Associate elements in non-integral domains. [duplicate]
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This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
Elements $a$ and $b$ of an integral domain are associates if
$amathrel{vdots}b$ and $bmathrel{vdots}a$
I have proved this fact. Then I tried to find out $a$ and $b$ which divide each other and does not associate. This may happen only in non-integral domains of course. But I couldn't find an example. Can someone suggest one?
Definition of associates.
$a$ and $b$ are associates if $a=bepsilon$ where $epsilon$ is
invertible element of the ring.
$amathrel{vdots}b$ means $a$ is divided by $b$. ($a=bc$). It is the
same as $b|a$.
abstract-algebra ring-theory integral-domain
asked Dec 1 '18 at 9:00
AshotAshot
2,61432044
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marked as duplicate by rschwieb abstract-algebra
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Dec 1 '18 at 13:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
Elements $a$ and $b$ of an integral domain are associates if
$amathrel{vdots}b$ and $bmathrel{vdots}a$
I have proved this fact. Then I tried to find out $a$ and $b$ which divide each other and does not associate. This may happen only in non-integral domains of course. But I couldn't find an example. Can someone suggest one?
Definition of associates.
$a$ and $b$ are associates if $a=bepsilon$ where $epsilon$ is
invertible element of the ring.
$amathrel{vdots}b$ means $a$ is divided by $b$. ($a=bc$). It is the
same as $b|a$.
abstract-algebra ring-theory integral-domain
asked Dec 1 '18 at 9:00
AshotAshot
2,61432044
$endgroup$
marked as duplicate by rschwieb abstract-algebra
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Dec 1 '18 at 13:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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What is your definition of associates, if not in terms of each a divisor of the other? Also, what is the definition of a (vert dots) b or is that same as a | b?
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– coffeemath
Dec 1 '18 at 9:03
1
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@coffeemath I updated the question.
$endgroup$
– Ashot
Dec 1 '18 at 9:06
1
$begingroup$
Just deleted my "answer" and now I think the question is interesting! [already upvoted]
$endgroup$
– coffeemath
Dec 1 '18 at 9:24
$begingroup$
Just to be clear, do you need a commutative ring in an example? And I assume you require a ring having a unity (a 1 for multiplication a(1)=(1)a=a all a) so that a unit would be a divisor of 1? If not how is a unit defined?
$endgroup$
– coffeemath
Dec 1 '18 at 9:54
1
$begingroup$
I think if we found such elements in any ideal it would be counterexample of the theorem, but it would be nice to do it in a commutative ring with identity.
$endgroup$
– Ashot
Dec 1 '18 at 10:23
add a comment |
$begingroup$
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
Elements $a$ and $b$ of an integral domain are associates if
$amathrel{vdots}b$ and $bmathrel{vdots}a$
I have proved this fact. Then I tried to find out $a$ and $b$ which divide each other and does not associate. This may happen only in non-integral domains of course. But I couldn't find an example. Can someone suggest one?
Definition of associates.
$a$ and $b$ are associates if $a=bepsilon$ where $epsilon$ is
invertible element of the ring.
$amathrel{vdots}b$ means $a$ is divided by $b$. ($a=bc$). It is the
same as $b|a$.
abstract-algebra ring-theory integral-domain
asked Dec 1 '18 at 9:00
AshotAshot
2,61432044
$endgroup$
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
Elements $a$ and $b$ of an integral domain are associates if
$amathrel{vdots}b$ and $bmathrel{vdots}a$
I have proved this fact. Then I tried to find out $a$ and $b$ which divide each other and does not associate. This may happen only in non-integral domains of course. But I couldn't find an example. Can someone suggest one?
Definition of associates.
$a$ and $b$ are associates if $a=bepsilon$ where $epsilon$ is
invertible element of the ring.
$amathrel{vdots}b$ means $a$ is divided by $b$. ($a=bc$). It is the
same as $b|a$.
