Mixing
Show that the assignment $f(x) to f(cx+b)$ is one-to-one
$begingroup$
I'm trying to prove that,
For a commutative ring $R$ with identity, the assignment $phi: f(x) to f(cx+b)$ on $R[x]$, where $c,bin R$ and $c$ is a unit, is a one-to-one map.
To show this, I tried to show that the kernel is zero, so given $f = sum_k^n a_k x^k in Ker phi$,
$$f to sum_{k=0}^n sum_{t=0}^k binom{k}{t} a_k (cx)^t b^{k-t} = 0,$$
but this implies
$$a_r (1+b+b^2+...+b^{n-r}) = 0 quad forall r=0,1,..,n.$$
However, I stuck at this point; since $R$ is not a integral domain in general, I don't know how to proceed.
I would appreciate any help or hint.
abstract-algebra polynomials ring-theory
asked Jan 11 at 16:23
onurcanbektasonurcanbektas
3,36511036
$endgroup$
|
show 2 more comments
$begingroup$
I'm trying to prove that,
For a commutative ring $R$ with identity, the assignment $phi: f(x) to f(cx+b)$ on $R[x]$, where $c,bin R$ and $c$ is a unit, is a one-to-one map.
To show this, I tried to show that the kernel is zero, so given $f = sum_k^n a_k x^k in Ker phi$,
$$f to sum_{k=0}^n sum_{t=0}^k binom{k}{t} a_k (cx)^t b^{k-t} = 0,$$
but this implies
$$a_r (1+b+b^2+...+b^{n-r}) = 0 quad forall r=0,1,..,n.$$
However, I stuck at this point; since $R$ is not a integral domain in general, I don't know how to proceed.
I would appreciate any help or hint.
abstract-algebra polynomials ring-theory
asked Jan 11 at 16:23
onurcanbektasonurcanbektas
3,36511036
$endgroup$
$begingroup$
When you write, "... and $c$ is unit...", do you mean $c$ is the unit of $R?$ Or do you simply mean $R$ is a ring with unit (and there was a typo)?
$endgroup$
– Adrian Keister
Jan 11 at 16:26
1
$begingroup$
@AdrianKeister it means $c$ has an inverse; so it is a unit, not the unity.
$endgroup$
– onurcanbektas
Jan 11 at 16:27
4
$begingroup$
It's probably just easier to prove that $x mapsto c^{-1}(x-b)$ is the inverse, thereby getting injectivity and surjectivity at once.
$endgroup$
– Randall
Jan 11 at 16:35
$begingroup$
In that case, it is better English to say that $c$ is a unit.
$endgroup$
– Adrian Keister
Jan 11 at 16:35
1
$begingroup$
Yep, no problem. It's fun.
$endgroup$
– Randall
Jan 11 at 16:55
|
show 2 more comments
$begingroup$
I'm trying to prove that,
For a commutative ring $R$ with identity, the assignment $phi: f(x) to f(cx+b)$ on $R[x]$, where $c,bin R$ and $c$ is a unit, is a one-to-one map.
To show this, I tried to show that the kernel is zero, so given $f = sum_k^n a_k x^k in Ker phi$,
$$f to sum_{k=0}^n sum_{t=0}^k binom{k}{t} a_k (cx)^t b^{k-t} = 0,$$
but this implies
$$a_r (1+b+b^2+...+b^{n-r}) = 0 quad forall r=0,1,..,n.$$
However, I stuck at this point; since $R$ is not a integral domain in general, I don't know how to proceed.
I would appreciate any help or hint.
abstract-algebra polynomials ring-theory
asked Jan 11 at 16:23
onurcanbektasonurcanbektas
3,36511036
$endgroup$
I'm trying to prove that,
For a commutative ring $R$ with identity, the assignment $phi: f(x) to f(cx+b)$ on $R[x]$, where $c,bin R$ and $c$ is a unit, is a one-to-one map.
To show this, I tried to show that the kernel is zero, so given $f = sum_k^n a_k x^k in Ker phi$,
$$f to sum_{k=0}^n sum_{t=0}^k binom{k}{t} a_k (cx)^t b^{k-t} = 0,$$
but this implies
$$a_r (1+b+b^2+...+b^{n-r}) = 0 quad forall r=0,1,..,n.$$
However, I stuck at this point; since $R$ is not a integral domain in general, I don't know how to proceed.
