In a Euclidean domain's division with remainders, is $r=0$ if $t$ divides $s$?
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A Euclidean domain $D$ has a function $F$ from nonzero elements of $D$ to nonnegative integers that makes division with remainder possible in $D$: Let $D$ have elements $t$ and $s$, and $t$ is nonzero. Then $D$ has elements $q$ and $r$ for which $s = tq + r$, and if the remainder $r$ is nonzero too, then $F(t) > F(r)$. If $r$ happens to be zero, then $t$ divides $s$ because $s=tq$. If $t$ happens to divides $s$, then is $r$ zero? If $t$ divides $s$, then $s = ut$ for an element $u$ in $D$. Then $r=t(u-q)$ is an element of the ideal generated by $t$, so the ideal generated by $r$ is contained in the ideal $(t)$.
Do any of these mean $r=0$?
We have $t$ divides $r$, which is unusual for nonzero remainders, but, unlike with the Euclidean domain of the integers $mathbb Z$, I don't see the precise contradiction for a general Euclidean domain $D$. Bernard poses a good question. Usually, the $F$ that allows us to deduce $r=0$.
Something that might help: Can we deduce the existence of a function $G$ that satisfies $G(p) le G(pv)$ and $G(p)=0$ if and only if $p=0$?
Thanks, and have a happy new year!
abstract-algebra euclidean-domain
edited Jan 20 at 9:27
user26857
39.3k124183
asked Dec 31 '18 at 16:53
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
|
show 5 more comments
$begingroup$
A Euclidean domain $D$ has a function $F$ from nonzero elements of $D$ to nonnegative integers that makes division with remainder possible in $D$: Let $D$ have elements $t$ and $s$, and $t$ is nonzero. Then $D$ has elements $q$ and $r$ for which $s = tq + r$, and if the remainder $r$ is nonzero too, then $F(t) > F(r)$. If $r$ happens to be zero, then $t$ divides $s$ because $s=tq$. If $t$ happens to divides $s$, then is $r$ zero? If $t$ divides $s$, then $s = ut$ for an element $u$ in $D$. Then $r=t(u-q)$ is an element of the ideal generated by $t$, so the ideal generated by $r$ is contained in the ideal $(t)$.
Do any of these mean $r=0$?
We have $t$ divides $r$, which is unusual for nonzero remainders, but, unlike with the Euclidean domain of the integers $mathbb Z$, I don't see the precise contradiction for a general Euclidean domain $D$. Bernard poses a good question. Usually, the $F$ that allows us to deduce $r=0$.
Something that might help: Can we deduce the existence of a function $G$ that satisfies $G(p) le G(pv)$ and $G(p)=0$ if and only if $p=0$?
Thanks, and have a happy new year!
abstract-algebra euclidean-domain
edited Jan 20 at 9:27
user26857
39.3k124183
asked Dec 31 '18 at 16:53
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
$begingroup$
Do you suppose that you will stop editing your question eventually?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 16:55
$begingroup$
@JoséCarlosSantos It wasn't working. I stopped editing already.
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 16:56
$begingroup$
What is $F$ in this context?
$endgroup$
– Bernard
Dec 31 '18 at 17:25
$begingroup$
@Bernard any function from nonzero elements of D to nonnegative integers for which the division with remainder is possible. What's the difference if I use the other definition of F, for which $F(a) le F(ab)$?
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 17:31
$begingroup$
Any function, really?
$endgroup$
– Bernard
Dec 31 '18 at 17:32
|
show 5 more comments
$begingroup$
A Euclidean domain $D$ has a function $F$ from nonzero elements of $D$ to nonnegative integers that makes division with remainder possible in $D$: Let $D$ have elements $t$ and $s$, and $t$ is nonzero. Then $D$ has elements $q$ and $r$ for which $s = tq + r$, and if the remainder $r$ is nonzero too, then $F(t) > F(r)$. If $r$ happens to be zero, then $t$ divides $s$ because $s=tq$. If $t$ happens to divides $s$, then is $r$ zero? If $t$ divides $s$, then $s = ut$ for an element $u$ in $D$. Then $r=t(u-q)$ is an element of the ideal generated by $t$, so the ideal generated by $r$ is contained in the ideal $(t)$.
Do any of these mean $r=0$?
We have $t$ divides $r$, which is unusual for nonzero remainders, but, unlike with the Euclidean domain of the integers $mathbb Z$, I don't see the precise contradiction for a general Euclidean domain $D$. Bernard poses a good question. Usually, the $F$ that allows us to deduce $r=0$.
Something that might help: Can we deduce the existence of a function $G$ that satisfies $G(p) le G(pv)$ and $G(p)=0$ if and only if $p=0$?
