Mixing
How do we prove or visualize $[x+(x-2)]=[2+(x-2)]$ the same way we prove or visualize $0+5mathbb Z = 5 +...
Denote $langle x-2rangle$ as the principal ideal generated by $x-2$ in the polynomial ring $mathbb R[x]$.
$[x+langle x-2rangle]$ and $[2+langle x-2rangle]$ are elements of the quotient ring $mathbb R[x]/langle x-2rangle$, which happens to be a field because $x-2$ is monic irreducible in $mathbb R[x]$ (and the Proposition here).
Here is what I tried:
Let us take an element in one side and show it is in the other side. One of the elements in $[x+langle x-2rangle]$ is $x+x(x-2)$. Now we must find $q in mathbb R[x]$ such that
$$x+x(x-2) = 2+q(x-2)$$
And it is $q(x)=x+1$ by solving for $q$.
In general for $x+r(x-2)$ and $r in mathbb R[x]$, $q=r+1$.
Right to left is similar.
Is that correct?
- And then in general, to show
$$[a+langle x-2rangle] = [b+langle x-2rangle]$$ for elements $overline a=overline b$ in $mathbb R[x]/langle x-2rangle$, to show an element on the left hand side is on the right hand side, we are given and $r$ and must find $q$ such that
$$a+r(x-2)=b+q(x-2)$$
and we solve for $q$:
$$a+r(x-2)=b+q(x-2)$$
$$iff a-b+r(x-2)= q(x-2)$$
$$iff c(x-2)+r(x-2)= q(x-2), c in mathbb R[x]$$
$$iff (c+r)(x-2)= q(x-2), c in mathbb R[x]$$
Therefore $q=c+r$ where $c$ exists as a polynomial with coefficients in $mathbb R$ by definition of "$overline a=overline b$ in $mathbb R[x]/langle x-2rangle$", which is that as
$overline a=overline b$ in $mathbb Z/langle n rangle$" means that $a-b=cn$ for some $c in mathbb Z$,
$overline a=overline b$ in $mathbb R[x]/langle p rangle$" means that $a-b=cp$ for some $c in mathbb R[x]$.
Is that correct?
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
|
show 6 more comments
Denote $langle x-2rangle$ as the principal ideal generated by $x-2$ in the polynomial ring $mathbb R[x]$.
$[x+langle x-2rangle]$ and $[2+langle x-2rangle]$ are elements of the quotient ring $mathbb R[x]/langle x-2rangle$, which happens to be a field because $x-2$ is monic irreducible in $mathbb R[x]$ (and the Proposition here).
Here is what I tried:
Let us take an element in one side and show it is in the other side. One of the elements in $[x+langle x-2rangle]$ is $x+x(x-2)$. Now we must find $q in mathbb R[x]$ such that
$$x+x(x-2) = 2+q(x-2)$$
And it is $q(x)=x+1$ by solving for $q$.
In general for $x+r(x-2)$ and $r in mathbb R[x]$, $q=r+1$.
Right to left is similar.
Is that correct?
- And then in general, to show
$$[a+langle x-2rangle] = [b+langle x-2rangle]$$ for elements $overline a=overline b$ in $mathbb R[x]/langle x-2rangle$, to show an element on the left hand side is on the right hand side, we are given and $r$ and must find $q$ such that
$$a+r(x-2)=b+q(x-2)$$
and we solve for $q$:
$$a+r(x-2)=b+q(x-2)$$
$$iff a-b+r(x-2)= q(x-2)$$
$$iff c(x-2)+r(x-2)= q(x-2), c in mathbb R[x]$$
$$iff (c+r)(x-2)= q(x-2), c in mathbb R[x]$$
Therefore $q=c+r$ where $c$ exists as a polynomial with coefficients in $mathbb R$ by definition of "$overline a=overline b$ in $mathbb R[x]/langle x-2rangle$", which is that as
$overline a=overline b$ in $mathbb Z/langle n rangle$" means that $a-b=cn$ for some $c in mathbb Z$,
$overline a=overline b$ in $mathbb R[x]/langle p rangle$" means that $a-b=cp$ for some $c in mathbb R[x]$.
Is that correct?
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
1
In general, if $I$ is an ideal then $a + I = b + I$ iff $a-bin I$. That is often (such as here) a lot easier to work with.
