Natural Euclidean Function Not Satisfying the $d$-inequality
$begingroup$
Let me provide some background before I begin (although I feel as though it's hardly needed):
Let $R$ be an integral domain. I call a function $d:Rsetminus {0}tomathbb{N}cup{0}$ a
Euclidean function if for every $a,bin R$, $ane0$, there exists $q,rin R$ with $b=aq+r$ and
either $r=0$ or $d(r)<d(a)$.
I say that a Euclidean function satisfies the $d$-inequality
if $xmid y$ implies $d(x)leqslant d(y)$.
It is often assumed that Euclidean functions for example, in the context of Euclidean domains, always satisfy the $d$-inequality since
given a Euclidean function $d$ the function $displaystyle widetilde{d}(x)=min_{yne 0}d(xy)$ is a Euclidean function satisfying the $d$-inequality. Thus, Euclidean domains (i.e. rings for which there exists a Euclidean function on) are precisely the same as the rings that admit Euclidean functions satisfying the $d$-inequality.
Now, while, as I said above, the study of such rings (where the only key is the existence of such functions) is no different, practically it's much nicer to have Euclidean functions satisfying the $d$-inequality since they enjoy such benefits as $ain R^times$ if and only if $d(a)=d(1)$.
The strange thing is, most naturally occurring Euclidean functions satisfy the $d$-inequality (e.g. the degree function on $F[x]$, field norms, etc.) And, for the life of me, I can't think of a non-contrived example of a Euclidean function that does not satisfy the $d$-inequality. So, what are some? Moreover, there will undoubtedly be some trivial, common one that I've overlooked, then I would still love to hear more obscure ones that arise naturally in more advanced contexts.
Thanks for your time!
abstract-algebra euclidean-domain
edited Jan 2 at 22:14
user26857
39.3k124183
asked Oct 14 '11 at 7:25
MathEnthusiastMathEnthusiast
361
$endgroup$
add a comment |
$begingroup$
Let me provide some background before I begin (although I feel as though it's hardly needed):
Let $R$ be an integral domain. I call a function $d:Rsetminus {0}tomathbb{N}cup{0}$ a
Euclidean function if for every $a,bin R$, $ane0$, there exists $q,rin R$ with $b=aq+r$ and
either $r=0$ or $d(r)<d(a)$.
I say that a Euclidean function satisfies the $d$-inequality
if $xmid y$ implies $d(x)leqslant d(y)$.
It is often assumed that Euclidean functions for example, in the context of Euclidean domains, always satisfy the $d$-inequality since
given a Euclidean function $d$ the function $displaystyle widetilde{d}(x)=min_{yne 0}d(xy)$ is a Euclidean function satisfying the $d$-inequality. Thus, Euclidean domains (i.e. rings for which there exists a Euclidean function on) are precisely the same as the rings that admit Euclidean functions satisfying the $d$-inequality.
Now, while, as I said above, the study of such rings (where the only key is the existence of such functions) is no different, practically it's much nicer to have Euclidean functions satisfying the $d$-inequality since they enjoy such benefits as $ain R^times$ if and only if $d(a)=d(1)$.
The strange thing is, most naturally occurring Euclidean functions satisfy the $d$-inequality (e.g. the degree function on $F[x]$, field norms, etc.) And, for the life of me, I can't think of a non-contrived example of a Euclidean function that does not satisfy the $d$-inequality. So, what are some? Moreover, there will undoubtedly be some trivial, common one that I've overlooked, then I would still love to hear more obscure ones that arise naturally in more advanced contexts.
Thanks for your time!
abstract-algebra euclidean-domain
edited Jan 2 at 22:14
user26857
39.3k124183
asked Oct 14 '11 at 7:25
MathEnthusiastMathEnthusiast
361
$endgroup$
1
$begingroup$
I cannot think of a non-trivial one, but here is a trivial example: Let $k$ be a field with at least three elements. Then define the norm $d: k - {0} longrightarrow mathbb{N} cup {0}$ by $d(1) = 1$ and $d(x)=0$ if $x neq 1$. Then this is a norm, because for any $a,b in k$ such that $b neq 0$ we have $a = ab^{-1}b$, and $1|x$ for all nonzero $x$, but $d(1) > d(x)$. However, as this is in some sense an artificial example, this does not answer your question.
$endgroup$
– Rankeya
Oct 14 '11 at 7:47
$begingroup$
Rankeya, thank you, I appreciate your response. This is precisely the example I had come up with--so of course we know now that there is no implication that every Euclidean function satisfies the $d$-inequality, but this example isn't "natural"--not that you ever said it was.
$endgroup$
– MathEnthusiast
Oct 14 '11 at 7:52
$begingroup$
Yes, precisely why I did not post this as an answer, but rather chose to just put it as a comment.
