Units in $R[X]$, where $R=Bbb Z/p^2qBbb Z$ [closed]
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Let $p$ and $q$ be two prime numbers and let $R=Bbb Z/p^2qBbb Z$. find units in $R[X]$. i am not getting how to do such type of problems....
abstract-algebra group-theory
edited Nov 13 at 18:35
cansomeonehelpmeout
6,4583834
asked Nov 13 at 18:29
Ravi Agarwal
161
closed as off-topic by Morgan Rodgers, A. Pongrácz, Alexander Gruber♦ Nov 13 at 19:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, A. Pongrácz, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
Let $p$ and $q$ be two prime numbers and let $R=Bbb Z/p^2qBbb Z$. find units in $R[X]$. i am not getting how to do such type of problems....
abstract-algebra group-theory
edited Nov 13 at 18:35
cansomeonehelpmeout
6,4583834
asked Nov 13 at 18:29
Ravi Agarwal
161
closed as off-topic by Morgan Rodgers, A. Pongrácz, Alexander Gruber♦ Nov 13 at 19:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, A. Pongrácz, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
What have you tried? What do you know about the problem?
– faith_in_facts
Nov 13 at 18:34
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $p$ and $q$ be two prime numbers and let $R=Bbb Z/p^2qBbb Z$. find units in $R[X]$. i am not getting how to do such type of problems....
abstract-algebra group-theory
edited Nov 13 at 18:35
cansomeonehelpmeout
6,4583834
asked Nov 13 at 18:29
Ravi Agarwal
161
Let $p$ and $q$ be two prime numbers and let $R=Bbb Z/p^2qBbb Z$. find units in $R[X]$. i am not getting how to do such type of problems....
abstract-algebra group-theory
abstract-algebra group-theory
edited Nov 13 at 18:35
cansomeonehelpmeout
6,4583834
asked Nov 13 at 18:29
Ravi Agarwal
161
edited Nov 13 at 18:35
cansomeonehelpmeout
6,4583834
asked Nov 13 at 18:29
Ravi Agarwal
161
edited Nov 13 at 18:35
cansomeonehelpmeout
6,4583834
edited Nov 13 at 18:35
cansomeonehelpmeout
6,4583834
edited Nov 13 at 18:35
cansomeonehelpmeout
6,4583834
6,4583834
asked Nov 13 at 18:29
Ravi Agarwal
161
asked Nov 13 at 18:29
Ravi Agarwal
161
asked Nov 13 at 18:29
Ravi Agarwal
161
161
closed as off-topic by Morgan Rodgers, A. Pongrácz, Alexander Gruber♦ Nov 13 at 19:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, A. Pongrácz, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Morgan Rodgers, A. Pongrácz, Alexander Gruber♦ Nov 13 at 19:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, A. Pongrácz, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
What have you tried? What do you know about the problem?
– faith_in_facts
Nov 13 at 18:34
add a comment |
What have you tried? What do you know about the problem?
– faith_in_facts
Nov 13 at 18:34
What have you tried? What do you know about the problem?
– faith_in_facts
Nov 13 at 18:34
What have you tried? What do you know about the problem?
– faith_in_facts
Nov 13 at 18:34
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
Hint
All units of $R$ will definitely be there as units in $R[x]$. But the presence of nilpotent elements in $R=Bbb{Z}/p^2qBbb{Z}$ can give more units. In fact, an element like $1+ax$, where $a$ is a nilpotent element in $R$ is a unit in $R[x]$.
For example:
$a=pq$ is nilpotent in $R$ (because $a^2=0$). So $1+ax$ is a unit of $R[x]$. The reason being
$$(1+pqx)(1-pqx)=1-(pqx)^2=1.$$
General idea: unit + nilpotent=unit.
answered Nov 13 at 18:43
Anurag A
24.7k12249
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
1
@RaviAgarwal In this particular ring: you have $0,1,pq,kpq$ (where $k in {2,3,4, ldots p-1}$) are all nilpotent elements. Any nilpotent element (apart from $0$ and $1$) in this ring should have both $p$ and $q$ as prime divisors because you want $a^n equiv 0 pmod{p^2q}$ for some $n geq 1$.
