5-adic order of $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$












1












$begingroup$


Let $pi:$(the 5-adic integers)$to mathbb{Z}/5mathbb{Z}$ be the reduction map.



Let $f:mathbb{Z}/5 mathbb{Z} to $ (the 5-adic integers) have the following properties $forall x,yin mathbb{Z}/5mathbb{Z}$




  1. $f(x)f(y)=f(xy)$


  2. $f(x)^{p-1}=1$ unless $f(x)=0$


  3. $pi(f(x))=x$



Also, let us adjoin a primitive $5$th root of unity, call it $zeta_5=zeta$, to the 5-adic integers.




I want to find the 5-adic order of $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$




I see that the order of $1-zeta^i$ is $frac 1 4$ for $inotequiv0$ (mod 5).



I also see that $f(overline 4)=-1$ and that therefore $f(overline 2)^2=-1$ and $f(overline 3)^2=-1$. Perhaps this can be used to rewrite the sum $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$ in a "better" way.






abstract-algebra field-theory prime-numbers roots-of-unity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I would certainly write $f(overline 2)$ as good-old $i$, square root of $-1$. Then your expression is $zeta+i^3zeta^2+izeta^3 +zeta^4$.
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:19












  • $begingroup$
    @Lubin Certainly makes notation a lot easier to work with. Thanks for the idea. I feel like multiplying by some $i^aζ^b$ term would help.
    $endgroup$
    – Pascal's Wager
    Dec 18 '18 at 2:25












  • $begingroup$
    Using a symbolic algebra package, I seem to find that your number is a unit in its ring. Does that seem right, or wrong to you?
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:33










  • $begingroup$
    @Lubin I think that's right.
    $endgroup$
    – Pascal's Wager
    Dec 18 '18 at 2:36










  • $begingroup$
    Then I suggest defining $sigma$ in the Galois group over $Bbb Q_5$, $sigma(zeta)=zeta^2$, it’s a generator, and taking the Norm of your element, which just means replacing $zeta$ by $zeta^2$, $zeta^3$, and $zeta^4$ in your expression, and multiplying all four four-term factors together. You’ll probably get a lot of collapsing, to make the result obvious. (I’m displaying my ignorance of group-representation theory here, of course. To those who are not so ignorant, your result is probably obvious.)
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:41
















1












$begingroup$


Let $pi:$(the 5-adic integers)$to mathbb{Z}/5mathbb{Z}$ be the reduction map.



Let $f:mathbb{Z}/5 mathbb{Z} to $ (the 5-adic integers) have the following properties $forall x,yin mathbb{Z}/5mathbb{Z}$




  1. $f(x)f(y)=f(xy)$


  2. $f(x)^{p-1}=1$ unless $f(x)=0$


  3. $pi(f(x))=x$



Also, let us adjoin a primitive $5$th root of unity, call it $zeta_5=zeta$, to the 5-adic integers.




I want to find the 5-adic order of $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$




I see that the order of $1-zeta^i$ is $frac 1 4$ for $inotequiv0$ (mod 5).



I also see that $f(overline 4)=-1$ and that therefore $f(overline 2)^2=-1$ and $f(overline 3)^2=-1$. Perhaps this can be used to rewrite the sum $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$ in a "better" way.






abstract-algebra field-theory prime-numbers roots-of-unity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I would certainly write $f(overline 2)$ as good-old $i$, square root of $-1$. Then your expression is $zeta+i^3zeta^2+izeta^3 +zeta^4$.
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:19












  • $begingroup$
    @Lubin Certainly makes notation a lot easier to work with. Thanks for the idea. I feel like multiplying by some $i^aζ^b$ term would help.
    $endgroup$
    – Pascal's Wager
    Dec 18 '18 at 2:25












  • $begingroup$
    Using a symbolic algebra package, I seem to find that your number is a unit in its ring. Does that seem right, or wrong to you?
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:33










  • $begingroup$
    @Lubin I think that's right.
    $endgroup$
    – Pascal's Wager
    Dec 18 '18 at 2:36










  • $begingroup$
    Then I suggest defining $sigma$ in the Galois group over $Bbb Q_5$, $sigma(zeta)=zeta^2$, it’s a generator, and taking the Norm of your element, which just means replacing $zeta$ by $zeta^2$, $zeta^3$, and $zeta^4$ in your expression, and multiplying all four four-term factors together. You’ll probably get a lot of collapsing, to make the result obvious. (I’m displaying my ignorance of group-representation theory here, of course. To those who are not so ignorant, your result is probably obvious.)
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:41














1












1








1





$begingroup$


Let $pi:$(the 5-adic integers)$to mathbb{Z}/5mathbb{Z}$ be the reduction map.



