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It is in fact that easy. You can make it even simpler by noting you can conclude $\mathbb Q \subseteq \langle \frac{1}{c} \rangle$ from your assumption. Since $\frac{1}{2c}\in\mathbb Q$ we have $\frac{1}{2c}=\frac{k}{c}$ for some $k\in\mathbb Z$, but there is no $k\in\mathbb Z$ such that $2k=1$.

3

Your claim is NOT true. $M = \mathbb{Z}, N = \mathbb{2Z}, M' = \mathbb{8Z}, N' = \mathbb{4Z}.$ All are considered modules over $\mathbb{Z}.$ Then $\frac{M}{M'} \oplus \frac{N}{N'} \cong \mathbb{Z}_8 \oplus \mathbb{Z}_2.$ On the other hand $\frac{M}{N'} \oplus \frac{N}{M'} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_4.$ So this two are not isomorphic. The mistake ...

0

Define a homomorphism $\phi: \bigoplus_{i \in I} Hom(M,N_i) \rightarrow Hom(M, \bigoplus_{i \in I} N_i)$ by sending $f = (f_k)_{k \in I^*}$ to $\phi(f) = \sum_{k \in I^*} f_k$, where $I^*$ is a finite subset of $I$. Clearly $\phi$ is injective. For surjectivity, take $f \in Hom(M, \bigoplus_{i \in I} N_i)$ and let $x_1,\dots,x_n$ be the generators of $M$. ...

3

Given a ring $R$ that's Noetherian as a $\Bbb Z$-module, it's automatically a (left- or right- as desired) Noetherian ring, as ideals are by definition closed under sum of elements; so every increasing chain of ideals is an increasing chain of $\Bbb Z$-submodules of $R$, and thus stabilizes. $\Bbb Q$ is indeed a Noetherian ring (it only has two ideals) ...

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A ring $A$ is said to have (right) invariant basis property (IBP) if for any integers $s, t \geq 0$, $$A^{\oplus s} \cong A^{\oplus t} \Rightarrow s = t$$ where the above isomorphism is as right $A$ modules. Any non-zero commutative ring, left Noetherian ring, semi-local ring satisfies IBP. The following link might be useful: Cohn, P. M. - Some remarks ...

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No, if the ring is commutative, yes, if the ring is noncommutative. The easiest counterexample is the endomorphism ring $R$ of $V=F^{(\omega)}$, where $F$ is a field and $F^{(\omega)}$ denotes a direct sum of countably many copies of $F$ (as vector space). Let homomorphisms act on the left, so $V$ becomes a left $R$-module. Then $R\cong R^2$ as left ...

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Here is a more fundamental approach to showing that $\mathbb{Q}$ is not finitely generated, based on a general method. Here's the general method. In a group $G$ with a finite generating set $\{g_1,…,g_K\}$, for any increasing sequence of subgroups $H_1 \subset H_2 \subset \cdots$ whose union $\cup_i H_i$ equals $G$, for sufficiently large $i$ the subgroup ...

1

$S^{-1}B$ and $f(S)^{-1} B$ are in fact isomorphic, simply because the action of $A$ on $B$ is defined via $f: A \rightarrow B$. To see this explicitly, note that when you gave the definition of $S^{-1}B$, you defined $(b,s)$ ~ $(b',s')$ if there exists $t \in S$ such that $t (s b' - s' b)=0$. But what does $s b'$ mean? What is the action of $s$ on $B$? This ...

3

If an abelian group $G\ne\{0\}$ is free, then it is isomorphic to $\mathbb{Z}^{(X)}$ (direct sum of copies of $\mathbb{Z}$), for some set non empty set $X$. Then $\mathbb{Z}$ is an epimorphic image of $G$. Since $\mathbb{Z}$ is not divisible, $G$ can't be divisible. But $\mathbb{Q}$ is divisible. The group $\mathbb{Q}$ is not finitely generated: if ...

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For the non-free part: Take any two nonzero elements $x, y ∈ ℚ$ and show they satisfy $λx + μy = 0$ for some nonzero $λ, μ ∈ ℤ$, hence any two elements are linearly dependent. Thus, since $ℚ$ is not cyclic, it cannot have a basis. For the non-finitely-generated part: If $ℚ$ was finitely generated then without loss of generality (by finding the common ...

