Do $f(x)=3x+2^nu_2(x)$ and $g(x)=3x+2^nu_2(x)cdot3^nu_3(x)$ have compact support in $Bbb Z_2$? [on hold]

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Do $f(x)=3x+2^nu_2(x)$ and $g(x)=3x+2^nu_2(x)cdot3^nu_3(x)$ have compact support in $Bbb Z_2$?

$f:Bbb Z_2toBbb Z_2$ and $g:Bbb Z_2toBbb Z_2$

$Bbb Z_p$ is the p-adic integers and I understand the topology of any set $Bbb Z_p$ is that of a Cantor set.

$q^nu_q(x)$ is simply the largest power of some prime $q$ that divides $x$, e.g. $2^nu_2(24)=8$

---

I understand compact support to mean the closure of the set of arguments for which the function is non-zero, is a compact set, so I'm assessing compact support in $Bbb Z_2$ under the usual metric.

---

The values of $x$ for which $f(x)=0$ are $-frac2^m3$

So my thinking is that none of these is in $Bbb Z$ and therefore the closure of the set of arguments yielding nonzero $f$ still contains all of $Bbb Z$, whose closure in $lvertcdotrvert_2$ is still $Bbb Z_2$ which is compact, so $f(x)$ has compact support.

As for $g(x)$, and I could possibly do with some help showing this for sure but I think this function is zero nowhere because:

Let $x=2^m3^ry$

Then $g(2^m3^ry)=2^m3^r+1y+2^m3^r=2^m3^r(3y+1)$

This can only be zero for $y=-frac13$, which $y$ cannot be since we have factored its powers of $3$ into $3^r$ therefore $g$ is nonzero for all arguments in $Bbb Z_2$ which is compact so $g$ has compact support too.

Is that correct?

On a related note there seems to be freedom to define $q^nu_q(x)$ as I see fit for $x=0$, although I should probably make some explicit choice as without one, it doesn't seem well-defined. Implicit in the above, is that I have taken it to be $q^nu_q(0)=1$. How should $q^nu_p(0)$ be best chosen? Is there some common or more logical choice, e.g. that best-preserves continuity?

general-topology metric-spaces continuity p-adic-number-theory

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edited Jul 30 at 20:35

asked Jul 29 at 19:11

Robert Frost

3,883936

put on hold as off-topic by Did, user21820, amWhy, Shailesh, José Carlos Santos Aug 5 at 17:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – user21820, amWhy, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-5
down vote

favorite

Do $f(x)=3x+2^nu_2(x)$ and $g(x)=3x+2^nu_2(x)cdot3^nu_3(x)$ have compact support in $Bbb Z_2$?

$f:Bbb Z_2toBbb Z_2$ and $g:Bbb Z_2toBbb Z_2$

$Bbb Z_p$ is the p-adic integers and I understand the topology of any set $Bbb Z_p$ is that of a Cantor set.

$q^nu_q(x)$ is simply the largest power of some prime $q$ that divides $x$, e.g. $2^nu_2(24)=8$

---

I understand compact support to mean the closure of the set of arguments for which the function is non-zero, is a compact set, so I'm assessing compact support in $Bbb Z_2$ under the usual metric.

---

The values of $x$ for which $f(x)=0$ are $-frac2^m3$

So my thinking is that none of these is in $Bbb Z$ and therefore the closure of the set of arguments yielding nonzero $f$ still contains all of $Bbb Z$, whose closure in $lvertcdotrvert_2$ is still $Bbb Z_2$ which is compact, so $f(x)$ has compact support.

As for $g(x)$, and I could possibly do with some help showing this for sure but I think this function is zero nowhere because:

Let $x=2^m3^ry$

Then $g(2^m3^ry)=2^m3^r+1y+2^m3^r=2^m3^r(3y+1)$

This can only be zero for $y=-frac13$, which $y$ cannot be since we have factored its powers of $3$ into $3^r$ therefore $g$ is nonzero for all arguments in $Bbb Z_2$ which is compact so $g$ has compact support too.

