Mixing
Problem of modules over PID, Exercise $9.13(2)$ in Rotman's Advanced Modern Algebra $2$nd Edition
$begingroup$
Rotman's Advanced Modern Algebra ($3$rd Edition), Exercise B-$3.24(2)$, also Exercise $9.13(2)$ in $2$nd Edition.
$R$ is PID, $M$ is $P$-primary $R$-module. $P=(p)$ is non-zero prime ideal. $k_P:=R/P$ is field.
Define $M[p]:={min M, pm=0 }$ and $V_P(n,M):=text{dim}_{k_P}frac{p^nM cap M[p]}{p^{n+1}M cap M[p]}.$
Suppose $M$ is direct sum of cyclic modules $C_i$, prove the number of summands $C_i$ having order ideal/annihilator $color{blue}{(p^{n+1})}$ is $V_P(n,M)$.
Attention: The book ($3$rd Edition) originally writes "$ldots$$C_i$ having order ideal/annihilator $(p^{n})$ is $V_P(n,M)$", but that's wrong and should be $color{red}{(p^{n+1})}$ instead.
I wrote my thoughts in my answer below.
abstract-algebra modules homological-algebra
asked Dec 31 '18 at 14:41
AndrewsAndrews
3591317
$endgroup$
add a comment |
$begingroup$
Rotman's Advanced Modern Algebra ($3$rd Edition), Exercise B-$3.24(2)$, also Exercise $9.13(2)$ in $2$nd Edition.
$R$ is PID, $M$ is $P$-primary $R$-module. $P=(p)$ is non-zero prime ideal. $k_P:=R/P$ is field.
Define $M[p]:={min M, pm=0 }$ and $V_P(n,M):=text{dim}_{k_P}frac{p^nM cap M[p]}{p^{n+1}M cap M[p]}.$
Suppose $M$ is direct sum of cyclic modules $C_i$, prove the number of summands $C_i$ having order ideal/annihilator $color{blue}{(p^{n+1})}$ is $V_P(n,M)$.
Attention: The book ($3$rd Edition) originally writes "$ldots$$C_i$ having order ideal/annihilator $(p^{n})$ is $V_P(n,M)$", but that's wrong and should be $color{red}{(p^{n+1})}$ instead.
I wrote my thoughts in my answer below.
abstract-algebra modules homological-algebra
asked Dec 31 '18 at 14:41
AndrewsAndrews
3591317
$endgroup$
$begingroup$
What is (1) that you're referencing?
$endgroup$
– jgon
Dec 31 '18 at 19:23
$begingroup$
Ok, found a copy of the textbook, but I'm not sure where the notation $U_P(n,M)$ is from. That said, I think I have a better idea of what you're asking.
$endgroup$
– jgon
Dec 31 '18 at 19:29
$begingroup$
@jgon Sorry for not showing what $U_P(n,M)$ means. I found some of my thoughts are entirely wrong and some topos, now they're deleted or corrected.
$endgroup$
– Andrews
Dec 31 '18 at 19:50
add a comment |
$begingroup$
Rotman's Advanced Modern Algebra ($3$rd Edition), Exercise B-$3.24(2)$, also Exercise $9.13(2)$ in $2$nd Edition.
$R$ is PID, $M$ is $P$-primary $R$-module. $P=(p)$ is non-zero prime ideal. $k_P:=R/P$ is field.
Define $M[p]:={min M, pm=0 }$ and $V_P(n,M):=text{dim}_{k_P}frac{p^nM cap M[p]}{p^{n+1}M cap M[p]}.$
Suppose $M$ is direct sum of cyclic modules $C_i$, prove the number of summands $C_i$ having order ideal/annihilator $color{blue}{(p^{n+1})}$ is $V_P(n,M)$.
Attention: The book ($3$rd Edition) originally writes "$ldots$$C_i$ having order ideal/annihilator $(p^{n})$ is $V_P(n,M)$", but that's wrong and should be $color{red}{(p^{n+1})}$ instead.
I wrote my thoughts in my answer below.
abstract-algebra modules homological-algebra
asked Dec 31 '18 at 14:41
AndrewsAndrews
3591317
$endgroup$
Rotman's Advanced Modern Algebra ($3$rd Edition), Exercise B-$3.24(2)$, also Exercise $9.13(2)$ in $2$nd Edition.
