Mixing
Finding Annihilator of an Ideal in $mathscr{C} left[ 0, 1 right]$.
I am trying to construct examples of rings, ideals and annihilators, and have taken the example of $left( mathscr{C} left[ 0, 1 right], +, cdot right)$, where $mathscr{C} left[ 0, 1 right]$ denote the set of all continuous functions from the interval $left[ 0, 1 right]$ to $mathbb{R}$, $+$ denotes point - wise addition and $cdot$ denotes point - wise multiplication.
One can easily show that this forms a commutative ring.
The next thing I tried to obtain is an ideal $I_{x_0} = leftlbrace f in mathscr{C} left[ 0, 1 right] | f left( x_0 right) = 0 rightrbrace$. Again, it is easy to prove that this is an ideal of $mathscr{C} left[ 0, 1 right]$.
However, when I tried to find out annihilator of $I_{x_0}$, I got stuck (rather confused). As such, my guess is that the annihilator should be $leftlbrace 0 rightrbrace$, where $0: left[ 0, 1 right] rightarrow mathbb{R}$ is the function $forall x in mathbb{R}, 0 left( x right) = 0$ and is the addtivie identity of the ring $mathscr{C} left[ 0, 1 right]$.
Although this guess is completely intuitive, I am not able to prove it! I would like help in proving the guess or disproving it!
abstract-algebra ring-theory ideals
asked Nov 20 '18 at 5:09
Aniruddha Deshmukh
899418
add a comment |
I am trying to construct examples of rings, ideals and annihilators, and have taken the example of $left( mathscr{C} left[ 0, 1 right], +, cdot right)$, where $mathscr{C} left[ 0, 1 right]$ denote the set of all continuous functions from the interval $left[ 0, 1 right]$ to $mathbb{R}$, $+$ denotes point - wise addition and $cdot$ denotes point - wise multiplication.
One can easily show that this forms a commutative ring.
The next thing I tried to obtain is an ideal $I_{x_0} = leftlbrace f in mathscr{C} left[ 0, 1 right] | f left( x_0 right) = 0 rightrbrace$. Again, it is easy to prove that this is an ideal of $mathscr{C} left[ 0, 1 right]$.
However, when I tried to find out annihilator of $I_{x_0}$, I got stuck (rather confused). As such, my guess is that the annihilator should be $leftlbrace 0 rightrbrace$, where $0: left[ 0, 1 right] rightarrow mathbb{R}$ is the function $forall x in mathbb{R}, 0 left( x right) = 0$ and is the addtivie identity of the ring $mathscr{C} left[ 0, 1 right]$.
Although this guess is completely intuitive, I am not able to prove it! I would like help in proving the guess or disproving it!
abstract-algebra ring-theory ideals
asked Nov 20 '18 at 5:09
Aniruddha Deshmukh
899418
Annihilator of what? Any module structure here? Is it the module over itself?
– xbh
Nov 20 '18 at 5:25
Annihilator of an ideal will be all those elements of the ring which take all elements of the ideal to the additive identity of the ring. In this case it will be $A left( I_{x_0} right) = leftlbrace h in mathscr{C} left[ 0, 1 right] | h cdot f = 0 rightrbrace$.
– Aniruddha Deshmukh
Nov 20 '18 at 5:28
Then your guess could be true. You could prove it by taking specific $f in I_{x_0}$ to deduce that $h = 0$ at every point except possibly $x_0$, then use the continuity of $h$.
– xbh
Nov 20 '18 at 5:33
@xbh The ideal is a module over the ring.
– Praneet Srivastava
Nov 20 '18 at 5:34
add a comment |
I am trying to construct examples of rings, ideals and annihilators, and have taken the example of $left( mathscr{C} left[ 0, 1 right], +, cdot right)$, where $mathscr{C} left[ 0, 1 right]$ denote the set of all continuous functions from the interval $left[ 0, 1 right]$ to $mathbb{R}$, $+$ denotes point - wise addition and $cdot$ denotes point - wise multiplication.
One can easily show that this forms a commutative ring.
