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5

$[R^{op}, \text{Ab}]$ is the category of right $R$-modules, which is equivalently the category of left $R^{op}$-modules. The reason to prefer taking right modules here is the same reason why presheaves are contravariant functors and not covariant functors: it's so that the Yoneda embedding, which in this case is $R \to [R^{op}, \text{Ab}]$, is covariant. In ...


4

Hint: Let $M_i = e_i M$, where $e_i = (0,\dotsc,1,\dotsc,0) \in R$ with $1$ in the $i$th entry. Use that $e_i$ are pairwise orthogonal idempotents with $\sum_i e_i = 1$ to show $M = \oplus_i M_i$.


3

By definition, your ideal is isomorphic to $R^n$ but this isomorphism is quite abstract and your proof shows nothing. Counter-example : the ideal $2 \mathbb{Z}$ in $\mathbb{Z}$. Exercise : show that your property is true if and only if $R$ is a field.


2

There are two ways to see this. First, one characterization of the socle of a module $N$ is that it is the sum of the simple submodules of $N$. The simple submodules of $K$ are exactly the simple submodules of $M$ that lie in $K$. Second, using the theorem you stated in the comments let $f\colon K \to M$ be the inclusion map. Then that theorem gives ...


1

There are rings for which every right ideal is free, and it's even possible for them to have rank more than 1. If such a ring is also commutative, it is a principal ideal domain, and all the ideals are isomorphic (as modules) to the ring.


1

@Joseph: Isomorphisms are not same as equality. If the natural inclusion map is an isomorphism, then you can say that it is actually an equality. Otherwise not in general. For example, take the ring $A := k[x_1, x_2, x_3, \cdots],$ k is a field and let $B := k[x_2, x_3, x_4, \cdots].$ Then $B$ is a subring of $A$, and it is also isomorphic to $A$ (why?). But ...


1

It's easier with just two rings and then you can do induction. So consider $R=R_1\times R_2$ and the two idempotents $e_1=(1,0)$, $e_2=(0,1)$. If $M$ is an $R$-module, you can consider $M_1=e_1M$ and $M_2=e_2M$. For all $r\in R$, $e_1r=re_1$, and similarly for $e_2$, so $M_1$ and $M_2$ are $R$-submodules of $M$ and $M=M_1\oplus M_2$ is clear: $$ ...


1

$\ker(f) \otimes N$ doesn't have to be a submodule of $\ker(f \otimes g)$. But there is a canonical homomorphism $\ker(f) \otimes N \to \ker(f \otimes g)$. Similarly, we get a canonical homomorphism on $M \otimes \ker(g)$. Hence, we get a canonical homomorphism $\ker(f) \otimes N \oplus M \otimes \ker(g) \to \ker(f \otimes g)$. This turns out to be ...


1

First of all, setting $r(m\otimes n)=mr\otimes n$ would have no chance in general of giving us a left $R$-module structure, since $M$ is a right $R$-module, so let's try defining $r(m\otimes n) = m\otimes rn$. In order for this to give a left $R$-module structure to $M\otimes N$, we need (at least) that the maps $\theta_r \colon M\times N \to M\otimes N$ ...


1

If I understood you correctly, we have (as $\;\Bbb Z$- modules = abelian groups) for any basic tensor $\;a\otimes b\in \Bbb Z_2\otimes\Bbb Z_2\;$: $$f\otimes1(a\otimes b):=f(a)\otimes b:=(2a)\times b=2(a\otimes b)=a\otimes (2b)=\ldots$$


1

Here is a way of going about it without the hint: Let $V$ be a finite dimensional $\mathbb{k}G$-module and $W \leq V$. Choose an idempotent $e \in \text{End}_\mathbb{k}(V)$ such that $eV=W$. Let $f=\frac{1}{|G|}\sum_{g \in G} geg^{-1}$. Then if $h \in G$, what is $hf$? Then what can we say about $f$? Now as $W \leq V$, we know what about $fV$ and ...


1

No, this is not true, but it is quite non-trivial to find examples. At least at the top of my head I only know the one below - does anyone know a simpler one? The entries of the vector $a_1 := (r_1,...,r_n)$ being coprime means that the map $$\alpha: R^n\xrightarrow{(r_1,...,r_n)} R$$ is surjective, so $P := \text{ker}(\alpha)$ is a rank projective ...


1

Notice that $g_M(y_1 - y_2) = 0$ so $y_1 - y_2$ has a preimage in $M_1$. This preimage maps onto $z^\ast$.


1

Let $A$ be a commutative ring with identity and $$ 0 \rightarrow N \xrightarrow{\alpha} M \xrightarrow{\beta} P \rightarrow 0 $$ be an exact sequence of $A$-modules and homomorphisms. Then the followings are equivalent: (i) There is an isomorphism of sequences with the sequence $$ 0 \rightarrow N \xrightarrow{i} N\oplus P \xrightarrow{\pi} P \rightarrow 0 ...


1

One direction is trivial: if $M=N\oplus N'$, then by definition any element $m\in M$ can be written in one and only one way as $m=x+y$, with $x\in N$ and $y\in N'$; verifying that $r\colon m\mapsto x$ is the homomorphism you need is very easy. Conversely, let $N'=\ker r$. Try to prove that $M=N+N'$; hint: write $m=r(m)+(m-r(m))$. $N\cap N'=\{0\}$. If ...



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