Mixing
Primitive roots of unity and $I$-adically separated rings.
Clash Royale CLAN TAG#URR8PPP
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Let $R$ be an integral domain with $char(R) = 0$ and $zeta, zeta'$ two primitive roots of unity in $R$. The following are equivalent.
(1) $(q-zeta)^m in (q-zeta') + I[q]$ for som $m geq 0$ and an ideal $I subset R$ such that $R$ is $I$-adically separated (note: $(q-zeta')$ on the right side denotes the ideal generated by $q- zeta'$).
(2) $R$ is $(zeta-zeta')$-adically separated.
(3) $ord(zeta^-1zeta') = p^k$ for some prime $p$ and $j in mathbbZsetminus 0$ such that $R$ is $(p)$-adically separated.
I believe I have just the implication (2) $implies$ (1). If $R$ is $(zeta-zeta')$-adically separated, then $cap_jgeq 0(zeta-zeta')^j = (0)$. We have to show that $(q-zeta)^m in (q-zeta') + I[q]$ for some $m geq 0$ and an ideal $I subset R$ such that $R$ is $I$-adically separated. But
$$q-zeta in (q-zeta') + (zeta-zeta')[q]$$
since $q-zeta = (q-zeta') + (zeta'-zeta)$.
I sort of "have" the proof of the reciprocal, (1)$implies$ (2), but I don't understand a step. The author says that since $(q-zeta)^m in (q-zeta') + I[q]$ for some $m geq$ and $I subset R$ such that $R$ is $I$-adically separated, then $(zeta-zeta')^m in I$. This is what I don't find obvious at all. From that I believe the result follows since the ideal generated by (zeta - zeta')^m would be then contained on $I$, so
$$bigcap_jgeq 0(zeta-zeta')^j subset bigcap_jgeq 0I^j = (0)$$
and hence $R$ is $(zeta-zeta')$-adically separated.
For any of the implications (2)$implies$(3) or (3)$implies$(2) I literally have no clue. Any hero would be nice. Thank you!
abstract-algebra ring-theory commutative-algebra ideals roots-of-unity
asked Jul 15 at 18:35
user313212
622519
add a comment |Â
up vote
3
down vote
favorite
Let $R$ be an integral domain with $char(R) = 0$ and $zeta, zeta'$ two primitive roots of unity in $R$. The following are equivalent.
(1) $(q-zeta)^m in (q-zeta') + I[q]$ for som $m geq 0$ and an ideal $I subset R$ such that $R$ is $I$-adically separated (note: $(q-zeta')$ on the right side denotes the ideal generated by $q- zeta'$).
(2) $R$ is $(zeta-zeta')$-adically separated.
(3) $ord(zeta^-1zeta') = p^k$ for some prime $p$ and $j in mathbbZsetminus 0$ such that $R$ is $(p)$-adically separated.
I believe I have just the implication (2) $implies$ (1). If $R$ is $(zeta-zeta')$-adically separated, then $cap_jgeq 0(zeta-zeta')^j = (0)$. We have to show that $(q-zeta)^m in (q-zeta') + I[q]$ for some $m geq 0$ and an ideal $I subset R$ such that $R$ is $I$-adically separated. But
$$q-zeta in (q-zeta') + (zeta-zeta')[q]$$
since $q-zeta = (q-zeta') + (zeta'-zeta)$.
I sort of "have" the proof of the reciprocal, (1)$implies$ (2), but I don't understand a step. The author says that since $(q-zeta)^m in (q-zeta') + I[q]$ for some $m geq$ and $I subset R$ such that $R$ is $I$-adically separated, then $(zeta-zeta')^m in I$. This is what I don't find obvious at all. From that I believe the result follows since the ideal generated by (zeta - zeta')^m would be then contained on $I$, so
$$bigcap_jgeq 0(zeta-zeta')^j subset bigcap_jgeq 0I^j = (0)$$
and hence $R$ is $(zeta-zeta')$-adically separated.
For any of the implications (2)$implies$(3) or (3)$implies$(2) I literally have no clue. Any hero would be nice. Thank you!
abstract-algebra ring-theory commutative-algebra ideals roots-of-unity
asked Jul 15 at 18:35
user313212
622519
What is $q$? And $I[q]$?
– xarles
Jul 20 at 22:50
@xarles sorry I couldn't answer earlier. $q$ is a variable, just like $s$ or $t$. $I[q]$ is then the set of polynomials with coefficients in $I$.
– user313212
Aug 6 at 17:45
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $R$ be an integral domain with $char(R) = 0$ and $zeta, zeta'$ two primitive roots of unity in $R$. The following are equivalent.
(1) $(q-zeta)^m in (q-zeta') + I[q]$ for som $m geq 0$ and an ideal $I subset R$ such that $R$ is $I$-adically separated (note: $(q-zeta')$ on the right side denotes the ideal generated by $q- zeta'$).
(2) $R$ is $(zeta-zeta')$-adically separated.
(3) $ord(zeta^-1zeta') = p^k$ for some prime $p$ and $j in mathbbZsetminus 0$ such that $R$ is $(p)$-adically separated.
