# Tag Info

0

If $R^n\simeq R^{n+1}$ then $R^{n+1}\simeq R^{n+2}$ (why?), so $R^n\simeq R^{n+2}$ and so on.

1

In $K \otimes_R C$ all tensors are elementary, i.e., they all have the form $x \otimes c$. That comes from using a common denominator for any finite list of elements in $K$ (as ratios of elements of $R$). The kernel of the $R$-linear mapping $C \rightarrow C \otimes_R K$ given by $c \mapsto 1 \otimes c$ is precisely the torsion submodule $C_{\rm tor}$, so ...

0

Hint on well-definedness. define $\Phi: C \times R \to C \otimes_R K$ by $c \times r \mapsto c \otimes r$ , then by the universal property of tensor $\Phi$ induces a $R$-hom $C \otimes_{R} R \to C \otimes_R K$ with $c \otimes r \mapsto c \otimes r$, of course, it is well-defined.

5

Hint: $R=\mathbb{Z}$, $C=\mathbb{Z}/2\mathbb{Z}$. There is no problem in $\phi$ being well defined; there's some problem in showing it's injective. ;-) The map is well defined, because such is the map $\tilde\phi\colon C\to C\otimes_RK$ defined by $$x\mapsto x\otimes1$$ and composing with the natural isomorphism $C\otimes_RR\to C$ gives your $\phi$.

0

Consider a domain $R$ with a nontrivial maximal ideal $I$. All nonzero modules of $R/I$ are injective. It looks like you are expecting that $M$ being $R/I$-injective implies $M$ is $R$-divisible. But that isn't true, considering you can take $M$ to be any nondivisible $R$ module and then look at $R/I$ for any maximal ideal to get an injective $R/I$ module.

0

Try to show any proper subgroups of $\Bbb Z(p^\infty)$ is finite (by showing the contrapositive).

0

So given $M$ an injective torsion-free module, we have that $\varphi_r$ is an isomorphism. Note that we can view $M$ as a $K(R)$ module (the field of fractions) in the following manner: Define $(a/b)\cdot m$ to be $\varphi^{-1}_b(am)$, as $\varphi_b$ is an isomorphism this is well defined. Hence, we have that $M$ is a $K(R)$-module, that is, a ...

2

The answer is that such matrices are always similar over $\def\Z{\Bbb Z}\Z$ (conjugate in $GL_2(\Bbb Z)$). The question is deeper however than it might look at first, and as far as I can see any solution requires some somewhat subtle arithmetic considerations. A few things are easy: $A,B$ always have determinant$~1$ (from the constant coefficients of the ...

2

By the snake lemma there is an exact sequence $$0\rightarrow\mathrm{Ker} \ \alpha\rightarrow 0 \rightarrow \mathrm{Ker} \ \gamma \rightarrow \mathrm{Coker} \ \alpha\rightarrow 0\rightarrow\mathrm{Coker} \ \gamma\rightarrow 0,$$ whence $\mathrm{Ker} \ \alpha=0$ and $\mathrm{Coker} \ \gamma=0$; also $\mathrm{Ker} \ \gamma\cong\mathrm{Coker} \ \alpha$ so that ...

0

If $U : \mathsf{Mod}(R) \to \mathsf{Set}$ is the forgetful functor, then there is a natural inclusion map $\mathrm{End}(1_{\mathsf{Mod}(R)}) \hookrightarrow \mathrm{End}(U)$. But $U$ is represented by $R$, hence $\mathsf{End}(U) \cong \mathsf{End}(R) \cong Z(R)$ by Yoneda. The map $Z(R) \cong \mathrm{End}(U)$ is given by $c \mapsto (m \mapsto cm)$ (by the ...

2

Take a set of basis $e_1,\ldots,e_n$ of $V$ and let $x=\sum_ {i}x^ie_i,~y=\sum_ {j}y^je_j$, then $$x\otimes y=\sum_ {i,j}(x^ie_i)\otimes(y^je_j)=\sum_ {i,j}x^iy^je_i\otimes e_j$$ $$y\otimes x=\sum_ {i,j}y^jx^ie_j\otimes e_i$$ The symmetry implies $$x^iy^j=x^jy^i$$ That is, $$\frac{x^i}{y^i}=\frac{x^j}{y^j}=a$$ for some constant $a$.

3

So for such $x,y\neq 0$, you want to show $x\otimes y=y\otimes x$ in $V\otimes_F V$ if $x=ay$ for some $a\in F$. Then $x\otimes y=ay\otimes y=y\otimes ay=y\otimes x$. Since this seems by far to be the easier half of the problem, I am beginning to wonder if you meant to ask about the other direction. Suppose $x,y$ are linearly independent. As such, this ...