The next thing I tried to obtain is an ideal $I_{x_0} = leftlbrace f in mathscr{C} left[ 0, 1 right] | f left( x_0 right) = 0 rightrbrace$. Again, it is easy to prove that this is an ideal of $mathscr{C} left[ 0, 1 right]$.
However, when I tried to find out annihilator of $I_{x_0}$, I got stuck (rather confused). As such, my guess is that the annihilator should be $leftlbrace 0 rightrbrace$, where $0: left[ 0, 1 right] rightarrow mathbb{R}$ is the function $forall x in mathbb{R}, 0 left( x right) = 0$ and is the addtivie identity of the ring $mathscr{C} left[ 0, 1 right]$.
Although this guess is completely intuitive, I am not able to prove it! I would like help in proving the guess or disproving it!
abstract-algebra ring-theory ideals
asked Nov 20 '18 at 5:09
Aniruddha Deshmukh
899418
I am trying to construct examples of rings, ideals and annihilators, and have taken the example of $left( mathscr{C} left[ 0, 1 right], +, cdot right)$, where $mathscr{C} left[ 0, 1 right]$ denote the set of all continuous functions from the interval $left[ 0, 1 right]$ to $mathbb{R}$, $+$ denotes point - wise addition and $cdot$ denotes point - wise multiplication.
One can easily show that this forms a commutative ring.
The next thing I tried to obtain is an ideal $I_{x_0} = leftlbrace f in mathscr{C} left[ 0, 1 right] | f left( x_0 right) = 0 rightrbrace$. Again, it is easy to prove that this is an ideal of $mathscr{C} left[ 0, 1 right]$.
However, when I tried to find out annihilator of $I_{x_0}$, I got stuck (rather confused). As such, my guess is that the annihilator should be $leftlbrace 0 rightrbrace$, where $0: left[ 0, 1 right] rightarrow mathbb{R}$ is the function $forall x in mathbb{R}, 0 left( x right) = 0$ and is the addtivie identity of the ring $mathscr{C} left[ 0, 1 right]$.
Although this guess is completely intuitive, I am not able to prove it! I would like help in proving the guess or disproving it!
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
asked Nov 20 '18 at 5:09
Aniruddha Deshmukh
899418
asked Nov 20 '18 at 5:09
Aniruddha Deshmukh
899418
asked Nov 20 '18 at 5:09
Aniruddha Deshmukh
899418
asked Nov 20 '18 at 5:09
Aniruddha Deshmukh
899418
asked Nov 20 '18 at 5:09
Aniruddha Deshmukh
899418
899418
Annihilator of what? Any module structure here? Is it the module over itself?
– xbh
Nov 20 '18 at 5:25
Annihilator of an ideal will be all those elements of the ring which take all elements of the ideal to the additive identity of the ring. In this case it will be $A left( I_{x_0} right) = leftlbrace h in mathscr{C} left[ 0, 1 right] | h cdot f = 0 rightrbrace$.
– Aniruddha Deshmukh
Nov 20 '18 at 5:28
Then your guess could be true. You could prove it by taking specific $f in I_{x_0}$ to deduce that $h = 0$ at every point except possibly $x_0$, then use the continuity of $h$.
– xbh
Nov 20 '18 at 5:33
@xbh The ideal is a module over the ring.
– Praneet Srivastava
Nov 20 '18 at 5:34
add a comment |
Annihilator of what? Any module structure here? Is it the module over itself?
– xbh
Nov 20 '18 at 5:25
Annihilator of an ideal will be all those elements of the ring which take all elements of the ideal to the additive identity of the ring. In this case it will be $A left( I_{x_0} right) = leftlbrace h in mathscr{C} left[ 0, 1 right] | h cdot f = 0 rightrbrace$.
– Aniruddha Deshmukh
Nov 20 '18 at 5:28
Then your guess could be true. You could prove it by taking specific $f in I_{x_0}$ to deduce that $h = 0$ at every point except possibly $x_0$, then use the continuity of $h$.