$R$ is PID, $M$ is $P$-primary $R$-module. $P=(p)$ is non-zero prime ideal. $k_P:=R/P$ is field.
Define $M[p]:={min M, pm=0 }$ and $V_P(n,M):=text{dim}_{k_P}frac{p^nM cap M[p]}{p^{n+1}M cap M[p]}.$
Suppose $M$ is direct sum of cyclic modules $C_i$, prove the number of summands $C_i$ having order ideal/annihilator $color{blue}{(p^{n+1})}$ is $V_P(n,M)$.
Attention: The book ($3$rd Edition) originally writes "$ldots$$C_i$ having order ideal/annihilator $(p^{n})$ is $V_P(n,M)$", but that's wrong and should be $color{red}{(p^{n+1})}$ instead.
I wrote my thoughts in my answer below.
abstract-algebra modules homological-algebra
abstract-algebra modules homological-algebra
asked Dec 31 '18 at 14:41
AndrewsAndrews
3591317
asked Dec 31 '18 at 14:41
AndrewsAndrews
3591317
edited Jan 1 at 19:33
Andrews
asked Dec 31 '18 at 14:41
AndrewsAndrews
3591317
asked Dec 31 '18 at 14:41
AndrewsAndrews
3591317
asked Dec 31 '18 at 14:41
AndrewsAndrews
3591317
3591317
$begingroup$
What is (1) that you're referencing?
$endgroup$
– jgon
Dec 31 '18 at 19:23
$begingroup$
Ok, found a copy of the textbook, but I'm not sure where the notation $U_P(n,M)$ is from. That said, I think I have a better idea of what you're asking.
$endgroup$
– jgon
Dec 31 '18 at 19:29
$begingroup$
@jgon Sorry for not showing what $U_P(n,M)$ means. I found some of my thoughts are entirely wrong and some topos, now they're deleted or corrected.
$endgroup$
– Andrews
Dec 31 '18 at 19:50
add a comment |
$begingroup$
What is (1) that you're referencing?
$endgroup$
– jgon
Dec 31 '18 at 19:23
$begingroup$
Ok, found a copy of the textbook, but I'm not sure where the notation $U_P(n,M)$ is from. That said, I think I have a better idea of what you're asking.
$endgroup$
– jgon
Dec 31 '18 at 19:29
$begingroup$
@jgon Sorry for not showing what $U_P(n,M)$ means. I found some of my thoughts are entirely wrong and some topos, now they're deleted or corrected.
$endgroup$
– Andrews
Dec 31 '18 at 19:50
$begingroup$
What is (1) that you're referencing?
$endgroup$
– jgon
Dec 31 '18 at 19:23
$begingroup$
What is (1) that you're referencing?
$endgroup$
– jgon
Dec 31 '18 at 19:23
$begingroup$
Ok, found a copy of the textbook, but I'm not sure where the notation $U_P(n,M)$ is from. That said, I think I have a better idea of what you're asking.
$endgroup$
– jgon
Dec 31 '18 at 19:29
$begingroup$
Ok, found a copy of the textbook, but I'm not sure where the notation $U_P(n,M)$ is from. That said, I think I have a better idea of what you're asking.
$endgroup$
– jgon
Dec 31 '18 at 19:29
$begingroup$
@jgon Sorry for not showing what $U_P(n,M)$ means. I found some of my thoughts are entirely wrong and some topos, now they're deleted or corrected.
$endgroup$
– Andrews
Dec 31 '18 at 19:50
$begingroup$
@jgon Sorry for not showing what $U_P(n,M)$ means. I found some of my thoughts are entirely wrong and some topos, now they're deleted or corrected.
$endgroup$
– Andrews
Dec 31 '18 at 19:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A few things. I found the second edition online, so I'm working from that.
I found your question a little unclear, so the first part of this is me working through it trying to figure out precisely what you're asking and why. Edit It has since been clarified, and the contradiction resolved.
Regardless of your work, there is an obvious contradiction in the statements.
