$mathbb{Z}, mathbb{Z}_2$ and $mathbb{Q}[x]$ - rings
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I'm quite new in the theory of rings and hope some of you can help me in tackling some problems in understanding, which occured when I tried to solve a problem.
The problem is about the rings $mathbb{Z}_{2}, mathbb{Z}_{2}[x]$ and $mathbb{Q}[x] = {sum_{j=0}^{n}a_{j} x^{j} ~| ~a_{j} in mathbb{Q}, n in mathbb{N} }$
First I need to find the unities, so i need to find $ {e in R : e |1} $.
In $mathbb{Z}_2$ it is easy: $E[mathbb{Z}_2] = {-1,1}$. I guess $E[mathbb{Z}_2[x]] = {1}$ but I can't justify why (I did calculatios for polynomials up to degree 3, but this of course is no proof) . For the latter case, $E[mathbb{Q}[x]]$ I don't have a clue how to proof, that every polynomial with constant coefficient $a_0 = 1$ is a unit (I've found this on the Internet). I really can't imagine how every polynomial with constant $a_0$ should be a unit - at least I wasn't able to find even one pair of unities, which give 1 when multiplied.
Is there an easy way to show this two facts?
Furthermore I have to find out which of the given rings are isomorphic. According to a hint it has to do something with the unities. Does an isomorphism map unities to unities? If yes, non of them can be isomorphic, due to their number is different. Is that true?
Finally I should decide whether $mathbb{Q}[x]$ is a factorial ring. Here again I got a hint; Find $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$.
At these final point I really have no idea how to start, but maybe this might get better, when I understand the previous points?
I would really appreciate if some of you can help me understandig this task.
Thanks!
abstract-algebra ring-theory factoring
add a comment |
up vote
1
down vote
favorite
I'm quite new in the theory of rings and hope some of you can help me in tackling some problems in understanding, which occured when I tried to solve a problem.
The problem is about the rings $mathbb{Z}_{2}, mathbb{Z}_{2}[x]$ and $mathbb{Q}[x] = {sum_{j=0}^{n}a_{j} x^{j} ~| ~a_{j} in mathbb{Q}, n in mathbb{N} }$
First I need to find the unities, so i need to find $ {e in R : e |1} $.
In $mathbb{Z}_2$ it is easy: $E[mathbb{Z}_2] = {-1,1}$. I guess $E[mathbb{Z}_2[x]] = {1}$ but I can't justify why (I did calculatios for polynomials up to degree 3, but this of course is no proof) . For the latter case, $E[mathbb{Q}[x]]$ I don't have a clue how to proof, that every polynomial with constant coefficient $a_0 = 1$ is a unit (I've found this on the Internet). I really can't imagine how every polynomial with constant $a_0$ should be a unit - at least I wasn't able to find even one pair of unities, which give 1 when multiplied.
Is there an easy way to show this two facts?
Furthermore I have to find out which of the given rings are isomorphic. According to a hint it has to do something with the unities. Does an isomorphism map unities to unities? If yes, non of them can be isomorphic, due to their number is different. Is that true?
Finally I should decide whether $mathbb{Q}[x]$ is a factorial ring. Here again I got a hint; Find $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$.
At these final point I really have no idea how to start, but maybe this might get better, when I understand the previous points?
I would really appreciate if some of you can help me understandig this task.
Thanks!
abstract-algebra ring-theory factoring
You're confusing $mathbb{Q}[x]$ and $mathbb{Q}[[x]]$. The former consists of polynomials, the latter power series. In $mathbb{Q}[[x]]$ every polynomial with nonzero constant term is a unit, but in $mathbb{Q}[x]$ only the nonzero constant polynomials are units.
– Slade
Dec 1 at 16:24
Hi Slade! You're right - I'm a little bit confused, because I also thought, that the latter one denotes power series, but the problem description defines it as $mathbb{Q}[[x]] = {sum_{j=0}^{n} a_j x^{j} }$, so something I would call polynomials...