Thanks, and have a happy new year!
abstract-algebra euclidean-domain
edited Jan 20 at 9:27
user26857
39.3k124183
asked Dec 31 '18 at 16:53
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
$endgroup$
A Euclidean domain $D$ has a function $F$ from nonzero elements of $D$ to nonnegative integers that makes division with remainder possible in $D$: Let $D$ have elements $t$ and $s$, and $t$ is nonzero. Then $D$ has elements $q$ and $r$ for which $s = tq + r$, and if the remainder $r$ is nonzero too, then $F(t) > F(r)$. If $r$ happens to be zero, then $t$ divides $s$ because $s=tq$. If $t$ happens to divides $s$, then is $r$ zero? If $t$ divides $s$, then $s = ut$ for an element $u$ in $D$. Then $r=t(u-q)$ is an element of the ideal generated by $t$, so the ideal generated by $r$ is contained in the ideal $(t)$.
Do any of these mean $r=0$?
We have $t$ divides $r$, which is unusual for nonzero remainders, but, unlike with the Euclidean domain of the integers $mathbb Z$, I don't see the precise contradiction for a general Euclidean domain $D$. Bernard poses a good question. Usually, the $F$ that allows us to deduce $r=0$.
Something that might help: Can we deduce the existence of a function $G$ that satisfies $G(p) le G(pv)$ and $G(p)=0$ if and only if $p=0$?
Thanks, and have a happy new year!
abstract-algebra euclidean-domain
abstract-algebra euclidean-domain
edited Jan 20 at 9:27
user26857
39.3k124183
asked Dec 31 '18 at 16:53
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
edited Jan 20 at 9:27
user26857
39.3k124183
asked Dec 31 '18 at 16:53
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
edited Jan 20 at 9:27
user26857
39.3k124183
edited Jan 20 at 9:27
user26857
39.3k124183
edited Jan 20 at 9:27
user26857
39.3k124183
39.3k124183
asked Dec 31 '18 at 16:53
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Dec 31 '18 at 16:53
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
asked Dec 31 '18 at 16:53
Ekhin Taylor R. WilsonEkhin Taylor R. Wilson
558
558
$begingroup$
Do you suppose that you will stop editing your question eventually?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 16:55
$begingroup$
@JoséCarlosSantos It wasn't working. I stopped editing already.
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 16:56
$begingroup$
What is $F$ in this context?
$endgroup$
– Bernard
Dec 31 '18 at 17:25
$begingroup$
@Bernard any function from nonzero elements of D to nonnegative integers for which the division with remainder is possible. What's the difference if I use the other definition of F, for which $F(a) le F(ab)$?
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 17:31
$begingroup$
Any function, really?
$endgroup$
– Bernard
Dec 31 '18 at 17:32
|
show 5 more comments
$begingroup$
Do you suppose that you will stop editing your question eventually?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 16:55
$begingroup$
@JoséCarlosSantos It wasn't working. I stopped editing already.
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 16:56
$begingroup$
What is $F$ in this context?
$endgroup$
– Bernard
Dec 31 '18 at 17:25
$begingroup$
@Bernard any function from nonzero elements of D to nonnegative integers for which the division with remainder is possible. What's the difference if I use the other definition of F, for which $F(a) le F(ab)$?
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 17:31
$begingroup$
Any function, really?
$endgroup$
– Bernard
Dec 31 '18 at 17:32
$begingroup$
Do you suppose that you will stop editing your question eventually?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 16:55
$begingroup$
Do you suppose that you will stop editing your question eventually?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 16:55
$begingroup$
@JoséCarlosSantos It wasn't working. I stopped editing already.
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 16:56
$begingroup$
@JoséCarlosSantos It wasn't working. I stopped editing already.
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 16:56
$begingroup$
What is $F$ in this context?
$endgroup$
– Bernard
Dec 31 '18 at 17:25
$begingroup$
What is $F$ in this context?
$endgroup$
– Bernard
Dec 31 '18 at 17:25
$begingroup$
@Bernard any function from nonzero elements of D to nonnegative integers for which the division with remainder is possible. What's the difference if I use the other definition of F, for which $F(a) le F(ab)$?
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 17:31
$begingroup$
@Bernard any function from nonzero elements of D to nonnegative integers for which the division with remainder is possible. What's the difference if I use the other definition of F, for which $F(a) le F(ab)$?
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 17:31
$begingroup$
Any function, really?
$endgroup$
– Bernard
Dec 31 '18 at 17:32
$begingroup$
Any function, really?
$endgroup$
– Bernard
Dec 31 '18 at 17:32
|
show 5 more comments
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$begingroup$
Do you suppose that you will stop editing your question eventually?
$endgroup$
– José Carlos Santos
Dec 31 '18 at 16:55
$begingroup$
@JoséCarlosSantos It wasn't working. I stopped editing already.
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 16:56
$begingroup$
What is $F$ in this context?
$endgroup$
– Bernard
Dec 31 '18 at 17:25
$begingroup$
@Bernard any function from nonzero elements of D to nonnegative integers for which the division with remainder is possible. What's the difference if I use the other definition of F, for which $F(a) le F(ab)$?
$endgroup$
– Ekhin Taylor R. Wilson
Dec 31 '18 at 17:31
$begingroup$
Any function, really?
$endgroup$
– Bernard
Dec 31 '18 at 17:32