We have
$U_P(n,M)=#{text{cyclic summands w/ order $p^{n+1}$}}$ (Theorem 9.14)
$U_P(n,M)=V_P(n,M)$ for $M$ f.g. (Ex 9.13 (i))
$V_P(n,M)=#{text{cyclic summands w/ order $p^{n}$}}$ (Ex 9.14 (ii))
Now for a finitely generated module $M$, this presents a clear contradiction. This is I presume why you asked the question, you find yourself in the situation of having apparently proved a contradiction, so you need to resolve this situation.
Now as with most textbooks, minor typos are not uncommon. It's not likely for statement 2 to contain a typo, so let's see if we can find an example showing statements 1 or 3 to be false.
Definitions
We recall the definitions:
$U_P(n,M)=d(p^nM)-d(p^{n+1}M)$, where $d(M)=dim(M/pM)$ (definition immediately after corollary 9.13)- $$V_P(n,M)=dim frac{p^nMcap M[p]}{p^{n+1}Mcap M[p]}$$
Consider $M=R/P$. Then $d(p^nM)=delta_{0n}$, so $U_P(n,M)=d(p^nM)-d(p^{n+1}M)=delta_{0n}-delta_{0,n+1}=delta_{0n}$. Thus statement 1 holds.
On the other hand, $M[p]=M$, so $V_P(n,M)=dim(p^nM/p^{n+1}M)=d(p^nM)=d(p^nM)-d(p^{n+1}M)=delta_{0n}$, since $d(p^{n+1}M)=0$ for all $nge 0$. Thus statement 3 is violated.
How I would prove the corrected version of (3)
Note that everything is linear across direct sums, so it suffices to prove the statement for a cyclic module.
Consider $M=R/P^{e+1}$. We want to show that $V_P(n,M)=delta{ne}$.
Now $M[p]=P^e/P^{e+1}$, and $p^nM = P^n/P^{e+1}$ when $n le e$ and $0$ otherwise.
Then $p^nMcap M[p]= P^e/P^{e+1}$ for $nle e$, and $0$ otherwise.
Hence if $n < e$, $$p^nMcap M[p] = P^e/P^{e+1} = p^{n+1}Mcap M[p],$$
if $n > e$, then
$$p^nMcap M[p] = 0 = p^{n+1}Mcap M[p],$$
and if $n=e$, then
$$p^nMcap M[p] = P^e/P^{e+1},text{ but } p^{n+1}Mcap M[p]=0,$$
so $V_P(e,M)=1dim P^e/P^{e+1}=1$.
Thus $V_P(n,M)=delta_{ne}$ as desired.
answered Dec 31 '18 at 20:01
jgonjgon
13.5k22041
$endgroup$
$begingroup$
Thanks for your kindly explaination. As I mentioned, it should be $(p^{n+1})$ instead of $(p^n)$ in the book. I deleted it because I worked it out.
$endgroup$
– Andrews
Dec 31 '18 at 21:55
$begingroup$
@Andrews, yes I realized that, but I'd already written almost the entire answer by that point, so I edited in a comment and left the rest of my answer as is
$endgroup$
– jgon
Dec 31 '18 at 21:58
add a comment |
$begingroup$
Suppose $M=M_P cong bigoplus_{e_mgeq1} R/P^{e_m},$ then $M[p]cong bigoplus_{e_ngeq0}P^{e_n}/P^{e_n+1}.$
$p^nM = bigoplus_{e_mgeq1} P^n/P^{e_m}cong bigoplus_{e_mgeq n+1} P^n/P^{e_m}$.
$p^nM cap M[p]cong bigoplus_{e_mgeq n+1}P^{e_m-1}/P^{e_m}$.
$p^{n+1}M = bigoplus_{e_mgeq1} P^{n+1}/P^{e_m}cong bigoplus_{e_mgeq n+2} P^n/P^{e_m}$.
$p^{n+1}M cap M[p]cong bigoplus_{e_mgeq n+2}P^{e_m-1}/P^{e_m}$.
Thus $(p^nM cap M[p])/(p^{n+1}M cap M[p])congbigoplus_{e_m=n+1}P^{e_m-1}/P^{e_m}cong(R/P)^{oplus #{e_m|e_m=n+1}}$.
dim$(p^nM cap M[p])/(p^{n+1}M cap M[p]) = V_P(n,M)=#{e_m|e_m=n+1}$.
Thus there're $ V_P(n,M)$ $R/P^{n+1}$ as direct summands in $M_P$.
answered Dec 31 '18 at 22:23
AndrewsAndrews
3591317
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
A few things. I found the second edition online, so I'm working from that.
