Mixing
Showing $xto cx+b$ is a ring homomorphism
$begingroup$
In the book Algebra by Hungerford, at page 167, at question 14, it is asked that
Let $R$ be a commutative ring with identity, and $c,b in R$ with $c$
is unit.Show that the assignment $x to cx + b$ is a unique
automorphism of $R[x]$ that is the identity map on $R$.
However, consider $R = mathbb{Z}$, $c = 1$, $b = 2$. Then since this assignment is a homomorphism, it must be true that
$$x(x+1) to x^2 + x + 2 = (x+1)(x+3) = x^2 + 4x + 3,$$
which is clearly not true.Therefore, considering the fact that this question is even asked in math overflow, and got answer, what is wrong in my argument ?
abstract-algebra polynomials ring-theory
asked Jan 11 at 15:39
onurcanbektasonurcanbektas
3,36511036
$endgroup$
|
show 1 more comment
$begingroup$
In the book Algebra by Hungerford, at page 167, at question 14, it is asked that
Let $R$ be a commutative ring with identity, and $c,b in R$ with $c$
is unit.Show that the assignment $x to cx + b$ is a unique
automorphism of $R[x]$ that is the identity map on $R$.
However, consider $R = mathbb{Z}$, $c = 1$, $b = 2$. Then since this assignment is a homomorphism, it must be true that
$$x(x+1) to x^2 + x + 2 = (x+1)(x+3) = x^2 + 4x + 3,$$
which is clearly not true.Therefore, considering the fact that this question is even asked in math overflow, and got answer, what is wrong in my argument ?
abstract-algebra polynomials ring-theory
asked Jan 11 at 15:39
onurcanbektasonurcanbektas
3,36511036
$endgroup$
1
$begingroup$
In the first part, doesn't $x(x+1)$ map to $(x+2)(x+3)$?
$endgroup$
– Randall
Jan 11 at 15:41
$begingroup$
@Randall $x(x+1) = x^2 + x to x^2 + x + 2$; $c=1$ and $b = 2$, so we just add $2$ to every polynomial.
$endgroup$
– onurcanbektas
Jan 11 at 15:42
2
$begingroup$
Not so. You're not applying the transformation to each individual $x$, and you should be.
$endgroup$
– Randall
Jan 11 at 15:43
1
$begingroup$
Yes, that's what he said: replace $x$ by $cx+b$.
$endgroup$
– Randall
Jan 11 at 15:46
2
$begingroup$
Note that he is not defining the auto on ALL of the polynomial ring, just on its action by $x$. But this then dictates the entire thing by linearity and distribution.
$endgroup$
– Randall
Jan 11 at 15:47
|
show 1 more comment
$begingroup$
In the book Algebra by Hungerford, at page 167, at question 14, it is asked that
Let $R$ be a commutative ring with identity, and $c,b in R$ with $c$
is unit.Show that the assignment $x to cx + b$ is a unique
automorphism of $R[x]$ that is the identity map on $R$.
However, consider $R = mathbb{Z}$, $c = 1$, $b = 2$. Then since this assignment is a homomorphism, it must be true that
$$x(x+1) to x^2 + x + 2 = (x+1)(x+3) = x^2 + 4x + 3,$$
which is clearly not true.Therefore, considering the fact that this question is even asked in math overflow, and got answer, what is wrong in my argument ?
abstract-algebra polynomials ring-theory
asked Jan 11 at 15:39
onurcanbektasonurcanbektas
3,36511036
$endgroup$
In the book Algebra by Hungerford, at page 167, at question 14, it is asked that
Let $R$ be a commutative ring with identity, and $c,b in R$ with $c$
is unit.Show that the assignment $x to cx + b$ is a unique
automorphism of $R[x]$ that is the identity map on $R$.
However, consider $R = mathbb{Z}$, $c = 1$, $b = 2$. Then since this assignment is a homomorphism, it must be true that
$$x(x+1) to x^2 + x + 2 = (x+1)(x+3) = x^2 + 4x + 3,$$
which is clearly not true.Therefore, considering the fact that this question is even asked in math overflow, and got answer, what is wrong in my argument ?
abstract-algebra polynomials ring-theory
abstract-algebra polynomials ring-theory
asked Jan 11 at 15:39
onurcanbektasonurcanbektas
3,36511036
asked Jan 11 at 15:39
onurcanbektasonurcanbektas
3,36511036
asked Jan 11 at 15:39
onurcanbektasonurcanbektas
3,36511036
asked Jan 11 at 15:39
onurcanbektasonurcanbektas
3,36511036
asked Jan 11 at 15:39
onurcanbektasonurcanbektas
3,36511036
3,36511036
1
$begingroup$
In the first part, doesn't $x(x+1)$ map to $(x+2)(x+3)$?
