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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 $n$th root of unity and $K=\mathbb{Q}(\zeta_n)$ (this $K$ is known as the $n$th 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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