# Tag Info

1

Does this not just follow from the properties of semisimple modules? If $S$ and $M$ are simple, and $(\oplus_{i\in I} S)/K\cong M$, then $M'\oplus K=(\oplus_{i\in I} S)$ where $M'\cong M$. So $M$ must be isomorphic to a direct sum of copies of $S$, but since it is simple it is just isomorphic to $S$.

1

If I understand your question correctly, then the question is whether it follows that all simple modules are isomorphic provided that one has a simple module which is a generator. I think the answer is yes: Suppose $S$ is another simple module. Then there is an epi $T^{(I)}\twoheadrightarrow S$. However, one has the following isomorphism (using universal ...

0

Let $R=\mathbb Z$, and $A=\mathbb Z\oplus\mathbb Z\oplus\cdots$. Now set $B=\mathbb Z$ and $C=0$. Moreover, if you also want $A$ finitely generated, then there still are counterexamples, but maybe not that easily found; see e.g. here.

1

This is because, as $Rp$ is a free $R$-module, the short exact sequence: $$0\to N'_a\to N_a\to Rp\to 0$$ splits.

0

Hint. You can suppose that $B,C$ are torsion modules (why?). Then use their elementary divisors.

2

Since the submodule is torsion free, the mapping from $R$ to $\langle h \rangle$ by $r \mapsto rh$ is an isomorphism.

0

In order to get a handle on $\hom_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z}/m\mathbb{Z})$, consider that these are exactly the maps induced by those homomorphisms $\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ that vanish on $n\mathbb{Z} \subset \mathbb{Z}$. Such a map is completely determined by $f(1)$, and it vanishes on $n\mathbb{Z}$ exactly when the order ...

1

L.R. Vermani in the book: "An Elementary Approach to Homological Algebra" has: See also Rotman's book "Advanced Modern Algebra"

1

Ok, so $|x - 2| = p$, and so $|p - 3| > 1$. Then that means $p - 3 > 1$ or $p - 3 < -1$. Then that means $p > 4$ or $p < 2$. But $p = |x - 2|$, so $|x - 2| > 4$ or $|x - 2| < 2$. That means either $x - 2 > 4$ or $x - 2 < -4$ or $-2 < x - 2 < 2$. That means either $x > 6$ or $x < -2$ or $0 < x < 4$. Notice ...

2

Just do it with two simultaneous inequalities: $$||x-2| - 3| > 1 \Leftrightarrow |x-2|-3 > 1 \text{ or } |x-2|-3 < -1$$ $$\Leftrightarrow |x-2| > 4 \text{ or } |x-2| < 2$$ $$\Leftrightarrow x-2 > 4 \text{ or } x-2 < -4 \text{ or } -2 < x-2 < 2$$ $$\Leftrightarrow x>6 \text{ or } x < -2 \text{ or } 0 < x < 4$$ So ...

1

$||x−2|−3|>1$ then either $|x-2| - 3 > 1$ or $|x-2| - 3 < -1$; respectively $$|x-2| > 4 \ \text{ or } |x-2| < 2$$ The first of these is equivalent to $x - 2 > 4$ or $x - 2 < -4$; i.e., $$x > 6 \text{ or } x < -2$$ The second is equivalent to $-2 < x - 2 < 2$; i.e., $$0 < x < 4$$ Now put that all together.

0

No, it's not! Suppose $R$ is a commutative unitary ring and $M$ a free $R$-module of rank $n$. If $A\subset M$ is a linearly independent set, then $|A|\le n$. We can assume $M=R^n$. A linearly independent set $A\subset R^n$ with $|A|=m$ gives rise to an injective $R$-module homomorphism $\phi:R^m\to R^n$, and then $m\le n$. (For this you can find a ...

1

It's an easy exercise to show that Rank(Def2) is equal to $\dim_K(K\otimes_RM)$, where $K$ is the field of fractions of $R$. This shows that Rank(Def2) is well-defined, and this is by definition the rank of $M$ in such context. "Is every cardinality of a $R$-linearly independent subset of a finitely generated $R$-module finite?" Yes, it is. In fact its ...