I found your question a little unclear, so the first part of this is me working through it trying to figure out precisely what you're asking and why. Edit It has since been clarified, and the contradiction resolved.
Regardless of your work, there is an obvious contradiction in the statements.
We have
$U_P(n,M)=#{text{cyclic summands w/ order $p^{n+1}$}}$ (Theorem 9.14)
$U_P(n,M)=V_P(n,M)$ for $M$ f.g. (Ex 9.13 (i))
$V_P(n,M)=#{text{cyclic summands w/ order $p^{n}$}}$ (Ex 9.14 (ii))
Now for a finitely generated module $M$, this presents a clear contradiction. This is I presume why you asked the question, you find yourself in the situation of having apparently proved a contradiction, so you need to resolve this situation.
Now as with most textbooks, minor typos are not uncommon. It's not likely for statement 2 to contain a typo, so let's see if we can find an example showing statements 1 or 3 to be false.
Definitions
We recall the definitions:
$U_P(n,M)=d(p^nM)-d(p^{n+1}M)$, where $d(M)=dim(M/pM)$ (definition immediately after corollary 9.13)- $$V_P(n,M)=dim frac{p^nMcap M[p]}{p^{n+1}Mcap M[p]}$$
Consider $M=R/P$. Then $d(p^nM)=delta_{0n}$, so $U_P(n,M)=d(p^nM)-d(p^{n+1}M)=delta_{0n}-delta_{0,n+1}=delta_{0n}$. Thus statement 1 holds.
On the other hand, $M[p]=M$, so $V_P(n,M)=dim(p^nM/p^{n+1}M)=d(p^nM)=d(p^nM)-d(p^{n+1}M)=delta_{0n}$, since $d(p^{n+1}M)=0$ for all $nge 0$. Thus statement 3 is violated.
How I would prove the corrected version of (3)
Note that everything is linear across direct sums, so it suffices to prove the statement for a cyclic module.
Consider $M=R/P^{e+1}$. We want to show that $V_P(n,M)=delta{ne}$.
Now $M[p]=P^e/P^{e+1}$, and $p^nM = P^n/P^{e+1}$ when $n le e$ and $0$ otherwise.
Then $p^nMcap M[p]= P^e/P^{e+1}$ for $nle e$, and $0$ otherwise.
Hence if $n < e$, $$p^nMcap M[p] = P^e/P^{e+1} = p^{n+1}Mcap M[p],$$
if $n > e$, then
$$p^nMcap M[p] = 0 = p^{n+1}Mcap M[p],$$
and if $n=e$, then
$$p^nMcap M[p] = P^e/P^{e+1},text{ but } p^{n+1}Mcap M[p]=0,$$
so $V_P(e,M)=1dim P^e/P^{e+1}=1$.
Thus $V_P(n,M)=delta_{ne}$ as desired.
answered Dec 31 '18 at 20:01
jgonjgon
13.5k22041
$endgroup$
$begingroup$
Thanks for your kindly explaination. As I mentioned, it should be $(p^{n+1})$ instead of $(p^n)$ in the book. I deleted it because I worked it out.
$endgroup$
– Andrews
Dec 31 '18 at 21:55
$begingroup$
@Andrews, yes I realized that, but I'd already written almost the entire answer by that point, so I edited in a comment and left the rest of my answer as is
$endgroup$
– jgon
Dec 31 '18 at 21:58
add a comment |
$begingroup$
A few things. I found the second edition online, so I'm working from that.
I found your question a little unclear, so the first part of this is me working through it trying to figure out precisely what you're asking and why. Edit It has since been clarified, and the contradiction resolved.
Regardless of your work, there is an obvious contradiction in the statements.
We have
$U_P(n,M)=#{text{cyclic summands w/ order $p^{n+1}$}}$ (Theorem 9.14)
$U_P(n,M)=V_P(n,M)$ for $M$ f.g. (Ex 9.13 (i))
$V_P(n,M)=#{text{cyclic summands w/ order $p^{n}$}}$ (Ex 9.14 (ii))
Now for a finitely generated module $M$, this presents a clear contradiction. This is I presume why you asked the question, you find yourself in the situation of having apparently proved a contradiction, so you need to resolve this situation.