$endgroup$
– Rankeya
Oct 14 '11 at 7:56
add a comment |
$begingroup$
Let me provide some background before I begin (although I feel as though it's hardly needed):
Let $R$ be an integral domain. I call a function $d:Rsetminus {0}tomathbb{N}cup{0}$ a
Euclidean function if for every $a,bin R$, $ane0$, there exists $q,rin R$ with $b=aq+r$ and
either $r=0$ or $d(r)<d(a)$.
I say that a Euclidean function satisfies the $d$-inequality
if $xmid y$ implies $d(x)leqslant d(y)$.
It is often assumed that Euclidean functions for example, in the context of Euclidean domains, always satisfy the $d$-inequality since
given a Euclidean function $d$ the function $displaystyle widetilde{d}(x)=min_{yne 0}d(xy)$ is a Euclidean function satisfying the $d$-inequality. Thus, Euclidean domains (i.e. rings for which there exists a Euclidean function on) are precisely the same as the rings that admit Euclidean functions satisfying the $d$-inequality.
Now, while, as I said above, the study of such rings (where the only key is the existence of such functions) is no different, practically it's much nicer to have Euclidean functions satisfying the $d$-inequality since they enjoy such benefits as $ain R^times$ if and only if $d(a)=d(1)$.
The strange thing is, most naturally occurring Euclidean functions satisfy the $d$-inequality (e.g. the degree function on $F[x]$, field norms, etc.) And, for the life of me, I can't think of a non-contrived example of a Euclidean function that does not satisfy the $d$-inequality. So, what are some? Moreover, there will undoubtedly be some trivial, common one that I've overlooked, then I would still love to hear more obscure ones that arise naturally in more advanced contexts.
Thanks for your time!
abstract-algebra euclidean-domain
edited Jan 2 at 22:14
user26857
39.3k124183
asked Oct 14 '11 at 7:25
MathEnthusiastMathEnthusiast
361
$endgroup$
Let me provide some background before I begin (although I feel as though it's hardly needed):
Let $R$ be an integral domain. I call a function $d:Rsetminus {0}tomathbb{N}cup{0}$ a
Euclidean function if for every $a,bin R$, $ane0$, there exists $q,rin R$ with $b=aq+r$ and
either $r=0$ or $d(r)<d(a)$.
I say that a Euclidean function satisfies the $d$-inequality
if $xmid y$ implies $d(x)leqslant d(y)$.
It is often assumed that Euclidean functions for example, in the context of Euclidean domains, always satisfy the $d$-inequality since
given a Euclidean function $d$ the function $displaystyle widetilde{d}(x)=min_{yne 0}d(xy)$ is a Euclidean function satisfying the $d$-inequality. Thus, Euclidean domains (i.e. rings for which there exists a Euclidean function on) are precisely the same as the rings that admit Euclidean functions satisfying the $d$-inequality.
Now, while, as I said above, the study of such rings (where the only key is the existence of such functions) is no different, practically it's much nicer to have Euclidean functions satisfying the $d$-inequality since they enjoy such benefits as $ain R^times$ if and only if $d(a)=d(1)$.
The strange thing is, most naturally occurring Euclidean functions satisfy the $d$-inequality (e.g. the degree function on $F[x]$, field norms, etc.) And, for the life of me, I can't think of a non-contrived example of a Euclidean function that does not satisfy the $d$-inequality. So, what are some? Moreover, there will undoubtedly be some trivial, common one that I've overlooked, then I would still love to hear more obscure ones that arise naturally in more advanced contexts.
Thanks for your time!
abstract-algebra euclidean-domain
abstract-algebra euclidean-domain
edited Jan 2 at 22:14
user26857
39.3k124183
asked Oct 14 '11 at 7:25
MathEnthusiastMathEnthusiast
361
edited Jan 2 at 22:14
user26857
39.3k124183
asked Oct 14 '11 at 7:25
MathEnthusiastMathEnthusiast
361
edited Jan 2 at 22:14
user26857
39.3k124183
edited Jan 2 at 22:14
user26857
39.3k124183
edited Jan 2 at 22:14
user26857
39.3k124183
39.3k124183
asked Oct 14 '11 at 7:25
MathEnthusiastMathEnthusiast
361
asked Oct 14 '11 at 7:25
MathEnthusiastMathEnthusiast
361
asked Oct 14 '11 at 7:25
MathEnthusiastMathEnthusiast
361
361
1
$begingroup$
I cannot think of a non-trivial one, but here is a trivial example: Let $k$ be a field with at least three elements. Then define the norm $d: k - {0} longrightarrow mathbb{N} cup {0}$ by $d(1) = 1$ and $d(x)=0$ if $x neq 1$. Then this is a norm, because for any $a,b in k$ such that $b neq 0$ we have $a = ab^{-1}b$, and $1|x$ for all nonzero $x$, but $d(1) > d(x)$. However, as this is in some sense an artificial example, this does not answer your question.
