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4

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 ...



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