Now as with most textbooks, minor typos are not uncommon. It's not likely for statement 2 to contain a typo, so let's see if we can find an example showing statements 1 or 3 to be false.
Definitions
We recall the definitions:
$U_P(n,M)=d(p^nM)-d(p^{n+1}M)$, where $d(M)=dim(M/pM)$ (definition immediately after corollary 9.13)- $$V_P(n,M)=dim frac{p^nMcap M[p]}{p^{n+1}Mcap M[p]}$$
Consider $M=R/P$. Then $d(p^nM)=delta_{0n}$, so $U_P(n,M)=d(p^nM)-d(p^{n+1}M)=delta_{0n}-delta_{0,n+1}=delta_{0n}$. Thus statement 1 holds.
On the other hand, $M[p]=M$, so $V_P(n,M)=dim(p^nM/p^{n+1}M)=d(p^nM)=d(p^nM)-d(p^{n+1}M)=delta_{0n}$, since $d(p^{n+1}M)=0$ for all $nge 0$. Thus statement 3 is violated.
How I would prove the corrected version of (3)
Note that everything is linear across direct sums, so it suffices to prove the statement for a cyclic module.
Consider $M=R/P^{e+1}$. We want to show that $V_P(n,M)=delta{ne}$.
Now $M[p]=P^e/P^{e+1}$, and $p^nM = P^n/P^{e+1}$ when $n le e$ and $0$ otherwise.
Then $p^nMcap M[p]= P^e/P^{e+1}$ for $nle e$, and $0$ otherwise.
Hence if $n < e$, $$p^nMcap M[p] = P^e/P^{e+1} = p^{n+1}Mcap M[p],$$
if $n > e$, then
$$p^nMcap M[p] = 0 = p^{n+1}Mcap M[p],$$
and if $n=e$, then
$$p^nMcap M[p] = P^e/P^{e+1},text{ but } p^{n+1}Mcap M[p]=0,$$
so $V_P(e,M)=1dim P^e/P^{e+1}=1$.
Thus $V_P(n,M)=delta_{ne}$ as desired.
answered Dec 31 '18 at 20:01
jgonjgon
13.5k22041
$endgroup$
$begingroup$
Thanks for your kindly explaination. As I mentioned, it should be $(p^{n+1})$ instead of $(p^n)$ in the book. I deleted it because I worked it out.
$endgroup$
– Andrews
Dec 31 '18 at 21:55
$begingroup$
@Andrews, yes I realized that, but I'd already written almost the entire answer by that point, so I edited in a comment and left the rest of my answer as is
$endgroup$
– jgon
Dec 31 '18 at 21:58
add a comment |
$begingroup$
A few things. I found the second edition online, so I'm working from that.
I found your question a little unclear, so the first part of this is me working through it trying to figure out precisely what you're asking and why. Edit It has since been clarified, and the contradiction resolved.
Regardless of your work, there is an obvious contradiction in the statements.
We have
$U_P(n,M)=#{text{cyclic summands w/ order $p^{n+1}$}}$ (Theorem 9.14)
$U_P(n,M)=V_P(n,M)$ for $M$ f.g. (Ex 9.13 (i))
$V_P(n,M)=#{text{cyclic summands w/ order $p^{n}$}}$ (Ex 9.14 (ii))
Now for a finitely generated module $M$, this presents a clear contradiction. This is I presume why you asked the question, you find yourself in the situation of having apparently proved a contradiction, so you need to resolve this situation.
Now as with most textbooks, minor typos are not uncommon. It's not likely for statement 2 to contain a typo, so let's see if we can find an example showing statements 1 or 3 to be false.
Definitions
We recall the definitions:
$U_P(n,M)=d(p^nM)-d(p^{n+1}M)$, where $d(M)=dim(M/pM)$ (definition immediately after corollary 9.13)- $$V_P(n,M)=dim frac{p^nMcap M[p]}{p^{n+1}Mcap M[p]}$$
Consider $M=R/P$. Then $d(p^nM)=delta_{0n}$, so $U_P(n,M)=d(p^nM)-d(p^{n+1}M)=delta_{0n}-delta_{0,n+1}=delta_{0n}$. Thus statement 1 holds.
