(Milne CFT p. 36) Showing an $A$-module in an exact sequence is cyclic. The Next CEO of Stack OverflowQuestion about topology on $K^times$ in local CFTCharacterizing valuation rings related to a given principal ideal domain?Confusing application of power residue reciprocity in Milne's CFTA question on valuation overrings of a PIDProving A Discrete Valuation Ringsemi-direct product of subgroups of $D_6$local PID that is not a field is a DVRWhy is a discrete valuation ring's unique prime ideal generated by a non-nilpotent element?Hasse invariants under extension of scalarsequivalent characterizations of discrete valuation rings
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(Milne CFT p. 36) Showing an $A$-module in an exact sequence is cyclic.
The Next CEO of Stack OverflowQuestion about topology on $K^times$ in local CFTCharacterizing valuation rings related to a given principal ideal domain?Confusing application of power residue reciprocity in Milne's CFTA question on valuation overrings of a PIDProving A Discrete Valuation Ringsemi-direct product of subgroups of $D_6$local PID that is not a field is a DVRWhy is a discrete valuation ring's unique prime ideal generated by a non-nilpotent element?Hasse invariants under extension of scalarsequivalent characterizations of discrete valuation rings
$begingroup$
I am currently working on the following proof from Milne's Class Field Theory.
Here $K$ is a nonarchimedean local field, $A$ its valuation ring $A = alpha in K: $. So $A$ is a DVR and its maximal ideal $mathcalm = alpha$ is generated by some prime element $pi$. $A/ (pi)$ is a finite field with $q$ elements.
The highlighted text is the only part that is not clear to me. I can see that $M_n/M_1 simeq M_n-1$ by exactness, and by induction we may even assume that $M_1$ and $M_n-1$ are cyclic. $M_n$ decomposes into a direct sum of cyclic modules (since $A$ is a PID with a unique prime element), but why is it necessary that $M_n$ is cyclic? I feel like I am missing something very obvious and I would appreciate any help!
abstract-algebra algebraic-number-theory class-field-theory
asked Mar 19 at 12:20
vxnturevxnture
36910
$endgroup$
add a comment |
$begingroup$
I am currently working on the following proof from Milne's Class Field Theory.
Here $K$ is a nonarchimedean local field, $A$ its valuation ring $A = alpha in K: $. So $A$ is a DVR and its maximal ideal $mathcalm = alpha$ is generated by some prime element $pi$. $A/ (pi)$ is a finite field with $q$ elements.
The highlighted text is the only part that is not clear to me. I can see that $M_n/M_1 simeq M_n-1$ by exactness, and by induction we may even assume that $M_1$ and $M_n-1$ are cyclic. $M_n$ decomposes into a direct sum of cyclic modules (since $A$ is a PID with a unique prime element), but why is it necessary that $M_n$ is cyclic? I feel like I am missing something very obvious and I would appreciate any help!
abstract-algebra algebraic-number-theory class-field-theory
asked Mar 19 at 12:20
vxnturevxnture
36910
$endgroup$
2
$begingroup$
Write $M_nsimeq A/(pi^k_1)oplus A/(pi^k_2)opluscdots$ with $k_jge 1$. Then $$M_1 simeq (a_1pi^k_1-1,a_2pi^k_2 -1, cdots)mid a_jin A/(pi).$$ Agree? OK?
$endgroup$
– peter a g
Mar 19 at 12:41
$begingroup$
@peterag Oh, I see it! Thank you!
$endgroup$
– vxnture
Mar 19 at 12:49
$begingroup$
well, it's the idea. I see my notation was crummy ( namely $a_j in A/(pi)$)
$endgroup$
– peter a g
Mar 19 at 12:52
$begingroup$
@peterag Right it explains everything. What about thisone : let $M_infty = x in M, exists n, pi^n x = 0$. If it is non-trivial try to find some $y in M_infty, not in pi M$, $l = min l, pi^l y = 0$, if it exists and $M_infty supsetneq A y$ we have $|M_1| > q$. If such a $y$ doesn't exist then $M_infty$ contains a sequence $z_0 = 0, z_1 ne 0, z_n = pi z_n+1$ thus $N=lim_n to infty A z_n$ is isomorphic to $Frac(A)/A$ and if $M_infty supsetneq N$ then $|M_1| > q$. Thus $|M_1|=q$ means $M_infty cong A/(pi^l)$ or $M_infty cong Frac(A)/A$
$endgroup$
– reuns
Mar 19 at 14:07
$begingroup$
@reuns I think ok... You are doing away with the hypothesis that multiplication by $pi$ is onto (for $y$ to have a chance of existing)?
$endgroup$
– peter a g
Mar 19 at 16:07
add a comment |
$begingroup$
I am currently working on the following proof from Milne's Class Field Theory.
Here $K$ is a nonarchimedean local field, $A$ its valuation ring $A = alpha in K: $. So $A$ is a DVR and its maximal ideal $mathcalm = alpha$ is generated by some prime element $pi$. $A/ (pi)$ is a finite field with $q$ elements.