– Anurag A
Nov 14 at 4:45
1
@RaviAgarwal Just to clarify, the units need not be only degree $1$ polynomials. For example $1+pqx+pqx^2$ is also a unit because $(1+pqx+pqx^2)(1-pqx-pqx^2)=1$.
– Anurag A
Nov 14 at 4:56
add a comment |
up vote
1
down vote
Whenever you have to deal with $(mathbb Z / nmathbb Z)^times$, 'always' use the Chinese Remainder Theorem. I'm assuming $p,q$ are coprime. This reduces the problem to understanding units in $(mathbb Z/p^2 mathbb Z)^times$ and $(mathbb Z/q mathbb Z)^times$. Hence, you are after numbers that are coprime to $p$ and also to $q$.
answered Nov 13 at 18:47
Richard Martin
1,4618
OP is looking for units in the polynomial ring and not in the coefficient ring. So your answer doesn't help.
– Anurag A
Nov 13 at 18:48
Oh yes well spotted; but actually it does help. If it isn't a unit in the coefficient ring then it certainly won't be one in the polynomial ring either.
– Richard Martin
Nov 13 at 18:52
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint
All units of $R$ will definitely be there as units in $R[x]$. But the presence of nilpotent elements in $R=Bbb{Z}/p^2qBbb{Z}$ can give more units. In fact, an element like $1+ax$, where $a$ is a nilpotent element in $R$ is a unit in $R[x]$.
For example:
$a=pq$ is nilpotent in $R$ (because $a^2=0$). So $1+ax$ is a unit of $R[x]$. The reason being
$$(1+pqx)(1-pqx)=1-(pqx)^2=1.$$
General idea: unit + nilpotent=unit.
answered Nov 13 at 18:43
Anurag A
24.7k12249
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
1
@RaviAgarwal In this particular ring: you have $0,1,pq,kpq$ (where $k in {2,3,4, ldots p-1}$) are all nilpotent elements. Any nilpotent element (apart from $0$ and $1$) in this ring should have both $p$ and $q$ as prime divisors because you want $a^n equiv 0 pmod{p^2q}$ for some $n geq 1$.
– Anurag A
Nov 14 at 4:45
1
@RaviAgarwal Just to clarify, the units need not be only degree $1$ polynomials. For example $1+pqx+pqx^2$ is also a unit because $(1+pqx+pqx^2)(1-pqx-pqx^2)=1$.
– Anurag A
Nov 14 at 4:56
add a comment |
up vote
3
down vote
Hint
All units of $R$ will definitely be there as units in $R[x]$. But the presence of nilpotent elements in $R=Bbb{Z}/p^2qBbb{Z}$ can give more units. In fact, an element like $1+ax$, where $a$ is a nilpotent element in $R$ is a unit in $R[x]$.
For example:
$a=pq$ is nilpotent in $R$ (because $a^2=0$). So $1+ax$ is a unit of $R[x]$. The reason being
$$(1+pqx)(1-pqx)=1-(pqx)^2=1.$$
General idea: unit + nilpotent=unit.
answered Nov 13 at 18:43
Anurag A
24.7k12249
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
1
@RaviAgarwal In this particular ring: you have $0,1,pq,kpq$ (where $k in {2,3,4, ldots p-1}$) are all nilpotent elements. Any nilpotent element (apart from $0$ and $1$) in this ring should have both $p$ and $q$ as prime divisors because you want $a^n equiv 0 pmod{p^2q}$ for some $n geq 1$.
– Anurag A
Nov 14 at 4:45
1
@RaviAgarwal Just to clarify, the units need not be only degree $1$ polynomials. For example $1+pqx+pqx^2$ is also a unit because $(1+pqx+pqx^2)(1-pqx-pqx^2)=1$.
– Anurag A
Nov 14 at 4:56
add a comment |
up vote
3
down vote
up vote
3
down vote
Hint
All units of $R$ will definitely be there as units in $R[x]$. But the presence of nilpotent elements in $R=Bbb{Z}/p^2qBbb{Z}$ can give more units. In fact, an element like $1+ax$, where $a$ is a nilpotent element in $R$ is a unit in $R[x]$.