Is that correct?

On a related note there seems to be freedom to define $q^nu_q(x)$ as I see fit for $x=0$, although I should probably make some explicit choice as without one, it doesn't seem well-defined. Implicit in the above, is that I have taken it to be $q^nu_q(0)=1$. How should $q^nu_p(0)$ be best chosen? Is there some common or more logical choice, e.g. that best-preserves continuity?

general-topology metric-spaces continuity p-adic-number-theory

share|cite|improve this question

edited Jul 30 at 20:35

asked Jul 29 at 19:11

Robert Frost

3,883936

put on hold as off-topic by Did, user21820, amWhy, Shailesh, José Carlos Santos Aug 5 at 17:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – user21820, amWhy, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-5
down vote

favorite

up vote
-5
down vote

favorite

Do $f(x)=3x+2^nu_2(x)$ and $g(x)=3x+2^nu_2(x)cdot3^nu_3(x)$ have compact support in $Bbb Z_2$?

$f:Bbb Z_2toBbb Z_2$ and $g:Bbb Z_2toBbb Z_2$

$Bbb Z_p$ is the p-adic integers and I understand the topology of any set $Bbb Z_p$ is that of a Cantor set.

$q^nu_q(x)$ is simply the largest power of some prime $q$ that divides $x$, e.g. $2^nu_2(24)=8$

---

I understand compact support to mean the closure of the set of arguments for which the function is non-zero, is a compact set, so I'm assessing compact support in $Bbb Z_2$ under the usual metric.

---

The values of $x$ for which $f(x)=0$ are $-frac2^m3$

So my thinking is that none of these is in $Bbb Z$ and therefore the closure of the set of arguments yielding nonzero $f$ still contains all of $Bbb Z$, whose closure in $lvertcdotrvert_2$ is still $Bbb Z_2$ which is compact, so $f(x)$ has compact support.

As for $g(x)$, and I could possibly do with some help showing this for sure but I think this function is zero nowhere because:

Let $x=2^m3^ry$

Then $g(2^m3^ry)=2^m3^r+1y+2^m3^r=2^m3^r(3y+1)$

This can only be zero for $y=-frac13$, which $y$ cannot be since we have factored its powers of $3$ into $3^r$ therefore $g$ is nonzero for all arguments in $Bbb Z_2$ which is compact so $g$ has compact support too.

Is that correct?

On a related note there seems to be freedom to define $q^nu_q(x)$ as I see fit for $x=0$, although I should probably make some explicit choice as without one, it doesn't seem well-defined. Implicit in the above, is that I have taken it to be $q^nu_q(0)=1$. How should $q^nu_p(0)$ be best chosen? Is there some common or more logical choice, e.g. that best-preserves continuity?

general-topology metric-spaces continuity p-adic-number-theory

share|cite|improve this question

edited Jul 30 at 20:35

asked Jul 29 at 19:11

Robert Frost

3,883936

Do $f(x)=3x+2^nu_2(x)$ and $g(x)=3x+2^nu_2(x)cdot3^nu_3(x)$ have compact support in $Bbb Z_2$?

$f:Bbb Z_2toBbb Z_2$ and $g:Bbb Z_2toBbb Z_2$

$Bbb Z_p$ is the p-adic integers and I understand the topology of any set $Bbb Z_p$ is that of a Cantor set.

$q^nu_q(x)$ is simply the largest power of some prime $q$ that divides $x$, e.g. $2^nu_2(24)=8$

---

I understand compact support to mean the closure of the set of arguments for which the function is non-zero, is a compact set, so I'm assessing compact support in $Bbb Z_2$ under the usual metric.

---

The values of $x$ for which $f(x)=0$ are $-frac2^m3$

So my thinking is that none of these is in $Bbb Z$ and therefore the closure of the set of arguments yielding nonzero $f$ still contains all of $Bbb Z$, whose closure in $lvertcdotrvert_2$ is still $Bbb Z_2$ which is compact, so $f(x)$ has compact support.