2

Show any finitely generated subgroup of $\mathbf Q$ is cyclic. Since $\mathbf Q$ is not cyclic, it cannot be finitely generated. $(1)$ It cannot be free: it is not cyclic, so any putative basis has at least two elements. But if $x,y$ are elements of the basis, we know $\mathbf Z^2\simeq \langle x,y\rangle=\langle x'\rangle\simeq\bf Z$ which is impossible. ...

0

It suffices to show that $I^{n-1}/I^n$ is finitely generated as an $R$-module, because then it is also finitely generated, hence Noetherian (since $R/I$ is Noetherian), as an $R/I$-module, and finally being Noetherian over $R/I$ or over $R$ is the same.

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Let us assume that $g$ has the quoted property. Let $Y = N/\text{im}(g)$ and $k : N \rightarrow Y$ the natural projection. Then $k \circ g = 0$, so $k = 0$ by assumption. But $k$ is surjective, so we have to have $Y = 0$, hence $N = \text{im}(g)$ and so $g$ is surjective.

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First, if $\text{coker}\ f$ is Artinian, then so is $\text{coker}\ f^m$ for all $m\geq 1$: it suffices to check this for $m=2$, and in this case, note that there is an epimorphism $$\text{coker}\ f\twoheadrightarrow\text{im} f / \text{im} f^2$$ and a short exact sequence $$0\to\text{im}\ f / \text{im}\ f^2\to\text{coker}\ f^2\to \text{coker}\ f\to 0.$$It ...

