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5

You can find good information about the topic in Hungerford's book "Algebra", page 199 (specially theorem 4.1). If $A$ and $B$ are modules over a ring $R$, then $Hom_R(A,B)$ is the set of all $R$-module homomorphisms $f:A\to B.$ The homomorhisms are expressed in the book. Let A,B,C,D he modules over a ring R and $f:C\to A$ and $g:B\to D$, R-module ...


5

If I understand your question correctly, it seems to me that $X$ is the unique smooth projective curve with function field $K$. Your question then boils down to the following. Lemma. Let $X$ be a smooth projective curve, and let $x_1, \ldots, x_r \in X$ be distinct points. Set $U = X \setminus \{x_1, \ldots, x_r\}$. Then $\Gamma(U, \mathcal{O}_U)^\times/...


5

Each element of $F$ can be written as a finite linear-combination of elements of $\Lambda$. Since there are finitely many elements in $F$, this gives rise to a finite subset $\Omega \subset \Lambda$ with $A\langle \Omega \rangle \supset A \langle F \rangle = X$.


4

Verify that the sequence $$0\longrightarrow \ker \phi\stackrel{}{\longrightarrow}M\stackrel{\phi}{\longrightarrow}N\stackrel{(Id-\phi \circ \psi)|^{\ker \psi}}{\longrightarrow}\ker \psi ~ {\longrightarrow} ~0$$ is well-defined and exact. Clarification: The sequence $$0\longrightarrow \ker \phi\stackrel{}{\longrightarrow}M\stackrel{\phi}{\...


4

There are plenty of cases where this happens. It's really only interesting when the module action interacts in an interesting way with the multiplication The most straightforward example, which works when $R$ is commutative, is the concept of an $R$-algebra, which I'll let you look up yourself. In this case it's easy to see what the action does. If you ...


4

$\operatorname{Hom}(M,N)$ refers to the set of $A$-module homomorphisms from $M$ to $N$. These form an abelian group under pointwise addition (define $f+g$ by $(f+g)(x)=f(x)+g(x)$), and if $A$ is commutative they in fact form an $A$-module by pointwise scalar multiplication (define $a\cdot f$ by $(af)(x)=a\cdot f(x)$). Given an $A$-module homomorphism $g:M\...


4

This follows at once if you show that $$R/I \otimes R/J \simeq R/(I+J)$$


4

No. For example, suppose $N=S$ is simple, so that $\operatorname{soc}N=S$, $M'$ is a non-split extension of $S$ by another simple module $T$, so that $\operatorname{soc}M'=T$ and there is an epimorphism $\alpha:M'\to N$ which is zero on $\operatorname{soc}M'$, and that $M=M'\oplus S$. Then the map $\begin{pmatrix}\alpha&\operatorname{id}_S\end{pmatrix}:...


4

Slup has given you the answer where it is mostly used, though it is not in general true that Trace maps $B$ to $A$, unless you assume something more (typically, one assumes that $A$ is integrally closed). One has standard counterexamples for general cases. For example take a ring $A$ which has a non-free projective module $P$ such that $A\oplus P$ is free. (...


3

You have the right idea. Note that $\gcd(3,q)=1$, by Euler's theorem $3^{\phi(q)}\equiv 1(\mod q)$, and hence the order of $3$ module $q$ is a divisor of $\phi(q)$. Hence, if we can prove $q-1$ is the order of $3$ modulo $q$, you can use the statement you mentioned. Now, assume $h=$ the order of $3$ modulo $q$. Since $3^{\frac{q-1}{2}}\equiv -1(\mod q)$, we ...


3

First show that if $A$ is any ring, $I\subseteq A$ a left ideal and $M$ a right $A$-module, there is a (natural) isomorphism $$ \eta_M : M\otimes_A A/I\longrightarrow M/IM$$ that sends $m\otimes a$ to the class of $ma$. Conclude that in particular $A/J\otimes_A A/I=A/(I+J)$ when $A$ is a commutative ring and finally consider the case when $I,J$ are ...


3

I think these problems should be tackled by the universal property of exterior powers and then your question will have a natural answer even if it were just projective modules (not necessarily free) and even for vector bundles (when the sequence may no longer split). Giving an $R$-module homomorphism $\wedge^r M\to N$ where $M, N$ are $R$-modules is ...


3

A structure of (unitary left) $R$-module over an abelian group $G$ (written additively) is determined by a (unitary) ring homomorphism $R\to\operatorname{End}(G)$, where $\operatorname{End}(G)$ consists of the endomorphisms of $G$ under the standard sum operation and map composition. To see why, suppose $G$ is an $R$-module. For $r\in R$, define $\lambda_r\...


3

Newman says that Smith form can be accomplished with elementary row and column operations as long as the coefficient ring is Euclidean, as here. Since we are not going to change the determinant, this means diagonal $(1,1,6).$ Newman's assurance means that we can accomplish this for the two by two square with entries $(2,3).$ Take distinct positive integers $...


3

Ok Slup, here goes. Let $R$ be any commutative ring and let $A$ be a polynomial ring over $R$. Let $P$ be any projective module over $R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal $\mathfrak{m}$ of $R$, $P_{\mathfrak{m}}$ is of the form $Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$ for some projective $...


3

This is a very general question and can not be answered in a few words. So, may be let me describe one aspect. If $R$ is a commutative ring and $M$ an $R$-module, giving an $R$-algebra homomorphism $A\to \operatorname{End}_R M$, where $A$ is an $R$-algebra makes $M$ into an $A$-module. In the example you write above, $R=\mathbb{Z}$. Now, let me look at a ...