As for $g(x)$, and I could possibly do with some help showing this for sure but I think this function is zero nowhere because:

Let $x=2^m3^ry$

Then $g(2^m3^ry)=2^m3^r+1y+2^m3^r=2^m3^r(3y+1)$

This can only be zero for $y=-frac13$, which $y$ cannot be since we have factored its powers of $3$ into $3^r$ therefore $g$ is nonzero for all arguments in $Bbb Z_2$ which is compact so $g$ has compact support too.

Is that correct?

On a related note there seems to be freedom to define $q^nu_q(x)$ as I see fit for $x=0$, although I should probably make some explicit choice as without one, it doesn't seem well-defined. Implicit in the above, is that I have taken it to be $q^nu_q(0)=1$. How should $q^nu_p(0)$ be best chosen? Is there some common or more logical choice, e.g. that best-preserves continuity?

general-topology metric-spaces continuity p-adic-number-theory

share|cite|improve this question

edited Jul 30 at 20:35

asked Jul 29 at 19:11

Robert Frost

3,883936

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 20:35

edited Jul 30 at 20:35

edited Jul 30 at 20:35

asked Jul 29 at 19:11

Robert Frost

3,883936

asked Jul 29 at 19:11

Robert Frost

3,883936

asked Jul 29 at 19:11

Robert Frost

3,883936

3,883936

put on hold as off-topic by Did, user21820, amWhy, Shailesh, José Carlos Santos Aug 5 at 17:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – user21820, amWhy, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Did, user21820, amWhy, Shailesh, José Carlos Santos Aug 5 at 17:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – user21820, amWhy, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

$mathbbZ_2$ is compact, so every closed subset of it is compact. So every function on $mathbbZ_2$ has compact support.

(I would remark, though, that your definition of $g$ is meaningless since $3$ is a unit in $mathbbZ_2$ so $nu_3(x)$ has no meaning for an element of $mathbbZ_2$. The obvious definition of $2^nu_2(0)$ would be $0$, since $2^nu_2(x)$ approaches $0$ as $x$ approaches $0$.)

share|cite|improve this answer

edited Jul 30 at 20:19

answered Jul 30 at 20:11

Eric Wofsey

162k12188298

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank-you in particular for $2^nu_2(0)=0$ as I had come to the same conclusion but was a little uncomfortable with it. On your other point, I want to conclude $g(x)$ is continuous everywhere, but your statement about $nu_3(x)$ is a real fly in the ointment. I was careless to use $p$. Is it now meaningful if I rewrite as I have done using $q$, which may or may not coincide with $p$?
  – Robert Frost
  Jul 30 at 20:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That doesn't change the fact that $nu_3(x)$ is meaningless for $xinmathbbZ_2$...
  – Eric Wofsey
  Jul 30 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need a function on $Bbb Z_2$ with exactly the properties of $3^nu_3(x)$ because I need to preserve the properties of $g(x)$ on all numbers $forall zin Z:zequiv1,2pmod3$ but for it to map multiples of $3$ into $Z$, and for it to commute with $3x$ in the sense that $3cdot g(x)=g(3x)$. I can see why $nu_p(x)$ has special significance in $Bbb Z_p$, a significance which is not there for $nu_q(x):qneq p$, which I think is the thrust of what you're saying. Is that the full extent of what you mean saying the function is meaningless?
  – Robert Frost
  Jul 30 at 20:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Every element of $mathbbZ_2$ is a multiple of $3$, so I have no idea what you are hoping for your function to accomplish.
  – Eric Wofsey
  Jul 30 at 20:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Right, so there's something abstract going on I hadn't appreciated. I need a function $h(x)$ having the properties $3^h(18)=9$ etc.
  – Robert Frost
  Jul 30 at 20:51
  

  
  
  
  

 | 
show 2 more comments

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

$mathbbZ_2$ is compact, so every closed subset of it is compact. So every function on $mathbbZ_2$ has compact support.