On the other hand, $M[p]=M$, so $V_P(n,M)=dim(p^nM/p^{n+1}M)=d(p^nM)=d(p^nM)-d(p^{n+1}M)=delta_{0n}$, since $d(p^{n+1}M)=0$ for all $nge 0$. Thus statement 3 is violated.
How I would prove the corrected version of (3)
Note that everything is linear across direct sums, so it suffices to prove the statement for a cyclic module.
Consider $M=R/P^{e+1}$. We want to show that $V_P(n,M)=delta{ne}$.
Now $M[p]=P^e/P^{e+1}$, and $p^nM = P^n/P^{e+1}$ when $n le e$ and $0$ otherwise.
Then $p^nMcap M[p]= P^e/P^{e+1}$ for $nle e$, and $0$ otherwise.
Hence if $n < e$, $$p^nMcap M[p] = P^e/P^{e+1} = p^{n+1}Mcap M[p],$$
if $n > e$, then
$$p^nMcap M[p] = 0 = p^{n+1}Mcap M[p],$$
and if $n=e$, then
$$p^nMcap M[p] = P^e/P^{e+1},text{ but } p^{n+1}Mcap M[p]=0,$$
so $V_P(e,M)=1dim P^e/P^{e+1}=1$.
Thus $V_P(n,M)=delta_{ne}$ as desired.
answered Dec 31 '18 at 20:01
jgonjgon
13.5k22041
$endgroup$
A few things. I found the second edition online, so I'm working from that.
I found your question a little unclear, so the first part of this is me working through it trying to figure out precisely what you're asking and why. Edit It has since been clarified, and the contradiction resolved.
Regardless of your work, there is an obvious contradiction in the statements.
We have
$U_P(n,M)=#{text{cyclic summands w/ order $p^{n+1}$}}$ (Theorem 9.14)
$U_P(n,M)=V_P(n,M)$ for $M$ f.g. (Ex 9.13 (i))
$V_P(n,M)=#{text{cyclic summands w/ order $p^{n}$}}$ (Ex 9.14 (ii))
Now for a finitely generated module $M$, this presents a clear contradiction. This is I presume why you asked the question, you find yourself in the situation of having apparently proved a contradiction, so you need to resolve this situation.
Now as with most textbooks, minor typos are not uncommon. It's not likely for statement 2 to contain a typo, so let's see if we can find an example showing statements 1 or 3 to be false.
Definitions
We recall the definitions:
$U_P(n,M)=d(p^nM)-d(p^{n+1}M)$, where $d(M)=dim(M/pM)$ (definition immediately after corollary 9.13)- $$V_P(n,M)=dim frac{p^nMcap M[p]}{p^{n+1}Mcap M[p]}$$
Consider $M=R/P$. Then $d(p^nM)=delta_{0n}$, so $U_P(n,M)=d(p^nM)-d(p^{n+1}M)=delta_{0n}-delta_{0,n+1}=delta_{0n}$. Thus statement 1 holds.
On the other hand, $M[p]=M$, so $V_P(n,M)=dim(p^nM/p^{n+1}M)=d(p^nM)=d(p^nM)-d(p^{n+1}M)=delta_{0n}$, since $d(p^{n+1}M)=0$ for all $nge 0$. Thus statement 3 is violated.
How I would prove the corrected version of (3)
Note that everything is linear across direct sums, so it suffices to prove the statement for a cyclic module.
Consider $M=R/P^{e+1}$. We want to show that $V_P(n,M)=delta{ne}$.
Now $M[p]=P^e/P^{e+1}$, and $p^nM = P^n/P^{e+1}$ when $n le e$ and $0$ otherwise.
Then $p^nMcap M[p]= P^e/P^{e+1}$ for $nle e$, and $0$ otherwise.
Hence if $n < e$, $$p^nMcap M[p] = P^e/P^{e+1} = p^{n+1}Mcap M[p],$$
if $n > e$, then
$$p^nMcap M[p] = 0 = p^{n+1}Mcap M[p],$$
and if $n=e$, then
$$p^nMcap M[p] = P^e/P^{e+1},text{ but } p^{n+1}Mcap M[p]=0,$$
so $V_P(e,M)=1dim P^e/P^{e+1}=1$.
