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If you want $\operatorname{Hom}_R(P,R)\simeq P$ for $P$ a finitely generated projective $R$-module, then forget it. $\bullet$ If $R$ is an integral domain, and $I\subset R$ is a non-zero ideal, then $\operatorname{Hom}_R(I,R)\simeq I^{-1}$, where $I^{-1}=\{x\in Q(R):xI\subseteq R\}$. (Here $Q(R)$ stands for the field of fractions of $R$.) $\bullet$ If ...

3

Consider $R=\mathbb Z$, $M=\mathbb Z$ and $N=\mathbb Z/2\mathbb Z$.

3

An example of a difference: $\mathbb{Z}/3\mathbb{Z}$ is projective (and free) as a $\mathbb{Z}/3\mathbb{Z}$-module but is not projective (nor free) as a $\mathbb{Z}$-module. However things like simplicity and indecomposability will remain the same, which is easy to prove since the action of $A/\mathfrak a$ is given by the action of $A.$ The only thing ...

2

Hint: what exact sequence can you build up out of $M_1\oplus M_2, M_1 + M_2$ and $M_1\cap M_2$?

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The problem asks to find an $A$-module $N$ (eventually you already know from class) such that $M/S\simeq N$ (eventually by using a well known isomorphism theorem). Hints. Define $f:A^n\to A$ by $f(a_1,\dots,a_n)=\sum_{i=1}^na_i$. Define $f:A[X]\to A$ by $f(p(X))=p(1)$. Define $f:M_n(A)\to A^n$ by $f((a_{ij})_{i,j})=(a_{11},\dots,a_{n1})$.

2

This result is known as Schanuel's lemma. A quick proof is given by introducing the pullback of $f$ and $f'$. It is the submodule of $P\oplus P'$ given by $$X = \{(p,p')\in P\oplus P'\mid f(p) = f(p')\}.$$ Then, the following sequences are exact : $$0 \longrightarrow \ker(f')\simeq K' \longrightarrow X \longrightarrow P \longrightarrow 0$$ and 0 ... 1 You have confused k-vector space basis as R-module basis, hence this issue. ANy commutative ring is a free module overitself with basis the singleton \{1\}, consiting of the multiplicative identity. Given two elements a,b\in R, by commutativity we have a.b -b.a=0 and so a basis has to have less than 2 elements. 1 In the general case, there is a canonical isomorphism: \begin{align*}P/N\cap P &\longrightarrow(P+N)/N\\ x+N\cap P & \longmapsto x+N\end{align*} Also, there's a bijection between submodules of M/N and submodules of M that contain N, and P is not supposed to contain N. However, by the canonical homomorphism \,p\colon M\longrightarrow M/N, ... 1 If P is a finitely generated and projective R-module, then there is a canonical isomorphism\zeta:P^*\otimes_RM\simeq\operatorname{Hom}_R(P,M)$$given by \zeta(f\otimes m)(x)=f(x)m. (For a proof see Proposition 6 from these notes.) Now, coming back to the question we have \mathfrak a^*\otimes_RM\simeq\operatorname{Hom}_R(\mathfrak a,M). But ... 1 As you remarked, there are four non-trivial subgroups of \mathbb Z_{18} given by the divisors of 18 (others than 1 and 18). Now it's good to know what's the sum and the intersection of such subgroups. In general, in \mathbb Z_n we have \bar k\mathbb Z_n+\bar l\mathbb Z_n=\overline{\gcd(k,l)}\mathbb Z_n and \bar k\mathbb Z_n\cap\bar l\mathbb ... 1 Hint: Prove the statements below ( may assume I \ne 0 ): If I, J are ideals of an integral domain R with field of fractions K then every morphism of R-modules I \to J is given by the multiplication by an element in K. If I is a projective module then there exists an imbedding of I into a free R module i \colon I \hookrightarrow ... 1 I think you want I to be a right ideal and J a left ideal, not the other way round. You are almost done. Consider the canonical homomorphism \mathbf{m} :A\otimes_{A}A\rightarrow A of \mathbb{Z}-modules which sends every a\otimes b\in A\otimes_{A}A to ab\in A. It is well-known that \mathbf{m} is an isomorphism. Thus, \left( ... 1 To augment user26857's answer, note that: (a_1,a_2,\dots,a_n) + S = (a_1+a_2+\cdots+a_n,0,\dots,0) + (a_1-a_2-\cdots-a_n,a_2,\dots,a_n) + S = (a_1+a_2+\cdots+a_n,0,\dots,0) + S. If p(X) = q(X)(X - 1) + r (for r \in A), then p(X) + S = r + S. If B = \begin{bmatrix} x_{11} & x_{12} & x_{13} & \dots & x_{1n} \\ ... 1 \mathbb{C}[M] will be Laurent polynomials in x_1,\dots,x_n. Algebra homomorphisms to \mathbb{C} are thus determined by sending each x_i to an element of \mathbb{C}^\times and therefore the set of all such homomorphisms is associated with (\mathbb{C}^\times)^n More generally, the group algebra functor is left adjoint to the functor taking an ... 1 Let b_1+A,b_2+A,\dots,b_m+A be generators of B/A and let a_1,\dots,a_n be generators of A. If b\in B, then$$ b+A=\sum_{i=1}^m r_i(b_i+A)=\biggl(\sum_{i=1}^m r_ib_i\biggr)+A $$for some r_1,\dots,r_m\in R, which means$$ b-\biggl(\sum_{i=1}^m r_ib_i\biggr)=\sum_{j=1}^n s_ja_j $$for some s_1,\dots,s_n\in R, so$$ b=\sum_{i=1}^m ...

1

One should approch this more conceptually: The uniqueness is an immediate consequence of the surjectivity of $f_2$ (You can view this as a definition of the term surjective). The existence follows from the universal property of the cokernel, which is the same as the fundamental homomorphism theorem: The morphism $g_2 \circ \beta$ annihilates anything, ...

1

It is clear that the product is injective wrt itself and from a theorem that a product is injective iff each factor is injective the first part is settled. As for the second part we need to see that we cannot inject the product inside the direct sum since we have infinitely many factors, see Ribenbenboim : Rings and modules top of page 25.

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The inclusion $\bigoplus R_n \hookrightarrow \prod R_n$ is essential, hence $\bigoplus R_n$ is not injective.

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This is well-studied under the heading of "cancellability," and Lam's crash course on the topic is very nice. Are there any simple conditions on $R$-modules $M,A$ and $B$... The readiest one is that if $R$ has stable range 1 and $M$ is finitely generated and projective, then it cancels from $M\oplus A\cong M\oplus B$. You can find this, for example, in ...

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