Let $f:mathbb{Z}/5 mathbb{Z} to $ (the 5-adic integers) have the following properties $forall x,yin mathbb{Z}/5mathbb{Z}$




  1. $f(x)f(y)=f(xy)$


  2. $f(x)^{p-1}=1$ unless $f(x)=0$


  3. $pi(f(x))=x$



Also, let us adjoin a primitive $5$th root of unity, call it $zeta_5=zeta$, to the 5-adic integers.




I want to find the 5-adic order of $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$




I see that the order of $1-zeta^i$ is $frac 1 4$ for $inotequiv0$ (mod 5).



I also see that $f(overline 4)=-1$ and that therefore $f(overline 2)^2=-1$ and $f(overline 3)^2=-1$. Perhaps this can be used to rewrite the sum $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$ in a "better" way.






abstract-algebra field-theory prime-numbers roots-of-unity







share|cite|improve this question











$endgroup$




Let $pi:$(the 5-adic integers)$to mathbb{Z}/5mathbb{Z}$ be the reduction map.



Let $f:mathbb{Z}/5 mathbb{Z} to $ (the 5-adic integers) have the following properties $forall x,yin mathbb{Z}/5mathbb{Z}$




  1. $f(x)f(y)=f(xy)$


  2. $f(x)^{p-1}=1$ unless $f(x)=0$


  3. $pi(f(x))=x$



Also, let us adjoin a primitive $5$th root of unity, call it $zeta_5=zeta$, to the 5-adic integers.




I want to find the 5-adic order of $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$




I see that the order of $1-zeta^i$ is $frac 1 4$ for $inotequiv0$ (mod 5).



I also see that $f(overline 4)=-1$ and that therefore $f(overline 2)^2=-1$ and $f(overline 3)^2=-1$. Perhaps this can be used to rewrite the sum $zeta-f(overline 2)zeta^2+f(overline 2)zeta^3+zeta^4$ in a "better" way.






abstract-algebra field-theory prime-numbers roots-of-unity




abstract-algebra field-theory prime-numbers roots-of-unity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 2:37









Kemono Chen

3,1991844




3,1991844










asked Dec 18 '18 at 2:01









Pascal's WagerPascal's Wager

371315




371315








  • 1




    $begingroup$
    I would certainly write $f(overline 2)$ as good-old $i$, square root of $-1$. Then your expression is $zeta+i^3zeta^2+izeta^3 +zeta^4$.
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:19












  • $begingroup$
    @Lubin Certainly makes notation a lot easier to work with. Thanks for the idea. I feel like multiplying by some $i^aζ^b$ term would help.
    $endgroup$
    – Pascal's Wager
    Dec 18 '18 at 2:25












  • $begingroup$
    Using a symbolic algebra package, I seem to find that your number is a unit in its ring. Does that seem right, or wrong to you?
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:33










  • $begingroup$
    @Lubin I think that's right.
    $endgroup$
    – Pascal's Wager
    Dec 18 '18 at 2:36










  • $begingroup$
    Then I suggest defining $sigma$ in the Galois group over $Bbb Q_5$, $sigma(zeta)=zeta^2$, it’s a generator, and taking the Norm of your element, which just means replacing $zeta$ by $zeta^2$, $zeta^3$, and $zeta^4$ in your expression, and multiplying all four four-term factors together. You’ll probably get a lot of collapsing, to make the result obvious. (I’m displaying my ignorance of group-representation theory here, of course. To those who are not so ignorant, your result is probably obvious.)
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:41