5

Let $R$ be ring for which $F$ is a free module. Then we have isomorphisms $R^n \rightarrow F$ and $F \rightarrow R^{n+1}$ which gives us an isomorphism $R^n \rightarrow R^{n+1}$. Thus for any $m \geq n$ we have an isomorphism $R^m \cong R^n \oplus R^{m-n} \cong R^{n+1} \oplus R^{m-n} \cong R^{m+1}$. Composing these isomorphisms we get an isomorphism $R^n ... 1 Let$A$and$B$be ideals of the ring$R$. Then$A/BA$is a module over$R/B$in a natural way. Indeed, for$b\in B$and$a\in A$, the product$ba\in BA$, so $$b(a+AB)=0+AB$$ in the module$A/AB$. Thus$\operatorname{Ann}(A/BA)\supseteq B$. If$M$is an$R$-module and$\operatorname{Ann}(M)\supseteq B$, then$M$is a module over$R/B$by defining $$... 1 (a) Yes. (b) Yes. (c) Over a semisimple ring every non-zero module is semisimple (as a quotient of a free module which is a direct sum of copies of your semisimple ring, hence semisimple). 5 The answer is negative since A\subset B flat and B regular implies A regular; see Bruns and Herzog, Theorem 2.2.12. But in this case A\simeq k[a,b,c]/(ac-b^2), so A is not regular. Edit. A simpler approach: let I=(x^2,xy) and A/I\to A/I be the multiplication by y^2. Since A/I\simeq k[y^2] this is injective, but on A/I\otimes_AB\to ... -2 Thanks to the comments of Manos. I could figure out finally what was going on. Let I be an ideal of A. The following is a condition for flatness: An R module M is flat iff I\otimes_R M injects into M. Let I = <x^2, xy> be the ideal of A. Then note that the elements x^2 \otimes y and xy \otimes x of I \otimes_A B are mapped to the ... 2 Such a map can exist, but not for every noncommutative ring. For example, no such map can exist over a division ring. This is because the annihilator of the left tensor product module is a nonzero left ideal, hence the left tensor product is always trivial in that case. For an example where the map does exist, let T be the tensor algebra over a field, and ... 3 For an explicit counterexample, consider R = k[X,Y] for any field k, which is free over itself. Then the ideal \mathfrak m = RX+RY is not free. 2 You are asking if a submodule of a free module is necessarily free. This is not true in general. It is, however, true if R is a PID. A counter example is given by R=\mathbb{Z}/\mathbb4{Z} as a module over itself. The submodule 2\mathbb{Z}/4\mathbb{Z} is not free. 2 For Dedekind domains we have \operatorname{Pic}(R)\simeq\operatorname{Cl}(R), where \operatorname{Cl}(R) is the ideal class group of R. Your case is treated here in detail. 2 There's no need to consider J; just assume R is semisimple. Every (right) module over a semisimple ring is a direct sum of simple modules. Moreover, consider (S_i)_{i\in I}, a family of simple modules over R such that every simple R-module S is isomorphic to S_i, for some i\in I; if i\ne j, then S_i is not isomorphic to S_j. We can ... 3 You can proceed as usual by starting with F\stackrel{f}\to A\to 0 and H\stackrel{h}\to C\to 0, where F and H are free of finite rank. Then show that there is an exact sequence G=F\oplus H\stackrel{g}\to B\to 0. Now consider F'=\ker f and so on. You have a short exact sequence 0\to F'\to G'\to H'\to 0. Now use the result for finitely generated ... 3 If I understand your question correctly, consider that a \equiv b \pmod x if and only if a-b = kx for some k\in\mathbb{Z^+}. Then x=\frac{a-b}{k}, hence there is precisely one x for each divisor k\in\mathbb{Z^+}. 3 user26857's answer shows how to repair the reduction to the Noetherian case. Here is how to repair the proof of the Noetherian case: Let M be a noetherian A-module and let N \subseteq M be a submodule. Let f : N \to M be a surjective linear map. Then f is injective. Proof: Let n \geq 0. Although f^n is not a well-defined homomorphism this ... 3 Let A be a commutative ring. Let M be a finitely generated A-module and N be an A-submodule of M. Let f\colon N \rightarrow M be a surjective homomorphism of A-modules Then f is injective. Proof. Let 0 \neq x'_0 \in N. It suffices to prove f(x'_0) \neq 0. Set f(x'_0) = x_0. Let x_1, \dots, x_n be generators for M. Then ... 0 A covariant exact functor {\bf A}\to{\bf B} preserves finite limits and colimits, so a contravariant exact functor does it too as a (covariant) functor on the opposite category: F:{\bf A}^{op}\to {\bf B}. Here {\bf A}^{op} is again Abelian (as [at least one form of] the axioms are self dual). Now limits (in particular, kernels and products) in {\bf ... 2 Show that \hom_{R/I}(M/IM,-) \cong \hom_R(M,U(-)), where U is the forgetful functor from R/I-modules to R-modules. Hence, this is a composition of two exact functors, hence exact. 