Clarification on this submodule over $mathbb{Z}[t]$
I am having trouble making the following calculation consistent. Consider the map from a $mathbb{Z}[t]$-module to itself:
$$phi:mathbb{Z}[t]/Irightarrow mathbb{Z}[t]/I,\p(t)+Imapsto (p(t)+I)(t^2-1),$$
where $I=(t+1)(t-n+1)mathbb{Z}[t]$.
We know $phi$ is a homomorphism. Thus im $phi$ is a submodule.
Let $x=at+b+Iinmathbb{Z}[t]/I$. Then
begin{align}
phi(x) & = (at+b)(t^2-1)+I \
& = (at+b)(t+1)(t-n+1+n-2)+I \
& = (n-2)(at+b)(t+1)+(at+b)(t+1)(t-n+1)+I \
& = (n-2)(a(t-n+1)+a(n-1)+b)(t+1)+I\
& = (n-2)(a(n-1)+b)(t+1)+I
end{align}
Then I concluded that im $phi=(n-2)(t+1)mathbb{Z}[t]/I$.
However, every submodule of $mathbb{Z}[t]/I$ should be a quotient like $J/I$ such that $I$ is a submodule of $J$ and $J$ is a submodule of $mathbb{Z}[t]$. But in my calculation $I$ is not a submodule of $(n-2)(t+1)mathbb{Z}[t]$ for $n>3$, this quotient even doesn't make sense.
Where does the flaw in my reasoning occur? Any help will be appreciated.
abstract-algebra modules
asked Nov 26 at 23:38
J. Wang
524
add a comment |
I am having trouble making the following calculation consistent. Consider the map from a $mathbb{Z}[t]$-module to itself:
$$phi:mathbb{Z}[t]/Irightarrow mathbb{Z}[t]/I,\p(t)+Imapsto (p(t)+I)(t^2-1),$$
where $I=(t+1)(t-n+1)mathbb{Z}[t]$.
We know $phi$ is a homomorphism. Thus im $phi$ is a submodule.
Let $x=at+b+Iinmathbb{Z}[t]/I$. Then
begin{align}
phi(x) & = (at+b)(t^2-1)+I \
& = (at+b)(t+1)(t-n+1+n-2)+I \
& = (n-2)(at+b)(t+1)+(at+b)(t+1)(t-n+1)+I \
& = (n-2)(a(t-n+1)+a(n-1)+b)(t+1)+I\
& = (n-2)(a(n-1)+b)(t+1)+I
end{align}
Then I concluded that im $phi=(n-2)(t+1)mathbb{Z}[t]/I$.
However, every submodule of $mathbb{Z}[t]/I$ should be a quotient like $J/I$ such that $I$ is a submodule of $J$ and $J$ is a submodule of $mathbb{Z}[t]$. But in my calculation $I$ is not a submodule of $(n-2)(t+1)mathbb{Z}[t]$ for $n>3$, this quotient even doesn't make sense.
Where does the flaw in my reasoning occur? Any help will be appreciated.
abstract-algebra modules
asked Nov 26 at 23:38
J. Wang
524
add a comment |
I am having trouble making the following calculation consistent. Consider the map from a $mathbb{Z}[t]$-module to itself:
$$phi:mathbb{Z}[t]/Irightarrow mathbb{Z}[t]/I,\p(t)+Imapsto (p(t)+I)(t^2-1),$$
where $I=(t+1)(t-n+1)mathbb{Z}[t]$.
We know $phi$ is a homomorphism. Thus im $phi$ is a submodule.
Let $x=at+b+Iinmathbb{Z}[t]/I$. Then
begin{align}
phi(x) & = (at+b)(t^2-1)+I \
& = (at+b)(t+1)(t-n+1+n-2)+I \
& = (n-2)(at+b)(t+1)+(at+b)(t+1)(t-n+1)+I \
& = (n-2)(a(t-n+1)+a(n-1)+b)(t+1)+I\
& = (n-2)(a(n-1)+b)(t+1)+I
end{align}
Then I concluded that im $phi=(n-2)(t+1)mathbb{Z}[t]/I$.
However, every submodule of $mathbb{Z}[t]/I$ should be a quotient like $J/I$ such that $I$ is a submodule of $J$ and $J$ is a submodule of $mathbb{Z}[t]$. But in my calculation $I$ is not a submodule of $(n-2)(t+1)mathbb{Z}[t]$ for $n>3$, this quotient even doesn't make sense.