2

You have to prove that if $r\equiv s \mod I$, then $rm=sm$ for any $m\in$M$. That is equivalent to$(r-s)m=0$, which is by definition since$r\equiv s\mod I\iff r-s\in I\subseteq\operatorname{Ann}_AM$. 1 The point is exactly that$A$itself is a projective$A$-module, but you can understand the proof as long as you know that$A$is a free$A$-module with basis$\{1\}$. First the inclusion maps$I\subset A$and$J\subset A$give a homomorphism$\pi:I\oplus J\to A$. Since$I+J=A$,$\pi$is a epimorphism and there exist$i\in I$and$j\in J$such that ... 2 Consider the$A$-module (or$R$-module) epimorphism $$I\oplus J \rightarrow I+J=A, \;\; (i, j)\mapsto i+j \in A,$$ compute its kernel and use the first isomorphism theorem. Now, if you know that$A$is a projective$A$-module and know what properties projective modules have, this should give you the isomorphism you are looking for. If not, the isomorphism ... 0 If you look careful at$M=t(M)\oplus F$can notice that$F\simeq M/t(M)$. But finitely generated torsion-free modules over a PID are free, so$M/t(M)$is free and thus has a (finite) rank. That's why Lang calls this the rank of$M$which seems a reasonable choice. (Bourbaki also calls this the rank of$M$.) But as you can see there is no standard ... 0 I suspect what's happened is that Lang has used an odd choice of language and you've misread it. The definition is meant to be rank of$M :=$rank of any$F$s.t.$M \cong Tor(M) \oplus F$He isn't trying to define the rank of$F$, though I can see why you might read it that way. Essentially, we're trying to extend the definition of rank for free modules ... 0 Your statement about$I$maximal is true, and so we see that$R/I$is simple as an$R$-module. Also, note that if$J$is any non-maximal ideal, say$J\subset N\subset R$. Then$N/J$is a submodule of$R/J$, so$R/J$is not simple. Let$M$be a simple$R$-module. Let$x\in M$such that there exists$r\in R$such that$rx \neq 0$. This exists since$M$is ... 2 Over a local commutative ring, projective modules coincide with free modules, so the question is whether$\mathfrak m$is free. If it is free it must be of rank one, because two elements of$a,b\in A$are necessarily$A$-linearly dependent:$a\cdot b-b\cdot a=0$(duh!) Freeness of dimension one means that for some$m\in \mathfrak m$the$A$-linear map ... 2 It depends on the ring. For instance, if$R=\mathbb{Z}_{(p)}$is the localization of$\mathbb{Z}$at the prime ideal$p$, then the maximal ideal is principal, so isomorphic as a module to the ring itself, hence projective. In case the ring is$\mathbb{Z}/4\mathbb{Z}$, the maximal ideal is not projective. Added from comment. If your aim is to discuss ... 1 Here's another one. Let$R$be a ring which has any non-free finite projective module$A$(where here$A$finite projective means there exists$B$with$A \oplus B\cong R^k$for a finite$k$). Then if we denote$R^\infty$the countable direct sum of copies of$R$, we have$A\oplus R^{\infty}\cong R \oplus R^{\infty}\cong R^\infty$by the Eilenberg-Mazur ... 3 Without using any topology, you can take$R$the ring of endomorphisms of an infinite-dimensional vector space$V$, say over$\mathbb{R}$. Then as a (left) module over itself,$R=R\oplus 0$is isomorphic to$R\oplus R$. 4 Let$R$be the ring of smooth functions on$S^2$. Then$R^3 \cong \operatorname{Vect}(S^2)\oplus R$where$\operatorname{Vect}(S^2)$is the$R$-module of smooth vector fields on$S^2$.$\operatorname{Vect}(S^2)$is not isomorphic to$R^2$by the hairy-ball theorem, and the standard smooth embedding$S^2\rightarrow \Bbb R^3$gives the first isomorphism. ... 