3

The straightforward answer for "what is a basis for $\mathbb{C}/\mathbb{Q}$" is that we don't know. The sneaky answer is that we do know there is one, because any maximal linearly independent set is a basis, and exists by Zorn's Lemma. This shows that $\mathbb{C}$ is a free $\mathbb{Q}$-module, since that concept is literally equivalent to the existence of ...

-3

Off course given set generates C(Q) but is not L.I. as you may notice 2^(1/2) and all its rational multiples are there. In fact you can't write precisely what will be the basis. The basis can be written iff vector space is finite dimensional.

5

Note first that for any ring $R$ and any $R$-module $M$, if the cardinality $\lvert M\rvert$ is infinite and greater than $\lvert R \rvert$, then any generating set of $M$ (in particular, any basis) must have cardinality equal to $\lvert M\rvert$ (this is simple combinatorics). In particular, any basis of $\bf C$ over $\bf Q$ must have cardinality of the ...

0

If $F(x^2)$ is not $F(x)$ then $[F(x) : F(x^2)] = 2$, and $[F(x) : F] = [F(x) : F(x^2)] [F(x^2) : F]$ is even.

2

Any basis for $\Bbb C$ as a $\Bbb Q$ vector space must be infinite, since any finite dimensional $\Bbb Q$ vector space is countable, but $\Bbb C$ is not. Constructing such a basis requires the axiom of choice.

2

Consider the ring $R=F_2[Z]/(Z^2)=M$ where $F_2$ is the field of two elements. This is a self-injective ring, so $M$ is an injective $R$-module. But now consider $x=1$ and $r=Z$, where I abuse notation for the images of $1$ and $Z$ in this ring. Saying that there exists $y\in R$ such that $yZ=1$ implies that $Z$ is a unit, but it is clearly not since it is ...

1

Edit: I misread the question. If you only want to determine whether some particular pair of matrices $(A,B)$ are similar to each other, you should solve the two equations $PA=BP$ and $|\det P|=1$. You may find Dario Alpern's online solver useful. (My original answer.) $x^2+x+1$ is the characteristic polynomial of $A=\pmatrix{a&b\\ c&d}$ if and only ...

3

Suppose $\eta$ is a natural tranformation. Its value at the left $R$-module $R$ is a map $\eta_R:R\to R$ of left $R$-modules. You can easily check that $c=\eta_R(1)$, the image of $1\in R$, belongs to the center $Z(R)$ of $R$. If now $M$ is any $R$-module, and $m\in M$, there is a unique map $f:R\to M$ such that $f(1)=m$. Using naturality of $\eta$, we see ...

1

Well, if you know the characteristic polynomial, you know the trace ($-1$) and the determinant ($1$), so you should be able to write the general matrix using two (integer) unknowns [for example, if the $1, 1$ entry is $a$ then the $2, 2$ entry is $-1-a$). Now, you have your two possible matrices $M_1$ and $M_2.$ Being similar, means that there exists a ...

4

Nope. Consider two $4 \times 4$ matrices: $$\begin{pmatrix}0 & & & \\ & 0 & & \\ & & 0 & 1\\&&0&0\end{pmatrix}$$ $$\begin{pmatrix}0 & 1 & & \\ 0 & 0 & & \\ & & 0 & 1\\&&0&0\end{pmatrix}$$ (where I've written the matrices in Jordan block form; ...

1

I think the author mean that we define an addition on $M\oplus N$, but not on $M\times N$. Moreover we consider $M\times N$ as a generating set for formal sums $(a,b)+(c,d)$. As to second question I think this expression is not nice and explains nothing.

0

I'm assuming you mean $M$ to be a submodule of $V$. By definition, $KM$ is the set of $K$-linear combinations of vectors in $M$. This is a $K$-subspace of $V$. In fact, since $M$ itself is finitely generated over $R$, the set $KM$ is equal to the set of $K$-linear combinations of a set of generators $x_1,\ldots,x_n\in M$ for $M$ as an $R$-module. In general, ...

0

Hint: What you're talking about are the induced isomorphisms of a non-degenerate (not necessarily perfect) pairing. The definition of $P'$ is, however, a bit different. Here is the right one: $$P': A \to B^*, a \mapsto P(a,-)$$ So the image of an $a \in A$ is really a dual map $B \to R$. Are those modules finitely generated over $R$?