The highlighted text is the only part that is not clear to me. I can see that $M_n/M_1 simeq M_n-1$ by exactness, and by induction we may even assume that $M_1$ and $M_n-1$ are cyclic. $M_n$ decomposes into a direct sum of cyclic modules (since $A$ is a PID with a unique prime element), but why is it necessary that $M_n$ is cyclic? I feel like I am missing something very obvious and I would appreciate any help!
abstract-algebra algebraic-number-theory class-field-theory
asked Mar 19 at 12:20
vxnturevxnture
36910
$endgroup$
I am currently working on the following proof from Milne's Class Field Theory.
Here $K$ is a nonarchimedean local field, $A$ its valuation ring $A = alpha in K: $. So $A$ is a DVR and its maximal ideal $mathcalm = alpha$ is generated by some prime element $pi$. $A/ (pi)$ is a finite field with $q$ elements.
The highlighted text is the only part that is not clear to me. I can see that $M_n/M_1 simeq M_n-1$ by exactness, and by induction we may even assume that $M_1$ and $M_n-1$ are cyclic. $M_n$ decomposes into a direct sum of cyclic modules (since $A$ is a PID with a unique prime element), but why is it necessary that $M_n$ is cyclic? I feel like I am missing something very obvious and I would appreciate any help!
abstract-algebra algebraic-number-theory class-field-theory
abstract-algebra algebraic-number-theory class-field-theory
asked Mar 19 at 12:20
vxnturevxnture
36910
asked Mar 19 at 12:20
vxnturevxnture
36910
edited Mar 19 at 12:38
vxnture
asked Mar 19 at 12:20
vxnturevxnture
36910
asked Mar 19 at 12:20
vxnturevxnture
36910
asked Mar 19 at 12:20
vxnturevxnture
36910
36910
2
$begingroup$
Write $M_nsimeq A/(pi^k_1)oplus A/(pi^k_2)opluscdots$ with $k_jge 1$. Then $$M_1 simeq (a_1pi^k_1-1,a_2pi^k_2 -1, cdots)mid a_jin A/(pi).$$ Agree? OK?
$endgroup$
– peter a g
Mar 19 at 12:41
$begingroup$
@peterag Oh, I see it! Thank you!
$endgroup$
– vxnture
Mar 19 at 12:49
$begingroup$
well, it's the idea. I see my notation was crummy ( namely $a_j in A/(pi)$)
$endgroup$
– peter a g
Mar 19 at 12:52
$begingroup$
@peterag Right it explains everything. What about thisone : let $M_infty = x in M, exists n, pi^n x = 0$. If it is non-trivial try to find some $y in M_infty, not in pi M$, $l = min l, pi^l y = 0$, if it exists and $M_infty supsetneq A y$ we have $|M_1| > q$. If such a $y$ doesn't exist then $M_infty$ contains a sequence $z_0 = 0, z_1 ne 0, z_n = pi z_n+1$ thus $N=lim_n to infty A z_n$ is isomorphic to $Frac(A)/A$ and if $M_infty supsetneq N$ then $|M_1| > q$. Thus $|M_1|=q$ means $M_infty cong A/(pi^l)$ or $M_infty cong Frac(A)/A$
$endgroup$
– reuns
Mar 19 at 14:07
$begingroup$
@reuns I think ok... You are doing away with the hypothesis that multiplication by $pi$ is onto (for $y$ to have a chance of existing)?
$endgroup$
– peter a g
Mar 19 at 16:07
add a comment |
2
$begingroup$
Write $M_nsimeq A/(pi^k_1)oplus A/(pi^k_2)opluscdots$ with $k_jge 1$. Then $$M_1 simeq (a_1pi^k_1-1,a_2pi^k_2 -1, cdots)mid a_jin A/(pi).$$ Agree? OK?
$endgroup$
– peter a g
Mar 19 at 12:41
$begingroup$
@peterag Oh, I see it! Thank you!
$endgroup$
– vxnture
Mar 19 at 12:49
$begingroup$
well, it's the idea. I see my notation was crummy ( namely $a_j in A/(pi)$)
$endgroup$
– peter a g
Mar 19 at 12:52
$begingroup$
@peterag Right it explains everything. What about thisone : let $M_infty = x in M, exists n, pi^n x = 0$. If it is non-trivial try to find some $y in M_infty, not in pi M$, $l = min l, pi^l y = 0$, if it exists and $M_infty supsetneq A y$ we have $|M_1| > q$. If such a $y$ doesn't exist then $M_infty$ contains a sequence $z_0 = 0, z_1 ne 0, z_n = pi z_n+1$ thus $N=lim_n to infty A z_n$ is isomorphic to $Frac(A)/A$ and if $M_infty supsetneq N$ then $|M_1| > q$. Thus $|M_1|=q$ means $M_infty cong A/(pi^l)$ or $M_infty cong Frac(A)/A$
$endgroup$
– reuns
Mar 19 at 14:07
$begingroup$
@reuns I think ok... You are doing away with the hypothesis that multiplication by $pi$ is onto (for $y$ to have a chance of existing)?