– xbh
Nov 20 '18 at 5:33
@xbh The ideal is a module over the ring.
– Praneet Srivastava
Nov 20 '18 at 5:34
Annihilator of what? Any module structure here? Is it the module over itself?
– xbh
Nov 20 '18 at 5:25
Annihilator of what? Any module structure here? Is it the module over itself?
– xbh
Nov 20 '18 at 5:25
Annihilator of an ideal will be all those elements of the ring which take all elements of the ideal to the additive identity of the ring. In this case it will be $A left( I_{x_0} right) = leftlbrace h in mathscr{C} left[ 0, 1 right] | h cdot f = 0 rightrbrace$.
– Aniruddha Deshmukh
Nov 20 '18 at 5:28
Annihilator of an ideal will be all those elements of the ring which take all elements of the ideal to the additive identity of the ring. In this case it will be $A left( I_{x_0} right) = leftlbrace h in mathscr{C} left[ 0, 1 right] | h cdot f = 0 rightrbrace$.
– Aniruddha Deshmukh
Nov 20 '18 at 5:28
Then your guess could be true. You could prove it by taking specific $f in I_{x_0}$ to deduce that $h = 0$ at every point except possibly $x_0$, then use the continuity of $h$.
– xbh
Nov 20 '18 at 5:33
Then your guess could be true. You could prove it by taking specific $f in I_{x_0}$ to deduce that $h = 0$ at every point except possibly $x_0$, then use the continuity of $h$.
– xbh
Nov 20 '18 at 5:33
@xbh The ideal is a module over the ring.
– Praneet Srivastava
Nov 20 '18 at 5:34
@xbh The ideal is a module over the ring.
– Praneet Srivastava
Nov 20 '18 at 5:34
add a comment |
2 Answers
2
active
oldest
votes
Suppose $f in I_{x_0}$. Especially take
$$
f(x) = |x - x_0|,
$$
then since $h cdot f = 0$, $h(x) = 0$ except $x_0$ [since $f(x)neq 0$ whenever $x neq x_0$]. Since $h in mathcal C[0,1]$, $h(x_0) = lim_{xto x_0} h(x) = 0$, so the annihilator is just $ {0} $.
answered Nov 20 '18 at 5:38
xbh
5,6551522
add a comment |
Take $f(x) = (x-x_0)$ on $[0,1]$. Clearly, $f in I_{x_0}$. Now consider $h in Ann(I_{x_0})$. For $x neq x_0$, $h(x)$ must be $0$ for $h(x)f(x)$ to be $0$.
So $h$ is a continuous functions such that $h(x) = 0$ for any $x neq x_0$. By the continuity of $h$, we then have $h(x) = 0$ for any $x in [0,1]$
answered Nov 20 '18 at 5:38
Praneet Srivastava
762516
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
Suppose $f in I_{x_0}$. Especially take
$$
f(x) = |x - x_0|,
$$
then since $h cdot f = 0$, $h(x) = 0$ except $x_0$ [since $f(x)neq 0$ whenever $x neq x_0$]. Since $h in mathcal C[0,1]$, $h(x_0) = lim_{xto x_0} h(x) = 0$, so the annihilator is just $ {0} $.
answered Nov 20 '18 at 5:38
xbh
5,6551522
add a comment |
Suppose $f in I_{x_0}$. Especially take
$$
f(x) = |x - x_0|,
$$
then since $h cdot f = 0$, $h(x) = 0$ except $x_0$ [since $f(x)neq 0$ whenever $x neq x_0$]. Since $h in mathcal C[0,1]$, $h(x_0) = lim_{xto x_0} h(x) = 0$, so the annihilator is just $ {0} $.
answered Nov 20 '18 at 5:38
xbh
5,6551522
add a comment |
Suppose $f in I_{x_0}$. Especially take
$$
f(x) = |x - x_0|,
$$
then since $h cdot f = 0$, $h(x) = 0$ except $x_0$ [since $f(x)neq 0$ whenever $x neq x_0$]. Since $h in mathcal C[0,1]$, $h(x_0) = lim_{xto x_0} h(x) = 0$, so the annihilator is just $ {0} $.