1

The element $xab^{-1}$ is an element in the localization $S^{-1}M$ where $S = R \setminus \{0\}$. There is a natural isomorphism $M \otimes_R S^{-1}R \simeq S^{-1}M$ defined by $m \otimes \frac{a}{b} \mapsto \frac{ma}{b}$. Here $S^{-1}R$ is the ring $K$. In general writing $x_1 \otimes x_2 = x_1x_2$ doesn't make sense unless you are invoking some type of ...

2

The notation $R \cap \mathfrak m$ is sometimes used to denote the inverse image of $\mathfrak m$ in $R$. It is a prime ideal $\mathfrak p$ of $A$. There is a canonical morphism $R_{\mathfrak p} \to A_{\mathfrak m}$, and the $A_{\mathfrak m}$-module $M_{\mathfrak m}$ acquires an $R_{\mathfrak p}$-module structure via this map.

9

You just have to work with the definitions (i.e. universal properties) in order to answer this question. It is not an extra convention or something like that (unfortunately, many mathematicians believe this). If $(E_i)_{i \in I}$ is a family of $R$-modules with underlying sets $|E_i|$, and $F$ is some $R$-module, a map $\prod_i |E_i| \to |F|$ is called ...

1

First, let's settle the issue of the empty tensor product. Consider the identity map $id: W\to W$ for an arbitrary $R$-module $W$. This is an $R$-balanced map out of the collection $\{W\}\cup \emptyset$ and so descends to a map $W\otimes E\to W$. But we know that $(W,id)$ satisfies the universal property of the tensor product of $W$, so $W\cong W\otimes E$ ...

3

$E^{\otimes 0}$ should be the identity of $\otimes$, i.e. the base ring $R$.

5

For ease of notation, let $A = \operatorname{im}\varphi_1 \subset N$, $T = \operatorname{im}\varphi_2 \subset S$, and $Q = T/\operatorname{im} (\varphi_2\circ\varphi_1) = T/\varphi_2(A)$. Consider the map $\psi = \pi \circ \varphi_2 \colon N \to Q$, where $\pi \colon T \to Q$ is the canonical projection. The kernel of $\psi$ is $$\ker \psi = \psi^{-1}(0) = ... 0 The proof does not exclude the possibility that the restriction of V to N is irreducible. Then U is necessarily equal to V. For example V could be 1-dimensional. 2 We are given N and that will give us the prime factors p and q as:$$N = 91 = p \times q = 7 \times 13$$We need the Euler Totient Function of the modulus, hence we get:$$\varphi(N) = \varphi(91) = (p-1)(q-1) = 6 \times 12 = 72$$Now, we choose an encryption exponent 1 \lt e \lt \varphi(N) = 72. We were told to pick an an e \lt 6, so lets ... 3 As you remarked, injective modules are divisible, that is, rM=M for all r\in R, r\ne 0. The key step is to show that every non-zero homomorphism f:M\to R is surjective. Let x\in M such that f(x)\ne 0. Set r=f(x). Since rM=M there exists y\in M such that ry=x. Then rf(y)=r, so f(y)=1, and this is enough. Now use that M is ... 0 \mathbb{Z}G is a free module over \mathbb{Z}, with G as basis. Try proving that F is (isomorphic to) a free \mathbb{Z}-module over X\times G. 4 As P is f. g. we have an exact sequence 0\rightarrow Q\rightarrow A^n\rightarrow P\rightarrow 0, Q denoting the kernel of the map A^n\rightarrow P. As P is projective, this exact sequence splits, A^n\cong Q\oplus P. The exact sequence A^n\cong Q\oplus P\rightarrow A^n\rightarrow P\rightarrow 0 shows P to be f. p. (where Q\oplus P\rightarrow ... 0 F is sums of elements of X with coefficients in \mathbb{Z}G, which is the same as sums of \{gx:g\in G, x\in X\} with coefficients in \mathbb{Z}. F is not a free \mathbb{Z}-module over X (over \{1_G\cdot x:x\in X\}) because X does not generate F as a \mathbb{Z}-module. 1 Do you understand that \Bbb Z G\cong \Bbb Z^{|G|}\not\cong\Bbb Z as \Bbb Z-modules when |G|>1? 0 As Martin suggested, a nice reference is Eisenbud - Commutative Algebra with a view towards algebraic geometry The definition of Fitting ideals that can be found there is indeed in term of exterior algebra, and precisely is the following (it consists of 2 parts) Def: Let \varphi:F\to G be a map of free modules over a ring R. We define the ... 2 Just some hints: What you've just done is to make M into an R-module. You could equally look at M as an abelian group (= \Bbb Z-module) with an action of the ring R, defined by R \to Aut(M), but the main problem is, that not every image of \mu is an automorphism. So, in fact, you could have \varphi \in End(M) \setminus Aut(M) as an image of ... 0 Using formulae used here,$$\frac{386}{97}=3+\frac{95}{97}=3+\frac1{\frac{97}{95}}=3+\frac1{1+\frac2{95}}=3+\frac1{1+\frac1{\frac{95}2}}=3+\frac1{1+\frac1{47+\frac12}}$$So, the previous convergent of \displaystyle \frac{386}{97} is \displaystyle3+\frac1{1+\frac1{47}}=3+\frac{47}{48}=\frac{191}{48}$$\implies386\cdot48-191\cdot97=1\implies ...