Where does the flaw in my reasoning occur? Any help will be appreciated.
abstract-algebra modules
asked Nov 26 at 23:38
J. Wang
524
I am having trouble making the following calculation consistent. Consider the map from a $mathbb{Z}[t]$-module to itself:
$$phi:mathbb{Z}[t]/Irightarrow mathbb{Z}[t]/I,\p(t)+Imapsto (p(t)+I)(t^2-1),$$
where $I=(t+1)(t-n+1)mathbb{Z}[t]$.
We know $phi$ is a homomorphism. Thus im $phi$ is a submodule.
Let $x=at+b+Iinmathbb{Z}[t]/I$. Then
begin{align}
phi(x) & = (at+b)(t^2-1)+I \
& = (at+b)(t+1)(t-n+1+n-2)+I \
& = (n-2)(at+b)(t+1)+(at+b)(t+1)(t-n+1)+I \
& = (n-2)(a(t-n+1)+a(n-1)+b)(t+1)+I\
& = (n-2)(a(n-1)+b)(t+1)+I
end{align}
Then I concluded that im $phi=(n-2)(t+1)mathbb{Z}[t]/I$.
However, every submodule of $mathbb{Z}[t]/I$ should be a quotient like $J/I$ such that $I$ is a submodule of $J$ and $J$ is a submodule of $mathbb{Z}[t]$. But in my calculation $I$ is not a submodule of $(n-2)(t+1)mathbb{Z}[t]$ for $n>3$, this quotient even doesn't make sense.
Where does the flaw in my reasoning occur? Any help will be appreciated.
abstract-algebra modules
abstract-algebra modules
asked Nov 26 at 23:38
J. Wang
524
asked Nov 26 at 23:38
J. Wang
524
asked Nov 26 at 23:38
J. Wang
524
asked Nov 26 at 23:38
J. Wang
524
asked Nov 26 at 23:38
J. Wang
524
524
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The map can be written $phi(p(t)+I)=(t^2-1)p(t)+I$. Thus the image is
$$
bigl((t^2-1)mathbb{Z}[t]+Ibigr)big/I
$$
An element of the top submodule $J$ is of the form
$$
(t^2-1)f(t)+(t+1)(t-n+1)g(t)=
(t+1)bigl((t-1)f(t)+(t-n+1)g(t)bigr)
$$
Now
begin{align}
(t-1)f(t)+(t-n+1)g(t)
&=(t-n+1+n-2)f(t)+(t-n+1)g(t)\[4px]
&=(t-n+1)(f(t)+g(t))+(n-2)f(t)
end{align}
so $J=(t-n+1)mathbb{Z}[t]+(n-2)mathbb{Z}[t]$ is generated by $t-n+1$ and $n-2$.
answered Nov 27 at 0:11
egreg
178k1484200
add a comment |
Your first step fails; in fact
$$phi(x)=(at+b)(t^2-1)+I(t^2-1).$$
The map $phi$ is given by multiplication by $t^2-1$, hence its image consists of all multiples of $t^2-1$. That is to say $operatorname{im}phi=(t^2-1)Bbb{Z}[t]/I$.
answered Nov 26 at 23:43
Servaes
22.3k33793
Thank you for your reply. Your argument is quite straightforward. But I am not fully convinced, because $I$ is still not a submodule of $(t^2-1)mathbb{Z}[t]$.
– J. Wang
Nov 27 at 0:56
1
I should have been more clear, perhaps. I meant that $$operatorname{im}phi=(t^2-1)(Bbb{Z}[t]/I),$$ which is the same as $$operatorname{im}phi=((t^2-1)Bbb{Z}[t]+I)/I.$$ After all, the domain of $phi$ is $Bbb{Z}[t]/I$, so its image is the set of all multiples of $(t^2-1)$ in $Bbb{Z}[t]/I$.
– Servaes
Nov 27 at 1:46
That makes sense. Thanks!
– J. Wang
Nov 27 at 1:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The map can be written $phi(p(t)+I)=(t^2-1)p(t)+I$. Thus the image is
$$
bigl((t^2-1)mathbb{Z}[t]+Ibigr)big/I
$$
An element of the top submodule $J$ is of the form
$$
(t^2-1)f(t)+(t+1)(t-n+1)g(t)=
(t+1)bigl((t-1)f(t)+(t-n+1)g(t)bigr)
$$
Now
begin{align}
(t-1)f(t)+(t-n+1)g(t)
&=(t-n+1+n-2)f(t)+(t-n+1)g(t)\[4px]
&=(t-n+1)(f(t)+g(t))+(n-2)f(t)
end{align}
so $J=(t-n+1)mathbb{Z}[t]+(n-2)mathbb{Z}[t]$ is generated by $t-n+1$ and $n-2$.