– noNickname
2 days ago
But I guess they mean power series, because I really can't imagine, that finding $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$ is possible for polynomials...
– noNickname
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm quite new in the theory of rings and hope some of you can help me in tackling some problems in understanding, which occured when I tried to solve a problem.
The problem is about the rings $mathbb{Z}_{2}, mathbb{Z}_{2}[x]$ and $mathbb{Q}[x] = {sum_{j=0}^{n}a_{j} x^{j} ~| ~a_{j} in mathbb{Q}, n in mathbb{N} }$
First I need to find the unities, so i need to find $ {e in R : e |1} $.
In $mathbb{Z}_2$ it is easy: $E[mathbb{Z}_2] = {-1,1}$. I guess $E[mathbb{Z}_2[x]] = {1}$ but I can't justify why (I did calculatios for polynomials up to degree 3, but this of course is no proof) . For the latter case, $E[mathbb{Q}[x]]$ I don't have a clue how to proof, that every polynomial with constant coefficient $a_0 = 1$ is a unit (I've found this on the Internet). I really can't imagine how every polynomial with constant $a_0$ should be a unit - at least I wasn't able to find even one pair of unities, which give 1 when multiplied.
Is there an easy way to show this two facts?
Furthermore I have to find out which of the given rings are isomorphic. According to a hint it has to do something with the unities. Does an isomorphism map unities to unities? If yes, non of them can be isomorphic, due to their number is different. Is that true?
Finally I should decide whether $mathbb{Q}[x]$ is a factorial ring. Here again I got a hint; Find $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$.
At these final point I really have no idea how to start, but maybe this might get better, when I understand the previous points?
I would really appreciate if some of you can help me understandig this task.
Thanks!
abstract-algebra ring-theory factoring
I'm quite new in the theory of rings and hope some of you can help me in tackling some problems in understanding, which occured when I tried to solve a problem.
The problem is about the rings $mathbb{Z}_{2}, mathbb{Z}_{2}[x]$ and $mathbb{Q}[x] = {sum_{j=0}^{n}a_{j} x^{j} ~| ~a_{j} in mathbb{Q}, n in mathbb{N} }$
First I need to find the unities, so i need to find $ {e in R : e |1} $.
In $mathbb{Z}_2$ it is easy: $E[mathbb{Z}_2] = {-1,1}$. I guess $E[mathbb{Z}_2[x]] = {1}$ but I can't justify why (I did calculatios for polynomials up to degree 3, but this of course is no proof) . For the latter case, $E[mathbb{Q}[x]]$ I don't have a clue how to proof, that every polynomial with constant coefficient $a_0 = 1$ is a unit (I've found this on the Internet). I really can't imagine how every polynomial with constant $a_0$ should be a unit - at least I wasn't able to find even one pair of unities, which give 1 when multiplied.
Is there an easy way to show this two facts?
Furthermore I have to find out which of the given rings are isomorphic. According to a hint it has to do something with the unities. Does an isomorphism map unities to unities? If yes, non of them can be isomorphic, due to their number is different. Is that true?
Finally I should decide whether $mathbb{Q}[x]$ is a factorial ring. Here again I got a hint; Find $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$.
At these final point I really have no idea how to start, but maybe this might get better, when I understand the previous points?
I would really appreciate if some of you can help me understandig this task.
Thanks!
abstract-algebra ring-theory factoring
abstract-algebra ring-theory factoring
asked Dec 1 at 13:30
noNickname
353
353
You're confusing $mathbb{Q}[x]$ and $mathbb{Q}[[x]]$. The former consists of polynomials, the latter power series. In $mathbb{Q}[[x]]$ every polynomial with nonzero constant term is a unit, but in $mathbb{Q}[x]$ only the nonzero constant polynomials are units.
– Slade
Dec 1 at 16:24
Hi Slade! You're right - I'm a little bit confused, because I also thought, that the latter one denotes power series, but the problem description defines it as $mathbb{Q}[[x]] = {sum_{j=0}^{n} a_j x^{j} }$, so something I would call polynomials...