$endgroup$
– Randall
Jan 11 at 15:41
$begingroup$
@Randall $x(x+1) = x^2 + x to x^2 + x + 2$; $c=1$ and $b = 2$, so we just add $2$ to every polynomial.
$endgroup$
– onurcanbektas
Jan 11 at 15:42
2
$begingroup$
Not so. You're not applying the transformation to each individual $x$, and you should be.
$endgroup$
– Randall
Jan 11 at 15:43
1
$begingroup$
Yes, that's what he said: replace $x$ by $cx+b$.
$endgroup$
– Randall
Jan 11 at 15:46
2
$begingroup$
Note that he is not defining the auto on ALL of the polynomial ring, just on its action by $x$. But this then dictates the entire thing by linearity and distribution.
$endgroup$
– Randall
Jan 11 at 15:47
|
show 1 more comment
1
$begingroup$
In the first part, doesn't $x(x+1)$ map to $(x+2)(x+3)$?
$endgroup$
– Randall
Jan 11 at 15:41
$begingroup$
@Randall $x(x+1) = x^2 + x to x^2 + x + 2$; $c=1$ and $b = 2$, so we just add $2$ to every polynomial.
$endgroup$
– onurcanbektas
Jan 11 at 15:42
2
$begingroup$
Not so. You're not applying the transformation to each individual $x$, and you should be.
$endgroup$
– Randall
Jan 11 at 15:43
1
$begingroup$
Yes, that's what he said: replace $x$ by $cx+b$.
$endgroup$
– Randall
Jan 11 at 15:46
2
$begingroup$
Note that he is not defining the auto on ALL of the polynomial ring, just on its action by $x$. But this then dictates the entire thing by linearity and distribution.
$endgroup$
– Randall
Jan 11 at 15:47
1
1
$begingroup$
In the first part, doesn't $x(x+1)$ map to $(x+2)(x+3)$?
$endgroup$
– Randall
Jan 11 at 15:41
$begingroup$
In the first part, doesn't $x(x+1)$ map to $(x+2)(x+3)$?
$endgroup$
– Randall
Jan 11 at 15:41
$begingroup$
@Randall $x(x+1) = x^2 + x to x^2 + x + 2$; $c=1$ and $b = 2$, so we just add $2$ to every polynomial.
$endgroup$
– onurcanbektas
Jan 11 at 15:42
$begingroup$
@Randall $x(x+1) = x^2 + x to x^2 + x + 2$; $c=1$ and $b = 2$, so we just add $2$ to every polynomial.
$endgroup$
– onurcanbektas
Jan 11 at 15:42
2
2
$begingroup$
Not so. You're not applying the transformation to each individual $x$, and you should be.
$endgroup$
– Randall
Jan 11 at 15:43
$begingroup$
Not so. You're not applying the transformation to each individual $x$, and you should be.
$endgroup$
– Randall
Jan 11 at 15:43
1
1
$begingroup$
Yes, that's what he said: replace $x$ by $cx+b$.
$endgroup$
– Randall
Jan 11 at 15:46
$begingroup$
Yes, that's what he said: replace $x$ by $cx+b$.
$endgroup$
– Randall
Jan 11 at 15:46
2
2
$begingroup$
Note that he is not defining the auto on ALL of the polynomial ring, just on its action by $x$. But this then dictates the entire thing by linearity and distribution.
$endgroup$
– Randall
Jan 11 at 15:47
$begingroup$
Note that he is not defining the auto on ALL of the polynomial ring, just on its action by $x$. But this then dictates the entire thing by linearity and distribution.
$endgroup$
– Randall
Jan 11 at 15:47
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Everything is fine. You are transforming by $x mapsto x+2$ and extending by linearity and distribution. Call the resulting map $phi: mathbb{Z}[x] to mathbb{Z}[x]$.