1 It might be easier to approach this problem the following way: We have the following free resolution of$I$: $$0 \to R \to R \oplus R \to I \to 0,$$ where the first map is given by$1 \mapsto (-y,x)$and the second map is given by$(1,0) \mapsto x, (0,1) \mapsto y$. After tensoring with$R/I$we get the following exact sequence: $$0 \to Tor_1(I,R/I) \to ... 1 You have to distinguish between the free module spanned by one basis element and the ideal generated by an element of R. The free module spanned by an element e is just \{0,(1,0)e,(0,1)e,(1,1)e\} and is isomorphic to R as a module. The ideal RA generated by your A is indeed \{0,(1,0)\} and is NOT a free module over R. Ideals are free if and ... 2 A counterexample is \mathbb{Q} considered as a \mathbb{Z}-module. Definitely torsion-free but not free. 1 If \zeta_n is any primitive nth root of unity and K=\mathbb{Q}(\zeta_n) (this K is known as the nth cyclotomic field) then we have \mathcal{O}_K=\mathbb{Z}[\zeta_n] and [K:\mathbb{Q}]=\varphi(n), where \varphi is the Euler phi function. A good reference for this would be chapter 2 of Marcus' Number Fields. In particular, you're correct that ... 3 It might be ambiguous, but I would usually parse it as a set of generators as an R-algebra, i.e. that every element of A can be written as a polynomial in the elements of X with coefficients in R. For instance, \{x\} generates \Bbb Z[x] as a \Bbb Z-algebra, whereas you would need something like \{1, x, x^2, \ldots \} to get a set of ... 2 This is kind of a silly example: \mathbb{Z} \subseteq \mathbb{Q}, but the Krull dimension of \mathbb{Z} is one while the Krull dimension of \mathbb{Q} is 0. 1 Look at the answer to this StackExchange question: http://math.stackexchange.com/questions/132729/a-free-submodule-of-a-free-module-having-greater-rank-the-submodule (and also the MO discussion referenced therein). 0 I looked up some more facts about projective dimension which ultimately lead to me solving part of the problem. It turns out that the fact that F_0 and N have the same rank was a geometric implication, not an algebraic one so that probably can't be proved from the question as it is. (Simple) Fact: If pd(M)\leq n then given any exact sequence ... 3 More generally, for any PID R and every non-zero element e \in R, the ring R/(e) is self-injective: Baer's criterion implies that, if S is a commutative ring in which every ideal is principal, an S-module M is injective if and only if for all a \in S, m \in M with \mathrm{Ann}(a) \subseteq \mathrm{Ann}(m) we have m \in aM. This can be ... 0 Dividing the second row by 2 gives you$$x=-2y-3z$$while taking 2\cdot \mbox{second row} - \mbox{first row} gives you$$y=-10z$$so that x=17z. So your space solution is generated by (17,-10,1), which is also a \Bbb{Z}-basis of the space. 1 You have a map \phi:S^3\to S given by \phi(e_1)=x^2 and so on. (In this case e_1=(1, 0, 0) and so on.) The matrix of this map is (as in the vector spaces case) the following: (\phi(e_1)\ \phi(e_2)\ \phi(e_3)), so your guess is correct. 1 Each x\in X_1 can be written as a K-linear combination of elements of X_2. If we multiply away denominator, we see that cx\in X_2 for suitable 0\ne c\in A. If x_1,\ldots,x_n generate X_1 and lead to corresponding factors c_1,\ldots, c_n with c_ix_i\in X_2, then let c=c_1\cdots c_n and find that cX_1\subseteq X_2. Then X_1/X_2 is a ... 2 Since _SN is projective, there is N' such that N\oplus N'\cong S^{(I)} is a free module. Then we can tensor:$$ M\otimes_SS^{(I)}\cong (M\otimes_SN)\oplus(M\otimes_SN') $$as R-modules. Now$$ M\otimes_SS^{(I)}\cong M^{(I)} $$as R-modules. Since M is projective, every direct power of it is projective as well. So M\otimes_SN is a direct summand ... 1 We have$$\mathbb{Z}^{3}/M \mathbb{Z}^{3}\simeq(\mathbb{Z}\oplus\mathbb Z\oplus\mathbb Z)/(\mathbb Z\oplus\mathbb Z\oplus 32\mathbb Z)\simeq\mathbb Z/32\mathbb Z,$$so your guess is correct. 