  • 1




    $begingroup$
    I would certainly write $f(overline 2)$ as good-old $i$, square root of $-1$. Then your expression is $zeta+i^3zeta^2+izeta^3 +zeta^4$.
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:19












  • $begingroup$
    @Lubin Certainly makes notation a lot easier to work with. Thanks for the idea. I feel like multiplying by some $i^aζ^b$ term would help.
    $endgroup$
    – Pascal's Wager
    Dec 18 '18 at 2:25












  • $begingroup$
    Using a symbolic algebra package, I seem to find that your number is a unit in its ring. Does that seem right, or wrong to you?
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:33










  • $begingroup$
    @Lubin I think that's right.
    $endgroup$
    – Pascal's Wager
    Dec 18 '18 at 2:36










  • $begingroup$
    Then I suggest defining $sigma$ in the Galois group over $Bbb Q_5$, $sigma(zeta)=zeta^2$, it’s a generator, and taking the Norm of your element, which just means replacing $zeta$ by $zeta^2$, $zeta^3$, and $zeta^4$ in your expression, and multiplying all four four-term factors together. You’ll probably get a lot of collapsing, to make the result obvious. (I’m displaying my ignorance of group-representation theory here, of course. To those who are not so ignorant, your result is probably obvious.)
    $endgroup$
    – Lubin
    Dec 18 '18 at 2:41








1




1




$begingroup$
I would certainly write $f(overline 2)$ as good-old $i$, square root of $-1$. Then your expression is $zeta+i^3zeta^2+izeta^3 +zeta^4$.
$endgroup$
– Lubin
Dec 18 '18 at 2:19






$begingroup$
I would certainly write $f(overline 2)$ as good-old $i$, square root of $-1$. Then your expression is $zeta+i^3zeta^2+izeta^3 +zeta^4$.
$endgroup$
– Lubin
Dec 18 '18 at 2:19














$begingroup$
@Lubin Certainly makes notation a lot easier to work with. Thanks for the idea. I feel like multiplying by some $i^aζ^b$ term would help.
$endgroup$
– Pascal's Wager
Dec 18 '18 at 2:25






$begingroup$
@Lubin Certainly makes notation a lot easier to work with. Thanks for the idea. I feel like multiplying by some $i^aζ^b$ term would help.
$endgroup$
– Pascal's Wager
Dec 18 '18 at 2:25














$begingroup$
Using a symbolic algebra package, I seem to find that your number is a unit in its ring. Does that seem right, or wrong to you?
$endgroup$
– Lubin
Dec 18 '18 at 2:33




$begingroup$
Using a symbolic algebra package, I seem to find that your number is a unit in its ring. Does that seem right, or wrong to you?
$endgroup$
– Lubin
Dec 18 '18 at 2:33












$begingroup$
@Lubin I think that's right.
$endgroup$
– Pascal's Wager
Dec 18 '18 at 2:36




$begingroup$
@Lubin I think that's right.
$endgroup$
– Pascal's Wager
Dec 18 '18 at 2:36












$begingroup$
Then I suggest defining $sigma$ in the Galois group over $Bbb Q_5$, $sigma(zeta)=zeta^2$, it’s a generator, and taking the Norm of your element, which just means replacing $zeta$ by $zeta^2$, $zeta^3$, and $zeta^4$ in your expression, and multiplying all four four-term factors together. You’ll probably get a lot of collapsing, to make the result obvious. (I’m displaying my ignorance of group-representation theory here, of course. To those who are not so ignorant, your result is probably obvious.)
$endgroup$
– Lubin
Dec 18 '18 at 2:41




$begingroup$
Then I suggest defining $sigma$ in the Galois group over $Bbb Q_5$, $sigma(zeta)=zeta^2$, it’s a generator, and taking the Norm of your element, which just means replacing $zeta$ by $zeta^2$, $zeta^3$, and $zeta^4$ in your expression, and multiplying all four four-term factors together. You’ll probably get a lot of collapsing, to make the result obvious. (I’m displaying my ignorance of group-representation theory here, of course. To those who are not so ignorant, your result is probably obvious.)
$endgroup$
– Lubin
Dec 18 '18 at 2:41










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