Thus $V_P(n,M)=delta_{ne}$ as desired.
answered Dec 31 '18 at 20:01
jgonjgon
13.5k22041
answered Dec 31 '18 at 20:01
jgonjgon
13.5k22041
answered Dec 31 '18 at 20:01
jgonjgon
13.5k22041
answered Dec 31 '18 at 20:01
jgonjgon
13.5k22041
13.5k22041
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Thanks for your kindly explaination. As I mentioned, it should be $(p^{n+1})$ instead of $(p^n)$ in the book. I deleted it because I worked it out.
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– Andrews
Dec 31 '18 at 21:55
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@Andrews, yes I realized that, but I'd already written almost the entire answer by that point, so I edited in a comment and left the rest of my answer as is
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– jgon
Dec 31 '18 at 21:58
add a comment |
$begingroup$
Thanks for your kindly explaination. As I mentioned, it should be $(p^{n+1})$ instead of $(p^n)$ in the book. I deleted it because I worked it out.
$endgroup$
– Andrews
Dec 31 '18 at 21:55
$begingroup$
@Andrews, yes I realized that, but I'd already written almost the entire answer by that point, so I edited in a comment and left the rest of my answer as is
$endgroup$
– jgon
Dec 31 '18 at 21:58
$begingroup$
Thanks for your kindly explaination. As I mentioned, it should be $(p^{n+1})$ instead of $(p^n)$ in the book. I deleted it because I worked it out.
$endgroup$
– Andrews
Dec 31 '18 at 21:55
$begingroup$
Thanks for your kindly explaination. As I mentioned, it should be $(p^{n+1})$ instead of $(p^n)$ in the book. I deleted it because I worked it out.
$endgroup$
– Andrews
Dec 31 '18 at 21:55
$begingroup$
@Andrews, yes I realized that, but I'd already written almost the entire answer by that point, so I edited in a comment and left the rest of my answer as is
$endgroup$
– jgon
Dec 31 '18 at 21:58
$begingroup$
@Andrews, yes I realized that, but I'd already written almost the entire answer by that point, so I edited in a comment and left the rest of my answer as is
$endgroup$
– jgon
Dec 31 '18 at 21:58
add a comment |
$begingroup$
Suppose $M=M_P cong bigoplus_{e_mgeq1} R/P^{e_m},$ then $M[p]cong bigoplus_{e_ngeq0}P^{e_n}/P^{e_n+1}.$
$p^nM = bigoplus_{e_mgeq1} P^n/P^{e_m}cong bigoplus_{e_mgeq n+1} P^n/P^{e_m}$.
$p^nM cap M[p]cong bigoplus_{e_mgeq n+1}P^{e_m-1}/P^{e_m}$.
$p^{n+1}M = bigoplus_{e_mgeq1} P^{n+1}/P^{e_m}cong bigoplus_{e_mgeq n+2} P^n/P^{e_m}$.
$p^{n+1}M cap M[p]cong bigoplus_{e_mgeq n+2}P^{e_m-1}/P^{e_m}$.
Thus $(p^nM cap M[p])/(p^{n+1}M cap M[p])congbigoplus_{e_m=n+1}P^{e_m-1}/P^{e_m}cong(R/P)^{oplus #{e_m|e_m=n+1}}$.
dim$(p^nM cap M[p])/(p^{n+1}M cap M[p]) = V_P(n,M)=#{e_m|e_m=n+1}$.
Thus there're $ V_P(n,M)$ $R/P^{n+1}$ as direct summands in $M_P$.
answered Dec 31 '18 at 22:23
AndrewsAndrews
3591317
$endgroup$
add a comment |
$begingroup$
Suppose $M=M_P cong bigoplus_{e_mgeq1} R/P^{e_m},$ then $M[p]cong bigoplus_{e_ngeq0}P^{e_n}/P^{e_n+1}.$
$p^nM = bigoplus_{e_mgeq1} P^n/P^{e_m}cong bigoplus_{e_mgeq n+1} P^n/P^{e_m}$.
$p^nM cap M[p]cong bigoplus_{e_mgeq n+1}P^{e_m-1}/P^{e_m}$.
$p^{n+1}M = bigoplus_{e_mgeq1} P^{n+1}/P^{e_m}cong bigoplus_{e_mgeq n+2} P^n/P^{e_m}$.
$p^{n+1}M cap M[p]cong bigoplus_{e_mgeq n+2}P^{e_m-1}/P^{e_m}$.