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
abstract-algebra ring-theory integral-domain
abstract-algebra ring-theory integral-domain
asked Dec 1 '18 at 9:00
AshotAshot
2,61432044
asked Dec 1 '18 at 9:00
AshotAshot
2,61432044
edited Dec 1 '18 at 9:05
Ashot
asked Dec 1 '18 at 9:00
AshotAshot
2,61432044
asked Dec 1 '18 at 9:00
AshotAshot
2,61432044
asked Dec 1 '18 at 9:00
AshotAshot
2,61432044
2,61432044
marked as duplicate by rschwieb abstract-algebra
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Dec 1 '18 at 13:27
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Dec 1 '18 at 13:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
What is your definition of associates, if not in terms of each a divisor of the other? Also, what is the definition of a (vert dots) b or is that same as a | b?
$endgroup$
– coffeemath
Dec 1 '18 at 9:03
1
$begingroup$
@coffeemath I updated the question.
$endgroup$
– Ashot
Dec 1 '18 at 9:06
1
$begingroup$
Just deleted my "answer" and now I think the question is interesting! [already upvoted]
$endgroup$
– coffeemath
Dec 1 '18 at 9:24
$begingroup$
Just to be clear, do you need a commutative ring in an example? And I assume you require a ring having a unity (a 1 for multiplication a(1)=(1)a=a all a) so that a unit would be a divisor of 1? If not how is a unit defined?
$endgroup$
– coffeemath
Dec 1 '18 at 9:54
1
$begingroup$
I think if we found such elements in any ideal it would be counterexample of the theorem, but it would be nice to do it in a commutative ring with identity.
$endgroup$
– Ashot
Dec 1 '18 at 10:23
add a comment |
$begingroup$
What is your definition of associates, if not in terms of each a divisor of the other? Also, what is the definition of a (vert dots) b or is that same as a | b?
$endgroup$
– coffeemath
Dec 1 '18 at 9:03
1
$begingroup$
@coffeemath I updated the question.
$endgroup$
– Ashot
Dec 1 '18 at 9:06
1
$begingroup$
Just deleted my "answer" and now I think the question is interesting! [already upvoted]
$endgroup$
– coffeemath
Dec 1 '18 at 9:24
$begingroup$
Just to be clear, do you need a commutative ring in an example? And I assume you require a ring having a unity (a 1 for multiplication a(1)=(1)a=a all a) so that a unit would be a divisor of 1? If not how is a unit defined?
$endgroup$
– coffeemath
Dec 1 '18 at 9:54
1
$begingroup$
I think if we found such elements in any ideal it would be counterexample of the theorem, but it would be nice to do it in a commutative ring with identity.
$endgroup$
– Ashot
Dec 1 '18 at 10:23
$begingroup$
What is your definition of associates, if not in terms of each a divisor of the other? Also, what is the definition of a (vert dots) b or is that same as a | b?
$endgroup$
– coffeemath
Dec 1 '18 at 9:03
$begingroup$
What is your definition of associates, if not in terms of each a divisor of the other? Also, what is the definition of a (vert dots) b or is that same as a | b?
$endgroup$
– coffeemath
Dec 1 '18 at 9:03
1
1
$begingroup$
@coffeemath I updated the question.
$endgroup$
– Ashot
Dec 1 '18 at 9:06
$begingroup$
@coffeemath I updated the question.
$endgroup$
– Ashot
Dec 1 '18 at 9:06
1
1
$begingroup$
Just deleted my "answer" and now I think the question is interesting! [already upvoted]
$endgroup$
– coffeemath
Dec 1 '18 at 9:24
$begingroup$
Just deleted my "answer" and now I think the question is interesting! [already upvoted]
$endgroup$
– coffeemath
Dec 1 '18 at 9:24
$begingroup$
Just to be clear, do you need a commutative ring in an example? And I assume you require a ring having a unity (a 1 for multiplication a(1)=(1)a=a all a) so that a unit would be a divisor of 1? If not how is a unit defined?
$endgroup$
– coffeemath
Dec 1 '18 at 9:54
$begingroup$
Just to be clear, do you need a commutative ring in an example? And I assume you require a ring having a unity (a 1 for multiplication a(1)=(1)a=a all a) so that a unit would be a divisor of 1? If not how is a unit defined?