answered Nov 20 '18 at 5:38
xbh
5,6551522
Suppose $f in I_{x_0}$. Especially take
$$
f(x) = |x - x_0|,
$$
then since $h cdot f = 0$, $h(x) = 0$ except $x_0$ [since $f(x)neq 0$ whenever $x neq x_0$]. Since $h in mathcal C[0,1]$, $h(x_0) = lim_{xto x_0} h(x) = 0$, so the annihilator is just $ {0} $.
answered Nov 20 '18 at 5:38
xbh
5,6551522
answered Nov 20 '18 at 5:38
xbh
5,6551522
answered Nov 20 '18 at 5:38
xbh
5,6551522
answered Nov 20 '18 at 5:38
xbh
5,6551522
5,6551522
add a comment |
add a comment |
Take $f(x) = (x-x_0)$ on $[0,1]$. Clearly, $f in I_{x_0}$. Now consider $h in Ann(I_{x_0})$. For $x neq x_0$, $h(x)$ must be $0$ for $h(x)f(x)$ to be $0$.
So $h$ is a continuous functions such that $h(x) = 0$ for any $x neq x_0$. By the continuity of $h$, we then have $h(x) = 0$ for any $x in [0,1]$
answered Nov 20 '18 at 5:38
Praneet Srivastava
762516
add a comment |
Take $f(x) = (x-x_0)$ on $[0,1]$. Clearly, $f in I_{x_0}$. Now consider $h in Ann(I_{x_0})$. For $x neq x_0$, $h(x)$ must be $0$ for $h(x)f(x)$ to be $0$.
So $h$ is a continuous functions such that $h(x) = 0$ for any $x neq x_0$. By the continuity of $h$, we then have $h(x) = 0$ for any $x in [0,1]$
answered Nov 20 '18 at 5:38
Praneet Srivastava
762516
add a comment |
Take $f(x) = (x-x_0)$ on $[0,1]$. Clearly, $f in I_{x_0}$. Now consider $h in Ann(I_{x_0})$. For $x neq x_0$, $h(x)$ must be $0$ for $h(x)f(x)$ to be $0$.
So $h$ is a continuous functions such that $h(x) = 0$ for any $x neq x_0$. By the continuity of $h$, we then have $h(x) = 0$ for any $x in [0,1]$
answered Nov 20 '18 at 5:38
Praneet Srivastava
762516
Take $f(x) = (x-x_0)$ on $[0,1]$. Clearly, $f in I_{x_0}$. Now consider $h in Ann(I_{x_0})$. For $x neq x_0$, $h(x)$ must be $0$ for $h(x)f(x)$ to be $0$.
So $h$ is a continuous functions such that $h(x) = 0$ for any $x neq x_0$. By the continuity of $h$, we then have $h(x) = 0$ for any $x in [0,1]$
answered Nov 20 '18 at 5:38
Praneet Srivastava
762516
answered Nov 20 '18 at 5:38
Praneet Srivastava
762516
answered Nov 20 '18 at 5:38
Praneet Srivastava
762516
answered Nov 20 '18 at 5:38
Praneet Srivastava
762516
762516
add a comment |
add a comment |
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Annihilator of what? Any module structure here? Is it the module over itself?
– xbh
Nov 20 '18 at 5:25
Annihilator of an ideal will be all those elements of the ring which take all elements of the ideal to the additive identity of the ring. In this case it will be $A left( I_{x_0} right) = leftlbrace h in mathscr{C} left[ 0, 1 right] | h cdot f = 0 rightrbrace$.
– Aniruddha Deshmukh
Nov 20 '18 at 5:28
Then your guess could be true. You could prove it by taking specific $f in I_{x_0}$ to deduce that $h = 0$ at every point except possibly $x_0$, then use the continuity of $h$.
– xbh
Nov 20 '18 at 5:33
@xbh The ideal is a module over the ring.
– Praneet Srivastava
Nov 20 '18 at 5:34