$mathbb{Z}, mathbb{Z}_2$ and $mathbb{Q}[x]$ - rings
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
$begingroup$
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.
$endgroup$
– Slade
Dec 1 '18 at 16:24
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 9:47
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 10:00
add a comment |
$begingroup$
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
$endgroup$
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 '18 at 13:30
user606561
$begingroup$
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.
$endgroup$
– Slade
Dec 1 '18 at 16:24
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 9:47
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 10:00
add a comment |
$begingroup$
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.
$endgroup$
– Slade
Dec 1 '18 at 16:24
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 9:47
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 10:00
$begingroup$
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.
$endgroup$
– Slade
Dec 1 '18 at 16:24
$begingroup$
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.
$endgroup$
– Slade
Dec 1 '18 at 16:24
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 9:47
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 9:47
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 10:00
$begingroup$
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...
$endgroup$
– user606561
Dec 2 '18 at 10:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 15:14
WuestenfuxWuestenfux
4,5991413
$endgroup$
$begingroup$
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?
$endgroup$
– user606561
Dec 2 '18 at 9:44
$begingroup$
Units map to units in a ring isomorphism.
$endgroup$
– Wuestenfux
Dec 2 '18 at 9:44
$begingroup$
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 $mathbb{Q}$.
$endgroup$
– user606561
Dec 2 '18 at 10:07
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
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 '18 at 15:14
WuestenfuxWuestenfux
4,5991413
$endgroup$
$begingroup$
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?
$endgroup$
– user606561
Dec 2 '18 at 9:44
$begingroup$
Units map to units in a ring isomorphism.
$endgroup$
– Wuestenfux
Dec 2 '18 at 9:44
$begingroup$
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 $mathbb{Q}$.
$endgroup$
– user606561
Dec 2 '18 at 10:07
add a comment |
$begingroup$
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 '18 at 15:14
WuestenfuxWuestenfux
4,5991413
$endgroup$
$begingroup$
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?
$endgroup$
– user606561
Dec 2 '18 at 9:44
$begingroup$
Units map to units in a ring isomorphism.
$endgroup$
– Wuestenfux
Dec 2 '18 at 9:44
$begingroup$
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 $mathbb{Q}$.
$endgroup$
– user606561
Dec 2 '18 at 10:07
add a comment |
$begingroup$
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 '18 at 15:14
WuestenfuxWuestenfux
4,5991413
$endgroup$
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 '18 at 15:14
WuestenfuxWuestenfux
4,5991413
answered Dec 1 '18 at 15:14
WuestenfuxWuestenfux
4,5991413
answered Dec 1 '18 at 15:14
WuestenfuxWuestenfux
4,5991413
answered Dec 1 '18 at 15:14
WuestenfuxWuestenfux
4,5991413
4,5991413
$begingroup$
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?
$endgroup$
– user606561
Dec 2 '18 at 9:44
$begingroup$
Units map to units in a ring isomorphism.
$endgroup$
– Wuestenfux
Dec 2 '18 at 9:44
$begingroup$
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 $mathbb{Q}$.
$endgroup$
– user606561
Dec 2 '18 at 10:07
add a comment |
$begingroup$
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?
$endgroup$
– user606561
Dec 2 '18 at 9:44
$begingroup$
Units map to units in a ring isomorphism.
$endgroup$
– Wuestenfux
Dec 2 '18 at 9:44
$begingroup$
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 $mathbb{Q}$.
$endgroup$
– user606561
Dec 2 '18 at 10:07
$begingroup$
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?
$endgroup$
– user606561
Dec 2 '18 at 9:44
$begingroup$
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?
$endgroup$
– user606561
Dec 2 '18 at 9:44
$begingroup$
Units map to units in a ring isomorphism.
$endgroup$
– Wuestenfux
Dec 2 '18 at 9:44
$begingroup$
Units map to units in a ring isomorphism.
$endgroup$
– Wuestenfux
Dec 2 '18 at 9:44
$begingroup$
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 $mathbb{Q}$.
$endgroup$
– user606561
Dec 2 '18 at 10:07
$begingroup$
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 $mathbb{Q}$.
$endgroup$
– user606561
Dec 2 '18 at 10:07
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.
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– Slade
Dec 1 '18 at 16:24
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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...
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– user606561
Dec 2 '18 at 9:47
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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...
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– user606561
Dec 2 '18 at 10:00