answered Nov 27 at 0:11
egreg
178k1484200
add a comment |
The map can be written $phi(p(t)+I)=(t^2-1)p(t)+I$. Thus the image is
$$
bigl((t^2-1)mathbb{Z}[t]+Ibigr)big/I
$$
An element of the top submodule $J$ is of the form
$$
(t^2-1)f(t)+(t+1)(t-n+1)g(t)=
(t+1)bigl((t-1)f(t)+(t-n+1)g(t)bigr)
$$
Now
begin{align}
(t-1)f(t)+(t-n+1)g(t)
&=(t-n+1+n-2)f(t)+(t-n+1)g(t)\[4px]
&=(t-n+1)(f(t)+g(t))+(n-2)f(t)
end{align}
so $J=(t-n+1)mathbb{Z}[t]+(n-2)mathbb{Z}[t]$ is generated by $t-n+1$ and $n-2$.
answered Nov 27 at 0:11
egreg
178k1484200
add a comment |
The map can be written $phi(p(t)+I)=(t^2-1)p(t)+I$. Thus the image is
$$
bigl((t^2-1)mathbb{Z}[t]+Ibigr)big/I
$$
An element of the top submodule $J$ is of the form
$$
(t^2-1)f(t)+(t+1)(t-n+1)g(t)=
(t+1)bigl((t-1)f(t)+(t-n+1)g(t)bigr)
$$
Now
begin{align}
(t-1)f(t)+(t-n+1)g(t)
&=(t-n+1+n-2)f(t)+(t-n+1)g(t)\[4px]
&=(t-n+1)(f(t)+g(t))+(n-2)f(t)
end{align}
so $J=(t-n+1)mathbb{Z}[t]+(n-2)mathbb{Z}[t]$ is generated by $t-n+1$ and $n-2$.
answered Nov 27 at 0:11
egreg
178k1484200
The map can be written $phi(p(t)+I)=(t^2-1)p(t)+I$. Thus the image is
$$
bigl((t^2-1)mathbb{Z}[t]+Ibigr)big/I
$$
An element of the top submodule $J$ is of the form
$$
(t^2-1)f(t)+(t+1)(t-n+1)g(t)=
(t+1)bigl((t-1)f(t)+(t-n+1)g(t)bigr)
$$
Now
begin{align}
(t-1)f(t)+(t-n+1)g(t)
&=(t-n+1+n-2)f(t)+(t-n+1)g(t)\[4px]
&=(t-n+1)(f(t)+g(t))+(n-2)f(t)
end{align}
so $J=(t-n+1)mathbb{Z}[t]+(n-2)mathbb{Z}[t]$ is generated by $t-n+1$ and $n-2$.
answered Nov 27 at 0:11
egreg
178k1484200
answered Nov 27 at 0:11
egreg
178k1484200
answered Nov 27 at 0:11
egreg
178k1484200
answered Nov 27 at 0:11
egreg
178k1484200
178k1484200
add a comment |
add a comment |
Your first step fails; in fact
$$phi(x)=(at+b)(t^2-1)+I(t^2-1).$$
The map $phi$ is given by multiplication by $t^2-1$, hence its image consists of all multiples of $t^2-1$. That is to say $operatorname{im}phi=(t^2-1)Bbb{Z}[t]/I$.
answered Nov 26 at 23:43
Servaes
22.3k33793
Thank you for your reply. Your argument is quite straightforward. But I am not fully convinced, because $I$ is still not a submodule of $(t^2-1)mathbb{Z}[t]$.
– J. Wang
Nov 27 at 0:56
1
I should have been more clear, perhaps. I meant that $$operatorname{im}phi=(t^2-1)(Bbb{Z}[t]/I),$$ which is the same as $$operatorname{im}phi=((t^2-1)Bbb{Z}[t]+I)/I.$$ After all, the domain of $phi$ is $Bbb{Z}[t]/I$, so its image is the set of all multiples of $(t^2-1)$ in $Bbb{Z}[t]/I$.
– Servaes
Nov 27 at 1:46
That makes sense. Thanks!
– J. Wang
Nov 27 at 1:51
add a comment |
Your first step fails; in fact
$$phi(x)=(at+b)(t^2-1)+I(t^2-1).$$
The map $phi$ is given by multiplication by $t^2-1$, hence its image consists of all multiples of $t^2-1$. That is to say $operatorname{im}phi=(t^2-1)Bbb{Z}[t]/I$.
answered Nov 26 at 23:43
Servaes
22.3k33793
Thank you for your reply. Your argument is quite straightforward. But I am not fully convinced, because $I$ is still not a submodule of $(t^2-1)mathbb{Z}[t]$.
