# Tag Info

3

Provided the answer to Tobias' question is yes, i.e. you take a standard embedding ${\mathfrak g}{\mathfrak l}(2)\subset{\mathfrak g}{\mathfrak l}(3)$, Properties (1) and (2) are preserved, but (3) is in general not: First, note that under the assumption of Properties (1) and (2), Property (3) is equivalent to $W$ being finitely generated over the ...

3

Use the third isomorphism theorem. If $G$ is a group (or ring, or module) and $H$ and $K$ are normal subgroups (or ideals, or submodules, respectively) of $G$, with $H\subseteq K$, then there is a natural isomorphism $$(G/H)/(K/H)\cong G/K.$$

2

Hint: Consider $\ker f$: it's an $R$-submodule of $R_R$. But the $R$-submodules of $R$ are its ideals. Solution: Alternatively, we can use your argument: If $mr=0$ with $r\ne0$, then $m=m1=m(r(1/r))=(mr)(1/r)=0$, which means that $f$ is the zero homomorphism.

2

Claim: If $P,Q$ are two projective generators of $R\text{-Mod}$, then there exist sets $I,J$ such that $P^{(I)}\cong Q^{(J)}$ Proof: You can use the Eilenberg-Mazur Swindle here: Suppose $P,Q$ are projective generators. Then, to begin, using the generator property of $P$, there exists a surjection $P^{(I_0)}\twoheadrightarrow X_0:=Q$. By projectivity of ...

2

You should not try assigning a representative to each $a \in A^n$ independently. Instead, let $\left(e_1,e_2,\ldots,e_n\right)$ be the standard basis of the $A$-module $A^n$. For each $i\in\left\{1,2,\ldots,n\right\}$, choose some $g_i \in A^n$ such that $f\left(\pi\left(e_i\right)\right) = \pi\left(g_i\right)$. Then, let $\widehat f : A^n \to A^n$ be the ...

1

Consider first of all a nonzero irreducible subrepresentation $W \subset V^{\oplus n}$. Each projection map $\pi_j: W \to V$ is either zero or an isomorphism, by Schur's Lemma. Choose $j_0$ so that $\pi_{j_0}$ is an isomorphism. For every $1 \le k \le n$, the map$$\pi_k \circ \pi_{j_0}^{-1}: V \to V$$is, by Schur's Lemma, a scalar; call this scalar ...

1

Your notation has ambiguity. Use R for real numbers and $R$ for the group ring, R$[G]$. Now any R-module is also an R-vector space. As we are working with finite-dimensional ones how about that $R$-module that has the least dimension as an R-vector space ?

1

Although this is the usual property that defines the faithfully flatness, (one of the) Liu's definition(s) which is helpful here is the following: Let $M$ be a flat $A$-module. Then $M$ is faithfully flat iff for any $A$-module $N$ such that $M\otimes_AN=0$ we have $N=0$. Recall that $N'\stackrel{u}\to N\stackrel{v}\to N''$ is exact iff ...

1

Indeed $M/N$ is a simple module, i.e. has no nontrivial submodules; hence $M/N =Rm$ for any $m \neq 0$ in $M/N$. Hence, as an $R$-module, $M/N$ is isomorphic to $R/I$ where $I = (N:M)$. But an $R$-module of the form $R/I$ is a simple module if and only if $I$ is maximal.

1

First, let us remind ourselves that $\mathbb Z$-modules are exactly the same thing as abelian groups. So we're dealing with abelian groups here. To avoid lots of cumbersome notation, we write $$\mathbb Z/n\mathbb Z =: C_n$$ in the following. Regarding simplicity, you're right. $C_n$ is simple iff $n$ is prime. A somewhat more detailed explanation could ...

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