(I would remark, though, that your definition of $g$ is meaningless since $3$ is a unit in $mathbbZ_2$ so $nu_3(x)$ has no meaning for an element of $mathbbZ_2$. The obvious definition of $2^nu_2(0)$ would be $0$, since $2^nu_2(x)$ approaches $0$ as $x$ approaches $0$.)

share|cite|improve this answer

edited Jul 30 at 20:19

answered Jul 30 at 20:11

Eric Wofsey

162k12188298

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank-you in particular for $2^nu_2(0)=0$ as I had come to the same conclusion but was a little uncomfortable with it. On your other point, I want to conclude $g(x)$ is continuous everywhere, but your statement about $nu_3(x)$ is a real fly in the ointment. I was careless to use $p$. Is it now meaningful if I rewrite as I have done using $q$, which may or may not coincide with $p$?
  – Robert Frost
  Jul 30 at 20:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That doesn't change the fact that $nu_3(x)$ is meaningless for $xinmathbbZ_2$...
  – Eric Wofsey
  Jul 30 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need a function on $Bbb Z_2$ with exactly the properties of $3^nu_3(x)$ because I need to preserve the properties of $g(x)$ on all numbers $forall zin Z:zequiv1,2pmod3$ but for it to map multiples of $3$ into $Z$, and for it to commute with $3x$ in the sense that $3cdot g(x)=g(3x)$. I can see why $nu_p(x)$ has special significance in $Bbb Z_p$, a significance which is not there for $nu_q(x):qneq p$, which I think is the thrust of what you're saying. Is that the full extent of what you mean saying the function is meaningless?
  – Robert Frost
  Jul 30 at 20:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Every element of $mathbbZ_2$ is a multiple of $3$, so I have no idea what you are hoping for your function to accomplish.
  – Eric Wofsey
  Jul 30 at 20:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Right, so there's something abstract going on I hadn't appreciated. I need a function $h(x)$ having the properties $3^h(18)=9$ etc.
  – Robert Frost
  Jul 30 at 20:51
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote



accepted

$mathbbZ_2$ is compact, so every closed subset of it is compact. So every function on $mathbbZ_2$ has compact support.

(I would remark, though, that your definition of $g$ is meaningless since $3$ is a unit in $mathbbZ_2$ so $nu_3(x)$ has no meaning for an element of $mathbbZ_2$. The obvious definition of $2^nu_2(0)$ would be $0$, since $2^nu_2(x)$ approaches $0$ as $x$ approaches $0$.)

share|cite|improve this answer

edited Jul 30 at 20:19

answered Jul 30 at 20:11

Eric Wofsey

162k12188298

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank-you in particular for $2^nu_2(0)=0$ as I had come to the same conclusion but was a little uncomfortable with it. On your other point, I want to conclude $g(x)$ is continuous everywhere, but your statement about $nu_3(x)$ is a real fly in the ointment. I was careless to use $p$. Is it now meaningful if I rewrite as I have done using $q$, which may or may not coincide with $p$?
  – Robert Frost
  Jul 30 at 20:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That doesn't change the fact that $nu_3(x)$ is meaningless for $xinmathbbZ_2$...
  – Eric Wofsey
  Jul 30 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need a function on $Bbb Z_2$ with exactly the properties of $3^nu_3(x)$ because I need to preserve the properties of $g(x)$ on all numbers $forall zin Z:zequiv1,2pmod3$ but for it to map multiples of $3$ into $Z$, and for it to commute with $3x$ in the sense that $3cdot g(x)=g(3x)$. I can see why $nu_p(x)$ has special significance in $Bbb Z_p$, a significance which is not there for $nu_q(x):qneq p$, which I think is the thrust of what you're saying. Is that the full extent of what you mean saying the function is meaningless?
  – Robert Frost
  Jul 30 at 20:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Every element of $mathbbZ_2$ is a multiple of $3$, so I have no idea what you are hoping for your function to accomplish.
  – Eric Wofsey
  Jul 30 at 20:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Right, so there's something abstract going on I hadn't appreciated. I need a function $h(x)$ having the properties $3^h(18)=9$ etc.
  – Robert Frost
  Jul 30 at 20:51
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote



accepted

up vote
0
down vote



accepted

$mathbbZ_2$ is compact, so every closed subset of it is compact. So every function on $mathbbZ_2$ has compact support.