Thus $(p^nM cap M[p])/(p^{n+1}M cap M[p])congbigoplus_{e_m=n+1}P^{e_m-1}/P^{e_m}cong(R/P)^{oplus #{e_m|e_m=n+1}}$.
dim$(p^nM cap M[p])/(p^{n+1}M cap M[p]) = V_P(n,M)=#{e_m|e_m=n+1}$.
Thus there're $ V_P(n,M)$ $R/P^{n+1}$ as direct summands in $M_P$.
answered Dec 31 '18 at 22:23
AndrewsAndrews
3591317
$endgroup$
add a comment |
$begingroup$
Suppose $M=M_P cong bigoplus_{e_mgeq1} R/P^{e_m},$ then $M[p]cong bigoplus_{e_ngeq0}P^{e_n}/P^{e_n+1}.$
$p^nM = bigoplus_{e_mgeq1} P^n/P^{e_m}cong bigoplus_{e_mgeq n+1} P^n/P^{e_m}$.
$p^nM cap M[p]cong bigoplus_{e_mgeq n+1}P^{e_m-1}/P^{e_m}$.
$p^{n+1}M = bigoplus_{e_mgeq1} P^{n+1}/P^{e_m}cong bigoplus_{e_mgeq n+2} P^n/P^{e_m}$.
$p^{n+1}M cap M[p]cong bigoplus_{e_mgeq n+2}P^{e_m-1}/P^{e_m}$.
Thus $(p^nM cap M[p])/(p^{n+1}M cap M[p])congbigoplus_{e_m=n+1}P^{e_m-1}/P^{e_m}cong(R/P)^{oplus #{e_m|e_m=n+1}}$.
dim$(p^nM cap M[p])/(p^{n+1}M cap M[p]) = V_P(n,M)=#{e_m|e_m=n+1}$.
Thus there're $ V_P(n,M)$ $R/P^{n+1}$ as direct summands in $M_P$.
answered Dec 31 '18 at 22:23
AndrewsAndrews
3591317
$endgroup$
Suppose $M=M_P cong bigoplus_{e_mgeq1} R/P^{e_m},$ then $M[p]cong bigoplus_{e_ngeq0}P^{e_n}/P^{e_n+1}.$
$p^nM = bigoplus_{e_mgeq1} P^n/P^{e_m}cong bigoplus_{e_mgeq n+1} P^n/P^{e_m}$.
$p^nM cap M[p]cong bigoplus_{e_mgeq n+1}P^{e_m-1}/P^{e_m}$.
$p^{n+1}M = bigoplus_{e_mgeq1} P^{n+1}/P^{e_m}cong bigoplus_{e_mgeq n+2} P^n/P^{e_m}$.
$p^{n+1}M cap M[p]cong bigoplus_{e_mgeq n+2}P^{e_m-1}/P^{e_m}$.
Thus $(p^nM cap M[p])/(p^{n+1}M cap M[p])congbigoplus_{e_m=n+1}P^{e_m-1}/P^{e_m}cong(R/P)^{oplus #{e_m|e_m=n+1}}$.
dim$(p^nM cap M[p])/(p^{n+1}M cap M[p]) = V_P(n,M)=#{e_m|e_m=n+1}$.
Thus there're $ V_P(n,M)$ $R/P^{n+1}$ as direct summands in $M_P$.
answered Dec 31 '18 at 22:23
AndrewsAndrews
3591317
answered Dec 31 '18 at 22:23
AndrewsAndrews
3591317
answered Dec 31 '18 at 22:23
AndrewsAndrews
3591317
answered Dec 31 '18 at 22:23
AndrewsAndrews
3591317
3591317
add a comment |
add a comment |
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$begingroup$
What is (1) that you're referencing?
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– jgon
Dec 31 '18 at 19:23
$begingroup$
Ok, found a copy of the textbook, but I'm not sure where the notation $U_P(n,M)$ is from. That said, I think I have a better idea of what you're asking.
$endgroup$
– jgon
Dec 31 '18 at 19:29
$begingroup$
@jgon Sorry for not showing what $U_P(n,M)$ means. I found some of my thoughts are entirely wrong and some topos, now they're deleted or corrected.
$endgroup$
– Andrews
Dec 31 '18 at 19:50