$endgroup$
– coffeemath
Dec 1 '18 at 9:54
1
1
$begingroup$
I think if we found such elements in any ideal it would be counterexample of the theorem, but it would be nice to do it in a commutative ring with identity.
$endgroup$
– Ashot
Dec 1 '18 at 10:23
$begingroup$
I think if we found such elements in any ideal it would be counterexample of the theorem, but it would be nice to do it in a commutative ring with identity.
$endgroup$
– Ashot
Dec 1 '18 at 10:23
add a comment |
1 Answer
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Here is an example due to Kaplansky.
Let $R=C([0,3])$ be the ring of continuous funtions $f:[0,3]rightarrowmathbb{R}$. I claim that
$$R^{times}=leftlbrace f(t)in R hspace{2.5mm} | hspace{2.5mm} f(t)neq 0 hspace{2.5mm} forall tin[0,3]rightrbrace.$$
Indeed, if $f(t)g(t)=1$ for all $tin[0,3]$ then $f(t)neq 0$ because if there exists $sin [0,3]$ such that $f(s)=0$ then $1=f(s)g(s)=0cdot g(s)=0,$ a contradiction. Conversely, if $f(t)neq 0$ then $frac{1}{f(t)}neq 0$ is a continuous function.
Now define the three elements
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases},$$
$$b(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\2-t&tin[2,3]end{cases},$$
and
$$c(t)=begin{cases}1&tin[0,1]\3-2t&tin[1,2]\-1&tin[2,3]end{cases}.$$
Then $a(t),b(t),c(t)in R$ and it's clear that $c(t)a(t)=b(t)$ and $c(t)b(t)=a(t)$, so $a(t)mid b(t)$ and $b(t)mid a(t)$.
Let $u(t)in R^{times}$ be given. If $a(t)=b(t)u(t)$ then
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases}=begin{cases}(1-t)u(t)&tin[0,1]\0&tin[1,2]\(2-t)u(t)&tin[2,3]end{cases}=b(t)u(t),$$
which implies that
$$u(t)=begin{cases}1&tin[0,1]\star&tin[1,2]\-1&tin[2,3]end{cases}.$$
But due to the Intermediate Value Theorem we see that there exists $sin[0,3]$ such that $u(s)=0$ because it attains all the values between $-1$ and $1$. Thus we conclude that $u(t)$ is not a unit and therefore we see that $a(t)$ and $b(t)$ are not associates.
answered Dec 1 '18 at 12:24
YumekuiMathYumekuiMath
34114
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is an example due to Kaplansky.
Let $R=C([0,3])$ be the ring of continuous funtions $f:[0,3]rightarrowmathbb{R}$. I claim that
$$R^{times}=leftlbrace f(t)in R hspace{2.5mm} | hspace{2.5mm} f(t)neq 0 hspace{2.5mm} forall tin[0,3]rightrbrace.$$
Indeed, if $f(t)g(t)=1$ for all $tin[0,3]$ then $f(t)neq 0$ because if there exists $sin [0,3]$ such that $f(s)=0$ then $1=f(s)g(s)=0cdot g(s)=0,$ a contradiction. Conversely, if $f(t)neq 0$ then $frac{1}{f(t)}neq 0$ is a continuous function.
Now define the three elements
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases},$$
$$b(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\2-t&tin[2,3]end{cases},$$
and
$$c(t)=begin{cases}1&tin[0,1]\3-2t&tin[1,2]\-1&tin[2,3]end{cases}.$$
Then $a(t),b(t),c(t)in R$ and it's clear that $c(t)a(t)=b(t)$ and $c(t)b(t)=a(t)$, so $a(t)mid b(t)$ and $b(t)mid a(t)$.