– J. Wang
Nov 27 at 0:56
1
I should have been more clear, perhaps. I meant that $$operatorname{im}phi=(t^2-1)(Bbb{Z}[t]/I),$$ which is the same as $$operatorname{im}phi=((t^2-1)Bbb{Z}[t]+I)/I.$$ After all, the domain of $phi$ is $Bbb{Z}[t]/I$, so its image is the set of all multiples of $(t^2-1)$ in $Bbb{Z}[t]/I$.
– Servaes
Nov 27 at 1:46
That makes sense. Thanks!
– J. Wang
Nov 27 at 1:51
add a comment |
Your first step fails; in fact
$$phi(x)=(at+b)(t^2-1)+I(t^2-1).$$
The map $phi$ is given by multiplication by $t^2-1$, hence its image consists of all multiples of $t^2-1$. That is to say $operatorname{im}phi=(t^2-1)Bbb{Z}[t]/I$.
answered Nov 26 at 23:43
Servaes
22.3k33793
Your first step fails; in fact
$$phi(x)=(at+b)(t^2-1)+I(t^2-1).$$
The map $phi$ is given by multiplication by $t^2-1$, hence its image consists of all multiples of $t^2-1$. That is to say $operatorname{im}phi=(t^2-1)Bbb{Z}[t]/I$.
answered Nov 26 at 23:43
Servaes
22.3k33793
answered Nov 26 at 23:43
Servaes
22.3k33793
answered Nov 26 at 23:43
Servaes
22.3k33793
answered Nov 26 at 23:43
Servaes
22.3k33793
22.3k33793
Thank you for your reply. Your argument is quite straightforward. But I am not fully convinced, because $I$ is still not a submodule of $(t^2-1)mathbb{Z}[t]$.
– J. Wang
Nov 27 at 0:56
1
I should have been more clear, perhaps. I meant that $$operatorname{im}phi=(t^2-1)(Bbb{Z}[t]/I),$$ which is the same as $$operatorname{im}phi=((t^2-1)Bbb{Z}[t]+I)/I.$$ After all, the domain of $phi$ is $Bbb{Z}[t]/I$, so its image is the set of all multiples of $(t^2-1)$ in $Bbb{Z}[t]/I$.
– Servaes
Nov 27 at 1:46
That makes sense. Thanks!
– J. Wang
Nov 27 at 1:51
add a comment |
Thank you for your reply. Your argument is quite straightforward. But I am not fully convinced, because $I$ is still not a submodule of $(t^2-1)mathbb{Z}[t]$.
– J. Wang
Nov 27 at 0:56
1
I should have been more clear, perhaps. I meant that $$operatorname{im}phi=(t^2-1)(Bbb{Z}[t]/I),$$ which is the same as $$operatorname{im}phi=((t^2-1)Bbb{Z}[t]+I)/I.$$ After all, the domain of $phi$ is $Bbb{Z}[t]/I$, so its image is the set of all multiples of $(t^2-1)$ in $Bbb{Z}[t]/I$.
– Servaes
Nov 27 at 1:46
That makes sense. Thanks!
– J. Wang
Nov 27 at 1:51
Thank you for your reply. Your argument is quite straightforward. But I am not fully convinced, because $I$ is still not a submodule of $(t^2-1)mathbb{Z}[t]$.
– J. Wang
Nov 27 at 0:56
Thank you for your reply. Your argument is quite straightforward. But I am not fully convinced, because $I$ is still not a submodule of $(t^2-1)mathbb{Z}[t]$.
– J. Wang
Nov 27 at 0:56
1
1
I should have been more clear, perhaps. I meant that $$operatorname{im}phi=(t^2-1)(Bbb{Z}[t]/I),$$ which is the same as $$operatorname{im}phi=((t^2-1)Bbb{Z}[t]+I)/I.$$ After all, the domain of $phi$ is $Bbb{Z}[t]/I$, so its image is the set of all multiples of $(t^2-1)$ in $Bbb{Z}[t]/I$.
– Servaes
Nov 27 at 1:46
I should have been more clear, perhaps. I meant that $$operatorname{im}phi=(t^2-1)(Bbb{Z}[t]/I),$$ which is the same as $$operatorname{im}phi=((t^2-1)Bbb{Z}[t]+I)/I.$$ After all, the domain of $phi$ is $Bbb{Z}[t]/I$, so its image is the set of all multiples of $(t^2-1)$ in $Bbb{Z}[t]/I$.
– Servaes
Nov 27 at 1:46
That makes sense. Thanks!
– J. Wang
Nov 27 at 1:51
That makes sense. Thanks!
– J. Wang
Nov 27 at 1:51
add a comment |
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