(I would remark, though, that your definition of $g$ is meaningless since $3$ is a unit in $mathbbZ_2$ so $nu_3(x)$ has no meaning for an element of $mathbbZ_2$. The obvious definition of $2^nu_2(0)$ would be $0$, since $2^nu_2(x)$ approaches $0$ as $x$ approaches $0$.)

share|cite|improve this answer

edited Jul 30 at 20:19

answered Jul 30 at 20:11

Eric Wofsey

162k12188298

$mathbbZ_2$ is compact, so every closed subset of it is compact. So every function on $mathbbZ_2$ has compact support.

(I would remark, though, that your definition of $g$ is meaningless since $3$ is a unit in $mathbbZ_2$ so $nu_3(x)$ has no meaning for an element of $mathbbZ_2$. The obvious definition of $2^nu_2(0)$ would be $0$, since $2^nu_2(x)$ approaches $0$ as $x$ approaches $0$.)

share|cite|improve this answer

edited Jul 30 at 20:19

answered Jul 30 at 20:11

Eric Wofsey

162k12188298

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 20:19

edited Jul 30 at 20:19

edited Jul 30 at 20:19

answered Jul 30 at 20:11

Eric Wofsey

162k12188298

answered Jul 30 at 20:11

Eric Wofsey

162k12188298

answered Jul 30 at 20:11

Eric Wofsey

162k12188298

162k12188298

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank-you in particular for $2^nu_2(0)=0$ as I had come to the same conclusion but was a little uncomfortable with it. On your other point, I want to conclude $g(x)$ is continuous everywhere, but your statement about $nu_3(x)$ is a real fly in the ointment. I was careless to use $p$. Is it now meaningful if I rewrite as I have done using $q$, which may or may not coincide with $p$?
  – Robert Frost
  Jul 30 at 20:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That doesn't change the fact that $nu_3(x)$ is meaningless for $xinmathbbZ_2$...
  – Eric Wofsey
  Jul 30 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need a function on $Bbb Z_2$ with exactly the properties of $3^nu_3(x)$ because I need to preserve the properties of $g(x)$ on all numbers $forall zin Z:zequiv1,2pmod3$ but for it to map multiples of $3$ into $Z$, and for it to commute with $3x$ in the sense that $3cdot g(x)=g(3x)$. I can see why $nu_p(x)$ has special significance in $Bbb Z_p$, a significance which is not there for $nu_q(x):qneq p$, which I think is the thrust of what you're saying. Is that the full extent of what you mean saying the function is meaningless?
  – Robert Frost
  Jul 30 at 20:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Every element of $mathbbZ_2$ is a multiple of $3$, so I have no idea what you are hoping for your function to accomplish.
  – Eric Wofsey
  Jul 30 at 20:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Right, so there's something abstract going on I hadn't appreciated. I need a function $h(x)$ having the properties $3^h(18)=9$ etc.
  – Robert Frost
  Jul 30 at 20:51
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank-you in particular for $2^nu_2(0)=0$ as I had come to the same conclusion but was a little uncomfortable with it. On your other point, I want to conclude $g(x)$ is continuous everywhere, but your statement about $nu_3(x)$ is a real fly in the ointment. I was careless to use $p$. Is it now meaningful if I rewrite as I have done using $q$, which may or may not coincide with $p$?
  – Robert Frost
  Jul 30 at 20:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That doesn't change the fact that $nu_3(x)$ is meaningless for $xinmathbbZ_2$...
  – Eric Wofsey
  Jul 30 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need a function on $Bbb Z_2$ with exactly the properties of $3^nu_3(x)$ because I need to preserve the properties of $g(x)$ on all numbers $forall zin Z:zequiv1,2pmod3$ but for it to map multiples of $3$ into $Z$, and for it to commute with $3x$ in the sense that $3cdot g(x)=g(3x)$. I can see why $nu_p(x)$ has special significance in $Bbb Z_p$, a significance which is not there for $nu_q(x):qneq p$, which I think is the thrust of what you're saying. Is that the full extent of what you mean saying the function is meaningless?
  – Robert Frost
  Jul 30 at 20:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Every element of $mathbbZ_2$ is a multiple of $3$, so I have no idea what you are hoping for your function to accomplish.
  – Eric Wofsey
  Jul 30 at 20:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Right, so there's something abstract going on I hadn't appreciated. I need a function $h(x)$ having the properties $3^h(18)=9$ etc.
  – Robert Frost
  Jul 30 at 20:51
  