Let $u(t)in R^{times}$ be given. If $a(t)=b(t)u(t)$ then
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases}=begin{cases}(1-t)u(t)&tin[0,1]\0&tin[1,2]\(2-t)u(t)&tin[2,3]end{cases}=b(t)u(t),$$
which implies that
$$u(t)=begin{cases}1&tin[0,1]\star&tin[1,2]\-1&tin[2,3]end{cases}.$$
But due to the Intermediate Value Theorem we see that there exists $sin[0,3]$ such that $u(s)=0$ because it attains all the values between $-1$ and $1$. Thus we conclude that $u(t)$ is not a unit and therefore we see that $a(t)$ and $b(t)$ are not associates.
answered Dec 1 '18 at 12:24
YumekuiMathYumekuiMath
34114
$endgroup$
add a comment |
$begingroup$
Here is an example due to Kaplansky.
Let $R=C([0,3])$ be the ring of continuous funtions $f:[0,3]rightarrowmathbb{R}$. I claim that
$$R^{times}=leftlbrace f(t)in R hspace{2.5mm} | hspace{2.5mm} f(t)neq 0 hspace{2.5mm} forall tin[0,3]rightrbrace.$$
Indeed, if $f(t)g(t)=1$ for all $tin[0,3]$ then $f(t)neq 0$ because if there exists $sin [0,3]$ such that $f(s)=0$ then $1=f(s)g(s)=0cdot g(s)=0,$ a contradiction. Conversely, if $f(t)neq 0$ then $frac{1}{f(t)}neq 0$ is a continuous function.
Now define the three elements
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases},$$
$$b(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\2-t&tin[2,3]end{cases},$$
and
$$c(t)=begin{cases}1&tin[0,1]\3-2t&tin[1,2]\-1&tin[2,3]end{cases}.$$
Then $a(t),b(t),c(t)in R$ and it's clear that $c(t)a(t)=b(t)$ and $c(t)b(t)=a(t)$, so $a(t)mid b(t)$ and $b(t)mid a(t)$.
Let $u(t)in R^{times}$ be given. If $a(t)=b(t)u(t)$ then
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases}=begin{cases}(1-t)u(t)&tin[0,1]\0&tin[1,2]\(2-t)u(t)&tin[2,3]end{cases}=b(t)u(t),$$
which implies that
$$u(t)=begin{cases}1&tin[0,1]\star&tin[1,2]\-1&tin[2,3]end{cases}.$$
But due to the Intermediate Value Theorem we see that there exists $sin[0,3]$ such that $u(s)=0$ because it attains all the values between $-1$ and $1$. Thus we conclude that $u(t)$ is not a unit and therefore we see that $a(t)$ and $b(t)$ are not associates.
answered Dec 1 '18 at 12:24
YumekuiMathYumekuiMath
34114
$endgroup$
add a comment |
$begingroup$
Here is an example due to Kaplansky.
Let $R=C([0,3])$ be the ring of continuous funtions $f:[0,3]rightarrowmathbb{R}$. I claim that
$$R^{times}=leftlbrace f(t)in R hspace{2.5mm} | hspace{2.5mm} f(t)neq 0 hspace{2.5mm} forall tin[0,3]rightrbrace.$$
Indeed, if $f(t)g(t)=1$ for all $tin[0,3]$ then $f(t)neq 0$ because if there exists $sin [0,3]$ such that $f(s)=0$ then $1=f(s)g(s)=0cdot g(s)=0,$ a contradiction. Conversely, if $f(t)neq 0$ then $frac{1}{f(t)}neq 0$ is a continuous function.
Now define the three elements
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases},$$
$$b(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\2-t&tin[2,3]end{cases},$$
and
$$c(t)=begin{cases}1&tin[0,1]\3-2t&tin[1,2]\-1&tin[2,3]end{cases}.$$
Then $a(t),b(t),c(t)in R$ and it's clear that $c(t)a(t)=b(t)$ and $c(t)b(t)=a(t)$, so $a(t)mid b(t)$ and $b(t)mid a(t)$.