0

There is likely a typo in the question. Using the Euclidean algorithm, we have that: \begin{align*} 386 &= 3(97) + 95 \\ 97 &= 1(95) + 2 \\ 95 &= 47(2) + 1 \end{align*} Working backwards, we see that: \begin{align*} 1 &= 95 - 47(2) = 95 - 47(97 - 95) \\ &= -47(97) + 48(95) = -47(97) + 48(386 - 3(97))\\ &= 48(386) - 191(97)\\ ...

1

Using Extended Euclidean algorithm determine $p$ and $q$ such that $$97 \cdot p - 386\cdot q = 1$$ Then $97^{-1} \equiv p \mod 386$.

1

Hint. If $n=p^k$, $k\ge 1$, then $0=p^k\Bbb Z/p^k\Bbb Z<p^{k-1}\Bbb Z/p^k\Bbb Z<\cdots<\Bbb Z/p^k\Bbb Z$ is a series of composition of length $k$. Can you generalize this to the case $n=p_1^{k_1}\cdots p_r^{k_r}$, $k_i\ge 1$ and $r\ge 2$? Edit. Let's try the next step: $n=p^kq^l$. Then $$0<p^{k-1}q^l\Bbb Z/p^kq^l\Bbb Z<\cdots<pq^l\Bbb ... 5 The answer is yes. Hint: You can easily see that considering the kernel, image resp. of an R-homomorphism. Conversely, if assumptions ii) or iii) hold for such R-homomorphisms, then M has to be simple, where you can consider the possible submodules of M to check. 0 Let's denote by \times the “external” direct sum just to avoid confusion. Saying that$$ N_1+N_2+\dots+N_n=N_1\oplus N_2\oplus\dots\oplus N_n $$means that the obvious homomorphism$$ N_1\times N_2\times\dots\times N_n\to N_1+N_2+\dots+N_n $$is injective. Hint for the second part: you can define homomorphisms$$ M_i\to ...

1

Claim: $N = f(K) + f(L)$. Certainly, $N \supseteq f(K) + f(L)$. For the reverse containment, let $n \in N$. As $f$ is surjective, there exists $m \in M$ such that $f(m) = n$. Since $M = K + L$, there exist $k \in K$ and $\ell \in L$ such that $m = k + \ell$. Now, $f(k) + f(\ell) = n$, demonstrating that $N \subseteq f(K) + f(L)$. Claim: $N = f(K) ... 3 It's obvious (from surjectivity) that$f(K)+f(L)=N$. Now take$x\in f(K)\cap f(L)$. Then$x=f(y)=f(z)$with$y\in K$and$z\in L$. Since$y-z\in \ker f$we get$y\in L$, so$y\in K\cap L$and therefore$y\in\ker f$. Thus we get$x=f(y)=0$. 8 I'm assuming that$\mathbb Z_n$means$\mathbb Z/n\mathbb Z$. I'm also assuming that "subproduct" means "subgroup of the product". If this is the case then yes, the subgroup generated by$(1, 1, \ldots) \in \prod_{n > 1}\mathbb Z_n$is isomorphic to$\mathbb Z$. To see that this is the case note that we can always define a homomorphism out of$\mathbb ...

1

The second part follows at once from split short sequences' theorem: look at $$0\longrightarrow N\stackrel g\longrightarrow M\stackrel\pi\longrightarrow M/g(N)\longrightarrow 0$$ Prove the above is an exact sequence (i.e., $\;g\;$ is one-to-one, $\;\pi\;$ is onto and $\;\text{Im}\,g=\ker\pi\;$ , with $\;\pi\;$ the canonical projection). Well now, the ...

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