– Tobias Kildetoft
Nov 20 '18 at 11:24
1
@JackBauer The quotient itself is defined as the set of equivalence classes under the relation $a-b in I$, so checking that two equivalence classes are the same is equivalent to checking if the two given representatives are related or not. For example, just because $x - (+2) = x-2 in langle x-2rangle$, it follows that $x + langle x-2rangle = 2 + langle x-2rangle$.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 11:32
1
@JackBauer As good as, I would say.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 15:31
1
You are welcome! Please answer this question yourself and accept it, so that it can be closed.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 1:56
1
@JackBauer No issues at all. +1 for question and answer.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:43
|
show 6 more comments
Denote $langle x-2rangle$ as the principal ideal generated by $x-2$ in the polynomial ring $mathbb R[x]$.
$[x+langle x-2rangle]$ and $[2+langle x-2rangle]$ are elements of the quotient ring $mathbb R[x]/langle x-2rangle$, which happens to be a field because $x-2$ is monic irreducible in $mathbb R[x]$ (and the Proposition here).
Here is what I tried:
Let us take an element in one side and show it is in the other side. One of the elements in $[x+langle x-2rangle]$ is $x+x(x-2)$. Now we must find $q in mathbb R[x]$ such that
$$x+x(x-2) = 2+q(x-2)$$
And it is $q(x)=x+1$ by solving for $q$.
In general for $x+r(x-2)$ and $r in mathbb R[x]$, $q=r+1$.
Right to left is similar.
Is that correct?
- And then in general, to show
$$[a+langle x-2rangle] = [b+langle x-2rangle]$$ for elements $overline a=overline b$ in $mathbb R[x]/langle x-2rangle$, to show an element on the left hand side is on the right hand side, we are given and $r$ and must find $q$ such that
$$a+r(x-2)=b+q(x-2)$$
and we solve for $q$:
$$a+r(x-2)=b+q(x-2)$$
$$iff a-b+r(x-2)= q(x-2)$$
$$iff c(x-2)+r(x-2)= q(x-2), c in mathbb R[x]$$
$$iff (c+r)(x-2)= q(x-2), c in mathbb R[x]$$
Therefore $q=c+r$ where $c$ exists as a polynomial with coefficients in $mathbb R$ by definition of "$overline a=overline b$ in $mathbb R[x]/langle x-2rangle$", which is that as
$overline a=overline b$ in $mathbb Z/langle n rangle$" means that $a-b=cn$ for some $c in mathbb Z$,
$overline a=overline b$ in $mathbb R[x]/langle p rangle$" means that $a-b=cp$ for some $c in mathbb R[x]$.
Is that correct?
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
Denote $langle x-2rangle$ as the principal ideal generated by $x-2$ in the polynomial ring $mathbb R[x]$.
$[x+langle x-2rangle]$ and $[2+langle x-2rangle]$ are elements of the quotient ring $mathbb R[x]/langle x-2rangle$, which happens to be a field because $x-2$ is monic irreducible in $mathbb R[x]$ (and the Proposition here).
Here is what I tried:
Let us take an element in one side and show it is in the other side. One of the elements in $[x+langle x-2rangle]$ is $x+x(x-2)$. Now we must find $q in mathbb R[x]$ such that
$$x+x(x-2) = 2+q(x-2)$$
And it is $q(x)=x+1$ by solving for $q$.
In general for $x+r(x-2)$ and $r in mathbb R[x]$, $q=r+1$.
Right to left is similar.
Is that correct?
- And then in general, to show
$$[a+langle x-2rangle] = [b+langle x-2rangle]$$ for elements $overline a=overline b$ in $mathbb R[x]/langle x-2rangle$, to show an element on the left hand side is on the right hand side, we are given and $r$ and must find $q$ such that
$$a+r(x-2)=b+q(x-2)$$
and we solve for $q$:
$$a+r(x-2)=b+q(x-2)$$
$$iff a-b+r(x-2)= q(x-2)$$
$$iff c(x-2)+r(x-2)= q(x-2), c in mathbb R[x]$$
$$iff (c+r)(x-2)= q(x-2), c in mathbb R[x]$$
Therefore $q=c+r$ where $c$ exists as a polynomial with coefficients in $mathbb R$ by definition of "$overline a=overline b$ in $mathbb R[x]/langle x-2rangle$", which is that as
$overline a=overline b$ in $mathbb Z/langle n rangle$" means that $a-b=cn$ for some $c in mathbb Z$,
$overline a=overline b$ in $mathbb R[x]/langle p rangle$" means that $a-b=cp$ for some $c in mathbb R[x]$.