Let $u(t)in R^{times}$ be given. If $a(t)=b(t)u(t)$ then
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases}=begin{cases}(1-t)u(t)&tin[0,1]\0&tin[1,2]\(2-t)u(t)&tin[2,3]end{cases}=b(t)u(t),$$
which implies that
$$u(t)=begin{cases}1&tin[0,1]\star&tin[1,2]\-1&tin[2,3]end{cases}.$$
But due to the Intermediate Value Theorem we see that there exists $sin[0,3]$ such that $u(s)=0$ because it attains all the values between $-1$ and $1$. Thus we conclude that $u(t)$ is not a unit and therefore we see that $a(t)$ and $b(t)$ are not associates.
answered Dec 1 '18 at 12:24
YumekuiMathYumekuiMath
34114
$endgroup$
Here is an example due to Kaplansky.
Let $R=C([0,3])$ be the ring of continuous funtions $f:[0,3]rightarrowmathbb{R}$. I claim that
$$R^{times}=leftlbrace f(t)in R hspace{2.5mm} | hspace{2.5mm} f(t)neq 0 hspace{2.5mm} forall tin[0,3]rightrbrace.$$
Indeed, if $f(t)g(t)=1$ for all $tin[0,3]$ then $f(t)neq 0$ because if there exists $sin [0,3]$ such that $f(s)=0$ then $1=f(s)g(s)=0cdot g(s)=0,$ a contradiction. Conversely, if $f(t)neq 0$ then $frac{1}{f(t)}neq 0$ is a continuous function.
Now define the three elements
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases},$$
$$b(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\2-t&tin[2,3]end{cases},$$
and
$$c(t)=begin{cases}1&tin[0,1]\3-2t&tin[1,2]\-1&tin[2,3]end{cases}.$$
Then $a(t),b(t),c(t)in R$ and it's clear that $c(t)a(t)=b(t)$ and $c(t)b(t)=a(t)$, so $a(t)mid b(t)$ and $b(t)mid a(t)$.
Let $u(t)in R^{times}$ be given. If $a(t)=b(t)u(t)$ then
$$a(t)=begin{cases}1-t&tin[0,1]\0&tin[1,2]\t-2&tin[2,3]end{cases}=begin{cases}(1-t)u(t)&tin[0,1]\0&tin[1,2]\(2-t)u(t)&tin[2,3]end{cases}=b(t)u(t),$$
which implies that
$$u(t)=begin{cases}1&tin[0,1]\star&tin[1,2]\-1&tin[2,3]end{cases}.$$
But due to the Intermediate Value Theorem we see that there exists $sin[0,3]$ such that $u(s)=0$ because it attains all the values between $-1$ and $1$. Thus we conclude that $u(t)$ is not a unit and therefore we see that $a(t)$ and $b(t)$ are not associates.
answered Dec 1 '18 at 12:24
YumekuiMathYumekuiMath
34114
answered Dec 1 '18 at 12:24
YumekuiMathYumekuiMath
34114
answered Dec 1 '18 at 12:24
YumekuiMathYumekuiMath
34114
answered Dec 1 '18 at 12:24
YumekuiMathYumekuiMath
34114
34114
add a comment |
add a comment |
$begingroup$
What is your definition of associates, if not in terms of each a divisor of the other? Also, what is the definition of a (vert dots) b or is that same as a | b?
$endgroup$
– coffeemath
Dec 1 '18 at 9:03
1
$begingroup$
@coffeemath I updated the question.
$endgroup$
– Ashot
Dec 1 '18 at 9:06
1
$begingroup$
Just deleted my "answer" and now I think the question is interesting! [already upvoted]
$endgroup$
– coffeemath
Dec 1 '18 at 9:24
$begingroup$
Just to be clear, do you need a commutative ring in an example? And I assume you require a ring having a unity (a 1 for multiplication a(1)=(1)a=a all a) so that a unit would be a divisor of 1? If not how is a unit defined?
$endgroup$
– coffeemath
Dec 1 '18 at 9:54
1
$begingroup$
I think if we found such elements in any ideal it would be counterexample of the theorem, but it would be nice to do it in a commutative ring with identity.
$endgroup$
– Ashot
Dec 1 '18 at 10:23