3 Define$$\phi: R\to Rx\le M\;,\;\;\phi(r):=rx$$prove the above is a (left) \;R- module homomorphism, and now use the first isomorphism theorem. 2 For i=1, \dots, n+m call$$c_i = \left\{ \begin{matrix} a_i & \mbox{ if } &i \leq n \\ b_{i-n} & \mbox{ if } &i \geq n+1 \end{matrix} \right. $$and analogously$$w_i = \left\{ \begin{matrix} x_i & \mbox{ if } &i \leq n \\ y_{i-n} & \mbox{ if } &i \geq n+1 \end{matrix} \right.$$Then$$x+y = \sum_{i=1}^{n+m} c_iw_i \in ... 1 Let$z_1=x_1,\dots,z_n=x_n,z_{n+1}=y_1,\dots,z_{n+m}=y_m$, and let$c_1,\dots,c_n,c_{n+1},\dots,c_{n+m}$be defined similarly based on$a_i,b_j$. Then:$$x+y = \sum c_iz_i$$ 0 Do you know that the Jacobson radical of a semisimple ring is zero? If so, you can easily show that$n$would have to be a squarefree positive integer for$\Bbb Z_n$to have Jacobson radical zero. You've already mentioned you have the converse that$\Bbb Z_n$is a product of fields for a squarefree integer, so then you would have both directions. 0 You should just use the universal property, don't worry about exact sequences. For starters assume the$p_i$exist. You have to show that a family of maps$N_i\colon X_i \to Z$factors through a unique map$X \to Z$. Well using the$p_i$you get maps$X \overset{p_i}{\rightarrow} X_i \overset{N_i}{\rightarrow} Z$so sum this over$i$to get$X \to Z$. ... 0 Let$A=\Bbb Z_n$. I suppose the morphisms are the inclusion and multiplication by$s$. If$n=rs$, then multiplication by$r$induces an isomorphism,$rA\simeq \Bbb Z_s$. Similarly, multiplication by$s$induces an isomorphism$sA\simeq \Bbb Z_r$. What you have is a SES $$0\longrightarrow rA\longrightarrow A\longrightarrow sA\longrightarrow 0$$ If we ... 0 Remember that if$0 \to A \to B \to C \to 0$is a split exact sequence of modules, then$B \cong A \oplus C$. Suppose$(r,s)>1$. (Note that$r\Bbb Z_n \cong \Bbb Z_s$and vice versa.) Show that$\Bbb Z_n \not\cong \Bbb Z_r \oplus \Bbb Z_s$. In the other direction, you should be able to write down an explicit section$s \Bbb Z_n \to \Bbb Z_n$. 1 We can argue as follows. Pick an$h \in ker\ g_* \subseteq Hom_R(M,M_2).$Then we have$g\circ h = 0.$This means$im\ h \subseteq ker\ g = im\ f.$Since$f$is injective, we have the$R$-homomorphism$(f|_{im\ f})^{-1}:im\ f \rightarrow M_1.$So we can define$j := (f|_{im\ f})^{-1} \circ h: M \rightarrow M_1.$This is an$R$-homomorphism since it's a ... 3 Here is a hint: count the elements of order$p$in the group $$\mathbb Z/p^{i_1} \oplus\mathbb Z/p^{i_2} \oplus \dots \oplus \mathbb Z/p^{i_n}.$$ Here's the answer to the hint: You should find that there are exactly$p^n - 1$. Namely if you write down an element as$(x_1,\dots,x_n)$than this has order$p$if$(px_1,\dots,px_n) = (0,\dots,0)$, i.e. if ... 0 Your idea is not quite right: the product$M\times N$of two modules$M,N$is the pullback of$M\to 0\leftarrow N$, but in your problem you are interested in the pullback of${\mathbb Z}_{p^2}\to {\mathbb Z}_p\leftarrow {\mathbb Z}$, which can be realized as the submodule of${\mathbb Z}_{p^2}\times{\mathbb Z}$consisting of those pairs$(\overline{x},y)$... 4 As you say, there is no reason why this should be true. Consider on the one hand$R$regarded as an$(R, R)$-bimodule in the usual way and on the other hand$R$regarded as an$(R, R)$-bimodule where, say, the left$R$-module has been twisted by an automorphism$\varphi : R \to R$, which is to say that left multiplication now looks like$$L_r s = \varphi(r) ... 2 I suppose that you know that if we have a s.e.s.$0\to M'\to M\to M''\to 0$, then$M$is noetherian iff$M'$and$M''$are noetherian. (If not, take a look here.) Now split your exact sequence in two s.e.s.:$0\to X\to Y\to K\to 0$and$0\to K\to Z\to T\to 0$. (i)$Y$noetherian$\RightarrowK$noetherian... (ii)$Z$noetherian$\RightarrowK$... 4 One way to approach the problem: If$M$is a simple$R$-module, then$M \cong R/I$for some maximal left ideal$I ⊂ R$. What are the maximal left ideals in$R$? Elaborating: Show that a maximal left ideal$I ⊂ R $is already generated by any matrix of maximal rank in$I$and find that maximal rank. Hint: Use row reduction. Then show that any two matrices ... 6 Here's one (class of) example(s). Let us restrict our attention to$\mathbb Z$-modules, i.e. abelian groups. Note that two abelian groups of order$p$are always isomorphic, thus it is sufficient to find a group$P$of order$p^i$(for any$i$) with two subgroups$A$and$B$of order$p^{i-1}$such that$A\not\cong B$. For instance, if$P = \mathbf C_4 ...

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