$endgroup$
– peter a g
Mar 19 at 16:07
2
2
$begingroup$
Write $M_nsimeq A/(pi^k_1)oplus A/(pi^k_2)opluscdots$ with $k_jge 1$. Then $$M_1 simeq (a_1pi^k_1-1,a_2pi^k_2 -1, cdots)mid a_jin A/(pi).$$ Agree? OK?
$endgroup$
– peter a g
Mar 19 at 12:41
$begingroup$
Write $M_nsimeq A/(pi^k_1)oplus A/(pi^k_2)opluscdots$ with $k_jge 1$. Then $$M_1 simeq (a_1pi^k_1-1,a_2pi^k_2 -1, cdots)mid a_jin A/(pi).$$ Agree? OK?
$endgroup$
– peter a g
Mar 19 at 12:41
$begingroup$
@peterag Oh, I see it! Thank you!
$endgroup$
– vxnture
Mar 19 at 12:49
$begingroup$
@peterag Oh, I see it! Thank you!
$endgroup$
– vxnture
Mar 19 at 12:49
$begingroup$
well, it's the idea. I see my notation was crummy ( namely $a_j in A/(pi)$)
$endgroup$
– peter a g
Mar 19 at 12:52
$begingroup$
well, it's the idea. I see my notation was crummy ( namely $a_j in A/(pi)$)
$endgroup$
– peter a g
Mar 19 at 12:52
$begingroup$
@peterag Right it explains everything. What about thisone : let $M_infty = x in M, exists n, pi^n x = 0$. If it is non-trivial try to find some $y in M_infty, not in pi M$, $l = min l, pi^l y = 0$, if it exists and $M_infty supsetneq A y$ we have $|M_1| > q$. If such a $y$ doesn't exist then $M_infty$ contains a sequence $z_0 = 0, z_1 ne 0, z_n = pi z_n+1$ thus $N=lim_n to infty A z_n$ is isomorphic to $Frac(A)/A$ and if $M_infty supsetneq N$ then $|M_1| > q$. Thus $|M_1|=q$ means $M_infty cong A/(pi^l)$ or $M_infty cong Frac(A)/A$
$endgroup$
– reuns
Mar 19 at 14:07
$begingroup$
@peterag Right it explains everything. What about thisone : let $M_infty = x in M, exists n, pi^n x = 0$. If it is non-trivial try to find some $y in M_infty, not in pi M$, $l = min l, pi^l y = 0$, if it exists and $M_infty supsetneq A y$ we have $|M_1| > q$. If such a $y$ doesn't exist then $M_infty$ contains a sequence $z_0 = 0, z_1 ne 0, z_n = pi z_n+1$ thus $N=lim_n to infty A z_n$ is isomorphic to $Frac(A)/A$ and if $M_infty supsetneq N$ then $|M_1| > q$. Thus $|M_1|=q$ means $M_infty cong A/(pi^l)$ or $M_infty cong Frac(A)/A$
$endgroup$
– reuns
Mar 19 at 14:07
$begingroup$
@reuns I think ok... You are doing away with the hypothesis that multiplication by $pi$ is onto (for $y$ to have a chance of existing)?
$endgroup$
– peter a g
Mar 19 at 16:07
$begingroup$
@reuns I think ok... You are doing away with the hypothesis that multiplication by $pi$ is onto (for $y$ to have a chance of existing)?
$endgroup$
– peter a g
Mar 19 at 16:07
add a comment |
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2
$begingroup$
Write $M_nsimeq A/(pi^k_1)oplus A/(pi^k_2)opluscdots$ with $k_jge 1$. Then $$M_1 simeq (a_1pi^k_1-1,a_2pi^k_2 -1, cdots)mid a_jin A/(pi).$$ Agree? OK?
$endgroup$
– peter a g
Mar 19 at 12:41
$begingroup$
@peterag Oh, I see it! Thank you!
$endgroup$
– vxnture
Mar 19 at 12:49
$begingroup$
well, it's the idea. I see my notation was crummy ( namely $a_j in A/(pi)$)
$endgroup$
– peter a g
Mar 19 at 12:52
$begingroup$
@peterag Right it explains everything. What about thisone : let $M_infty = x in M, exists n, pi^n x = 0$. If it is non-trivial try to find some $y in M_infty, not in pi M$, $l = min l, pi^l y = 0$, if it exists and $M_infty supsetneq A y$ we have $|M_1| > q$. If such a $y$ doesn't exist then $M_infty$ contains a sequence $z_0 = 0, z_1 ne 0, z_n = pi z_n+1$ thus $N=lim_n to infty A z_n$ is isomorphic to $Frac(A)/A$ and if $M_infty supsetneq N$ then $|M_1| > q$. Thus $|M_1|=q$ means $M_infty cong A/(pi^l)$ or $M_infty cong Frac(A)/A$
$endgroup$
– reuns
Mar 19 at 14:07
$begingroup$
@reuns I think ok... You are doing away with the hypothesis that multiplication by $pi$ is onto (for $y$ to have a chance of existing)?
$endgroup$
– peter a g
Mar 19 at 16:07