– noNickname
2 days ago
But I guess they mean power series, because I really can't imagine, that finding $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$ is possible for polynomials...
– noNickname
2 days ago
add a comment |
You're confusing $mathbb{Q}[x]$ and $mathbb{Q}[[x]]$. The former consists of polynomials, the latter power series. In $mathbb{Q}[[x]]$ every polynomial with nonzero constant term is a unit, but in $mathbb{Q}[x]$ only the nonzero constant polynomials are units.
– Slade
Dec 1 at 16:24
Hi Slade! You're right - I'm a little bit confused, because I also thought, that the latter one denotes power series, but the problem description defines it as $mathbb{Q}[[x]] = {sum_{j=0}^{n} a_j x^{j} }$, so something I would call polynomials...
– noNickname
2 days ago
But I guess they mean power series, because I really can't imagine, that finding $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$ is possible for polynomials...
– noNickname
2 days ago
You're confusing $mathbb{Q}[x]$ and $mathbb{Q}[[x]]$. The former consists of polynomials, the latter power series. In $mathbb{Q}[[x]]$ every polynomial with nonzero constant term is a unit, but in $mathbb{Q}[x]$ only the nonzero constant polynomials are units.
– Slade
Dec 1 at 16:24
You're confusing $mathbb{Q}[x]$ and $mathbb{Q}[[x]]$. The former consists of polynomials, the latter power series. In $mathbb{Q}[[x]]$ every polynomial with nonzero constant term is a unit, but in $mathbb{Q}[x]$ only the nonzero constant polynomials are units.
– Slade
Dec 1 at 16:24
Hi Slade! You're right - I'm a little bit confused, because I also thought, that the latter one denotes power series, but the problem description defines it as $mathbb{Q}[[x]] = {sum_{j=0}^{n} a_j x^{j} }$, so something I would call polynomials...
– noNickname
2 days ago
Hi Slade! You're right - I'm a little bit confused, because I also thought, that the latter one denotes power series, but the problem description defines it as $mathbb{Q}[[x]] = {sum_{j=0}^{n} a_j x^{j} }$, so something I would call polynomials...
– noNickname
2 days ago
But I guess they mean power series, because I really can't imagine, that finding $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$ is possible for polynomials...
– noNickname
2 days ago
But I guess they mean power series, because I really can't imagine, that finding $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$ is possible for polynomials...
– noNickname
2 days ago
add a comment |
1 Answer
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1
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Let $K[x]$ be a polynomial ring over a field $K$.
Then a polynomial $fin K[x]$ is invertible if there is a polynomial $gin K[x]$ such that $fg=1$. Now the degree formula comes in: $${rm deg}(f) + {rm deg}(g) = {rm deg}(fg) = {rm deg}(1) = 0.$$ The first equality comes from the fact that the highest terms of $f$ and $g$ multiply while their degrees add; there is no cancellation of the highest terms since $K$ is a field.
Since the degrees of $f,g$ are non-negative and sum up to 0, both need to be zero and so $f,g$ are elements of $K$.
It follows that the unit group of $K[x]$ is $K^*=Ksetminus{0}$.
Hi Wuestenfux! Thanks for this hint! It's kind of obvious in retrospective, but I really didn't had the degree-law in mind, when I tried to solve this example. Is it possible to draw a conclusion about isomorphies between this three rings, with the knowlege about unities? So does the number of unities has to be the same, if the rings are isomorphic or something like this?
– noNickname
2 days ago
Units map to units in a ring isomorphism.
– Wuestenfux
2 days ago
So non of them can be isomorphic, due to the number of units differ? The unit in $mathbb{Z}_2[x]$ is only $1$, the units in $mathbb{Z}$ are ${-1,1}$ and the units in $mathbb{Q}[[x]]$ are infinitelly many, due the field is $mathbb{Q}$.
– noNickname
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $K[x]$ be a polynomial ring over a field $K$.