With your example you have
$$
phi(x(x+1)) = phi(x^2+x) = (x+2)^2+(x+2)= x^2+5x+6.
$$
On the other hand,
$$
phi(x)phi(x+1)= (x+2)(x+3) = x^2+5x+6
$$
so yes, they are equal.
answered Jan 11 at 15:45
RandallRandall
9,71111230
$endgroup$
add a comment |
$begingroup$
You should try to prove that $phi(x)phi(x+1) = phi(x(x+1))$.
Note that $phi(x) = x+2$, $phi(x+1) = x+3$, and their product is $x^2+5x+6$.
Next, $x(x+1) = x^2+x$, and $phi(x^2+x) = (x+2)^2 + x+2 = x^2+4x+4+x+2 = x^2+5x+6$.
What you did, instead, was find $x times x+1$, and compare this with $phi(x) times phi(x+1)$ (which was also computed wrongly).
Note that $x to cx+b$ means that for each $f in R[x]$ we have $f(x) to f(cx+b)$. In other words, $f$ is sent to $g$, the polynomial formed by substituting every occurence of $x$ in $f$, by $cx+b$.
The other way of saying this, is that any homomorphism from $R[x] to R[x]$ which is the identity on $R$, is specified uniquely by where $x$ goes, because of the above correspondence.
answered Jan 11 at 15:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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oldest
votes
$begingroup$
Everything is fine. You are transforming by $x mapsto x+2$ and extending by linearity and distribution. Call the resulting map $phi: mathbb{Z}[x] to mathbb{Z}[x]$.
With your example you have
$$
phi(x(x+1)) = phi(x^2+x) = (x+2)^2+(x+2)= x^2+5x+6.
$$
On the other hand,
$$
phi(x)phi(x+1)= (x+2)(x+3) = x^2+5x+6
$$
so yes, they are equal.
answered Jan 11 at 15:45
RandallRandall
9,71111230
$endgroup$
add a comment |
$begingroup$
Everything is fine. You are transforming by $x mapsto x+2$ and extending by linearity and distribution. Call the resulting map $phi: mathbb{Z}[x] to mathbb{Z}[x]$.
With your example you have
$$
phi(x(x+1)) = phi(x^2+x) = (x+2)^2+(x+2)= x^2+5x+6.
$$
On the other hand,
$$
phi(x)phi(x+1)= (x+2)(x+3) = x^2+5x+6
$$
so yes, they are equal.
answered Jan 11 at 15:45
RandallRandall
9,71111230
$endgroup$
add a comment |
$begingroup$
Everything is fine. You are transforming by $x mapsto x+2$ and extending by linearity and distribution. Call the resulting map $phi: mathbb{Z}[x] to mathbb{Z}[x]$.
With your example you have
$$
phi(x(x+1)) = phi(x^2+x) = (x+2)^2+(x+2)= x^2+5x+6.
$$
On the other hand,
$$
phi(x)phi(x+1)= (x+2)(x+3) = x^2+5x+6
$$
so yes, they are equal.
answered Jan 11 at 15:45
RandallRandall
9,71111230
$endgroup$
Everything is fine. You are transforming by $x mapsto x+2$ and extending by linearity and distribution. Call the resulting map $phi: mathbb{Z}[x] to mathbb{Z}[x]$.
With your example you have
$$
phi(x(x+1)) = phi(x^2+x) = (x+2)^2+(x+2)= x^2+5x+6.
$$
On the other hand,
$$
phi(x)phi(x+1)= (x+2)(x+3) = x^2+5x+6
$$
so yes, they are equal.
answered Jan 11 at 15:45
RandallRandall
9,71111230
answered Jan 11 at 15:45
RandallRandall
9,71111230
answered Jan 11 at 15:45
RandallRandall
9,71111230
answered Jan 11 at 15:45
RandallRandall
9,71111230
9,71111230
add a comment |
add a comment |
$begingroup$
You should try to prove that $phi(x)phi(x+1) = phi(x(x+1))$.
Note that $phi(x) = x+2$, $phi(x+1) = x+3$, and their product is $x^2+5x+6$.
Next, $x(x+1) = x^2+x$, and $phi(x^2+x) = (x+2)^2 + x+2 = x^2+4x+4+x+2 = x^2+5x+6$.
What you did, instead, was find $x times x+1$, and compare this with $phi(x) times phi(x+1)$ (which was also computed wrongly).