0 A cyclic module is a module, that is generated by one element. For sure R/AB is generated by the class of 1. AB is precisely the annihilator of \overline 1 = 1+AB \in R/AB. 2 The issue here is that you need to know what you mean by "unique". You can show that such an object L exists ; just take it to be the submodule of M of elements mapped to 0 under f, and check the two conditions. For "unicity", if you have two modules L_1,L_2 that satisfy property (i) and (ii), you cannot hope to show that L_1 = L_2, but you can ... 1 am=0\Rightarrow an_1+an_2=0\Rightarrow an_1=an_2=0 (why?). So \mathrm{Ann}(m)=\mathrm{Ann}(n_1)\cap\mathrm{Ann}(n_2). Moreover, N_i=Rn_i for i=1,2. We have R/\mathrm{Ann}(m)\simeq Rm, Rm=N_1\dotplus N_2, and N_1\dotplus N_2\simeq R/\mathrm{Ann}(n_1)\oplus R/\mathrm{Ann}(n_2), therefore$$R/\mathrm{Ann}(n_1)\cap\mathrm{Ann}(n_2)\simeq ... 1 An answer to your last question: If$\mathbb Z \to N$is onto, then any homomorphism$f:\mathbb Q \to N$is easily shown to be trivial. In particular we can always chose$h = 0$(the only homomorphism$\mathbb Q \to \mathbb Z$anyway) to satisfy$g \circ h = f$. This shows (a boring fact, since the involved maps are trivial, so nothing happens here) that ... 1 Since the map$\phi_m$is surjective we have$M=Rm$for all$m\in M-\{0\}$. Suppose$\mathrm{Ann}(m)$is not left maximal for some$m\in M-\{0\}$. Then there is$\mathrm{Ann}(m)\subsetneq I\subsetneq R$a left ideal. Let$a\in I-\mathrm{Ann}(m)$. Then$am\ne 0$, and$M=R(am)$. Since$M=Rm$we get$m\in R(am)$, so there is$r\in R$such that$m=ram$, that is, ... 1 Choose generators of$P$to get a surjection from a free module$F$onto$P$. This gives you a split (by your definition!) exact sequence$0 \to K \to F \to P \to 0$, showing$F = K \oplus P$. 1$R$is obviously generated by$1$as an$R$-module. The ideal (submodule)$(x_1,x_2,\ldots)$of$R$is not finitely generated. For otherwise$(x_1,x_2,\ldots)=(f_1,\ldots,f_k)$for some$f_1,\ldots,f_k\in R$, and every$f_i$involves only finitely many indeterminates. 3 Suppose$R=I\oplus J$, where$I$and$J$are ideals of$R$, with$I\cap J=\{0\}$. Then $$1=x+y$$ with$x\in I$and$y\in J$. It follows that$x=x(x+y)=x^2+xy$. Since$xy\in J$and$xy=x^2-x\in I$, we have$xy=0$. Since$R$is a domain, we have either$x=0$or$y=0$. In the first case$y=1$and$J=R$, in the second case$x=1$and$I=R$. A finite direct ... 2 Suppose that$R=M\oplus M'$with$M\neq 0\neq M'$. Then$(m,0)\cdot(0,m')=0$for$m\neq 0\neq m'$which is a contradiction to$R$being an integral domain. Let$e\in R$be idempotent, that is to say$e^2=e$and$e\neq 1$. (For example$(1,0)\in K^2$for any field$K$). Then$R=R(1-e)\oplus Re$(easy exercise). In our example this leads to$R=K\oplus K$as ... 1 Since$3^2\equiv -1\pmod {10}, one has \begin{align}5^{31}\cdot 2^{789}-23^{23}&\equiv0-3^{23}\\&\equiv -(3^2)^{11}\cdot 3\\&\equiv -(-1)^{11}\cdot 3\\&\equiv -(-1)\cdot 3\\&\equiv 3.\end{align} 1 Suppose\mathrm{Ann}(r_2rm)\not\subseteq Q$. Then there is$a\in \mathrm{Ann}(r_2rm)$,$a\notin Q$. Now$\mathrm{Ann}(arm)\subseteq Q$:$b\in\mathrm{Ann}(arm)\Rightarrow b(arm)=0\Rightarrow (ba)rm=0\Rightarrow ba\in P\Rightarrow ba\in Q\Rightarrow b\in Q$. Since$P\subseteq\mathrm{Ann}(arm)$and$P$is maximal with some properties that$\mathrm{Ann}(arm)$... 2 Your edit is basically the right idea but notationally there's a problem. You can't say every$m \in M$can be written as$m' + u$with$m' \in M/U$and$u \in U$because$M/U$is not a submodule of$M$. What you want to say is that$(v_1 + U, \ldots, v_m + U)$are generators for$M/U$so given$m \in M$you first push it$M/U$to get$m + U\$. Then write ...

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