong T \oplus T, then C=0. Proof: We have exterior powers in C with the usual properties. Hence, 0 = \Lambda^2 (T \oplus T) \cong (\Lambda^2 T) \oplus (\Lambda^1 T \otimes \Lambda^1 T) \oplus ... 4 I guess you really mean n>1, and that is rather a lot of questions. I'll try to hit them all. Well the nicest thing about \Bbb Z/n\Bbb Z is that it's a quasi-Frobnius ring meaning that it's Artinian and self-injective. This is in fact equivalent to the projective modules being precisely the injective modules. Actually, all of these are equivalent: ... 2 Let 0=M_0\subsetneq \cdots \subsetneq M_n=M be a composition series for M. Since f is an injective homomorphism, f(M_i) is a submodule for each M_i and strict inclusions are preserved, so f(M_0)\subsetneq \cdots\subsetneq f(M_n)\subseteq N, thus l(N)\geq n=l(M). 2 Because of the isomorphism theorem for modules there is an inclusion-preserving bijection between the submodules of M/N and the submodules of M containing N. Thus, because uA+N properly contains N and M/N is simple it has to be equal to M. 3 Your first part is correct. As for the second part, think bigger: show that any two elements of an ideal I in a commutative ring are linearly dependent. (Hint: this is just as easy as the first part!) As for the third part: in view of the second part, in any commutative ring the ideals which are free as R-modules are necessarily principal ideals. (In ... 3 Just for the fun of it, let me put this in a broader context: Let A be a ring (with 1), and let B be a ring (with 1) equipped with a morphism A \to B (respecting 1). If N is a left A-module, then Hom_A(B,N) (Hom of left A-modules) is naturally a left B_module. (We define b \cdot \phi, for b \in B and \phi in the Hom set, by (b ... 5 First check \bf C-linearity (this should be obvious). The only other step needed to check that \Gamma is a homomorphism of {\bf C}[G]-modules is to check that it is G-equivariant. We write out$$\quad \Gamma(h m)=\sum_{g\in G}\phi(g^{-1}h m)g $$Consider replacing g with hg (this is merely a substitution) so that the sum looks different. What ... 0 Following the comments, I left an answer as part of the following question: Understanding a paper concerning the a group-ring, an augmentation ideal and the Jacobson radical 0 In the reference you gave, it expresses M as$$M\cong\sum(\overline{\theta}-\theta)^{e_i}U_iR$$where the U_i are ideals of R_0. It then says "the class of \prod U_i is an invariant of M." My best guess is that this means that if there is another representation$$\sum(\overline{\theta}-\theta)^{e_i}V_iR for ideals $V_i$ in $R_0$, then $\prod ... 1 More generally, you can prove the following (called Baer's criterion): Let$R$be a ring. Assume that$P$is an$R$-module with the property that every homomorphism$I \to P$can be extended to a homomorphism$R \to P$, where$I$is a left ideal of$R$(when$R$is a PID, then this precisely means that$P$is divisible). Then this extension property actually ... 2 Not in general. You may be interested in looking up uniserial modules. These are modules that have only one composition series so the sequence of$X_i$'s is uniquely determined. 1 Every closed submodule is homeomorphic to$\Bbb Z^\Bbb Z$or to a$\Bbb Z^n$for some$n$: Set$M_0 = 0, M^0 = M, B_0 = \emptyset$.$M_0$is the closed submodule generated by$B_0$,$M = M_0 \oplus M^0$. If$M^n \neq \{0\}$, pick the smallest$l_{n+1} \in \omega$such that$\pi_{l_n}(M^n) = k\Bbb Z\neq \{0\}$, and pick an element$b_{n+1} \in M^n\$ such ...

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