Is that correct?
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals
edited Dec 29 '18 at 12:39
asked Nov 20 '18 at 11:06
user198044
1
In general, if $I$ is an ideal then $a + I = b + I$ iff $a-bin I$. That is often (such as here) a lot easier to work with.
– Tobias Kildetoft
Nov 20 '18 at 11:24
1
@JackBauer The quotient itself is defined as the set of equivalence classes under the relation $a-b in I$, so checking that two equivalence classes are the same is equivalent to checking if the two given representatives are related or not. For example, just because $x - (+2) = x-2 in langle x-2rangle$, it follows that $x + langle x-2rangle = 2 + langle x-2rangle$.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 11:32
1
@JackBauer As good as, I would say.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 15:31
1
You are welcome! Please answer this question yourself and accept it, so that it can be closed.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 1:56
1
@JackBauer No issues at all. +1 for question and answer.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:43
|
show 6 more comments
1
In general, if $I$ is an ideal then $a + I = b + I$ iff $a-bin I$. That is often (such as here) a lot easier to work with.
– Tobias Kildetoft
Nov 20 '18 at 11:24
1
@JackBauer The quotient itself is defined as the set of equivalence classes under the relation $a-b in I$, so checking that two equivalence classes are the same is equivalent to checking if the two given representatives are related or not. For example, just because $x - (+2) = x-2 in langle x-2rangle$, it follows that $x + langle x-2rangle = 2 + langle x-2rangle$.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 11:32
1
@JackBauer As good as, I would say.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 15:31
1
You are welcome! Please answer this question yourself and accept it, so that it can be closed.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 1:56
1
@JackBauer No issues at all. +1 for question and answer.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:43
1
1
In general, if $I$ is an ideal then $a + I = b + I$ iff $a-bin I$. That is often (such as here) a lot easier to work with.
– Tobias Kildetoft
Nov 20 '18 at 11:24
In general, if $I$ is an ideal then $a + I = b + I$ iff $a-bin I$. That is often (such as here) a lot easier to work with.
– Tobias Kildetoft
Nov 20 '18 at 11:24
1
1
@JackBauer The quotient itself is defined as the set of equivalence classes under the relation $a-b in I$, so checking that two equivalence classes are the same is equivalent to checking if the two given representatives are related or not. For example, just because $x - (+2) = x-2 in langle x-2rangle$, it follows that $x + langle x-2rangle = 2 + langle x-2rangle$.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 11:32
@JackBauer The quotient itself is defined as the set of equivalence classes under the relation $a-b in I$, so checking that two equivalence classes are the same is equivalent to checking if the two given representatives are related or not. For example, just because $x - (+2) = x-2 in langle x-2rangle$, it follows that $x + langle x-2rangle = 2 + langle x-2rangle$.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 11:32
1
1
@JackBauer As good as, I would say.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 15:31
@JackBauer As good as, I would say.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 15:31
1
1
You are welcome! Please answer this question yourself and accept it, so that it can be closed.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 1:56
You are welcome! Please answer this question yourself and accept it, so that it can be closed.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 1:56
1
1
@JackBauer No issues at all. +1 for question and answer.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:43