Then a polynomial $fin K[x]$ is invertible if there is a polynomial $gin K[x]$ such that $fg=1$. Now the degree formula comes in: $${rm deg}(f) + {rm deg}(g) = {rm deg}(fg) = {rm deg}(1) = 0.$$ The first equality comes from the fact that the highest terms of $f$ and $g$ multiply while their degrees add; there is no cancellation of the highest terms since $K$ is a field.
Since the degrees of $f,g$ are non-negative and sum up to 0, both need to be zero and so $f,g$ are elements of $K$.
It follows that the unit group of $K[x]$ is $K^*=Ksetminus{0}$.
Hi Wuestenfux! Thanks for this hint! It's kind of obvious in retrospective, but I really didn't had the degree-law in mind, when I tried to solve this example. Is it possible to draw a conclusion about isomorphies between this three rings, with the knowlege about unities? So does the number of unities has to be the same, if the rings are isomorphic or something like this?
– noNickname
2 days ago
Units map to units in a ring isomorphism.
– Wuestenfux
2 days ago
So non of them can be isomorphic, due to the number of units differ? The unit in $mathbb{Z}_2[x]$ is only $1$, the units in $mathbb{Z}$ are ${-1,1}$ and the units in $mathbb{Q}[[x]]$ are infinitelly many, due the field is $mathbb{Q}$.
– noNickname
2 days ago
add a comment |
up vote
1
down vote
Let $K[x]$ be a polynomial ring over a field $K$.
Then a polynomial $fin K[x]$ is invertible if there is a polynomial $gin K[x]$ such that $fg=1$. Now the degree formula comes in: $${rm deg}(f) + {rm deg}(g) = {rm deg}(fg) = {rm deg}(1) = 0.$$ The first equality comes from the fact that the highest terms of $f$ and $g$ multiply while their degrees add; there is no cancellation of the highest terms since $K$ is a field.
Since the degrees of $f,g$ are non-negative and sum up to 0, both need to be zero and so $f,g$ are elements of $K$.
It follows that the unit group of $K[x]$ is $K^*=Ksetminus{0}$.
Hi Wuestenfux! Thanks for this hint! It's kind of obvious in retrospective, but I really didn't had the degree-law in mind, when I tried to solve this example. Is it possible to draw a conclusion about isomorphies between this three rings, with the knowlege about unities? So does the number of unities has to be the same, if the rings are isomorphic or something like this?
– noNickname
2 days ago
Units map to units in a ring isomorphism.
– Wuestenfux
2 days ago
So non of them can be isomorphic, due to the number of units differ? The unit in $mathbb{Z}_2[x]$ is only $1$, the units in $mathbb{Z}$ are ${-1,1}$ and the units in $mathbb{Q}[[x]]$ are infinitelly many, due the field is $mathbb{Q}$.
– noNickname
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $K[x]$ be a polynomial ring over a field $K$.
Then a polynomial $fin K[x]$ is invertible if there is a polynomial $gin K[x]$ such that $fg=1$. Now the degree formula comes in: $${rm deg}(f) + {rm deg}(g) = {rm deg}(fg) = {rm deg}(1) = 0.$$ The first equality comes from the fact that the highest terms of $f$ and $g$ multiply while their degrees add; there is no cancellation of the highest terms since $K$ is a field.
Since the degrees of $f,g$ are non-negative and sum up to 0, both need to be zero and so $f,g$ are elements of $K$.
It follows that the unit group of $K[x]$ is $K^*=Ksetminus{0}$.
Let $K[x]$ be a polynomial ring over a field $K$.
Then a polynomial $fin K[x]$ is invertible if there is a polynomial $gin K[x]$ such that $fg=1$. Now the degree formula comes in: $${rm deg}(f) + {rm deg}(g) = {rm deg}(fg) = {rm deg}(1) = 0.$$ The first equality comes from the fact that the highest terms of $f$ and $g$ multiply while their degrees add; there is no cancellation of the highest terms since $K$ is a field.
Since the degrees of $f,g$ are non-negative and sum up to 0, both need to be zero and so $f,g$ are elements of $K$.
