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4

$\mathbb{Z}/6\mathbb{Z}$ is not a PID, because it's not a domain.


3

For a $b\in B$ consider the element $x=f (g(b)) -b\in B$. We claim that this $x$ is in the kernel of $g$. Then it will follow that $b= x+f(g(b))$ where in the rhs first element is in $\ker g$ and the second element is in image of $f$, that is what you wanted to be proved. To prove the claim apply $f$ to this element $x$ and use the condition that $g$ ...


3

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

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 $M_1\cap M_2$?


2

Your method is not going to work unless you're careful about the choice of the projective modules $P$ and $Q$ through which $gf-\text{id}_M$ and $fg-\text{id}_N$ factor. Indeed, you could replace $P$ by $P\oplus P'$ for any projective module $P'$, changing the isomorphism type of $M\oplus P$. Instead, use the fact that $gf-\text{id}_M$ factor through a ...


2

Without any assumptions about $R$, this is false. For example, consider $\mathbb{Z}$ as a $\mathbb{Z}$-module. It's finitely generated (the set $\{1\}$ is a generating set), but the collection of submodules $$\mathbb{Z}\supset 2\mathbb{Z}\supset 4\mathbb{Z}\supset 8\mathbb{Z}\supset\cdots$$ has no minimal element.


2

Embed $L$ in an injective module $E$ and consider the push-out diagram $$\require{AMScd} \begin{CD} {} @. {} @. 0 @. 0 \\ @. @. @VVV @VVV \\ 0 @>>> K @>>> L @>>> M @>>> 0 \\ @. @| @VVV @VVV \\ 0 @>>> K @>>> E @>>> N @>>> 0 \\ @. @. @VVV @VVV \\ {} @. {} @. E/L @= E/L \\ @. @. @VVV @VVV ...


2

We want to show ${\rm Ext}^1( P, N ) = 0$ for all $N$ (as this is equivalent to $P$ being projective). Accordingly, let $$ 0 \to N \to I^0 \mathop\to^{\delta^0} I^1 \mathop\to^{\delta^1} I^2 \to \cdots$$ be the beginning of an injective resolution for $N$. Suppose $f\colon P \to I^1$ is a cocycle, i.e., $\delta^1\circ f = 0$. Since the sequence is exact, ...


1

For a module such as $B$, with submodules $C$ and $D$, we of course have the submodule $C + D$ of $B$, which is, by definition, $C + D= \{ c + d \mid c \in C, d \in D \} \subset B; \tag{1}$ in the event that $C \cap D = \{ 0 \}$, the notation $C \oplus D$ may also be used; $C \oplus D$ is then referred to as the direct sum of $C$ and $D$. We observe that ...


1

It seems like your question is: prove that a finitely generated free module has a finite basis. Let $E$ be a basis for such a module $M$ over $R$. Then $E$ spans $M$. Now, $M$ is finitely generated so you can find a finite ste of elements that span $M$. Write each element as (finite) linear combinations of elements from $E$. Collect all these elements from ...


1

Write $M$ as the direct sum of its irreducible submodules. Each element of the generating set can be written as a sum of elements from finitely many of the irreducible submodules. Then $M$ is a direct sum of finitely many irreducible submodules. This implies $M$ is Artinian.


1

Hint: $$0\subseteq\ker(f)\subseteq\ker(f^2)\subseteq\cdots$$


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

I am surprised that it is not mentioned here- Example of a free module M which has bases having different cardinalities. Let $V$ be a vector space of countably infinite dimension over a division ring $D$. Let $R=End_D(V)$. We know that $R$ is free over $R$ with basis $\{1\}$. We claim that given a positive integer $n$, there is a $R$-basis $B_n=\{f_1,f_2, ...


1

In different categories (such as the category of rings or the category of modules) different substructures are important. Ideals are the natural substructures of rings, and submodules are the natural substructures of modules. These substructures are special because they are suitable for forming quotient objects in their categories (that is why I am not ...


1

From the artinian hypothesis you may deduce that the zero submodule is an intersection of finitely many maximal submodules. Therefore $M$ is isomorphic to a submodule of a finitely generated semisimple module, hence it is semisimple and finitely generated.


1

Let $\mathcal{M}$ be the set of maximal submodules of $M$. Lemma: There exist a finite subset $\mathcal{N}$ of $\mathcal{M}$ such that $\bigcap \mathcal{N} = 0$. Proof: Let $\mathcal{C}$ be the collection of the intersection finite number of maximal submodules. As $M$ is artinian, $\mathcal{C}$ must have a minimal element $N = \bigcap \mathcal{N}$. If it ...


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

When that definition says Let $R$ be a ring and $1_R$ its multiplicative identity... they are mentioning (implicitly) that they are requiring $R$ to be unital. However, it is certainly possible to define modules over a non-unital ring; just throw out statement 4 from that definition.


1

Some authors use the term "ring" to mean "ring with identity." This is neither standard nor nonstandard; there is just no consensus. Authors using this definition would use the term "rng" to denote a ring that possibly does not have an identity.


1

I don't have the book by Dummit and Foote, but the maps $$ R\to Ry_1,\quad r\mapsto ry_1\\ R\to Ra_1y_1,\quad r\mapsto ra_1y_1 $$ are surjective homomorphisms and they're also injective because $M$ is torsion-free. For the second one, note that $ra_1=0$ implies $r=0$, if $a_1\ne0$. If $a_1=0$, then $Ra_1y_1=\{0\}$ is free as well.



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