For example:
$a=pq$ is nilpotent in $R$ (because $a^2=0$). So $1+ax$ is a unit of $R[x]$. The reason being
$$(1+pqx)(1-pqx)=1-(pqx)^2=1.$$
General idea: unit + nilpotent=unit.
answered Nov 13 at 18:43
Anurag A
24.7k12249
Hint
All units of $R$ will definitely be there as units in $R[x]$. But the presence of nilpotent elements in $R=Bbb{Z}/p^2qBbb{Z}$ can give more units. In fact, an element like $1+ax$, where $a$ is a nilpotent element in $R$ is a unit in $R[x]$.
For example:
$a=pq$ is nilpotent in $R$ (because $a^2=0$). So $1+ax$ is a unit of $R[x]$. The reason being
$$(1+pqx)(1-pqx)=1-(pqx)^2=1.$$
General idea: unit + nilpotent=unit.
answered Nov 13 at 18:43
Anurag A
24.7k12249
edited Nov 13 at 18:59
answered Nov 13 at 18:43
Anurag A
24.7k12249
answered Nov 13 at 18:43
Anurag A
24.7k12249
answered Nov 13 at 18:43
Anurag A
24.7k12249
24.7k12249
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
1
@RaviAgarwal In this particular ring: you have $0,1,pq,kpq$ (where $k in {2,3,4, ldots p-1}$) are all nilpotent elements. Any nilpotent element (apart from $0$ and $1$) in this ring should have both $p$ and $q$ as prime divisors because you want $a^n equiv 0 pmod{p^2q}$ for some $n geq 1$.
– Anurag A
Nov 14 at 4:45
1
@RaviAgarwal Just to clarify, the units need not be only degree $1$ polynomials. For example $1+pqx+pqx^2$ is also a unit because $(1+pqx+pqx^2)(1-pqx-pqx^2)=1$.
– Anurag A
Nov 14 at 4:56
add a comment |
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
1
@RaviAgarwal In this particular ring: you have $0,1,pq,kpq$ (where $k in {2,3,4, ldots p-1}$) are all nilpotent elements. Any nilpotent element (apart from $0$ and $1$) in this ring should have both $p$ and $q$ as prime divisors because you want $a^n equiv 0 pmod{p^2q}$ for some $n geq 1$.
– Anurag A
Nov 14 at 4:45
1
@RaviAgarwal Just to clarify, the units need not be only degree $1$ polynomials. For example $1+pqx+pqx^2$ is also a unit because $(1+pqx+pqx^2)(1-pqx-pqx^2)=1$.
– Anurag A
Nov 14 at 4:56
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
is there any other nilpotent elements in R
– Ravi Agarwal
Nov 14 at 3:59
1
1
@RaviAgarwal In this particular ring: you have $0,1,pq,kpq$ (where $k in {2,3,4, ldots p-1}$) are all nilpotent elements. Any nilpotent element (apart from $0$ and $1$) in this ring should have both $p$ and $q$ as prime divisors because you want $a^n equiv 0 pmod{p^2q}$ for some $n geq 1$.
– Anurag A
Nov 14 at 4:45
@RaviAgarwal In this particular ring: you have $0,1,pq,kpq$ (where $k in {2,3,4, ldots p-1}$) are all nilpotent elements. Any nilpotent element (apart from $0$ and $1$) in this ring should have both $p$ and $q$ as prime divisors because you want $a^n equiv 0 pmod{p^2q}$ for some $n geq 1$.
– Anurag A
Nov 14 at 4:45
1
1
@RaviAgarwal Just to clarify, the units need not be only degree $1$ polynomials. For example $1+pqx+pqx^2$ is also a unit because $(1+pqx+pqx^2)(1-pqx-pqx^2)=1$.
– Anurag A
Nov 14 at 4:56
@RaviAgarwal Just to clarify, the units need not be only degree $1$ polynomials. For example $1+pqx+pqx^2$ is also a unit because $(1+pqx+pqx^2)(1-pqx-pqx^2)=1$.