It follows that the unit group of $K[x]$ is $K^*=Ksetminus{0}$.
answered Dec 1 at 15:14
Wuestenfux
2,6971410
2,6971410
Hi Wuestenfux! Thanks for this hint! It's kind of obvious in retrospective, but I really didn't had the degree-law in mind, when I tried to solve this example. Is it possible to draw a conclusion about isomorphies between this three rings, with the knowlege about unities? So does the number of unities has to be the same, if the rings are isomorphic or something like this?
– noNickname
2 days ago
Units map to units in a ring isomorphism.
– Wuestenfux
2 days ago
So non of them can be isomorphic, due to the number of units differ? The unit in $mathbb{Z}_2[x]$ is only $1$, the units in $mathbb{Z}$ are ${-1,1}$ and the units in $mathbb{Q}[[x]]$ are infinitelly many, due the field is $mathbb{Q}$.
– noNickname
2 days ago
add a comment |
Hi Wuestenfux! Thanks for this hint! It's kind of obvious in retrospective, but I really didn't had the degree-law in mind, when I tried to solve this example. Is it possible to draw a conclusion about isomorphies between this three rings, with the knowlege about unities? So does the number of unities has to be the same, if the rings are isomorphic or something like this?
– noNickname
2 days ago
Units map to units in a ring isomorphism.
– Wuestenfux
2 days ago
So non of them can be isomorphic, due to the number of units differ? The unit in $mathbb{Z}_2[x]$ is only $1$, the units in $mathbb{Z}$ are ${-1,1}$ and the units in $mathbb{Q}[[x]]$ are infinitelly many, due the field is $mathbb{Q}$.
– noNickname
2 days ago
Hi Wuestenfux! Thanks for this hint! It's kind of obvious in retrospective, but I really didn't had the degree-law in mind, when I tried to solve this example. Is it possible to draw a conclusion about isomorphies between this three rings, with the knowlege about unities? So does the number of unities has to be the same, if the rings are isomorphic or something like this?
– noNickname
2 days ago
Hi Wuestenfux! Thanks for this hint! It's kind of obvious in retrospective, but I really didn't had the degree-law in mind, when I tried to solve this example. Is it possible to draw a conclusion about isomorphies between this three rings, with the knowlege about unities? So does the number of unities has to be the same, if the rings are isomorphic or something like this?
– noNickname
2 days ago
Units map to units in a ring isomorphism.
– Wuestenfux
2 days ago
Units map to units in a ring isomorphism.
– Wuestenfux
2 days ago
So non of them can be isomorphic, due to the number of units differ? The unit in $mathbb{Z}_2[x]$ is only $1$, the units in $mathbb{Z}$ are ${-1,1}$ and the units in $mathbb{Q}[[x]]$ are infinitelly many, due the field is $mathbb{Q}$.
– noNickname
2 days ago
So non of them can be isomorphic, due to the number of units differ? The unit in $mathbb{Z}_2[x]$ is only $1$, the units in $mathbb{Z}$ are ${-1,1}$ and the units in $mathbb{Q}[[x]]$ are infinitelly many, due the field is $mathbb{Q}$.
– noNickname
2 days ago
add a comment |
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You're confusing $mathbb{Q}[x]$ and $mathbb{Q}[[x]]$. The former consists of polynomials, the latter power series. In $mathbb{Q}[[x]]$ every polynomial with nonzero constant term is a unit, but in $mathbb{Q}[x]$ only the nonzero constant polynomials are units.
– Slade
Dec 1 at 16:24
Hi Slade! You're right - I'm a little bit confused, because I also thought, that the latter one denotes power series, but the problem description defines it as $mathbb{Q}[[x]] = {sum_{j=0}^{n} a_j x^{j} }$, so something I would call polynomials...
– noNickname
2 days ago
But I guess they mean power series, because I really can't imagine, that finding $p in mathbb{Q}[x]$ such that $forall f in mathbb{Q}[x]: f=e p^k$ with $e in E[mathbb{Q}[x]]$ and $k in mathbb{N}$ is possible for polynomials...
– noNickname
2 days ago