I would appreciate any help or hint.
abstract-algebra polynomials ring-theory
abstract-algebra polynomials ring-theory
asked Jan 11 at 16:23
onurcanbektasonurcanbektas
3,36511036
asked Jan 11 at 16:23
onurcanbektasonurcanbektas
3,36511036
edited Jan 11 at 16:38
onurcanbektas
asked Jan 11 at 16:23
onurcanbektasonurcanbektas
3,36511036
asked Jan 11 at 16:23
onurcanbektasonurcanbektas
3,36511036
asked Jan 11 at 16:23
onurcanbektasonurcanbektas
3,36511036
3,36511036
$begingroup$
When you write, "... and $c$ is unit...", do you mean $c$ is the unit of $R?$ Or do you simply mean $R$ is a ring with unit (and there was a typo)?
$endgroup$
– Adrian Keister
Jan 11 at 16:26
1
$begingroup$
@AdrianKeister it means $c$ has an inverse; so it is a unit, not the unity.
$endgroup$
– onurcanbektas
Jan 11 at 16:27
4
$begingroup$
It's probably just easier to prove that $x mapsto c^{-1}(x-b)$ is the inverse, thereby getting injectivity and surjectivity at once.
$endgroup$
– Randall
Jan 11 at 16:35
$begingroup$
In that case, it is better English to say that $c$ is a unit.
$endgroup$
– Adrian Keister
Jan 11 at 16:35
1
$begingroup$
Yep, no problem. It's fun.
$endgroup$
– Randall
Jan 11 at 16:55
|
show 2 more comments
$begingroup$
When you write, "... and $c$ is unit...", do you mean $c$ is the unit of $R?$ Or do you simply mean $R$ is a ring with unit (and there was a typo)?
$endgroup$
– Adrian Keister
Jan 11 at 16:26
1
$begingroup$
@AdrianKeister it means $c$ has an inverse; so it is a unit, not the unity.
$endgroup$
– onurcanbektas
Jan 11 at 16:27
4
$begingroup$
It's probably just easier to prove that $x mapsto c^{-1}(x-b)$ is the inverse, thereby getting injectivity and surjectivity at once.
$endgroup$
– Randall
Jan 11 at 16:35
$begingroup$
In that case, it is better English to say that $c$ is a unit.
$endgroup$
– Adrian Keister
Jan 11 at 16:35
1
$begingroup$
Yep, no problem. It's fun.
$endgroup$
– Randall
Jan 11 at 16:55
$begingroup$
When you write, "... and $c$ is unit...", do you mean $c$ is the unit of $R?$ Or do you simply mean $R$ is a ring with unit (and there was a typo)?
$endgroup$
– Adrian Keister
Jan 11 at 16:26
$begingroup$
When you write, "... and $c$ is unit...", do you mean $c$ is the unit of $R?$ Or do you simply mean $R$ is a ring with unit (and there was a typo)?
$endgroup$
– Adrian Keister
Jan 11 at 16:26
1
1
$begingroup$
@AdrianKeister it means $c$ has an inverse; so it is a unit, not the unity.
$endgroup$
– onurcanbektas
Jan 11 at 16:27
$begingroup$
@AdrianKeister it means $c$ has an inverse; so it is a unit, not the unity.
$endgroup$
– onurcanbektas
Jan 11 at 16:27
4
4
$begingroup$
It's probably just easier to prove that $x mapsto c^{-1}(x-b)$ is the inverse, thereby getting injectivity and surjectivity at once.
$endgroup$
– Randall
Jan 11 at 16:35
$begingroup$
It's probably just easier to prove that $x mapsto c^{-1}(x-b)$ is the inverse, thereby getting injectivity and surjectivity at once.
$endgroup$
– Randall
Jan 11 at 16:35
$begingroup$
In that case, it is better English to say that $c$ is a unit.
$endgroup$
– Adrian Keister
Jan 11 at 16:35
$begingroup$
In that case, it is better English to say that $c$ is a unit.
$endgroup$
– Adrian Keister
Jan 11 at 16:35
1
1
$begingroup$
Yep, no problem. It's fun.
$endgroup$
– Randall
Jan 11 at 16:55
$begingroup$
Yep, no problem. It's fun.
$endgroup$
– Randall
Jan 11 at 16:55
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If we let $psi:g(x) mapsto g(c^{-1}(x-b))$, it follows that
$$
(psiphi)(f)(x)=psi(phi(f))(x)=phi(f)(c^{-1}(x-b))=f(cc^{-1}(x-b)+b)=f(x),
$$ and
$$(phipsi)(f)(x)=phi(psi(f))(x) = psi(f)(cx+b)=f(c^{-1}((cx+b)-b))=f(x).
$$ Hence, $psi = phi^{-1}$ and the bijectivity of $phi$ follows.