$endgroup$
– Rankeya
Oct 14 '11 at 7:47
$begingroup$
Rankeya, thank you, I appreciate your response. This is precisely the example I had come up with--so of course we know now that there is no implication that every Euclidean function satisfies the $d$-inequality, but this example isn't "natural"--not that you ever said it was.
$endgroup$
– MathEnthusiast
Oct 14 '11 at 7:52
$begingroup$
Yes, precisely why I did not post this as an answer, but rather chose to just put it as a comment.
$endgroup$
– Rankeya
Oct 14 '11 at 7:56
add a comment |
1
$begingroup$
I cannot think of a non-trivial one, but here is a trivial example: Let $k$ be a field with at least three elements. Then define the norm $d: k - {0} longrightarrow mathbb{N} cup {0}$ by $d(1) = 1$ and $d(x)=0$ if $x neq 1$. Then this is a norm, because for any $a,b in k$ such that $b neq 0$ we have $a = ab^{-1}b$, and $1|x$ for all nonzero $x$, but $d(1) > d(x)$. However, as this is in some sense an artificial example, this does not answer your question.
$endgroup$
– Rankeya
Oct 14 '11 at 7:47
$begingroup$
Rankeya, thank you, I appreciate your response. This is precisely the example I had come up with--so of course we know now that there is no implication that every Euclidean function satisfies the $d$-inequality, but this example isn't "natural"--not that you ever said it was.
$endgroup$
– MathEnthusiast
Oct 14 '11 at 7:52
$begingroup$
Yes, precisely why I did not post this as an answer, but rather chose to just put it as a comment.
$endgroup$
– Rankeya
Oct 14 '11 at 7:56
1
1
$begingroup$
I cannot think of a non-trivial one, but here is a trivial example: Let $k$ be a field with at least three elements. Then define the norm $d: k - {0} longrightarrow mathbb{N} cup {0}$ by $d(1) = 1$ and $d(x)=0$ if $x neq 1$. Then this is a norm, because for any $a,b in k$ such that $b neq 0$ we have $a = ab^{-1}b$, and $1|x$ for all nonzero $x$, but $d(1) > d(x)$. However, as this is in some sense an artificial example, this does not answer your question.
$endgroup$
– Rankeya
Oct 14 '11 at 7:47
$begingroup$
I cannot think of a non-trivial one, but here is a trivial example: Let $k$ be a field with at least three elements. Then define the norm $d: k - {0} longrightarrow mathbb{N} cup {0}$ by $d(1) = 1$ and $d(x)=0$ if $x neq 1$. Then this is a norm, because for any $a,b in k$ such that $b neq 0$ we have $a = ab^{-1}b$, and $1|x$ for all nonzero $x$, but $d(1) > d(x)$. However, as this is in some sense an artificial example, this does not answer your question.
$endgroup$
– Rankeya
Oct 14 '11 at 7:47
$begingroup$
Rankeya, thank you, I appreciate your response. This is precisely the example I had come up with--so of course we know now that there is no implication that every Euclidean function satisfies the $d$-inequality, but this example isn't "natural"--not that you ever said it was.
$endgroup$
– MathEnthusiast
Oct 14 '11 at 7:52
$begingroup$
Rankeya, thank you, I appreciate your response. This is precisely the example I had come up with--so of course we know now that there is no implication that every Euclidean function satisfies the $d$-inequality, but this example isn't "natural"--not that you ever said it was.
$endgroup$
– MathEnthusiast
Oct 14 '11 at 7:52
$begingroup$
Yes, precisely why I did not post this as an answer, but rather chose to just put it as a comment.
$endgroup$
– Rankeya
Oct 14 '11 at 7:56
$begingroup$
Yes, precisely why I did not post this as an answer, but rather chose to just put it as a comment.
$endgroup$
– Rankeya
Oct 14 '11 at 7:56
add a comment |
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$begingroup$
I cannot think of a non-trivial one, but here is a trivial example: Let $k$ be a field with at least three elements. Then define the norm $d: k - {0} longrightarrow mathbb{N} cup {0}$ by $d(1) = 1$ and $d(x)=0$ if $x neq 1$. Then this is a norm, because for any $a,b in k$ such that $b neq 0$ we have $a = ab^{-1}b$, and $1|x$ for all nonzero $x$, but $d(1) > d(x)$. However, as this is in some sense an artificial example, this does not answer your question.
$endgroup$
– Rankeya
Oct 14 '11 at 7:47
$begingroup$
Rankeya, thank you, I appreciate your response. This is precisely the example I had come up with--so of course we know now that there is no implication that every Euclidean function satisfies the $d$-inequality, but this example isn't "natural"--not that you ever said it was.
$endgroup$
– MathEnthusiast
Oct 14 '11 at 7:52
$begingroup$
Yes, precisely why I did not post this as an answer, but rather chose to just put it as a comment.
$endgroup$
– Rankeya
Oct 14 '11 at 7:56