– Anurag A
Nov 14 at 4:56
add a comment |
up vote
1
down vote
Whenever you have to deal with $(mathbb Z / nmathbb Z)^times$, 'always' use the Chinese Remainder Theorem. I'm assuming $p,q$ are coprime. This reduces the problem to understanding units in $(mathbb Z/p^2 mathbb Z)^times$ and $(mathbb Z/q mathbb Z)^times$. Hence, you are after numbers that are coprime to $p$ and also to $q$.
answered Nov 13 at 18:47
Richard Martin
1,4618
OP is looking for units in the polynomial ring and not in the coefficient ring. So your answer doesn't help.
– Anurag A
Nov 13 at 18:48
Oh yes well spotted; but actually it does help. If it isn't a unit in the coefficient ring then it certainly won't be one in the polynomial ring either.
– Richard Martin
Nov 13 at 18:52
add a comment |
up vote
1
down vote
Whenever you have to deal with $(mathbb Z / nmathbb Z)^times$, 'always' use the Chinese Remainder Theorem. I'm assuming $p,q$ are coprime. This reduces the problem to understanding units in $(mathbb Z/p^2 mathbb Z)^times$ and $(mathbb Z/q mathbb Z)^times$. Hence, you are after numbers that are coprime to $p$ and also to $q$.
answered Nov 13 at 18:47
Richard Martin
1,4618
OP is looking for units in the polynomial ring and not in the coefficient ring. So your answer doesn't help.
– Anurag A
Nov 13 at 18:48
Oh yes well spotted; but actually it does help. If it isn't a unit in the coefficient ring then it certainly won't be one in the polynomial ring either.
– Richard Martin
Nov 13 at 18:52
add a comment |
up vote
1
down vote
up vote
1
down vote
Whenever you have to deal with $(mathbb Z / nmathbb Z)^times$, 'always' use the Chinese Remainder Theorem. I'm assuming $p,q$ are coprime. This reduces the problem to understanding units in $(mathbb Z/p^2 mathbb Z)^times$ and $(mathbb Z/q mathbb Z)^times$. Hence, you are after numbers that are coprime to $p$ and also to $q$.
answered Nov 13 at 18:47
Richard Martin
1,4618
Whenever you have to deal with $(mathbb Z / nmathbb Z)^times$, 'always' use the Chinese Remainder Theorem. I'm assuming $p,q$ are coprime. This reduces the problem to understanding units in $(mathbb Z/p^2 mathbb Z)^times$ and $(mathbb Z/q mathbb Z)^times$. Hence, you are after numbers that are coprime to $p$ and also to $q$.
answered Nov 13 at 18:47
Richard Martin
1,4618
answered Nov 13 at 18:47
Richard Martin
1,4618
answered Nov 13 at 18:47
Richard Martin
1,4618
answered Nov 13 at 18:47
Richard Martin
1,4618
1,4618
OP is looking for units in the polynomial ring and not in the coefficient ring. So your answer doesn't help.
– Anurag A
Nov 13 at 18:48
Oh yes well spotted; but actually it does help. If it isn't a unit in the coefficient ring then it certainly won't be one in the polynomial ring either.
– Richard Martin
Nov 13 at 18:52
add a comment |
OP is looking for units in the polynomial ring and not in the coefficient ring. So your answer doesn't help.
– Anurag A
Nov 13 at 18:48
Oh yes well spotted; but actually it does help. If it isn't a unit in the coefficient ring then it certainly won't be one in the polynomial ring either.
– Richard Martin
Nov 13 at 18:52
OP is looking for units in the polynomial ring and not in the coefficient ring. So your answer doesn't help.
– Anurag A
Nov 13 at 18:48
OP is looking for units in the polynomial ring and not in the coefficient ring. So your answer doesn't help.
– Anurag A
Nov 13 at 18:48
Oh yes well spotted; but actually it does help. If it isn't a unit in the coefficient ring then it certainly won't be one in the polynomial ring either.
– Richard Martin
Nov 13 at 18:52
Oh yes well spotted; but actually it does help. If it isn't a unit in the coefficient ring then it certainly won't be one in the polynomial ring either.
– Richard Martin
Nov 13 at 18:52
add a comment |
What have you tried? What do you know about the problem?
– faith_in_facts
Nov 13 at 18:34