  
  
  
  

Thank-you in particular for $2^nu_2(0)=0$ as I had come to the same conclusion but was a little uncomfortable with it. On your other point, I want to conclude $g(x)$ is continuous everywhere, but your statement about $nu_3(x)$ is a real fly in the ointment. I was careless to use $p$. Is it now meaningful if I rewrite as I have done using $q$, which may or may not coincide with $p$?
– Robert Frost
Jul 30 at 20:35

Thank-you in particular for $2^nu_2(0)=0$ as I had come to the same conclusion but was a little uncomfortable with it. On your other point, I want to conclude $g(x)$ is continuous everywhere, but your statement about $nu_3(x)$ is a real fly in the ointment. I was careless to use $p$. Is it now meaningful if I rewrite as I have done using $q$, which may or may not coincide with $p$?
– Robert Frost
Jul 30 at 20:35

That doesn't change the fact that $nu_3(x)$ is meaningless for $xinmathbbZ_2$...
– Eric Wofsey
Jul 30 at 20:40

That doesn't change the fact that $nu_3(x)$ is meaningless for $xinmathbbZ_2$...
– Eric Wofsey
Jul 30 at 20:40

I need a function on $Bbb Z_2$ with exactly the properties of $3^nu_3(x)$ because I need to preserve the properties of $g(x)$ on all numbers $forall zin Z:zequiv1,2pmod3$ but for it to map multiples of $3$ into $Z$, and for it to commute with $3x$ in the sense that $3cdot g(x)=g(3x)$. I can see why $nu_p(x)$ has special significance in $Bbb Z_p$, a significance which is not there for $nu_q(x):qneq p$, which I think is the thrust of what you're saying. Is that the full extent of what you mean saying the function is meaningless?
– Robert Frost
Jul 30 at 20:47

I need a function on $Bbb Z_2$ with exactly the properties of $3^nu_3(x)$ because I need to preserve the properties of $g(x)$ on all numbers $forall zin Z:zequiv1,2pmod3$ but for it to map multiples of $3$ into $Z$, and for it to commute with $3x$ in the sense that $3cdot g(x)=g(3x)$. I can see why $nu_p(x)$ has special significance in $Bbb Z_p$, a significance which is not there for $nu_q(x):qneq p$, which I think is the thrust of what you're saying. Is that the full extent of what you mean saying the function is meaningless?
– Robert Frost
Jul 30 at 20:47

Every element of $mathbbZ_2$ is a multiple of $3$, so I have no idea what you are hoping for your function to accomplish.
– Eric Wofsey
Jul 30 at 20:49

Every element of $mathbbZ_2$ is a multiple of $3$, so I have no idea what you are hoping for your function to accomplish.
– Eric Wofsey
Jul 30 at 20:49

Right, so there's something abstract going on I hadn't appreciated. I need a function $h(x)$ having the properties $3^h(18)=9$ etc.
– Robert Frost
Jul 30 at 20:51

Right, so there's something abstract going on I hadn't appreciated. I need a function $h(x)$ having the properties $3^h(18)=9$ etc.
– Robert Frost
Jul 30 at 20:51

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