Note that $x to cx+b$ means that for each $f in R[x]$ we have $f(x) to f(cx+b)$. In other words, $f$ is sent to $g$, the polynomial formed by substituting every occurence of $x$ in $f$, by $cx+b$.
The other way of saying this, is that any homomorphism from $R[x] to R[x]$ which is the identity on $R$, is specified uniquely by where $x$ goes, because of the above correspondence.
answered Jan 11 at 15:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
$endgroup$
add a comment |
$begingroup$
You should try to prove that $phi(x)phi(x+1) = phi(x(x+1))$.
Note that $phi(x) = x+2$, $phi(x+1) = x+3$, and their product is $x^2+5x+6$.
Next, $x(x+1) = x^2+x$, and $phi(x^2+x) = (x+2)^2 + x+2 = x^2+4x+4+x+2 = x^2+5x+6$.
What you did, instead, was find $x times x+1$, and compare this with $phi(x) times phi(x+1)$ (which was also computed wrongly).
Note that $x to cx+b$ means that for each $f in R[x]$ we have $f(x) to f(cx+b)$. In other words, $f$ is sent to $g$, the polynomial formed by substituting every occurence of $x$ in $f$, by $cx+b$.
The other way of saying this, is that any homomorphism from $R[x] to R[x]$ which is the identity on $R$, is specified uniquely by where $x$ goes, because of the above correspondence.
answered Jan 11 at 15:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
$endgroup$
add a comment |
$begingroup$
You should try to prove that $phi(x)phi(x+1) = phi(x(x+1))$.
Note that $phi(x) = x+2$, $phi(x+1) = x+3$, and their product is $x^2+5x+6$.
Next, $x(x+1) = x^2+x$, and $phi(x^2+x) = (x+2)^2 + x+2 = x^2+4x+4+x+2 = x^2+5x+6$.
What you did, instead, was find $x times x+1$, and compare this with $phi(x) times phi(x+1)$ (which was also computed wrongly).
Note that $x to cx+b$ means that for each $f in R[x]$ we have $f(x) to f(cx+b)$. In other words, $f$ is sent to $g$, the polynomial formed by substituting every occurence of $x$ in $f$, by $cx+b$.
The other way of saying this, is that any homomorphism from $R[x] to R[x]$ which is the identity on $R$, is specified uniquely by where $x$ goes, because of the above correspondence.
answered Jan 11 at 15:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
$endgroup$
You should try to prove that $phi(x)phi(x+1) = phi(x(x+1))$.
Note that $phi(x) = x+2$, $phi(x+1) = x+3$, and their product is $x^2+5x+6$.
Next, $x(x+1) = x^2+x$, and $phi(x^2+x) = (x+2)^2 + x+2 = x^2+4x+4+x+2 = x^2+5x+6$.
What you did, instead, was find $x times x+1$, and compare this with $phi(x) times phi(x+1)$ (which was also computed wrongly).
Note that $x to cx+b$ means that for each $f in R[x]$ we have $f(x) to f(cx+b)$. In other words, $f$ is sent to $g$, the polynomial formed by substituting every occurence of $x$ in $f$, by $cx+b$.
The other way of saying this, is that any homomorphism from $R[x] to R[x]$ which is the identity on $R$, is specified uniquely by where $x$ goes, because of the above correspondence.
answered Jan 11 at 15:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
answered Jan 11 at 15:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
answered Jan 11 at 15:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
answered Jan 11 at 15:48
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.3k33376
38.3k33376
add a comment |
add a comment |
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1
$begingroup$
In the first part, doesn't $x(x+1)$ map to $(x+2)(x+3)$?
$endgroup$
– Randall
Jan 11 at 15:41
$begingroup$
@Randall $x(x+1) = x^2 + x to x^2 + x + 2$; $c=1$ and $b = 2$, so we just add $2$ to every polynomial.
$endgroup$
– onurcanbektas
Jan 11 at 15:42
2
$begingroup$
Not so. You're not applying the transformation to each individual $x$, and you should be.
$endgroup$
– Randall
Jan 11 at 15:43
1
$begingroup$
Yes, that's what he said: replace $x$ by $cx+b$.
$endgroup$
– Randall
Jan 11 at 15:46
2
$begingroup$
Note that he is not defining the auto on ALL of the polynomial ring, just on its action by $x$. But this then dictates the entire thing by linearity and distribution.
$endgroup$
– Randall
Jan 11 at 15:47