2

Note that $\Bbb{Z}/p^k\Bbb{Z}$ is a local ring with maximal ideal $(p)$, so the map $$(\Bbb{Z}/p^k\Bbb{Z})_{(p)}\ \longrightarrow\ \Bbb{Z}/p^k\Bbb{Z}:\ \tfrac{a}{b}\ \longmapsto\ ab^{-1},$$ is well defined, and in fact it is an isomorphism. There is no mistake in your reasoning on flatness; indeed $(\Bbb{Z}/p^k\Bbb{Z})_{(p)}$ is flat as a $\Bbb{Z}/p^k\Bbb{...


2

Both notations that bother you are defined and explained beginning at the top of page 18, right before the beginning of the subchapter "Submodules and quotient modules". Read carefully, don't skip paragraphs - the book is a classic written by two giants in that field, so you won't find serious omissions in it.


2

Since $\overline 6\in\operatorname{Ann}(M)$, there is an induced action of the ring $(\mathbb Z/36\mathbb Z)/(\overline 6) = (\mathbb Z/36\mathbb Z)/(6\mathbb Z/36\mathbb Z)\cong \mathbb Z/6\mathbb Z$ on $M$, i.e. $M$ may be considered as a $\mathbb Z/6\mathbb Z$-module, and under this action, $M$ has the same submodules as under the original action, so in ...


2

A start: The natural number $n$ divides $34(34x^2-42y+13y^2)$, which is equal to $$(34x-21y)^2+y^2.$$ Now use what you know about the prime factorization of numbers that are the sum of two squares. Added: We show that if $x$ and $y$ are relatively prime, then any prime divisor of $(34x-21y)^2+y^2$ is either $2$ or of the form $4k+1$. Since $2$ is a sum ...


2

First $(3)$ implies $(1)$. Suppose $M \cong R/I$, with $I$ a maximal ideal in $R$. Then as a ring $R/I$ is a field. Hence the only ideals of $R/I$ are the ring itself and $\{0\}$. But this means that the only subgroups of $R/I$ that are closed under multiplication by elements of $R$ are $R/I$ and $\{0\}$, and so $M$ is a simple module. For the other ...


2

An $R$-module structure on the abelian group $G$ is a ring homomorphism $R\to\operatorname{End}(G)$. Since $\operatorname{End}(\mathbb{Z}/5\mathbb{Z})\cong\mathbb{Z}/5\mathbb{Z}$ as rings, what you want is the number of ring homomorphisms $\mathbb{Z}[i]\to\mathbb{Z}/5\mathbb{Z}$. Such a homomorphism must send $1$ to $[1]$ (as we want unital modules, don't ...


2

I don't think this is true. Let $R$ be any ring with identity with maximal left ideal $\mathfrak M$ which is not a two sided ideal. Let $M$ be the left $R$-module $R/\mathfrak M$. Then $M$ is a simple left $R$-module. Now, $\textrm{Ann}_R(M)$ is the set of $r \in R$ for which $rs \in \mathfrak M$ for all $s \in R$. It is clearly a two-sided ideal. ...


2

The following is primarily first principles. It is not the most elegant way to answer the question if you have knowledge of short exact sequences, etc., but it seems to be more what you are looking for. Choose any $m\in f^{-1}(1)$. Then Let $L=Rm$. Let's verify that $M=\ker f \oplus L$. 1) If $f(rm)=0$, then $0=rf(m)=r1=r$, so $rm=0$, i.e. $\ker f \cap L=...


2

Assume $\varphi: M\times N \to P$ is a bilinear map, then the corresponding homomorphism is $\phi: M\to \hom_R(N,P)$, with $\phi(m)(n)=\varphi(m,n)$. We can check $\phi(m)$ is a homomorphism from $N$ to $P$ using the fact that $\varphi$ is bilinear. Convesely, if $\phi: M\to \hom_R(N,P)$ is a homomorphism, then the corresponding bilinear map is $\varphi: M\...


2

Hint: Take $f:M\rightarrow M/M'$ be the canonical projection.


2

Note that the image of your diagonal matrix is $\Bbb Z\oplus 2\Bbb Z\oplus 2\Bbb Z$. Hence the cokernel is $$ \frac{\Bbb Z\oplus\Bbb Z\oplus\Bbb Z}{\Bbb Z\oplus 2\Bbb Z\oplus 2\Bbb Z}\simeq(\Bbb Z/\Bbb Z)\oplus(\Bbb Z/2\Bbb Z)\oplus(\Bbb Z/2\Bbb Z)\simeq\Bbb Z/2\Bbb Z\oplus\Bbb Z/2\Bbb Z $$


2

Edit As @Batominovski and @TobiasKildetoft remarked, one should assume characteristic zero here. In positive characteristic $p$ and non-trivial $G$, the statement is indeed not true, at least if one takes the naive definition of characters: For example, the $p$-fold sum of the trivial representation satisfies the assumption but is not of the form $KG^{\oplus ...


2

I will prove that given a family $\{M_j\}_{j \in J}$ of $R$-modules then $Hom(\oplus M_{j},N) \cong \prod Hom(M_{j},N)$ for any given $R$-module. Then your question follows by setting $M_{j}=R$ for all $\thinspace$ $j \in J$. Assume that $i_{j} : M_{j} \rightarrow \oplus M_{j}$ is the canonical inclusion. Now define the above homomorphism $\phi: Hom(\oplus ...


2

The theorem is true more generally when the coefficient ring is a PID (see Quillen's original article). Quillen's original proof is different from the one given in Lang!



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