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Let $R$ be ring for which $F$ is a free module. Then we have isomorphisms $R^n \rightarrow F$ and $F \rightarrow R^{n+1}$ which gives us an isomorphism $R^n \rightarrow R^{n+1}$. Thus for any $m \geq n$ we have an isomorphism $R^m \cong R^n \oplus R^{m-n} \cong R^{n+1} \oplus R^{m-n} \cong R^{m+1}$. Composing these isomorphisms we get an isomorphism $R^n ... 3 You can proceed as usual by starting with$F\stackrel{f}\to A\to 0$and$H\stackrel{h}\to C\to 0$, where$F$and$H$are free of finite rank. Then show that there is an exact sequence$G=F\oplus H\stackrel{g}\to B\to 0$. Now consider$F'=\ker f$and so on. You have a short exact sequence$0\to F'\to G'\to H'\to 0$. Now use the result for finitely generated ... 3 For an explicit counterexample, consider$R = k[X,Y]$for any field$k$, which is free over itself. Then the ideal$\mathfrak m = RX+RY$is not free. 3 Define $$\phi: R\to Rx\le M\;,\;\;\phi(r):=rx$$ prove the above is a (left)$\;R$- module homomorphism, and now use the first isomorphism theorem. 3 If I understand your question correctly, consider that$a \equiv b \pmod x$if and only if$a-b = kx$for some$k\in\mathbb{Z^+}$. Then$x=\frac{a-b}{k}$, hence there is precisely one$x$for each divisor$k\in\mathbb{Z^+}$. 3 Let$A$be a commutative ring. Let$M$be a finitely generated$A$-module and$N$be an$A$-submodule of$M$. Let$f\colon N \rightarrow M$be a surjective homomorphism of$A$-modules Then$f$is injective. Proof. Let$0 \neq x'_0 \in N$. It suffices to prove$f(x'_0) \neq 0$. Set$f(x'_0) = x_0$. Let$x_1, \dots, x_n$be generators for$M$. Then ... 3 user26857's answer shows how to repair the reduction to the Noetherian case. Here is how to repair the proof of the Noetherian case: Let$M$be a noetherian$A$-module and let$N \subseteq M$be a submodule. Let$f : N \to M$be a surjective linear map. Then$f$is injective. Proof: Let$n \geq 0$. Although$f^n$is not a well-defined homomorphism this ... 2 For$i=1, \dots, n+m$call $$c_i = \left\{ \begin{matrix} a_i & \mbox{ if } &i \leq n \\ b_{i-n} & \mbox{ if } &i \geq n+1 \end{matrix} \right.$$ and analogously $$w_i = \left\{ \begin{matrix} x_i & \mbox{ if } &i \leq n \\ y_{i-n} & \mbox{ if } &i \geq n+1 \end{matrix} \right.$$ Then $$x+y = \sum_{i=1}^{n+m} c_iw_i \in ... 2 Show that \hom_{R/I}(M/IM,-) \cong \hom_R(M,U(-)), where U is the forgetful functor from R/I-modules to R-modules. Hence, this is a composition of two exact functors, hence exact. 2 There's no need to consider J; just assume R is semisimple. Every (right) module over a semisimple ring is a direct sum of simple modules. Moreover, consider (S_i)_{i\in I}, a family of simple modules over R such that every simple R-module S is isomorphic to S_i, for some i\in I; if i\ne j, then S_i is not isomorphic to S_j. We can ... 2 For Dedekind domains we have \operatorname{Pic}(R)\simeq\operatorname{Cl}(R), where \operatorname{Cl}(R) is the ideal class group of R. Your case is treated here in detail. 2 You are asking if a submodule of a free module is necessarily free. This is not true in general. It is, however, true if R is a PID. A counter example is given by R=\mathbb{Z}/\mathbb4{Z} as a module over itself. The submodule 2\mathbb{Z}/4\mathbb{Z} is not free. 2 Such a map can exist, but not for every noncommutative ring. For example, no such map can exist over a division ring. This is because the annihilator of the left tensor product module is a nonzero left ideal, hence the left tensor product is always trivial in that case. For an example where the map does exist, let T be the tensor algebra over a field, and ... 1 (a) Yes. (b) Yes. (c) Over a semisimple ring every non-zero module is semisimple (as a quotient of a free module which is a direct sum of copies of your semisimple ring, hence semisimple). 1 Let A and B be ideals of the ring R. Then A/BA is a module over R/B in a natural way. Indeed, for b\in B and a\in A, the product ba\in BA, so$$ b(a+AB)=0+AB $$in the module A/AB. Thus \operatorname{Ann}(A/BA)\supseteq B. If M is an R-module and \operatorname{Ann}(M)\supseteq B, then M is a module over R/B by defining$$ ... 1 Let$z_1=x_1,\dots,z_n=x_n,z_{n+1}=y_1,\dots,z_{n+m}=y_m$, and let$c_1,\dots,c_n,c_{n+1},\dots,c_{n+m}$be defined similarly based on$a_i,b_j$. Then:$$x+y = \sum c_iz_i$$ 1 We can argue as follows. Pick an$h \in ker\ g_* \subseteq Hom_R(M,M_2).$Then we have$g\circ h = 0.$This means$im\ h \subseteq ker\ g = im\ f.$Since$f$is injective, we have the$R$-homomorphism$(f|_{im\ f})^{-1}:im\ f \rightarrow M_1.$So we can define$j := (f|_{im\ f})^{-1} \circ h: M \rightarrow M_1.$This is an$R\$-homomorphism since it's a ...