I believe I have just the implication (2) $implies$ (1). If $R$ is $(zeta-zeta')$-adically separated, then $cap_jgeq 0(zeta-zeta')^j = (0)$. We have to show that $(q-zeta)^m in (q-zeta') + I[q]$ for some $m geq 0$ and an ideal $I subset R$ such that $R$ is $I$-adically separated. But
$$q-zeta in (q-zeta') + (zeta-zeta')[q]$$
since $q-zeta = (q-zeta') + (zeta'-zeta)$.
I sort of "have" the proof of the reciprocal, (1)$implies$ (2), but I don't understand a step. The author says that since $(q-zeta)^m in (q-zeta') + I[q]$ for some $m geq$ and $I subset R$ such that $R$ is $I$-adically separated, then $(zeta-zeta')^m in I$. This is what I don't find obvious at all. From that I believe the result follows since the ideal generated by (zeta - zeta')^m would be then contained on $I$, so
$$bigcap_jgeq 0(zeta-zeta')^j subset bigcap_jgeq 0I^j = (0)$$
and hence $R$ is $(zeta-zeta')$-adically separated.
For any of the implications (2)$implies$(3) or (3)$implies$(2) I literally have no clue. Any hero would be nice. Thank you!
abstract-algebra ring-theory commutative-algebra ideals roots-of-unity
asked Jul 15 at 18:35
user313212
622519
Let $R$ be an integral domain with $char(R) = 0$ and $zeta, zeta'$ two primitive roots of unity in $R$. The following are equivalent.
(1) $(q-zeta)^m in (q-zeta') + I[q]$ for som $m geq 0$ and an ideal $I subset R$ such that $R$ is $I$-adically separated (note: $(q-zeta')$ on the right side denotes the ideal generated by $q- zeta'$).
(2) $R$ is $(zeta-zeta')$-adically separated.
(3) $ord(zeta^-1zeta') = p^k$ for some prime $p$ and $j in mathbbZsetminus 0$ such that $R$ is $(p)$-adically separated.
I believe I have just the implication (2) $implies$ (1). If $R$ is $(zeta-zeta')$-adically separated, then $cap_jgeq 0(zeta-zeta')^j = (0)$. We have to show that $(q-zeta)^m in (q-zeta') + I[q]$ for some $m geq 0$ and an ideal $I subset R$ such that $R$ is $I$-adically separated. But
$$q-zeta in (q-zeta') + (zeta-zeta')[q]$$
since $q-zeta = (q-zeta') + (zeta'-zeta)$.
I sort of "have" the proof of the reciprocal, (1)$implies$ (2), but I don't understand a step. The author says that since $(q-zeta)^m in (q-zeta') + I[q]$ for some $m geq$ and $I subset R$ such that $R$ is $I$-adically separated, then $(zeta-zeta')^m in I$. This is what I don't find obvious at all. From that I believe the result follows since the ideal generated by (zeta - zeta')^m would be then contained on $I$, so
$$bigcap_jgeq 0(zeta-zeta')^j subset bigcap_jgeq 0I^j = (0)$$
and hence $R$ is $(zeta-zeta')$-adically separated.
For any of the implications (2)$implies$(3) or (3)$implies$(2) I literally have no clue. Any hero would be nice. Thank you!
abstract-algebra ring-theory commutative-algebra ideals roots-of-unity
asked Jul 15 at 18:35
user313212
622519
asked Jul 15 at 18:35
user313212
622519
asked Jul 15 at 18:35
user313212
622519
asked Jul 15 at 18:35
user313212
622519
622519
What is $q$? And $I[q]$?
– xarles
Jul 20 at 22:50
@xarles sorry I couldn't answer earlier. $q$ is a variable, just like $s$ or $t$. $I[q]$ is then the set of polynomials with coefficients in $I$.
– user313212
Aug 6 at 17:45
add a comment |Â
What is $q$? And $I[q]$?
– xarles
Jul 20 at 22:50
@xarles sorry I couldn't answer earlier. $q$ is a variable, just like $s$ or $t$. $I[q]$ is then the set of polynomials with coefficients in $I$.
– user313212
Aug 6 at 17:45
What is $q$? And $I[q]$?
– xarles
Jul 20 at 22:50
What is $q$? And $I[q]$?
– xarles
Jul 20 at 22:50
@xarles sorry I couldn't answer earlier. $q$ is a variable, just like $s$ or $t$. $I[q]$ is then the set of polynomials with coefficients in $I$.
– user313212
Aug 6 at 17:45
@xarles sorry I couldn't answer earlier. $q$ is a variable, just like $s$ or $t$. $I[q]$ is then the set of polynomials with coefficients in $I$.
– user313212
Aug 6 at 17:45
add a comment |Â
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What is $q$? And $I[q]$?
– xarles
Jul 20 at 22:50
@xarles sorry I couldn't answer earlier. $q$ is a variable, just like $s$ or $t$. $I[q]$ is then the set of polynomials with coefficients in $I$.
– user313212
Aug 6 at 17:45