But I couldn't see how you derived the condition $$
a_r(1+b+cdots b^{n-r})=0.
$$ Shouldn't it be $forall j=0,ldots,n$
$$
sum_{k=j}^n a_kbinom{k}{j}b^{k-j}=0?
$$ If it is the case, we can show inductively that $a_n=0$, $a_{n-1}=0$, ..., etc.
answered Jan 11 at 16:40
SongSong
12.2k630
$endgroup$
1
$begingroup$
Do you have $psi$ and $phi$ transposed?
$endgroup$
– Randall
Jan 11 at 16:42
$begingroup$
@Randall It looks correct. Check again.
$endgroup$
– David Hill
Jan 11 at 16:52
$begingroup$
Do you compose left-to-right?
$endgroup$
– Randall
Jan 11 at 16:53
$begingroup$
Oh wait, I see.
$endgroup$
– Randall
Jan 11 at 16:54
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If we let $psi:g(x) mapsto g(c^{-1}(x-b))$, it follows that
$$
(psiphi)(f)(x)=psi(phi(f))(x)=phi(f)(c^{-1}(x-b))=f(cc^{-1}(x-b)+b)=f(x),
$$ and
$$(phipsi)(f)(x)=phi(psi(f))(x) = psi(f)(cx+b)=f(c^{-1}((cx+b)-b))=f(x).
$$ Hence, $psi = phi^{-1}$ and the bijectivity of $phi$ follows.
But I couldn't see how you derived the condition $$
a_r(1+b+cdots b^{n-r})=0.
$$ Shouldn't it be $forall j=0,ldots,n$
$$
sum_{k=j}^n a_kbinom{k}{j}b^{k-j}=0?
$$ If it is the case, we can show inductively that $a_n=0$, $a_{n-1}=0$, ..., etc.
answered Jan 11 at 16:40
SongSong
12.2k630
$endgroup$
1
$begingroup$
Do you have $psi$ and $phi$ transposed?
$endgroup$
– Randall
Jan 11 at 16:42
$begingroup$
@Randall It looks correct. Check again.
$endgroup$
– David Hill
Jan 11 at 16:52
$begingroup$
Do you compose left-to-right?
$endgroup$
– Randall
Jan 11 at 16:53
$begingroup$
Oh wait, I see.
$endgroup$
– Randall
Jan 11 at 16:54
add a comment |
$begingroup$
If we let $psi:g(x) mapsto g(c^{-1}(x-b))$, it follows that
$$
(psiphi)(f)(x)=psi(phi(f))(x)=phi(f)(c^{-1}(x-b))=f(cc^{-1}(x-b)+b)=f(x),
$$ and
$$(phipsi)(f)(x)=phi(psi(f))(x) = psi(f)(cx+b)=f(c^{-1}((cx+b)-b))=f(x).
$$ Hence, $psi = phi^{-1}$ and the bijectivity of $phi$ follows.
But I couldn't see how you derived the condition $$
a_r(1+b+cdots b^{n-r})=0.
$$ Shouldn't it be $forall j=0,ldots,n$
$$
sum_{k=j}^n a_kbinom{k}{j}b^{k-j}=0?
$$ If it is the case, we can show inductively that $a_n=0$, $a_{n-1}=0$, ..., etc.
answered Jan 11 at 16:40
SongSong
12.2k630
$endgroup$
1
$begingroup$
Do you have $psi$ and $phi$ transposed?
$endgroup$
– Randall
Jan 11 at 16:42
$begingroup$
@Randall It looks correct. Check again.
$endgroup$
– David Hill
Jan 11 at 16:52
$begingroup$
Do you compose left-to-right?
$endgroup$
– Randall
Jan 11 at 16:53
$begingroup$
Oh wait, I see.
$endgroup$
– Randall
Jan 11 at 16:54
add a comment |
$begingroup$
If we let $psi:g(x) mapsto g(c^{-1}(x-b))$, it follows that
$$
(psiphi)(f)(x)=psi(phi(f))(x)=phi(f)(c^{-1}(x-b))=f(cc^{-1}(x-b)+b)=f(x),
$$ and
$$(phipsi)(f)(x)=phi(psi(f))(x) = psi(f)(cx+b)=f(c^{-1}((cx+b)-b))=f(x).
$$ Hence, $psi = phi^{-1}$ and the bijectivity of $phi$ follows.
