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Apr
14
awarded  Popular Question
Mar
28
comment If $K \leq H \leq G$, show that $[G:K] = [G:H][H:K]$.
$G = \cup_{a \in A} Ha = \cup_{a \in A, b \in B} (Kb)a = \cup_{a \in A, b \in B} Kba$, where A is a transversal for $H$ in $G$, and $B$ is a transversal for $K$ in $H$. Of course $|A| = [G:H]$ and $|B| = [H:K]$. You just need to show that the sets $Kba$ are distinct.
Mar
5
awarded  Popular Question
Oct
12
awarded  Yearling
Jul
21
accepted Proof details of Theorem 11.1 in Atiyah-Macdonald
Jul
20
comment Proof details of Theorem 11.1 in Atiyah-Macdonald
Thank you very much for these references. I was very confused, as for example Lang seems to give the same proof as AM and there $g(t)$ also appears. I'll go through the proofs in the links you posted in the morning and then accept your answer.
Jul
20
comment Proof details of Theorem 11.1 in Atiyah-Macdonald
There must be something simple that I am missing here. Given sequences $0 \rightarrow K_n \rightarrow M_n \xrightarrow{x_s} M_{n+k_s} \rightarrow L_{n+k_s} \rightarrow 0$ I would have that $L_{n+k_s} = M_{n+k_s}/x_sM_n$ by exactness. But then what do I take for $L_n$ where $n = 0, 1, \ldots, k_s-1$? I understood that $L$ is the cokernel of the map on $M$ given by multiplication by $x_s$.
Jul
19
comment Proof details of Theorem 11.1 in Atiyah-Macdonald
@YACP I am sorry, I do not really understand. Perhaps I am wrong in the assumption that $L_n = M_n$ for $n = 0, 1, \ldots, k_s-1$?
Jul
19
asked Proof details of Theorem 11.1 in Atiyah-Macdonald
Jun
15
accepted Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product.
Jun
15
comment Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product.
Oh, that was indeed very simple. Thanks.
Jun
15
asked Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product.
May
10
awarded  Nice Question
Aug
16
awarded  Yearling
Jul
11
comment Exercise on modules over PID involving injective modules, Baer's criterion.
@navigetor Exercise is from Hungerford. It is self-study, I complement it with Robert Ash's text Basic Abstract Algebra.
Jul
11
comment Exercise on modules over PID involving injective modules, Baer's criterion.
Thanks for reading it. I sort of thought it was too long for anybody to care enough to get through it. Indeed author leads the reader into a solution, but I need some confirmation I somewhat understood the material correctly, as similar concepts occur frequently later in the book.
Jul
10
revised Exercise on modules over PID involving injective modules, Baer's criterion.
deleted 39 characters in body
Jul
9
asked Exercise on modules over PID involving injective modules, Baer's criterion.
Jun
8
awarded  Caucus
Apr
3
accepted Infinite nilpotent group, any normal subgroup intersects the center nontrivially