@JackBauer No issues at all. +1 for question and answer.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:43
|
show 6 more comments
2 Answers
2
active
oldest
votes
From the comments:
@астонвіллаолофмэллбэрг So what I did is correct but reinventing the wheel? – Jack Bauer
@JackBauer As good as, I would say. – астон вілла олоф мэллбэрг
add a comment |
The analogy should be $5+5mathbb Z=0+5mathbb Z$ and $[x-2+langle x-2 rangle] = [0+langle x-2 rangle]$:
$$5+5mathbb Z = 5+{...,-5,0,5,...} = 0+{...,0,5,10,...}=0+5mathbb Z$$
or
$$5+5mathbb Z = 5+{5m} = 0+{5+5m}=0+{5(m+1)}=0+{5(n)}$$
Similarly,
$$x-2+langle x-2 rangle = x-2+{(x-2)(p)} = 0+{(x-2)(p+1)} = 0+{(x-2)(q)}$$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
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active
oldest
votes
From the comments:
@астонвіллаолофмэллбэрг So what I did is correct but reinventing the wheel? – Jack Bauer
@JackBauer As good as, I would say. – астон вілла олоф мэллбэрг
add a comment |
From the comments:
@астонвіллаолофмэллбэрг So what I did is correct but reinventing the wheel? – Jack Bauer
@JackBauer As good as, I would say. – астон вілла олоф мэллбэрг
add a comment |
From the comments:
@астонвіллаолофмэллбэрг So what I did is correct but reinventing the wheel? – Jack Bauer
@JackBauer As good as, I would say. – астон вілла олоф мэллбэрг
From the comments:
@астонвіллаолофмэллбэрг So what I did is correct but reinventing the wheel? – Jack Bauer
@JackBauer As good as, I would say. – астон вілла олоф мэллбэрг
answered Nov 22 '18 at 2:40
user198044
add a comment |
add a comment |
The analogy should be $5+5mathbb Z=0+5mathbb Z$ and $[x-2+langle x-2 rangle] = [0+langle x-2 rangle]$:
$$5+5mathbb Z = 5+{...,-5,0,5,...} = 0+{...,0,5,10,...}=0+5mathbb Z$$
or
$$5+5mathbb Z = 5+{5m} = 0+{5+5m}=0+{5(m+1)}=0+{5(n)}$$
Similarly,
$$x-2+langle x-2 rangle = x-2+{(x-2)(p)} = 0+{(x-2)(p+1)} = 0+{(x-2)(q)}$$
add a comment |
The analogy should be $5+5mathbb Z=0+5mathbb Z$ and $[x-2+langle x-2 rangle] = [0+langle x-2 rangle]$:
$$5+5mathbb Z = 5+{...,-5,0,5,...} = 0+{...,0,5,10,...}=0+5mathbb Z$$
or
$$5+5mathbb Z = 5+{5m} = 0+{5+5m}=0+{5(m+1)}=0+{5(n)}$$
Similarly,
$$x-2+langle x-2 rangle = x-2+{(x-2)(p)} = 0+{(x-2)(p+1)} = 0+{(x-2)(q)}$$
add a comment |
The analogy should be $5+5mathbb Z=0+5mathbb Z$ and $[x-2+langle x-2 rangle] = [0+langle x-2 rangle]$:
$$5+5mathbb Z = 5+{...,-5,0,5,...} = 0+{...,0,5,10,...}=0+5mathbb Z$$
or
$$5+5mathbb Z = 5+{5m} = 0+{5+5m}=0+{5(m+1)}=0+{5(n)}$$
Similarly,
$$x-2+langle x-2 rangle = x-2+{(x-2)(p)} = 0+{(x-2)(p+1)} = 0+{(x-2)(q)}$$
The analogy should be $5+5mathbb Z=0+5mathbb Z$ and $[x-2+langle x-2 rangle] = [0+langle x-2 rangle]$:
$$5+5mathbb Z = 5+{...,-5,0,5,...} = 0+{...,0,5,10,...}=0+5mathbb Z$$
or
$$5+5mathbb Z = 5+{5m} = 0+{5+5m}=0+{5(m+1)}=0+{5(n)}$$
Similarly,
$$x-2+langle x-2 rangle = x-2+{(x-2)(p)} = 0+{(x-2)(p+1)} = 0+{(x-2)(q)}$$
answered Dec 29 '18 at 13:03
user198044
add a comment |
add a comment |
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1
In general, if $I$ is an ideal then $a + I = b + I$ iff $a-bin I$. That is often (such as here) a lot easier to work with.
– Tobias Kildetoft
Nov 20 '18 at 11:24
1
@JackBauer The quotient itself is defined as the set of equivalence classes under the relation $a-b in I$, so checking that two equivalence classes are the same is equivalent to checking if the two given representatives are related or not. For example, just because $x - (+2) = x-2 in langle x-2rangle$, it follows that $x + langle x-2rangle = 2 + langle x-2rangle$.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 11:32
1
@JackBauer As good as, I would say.
– астон вілла олоф мэллбэрг
Nov 20 '18 at 15:31
1
You are welcome! Please answer this question yourself and accept it, so that it can be closed.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 1:56
1
@JackBauer No issues at all. +1 for question and answer.
– астон вілла олоф мэллбэрг
Nov 22 '18 at 2:43