But I couldn't see how you derived the condition $$
a_r(1+b+cdots b^{n-r})=0.
$$ Shouldn't it be $forall j=0,ldots,n$
$$
sum_{k=j}^n a_kbinom{k}{j}b^{k-j}=0?
$$ If it is the case, we can show inductively that $a_n=0$, $a_{n-1}=0$, ..., etc.
answered Jan 11 at 16:40
SongSong
12.2k630
$endgroup$
If we let $psi:g(x) mapsto g(c^{-1}(x-b))$, it follows that
$$
(psiphi)(f)(x)=psi(phi(f))(x)=phi(f)(c^{-1}(x-b))=f(cc^{-1}(x-b)+b)=f(x),
$$ and
$$(phipsi)(f)(x)=phi(psi(f))(x) = psi(f)(cx+b)=f(c^{-1}((cx+b)-b))=f(x).
$$ Hence, $psi = phi^{-1}$ and the bijectivity of $phi$ follows.
But I couldn't see how you derived the condition $$
a_r(1+b+cdots b^{n-r})=0.
$$ Shouldn't it be $forall j=0,ldots,n$
$$
sum_{k=j}^n a_kbinom{k}{j}b^{k-j}=0?
$$ If it is the case, we can show inductively that $a_n=0$, $a_{n-1}=0$, ..., etc.
answered Jan 11 at 16:40
SongSong
12.2k630
edited Jan 11 at 17:03
answered Jan 11 at 16:40
SongSong
12.2k630
answered Jan 11 at 16:40
SongSong
12.2k630
answered Jan 11 at 16:40
SongSong
12.2k630
12.2k630
1
$begingroup$
Do you have $psi$ and $phi$ transposed?
$endgroup$
– Randall
Jan 11 at 16:42
$begingroup$
@Randall It looks correct. Check again.
$endgroup$
– David Hill
Jan 11 at 16:52
$begingroup$
Do you compose left-to-right?
$endgroup$
– Randall
Jan 11 at 16:53
$begingroup$
Oh wait, I see.
$endgroup$
– Randall
Jan 11 at 16:54
add a comment |
1
$begingroup$
Do you have $psi$ and $phi$ transposed?
$endgroup$
– Randall
Jan 11 at 16:42
$begingroup$
@Randall It looks correct. Check again.
$endgroup$
– David Hill
Jan 11 at 16:52
$begingroup$
Do you compose left-to-right?
$endgroup$
– Randall
Jan 11 at 16:53
$begingroup$
Oh wait, I see.
$endgroup$
– Randall
Jan 11 at 16:54
1
1
$begingroup$
Do you have $psi$ and $phi$ transposed?
$endgroup$
– Randall
Jan 11 at 16:42
$begingroup$
Do you have $psi$ and $phi$ transposed?
$endgroup$
– Randall
Jan 11 at 16:42
$begingroup$
@Randall It looks correct. Check again.
$endgroup$
– David Hill
Jan 11 at 16:52
$begingroup$
@Randall It looks correct. Check again.
$endgroup$
– David Hill
Jan 11 at 16:52
$begingroup$
Do you compose left-to-right?
$endgroup$
– Randall
Jan 11 at 16:53
$begingroup$
Do you compose left-to-right?
$endgroup$
– Randall
Jan 11 at 16:53
$begingroup$
Oh wait, I see.
$endgroup$
– Randall
Jan 11 at 16:54
$begingroup$
Oh wait, I see.
$endgroup$
– Randall
Jan 11 at 16:54
add a comment |
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$begingroup$
When you write, "... and $c$ is unit...", do you mean $c$ is the unit of $R?$ Or do you simply mean $R$ is a ring with unit (and there was a typo)?
$endgroup$
– Adrian Keister
Jan 11 at 16:26
1
$begingroup$
@AdrianKeister it means $c$ has an inverse; so it is a unit, not the unity.
$endgroup$
– onurcanbektas
Jan 11 at 16:27
4
$begingroup$
It's probably just easier to prove that $x mapsto c^{-1}(x-b)$ is the inverse, thereby getting injectivity and surjectivity at once.
$endgroup$
– Randall
Jan 11 at 16:35
$begingroup$
In that case, it is better English to say that $c$ is a unit.
$endgroup$
– Adrian Keister
Jan 11 at 16:35
1
$begingroup$
Yep, no problem. It's fun.
$endgroup$
– Randall
Jan 11 at 16:55