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Hensel lemma - generalization?
Clash Royale CLAN TAG#URR8PPP
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Let $f in Bbb Z[X]$ be monic and assume that $f$ has a root $a_n$ modulo $p^n$ for every $n geq 1$ (where $p$ is a fixed prime). Does it follow that $f$ has a root in $Bbb Z_p$?
The problem is that we might have $f'(a_n) equiv 0 pmod p$ so that Hensel's lemma doesn't apply directly. If this is true, what reference/book gives a proof of this fact (stated as above)?
p-adic-number-theory hensels-lemma
asked 3 hours ago
Alphonse
2,017623
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up vote
1
down vote
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Let $f in Bbb Z[X]$ be monic and assume that $f$ has a root $a_n$ modulo $p^n$ for every $n geq 1$ (where $p$ is a fixed prime). Does it follow that $f$ has a root in $Bbb Z_p$?
The problem is that we might have $f'(a_n) equiv 0 pmod p$ so that Hensel's lemma doesn't apply directly. If this is true, what reference/book gives a proof of this fact (stated as above)?
p-adic-number-theory hensels-lemma
asked 3 hours ago
Alphonse
2,017623
If $a_n equiv a_n-1 pmodp^n-1$ for all $n$ then the answer is yes, so it feels to me this is more of a "definition of $mathbbZ_p$" problem rather than something to do with Hensel's Lemma. But I am not familiar with the set theoretic issues to say whether or not we can WLOG assume this in your question.
– dalbouvet
3 hours ago
@dalbouvet : thank you. But I precisely did not assume that condition (under which the statement is indeed trivial).
– Alphonse
3 hours ago
yes indeed, I would also be interested if someone else can shed some light on this issue (and thus answers the set theoretic issues I'm concerned about).
– dalbouvet
3 hours ago
Such an $f$ has a root in $mathbbZ_p$ if and only if there exists a compatible sequence as in @dalbouvet's answer. As they said, this is really a question about understanding the definition of $mathbbZ_p$, not Hensel's lemma.
– RghtHndSd
2 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f in Bbb Z[X]$ be monic and assume that $f$ has a root $a_n$ modulo $p^n$ for every $n geq 1$ (where $p$ is a fixed prime). Does it follow that $f$ has a root in $Bbb Z_p$?
The problem is that we might have $f'(a_n) equiv 0 pmod p$ so that Hensel's lemma doesn't apply directly. If this is true, what reference/book gives a proof of this fact (stated as above)?
p-adic-number-theory hensels-lemma
asked 3 hours ago
Alphonse
2,017623
Let $f in Bbb Z[X]$ be monic and assume that $f$ has a root $a_n$ modulo $p^n$ for every $n geq 1$ (where $p$ is a fixed prime). Does it follow that $f$ has a root in $Bbb Z_p$?
The problem is that we might have $f'(a_n) equiv 0 pmod p$ so that Hensel's lemma doesn't apply directly. If this is true, what reference/book gives a proof of this fact (stated as above)?
p-adic-number-theory hensels-lemma
p-adic-number-theory hensels-lemma
asked 3 hours ago
Alphonse
2,017623
asked 3 hours ago
Alphonse
2,017623
asked 3 hours ago
Alphonse
2,017623
asked 3 hours ago
Alphonse
2,017623
asked 3 hours ago
Alphonse
2,017623
2,017623
If $a_n equiv a_n-1 pmodp^n-1$ for all $n$ then the answer is yes, so it feels to me this is more of a "definition of $mathbbZ_p$" problem rather than something to do with Hensel's Lemma. But I am not familiar with the set theoretic issues to say whether or not we can WLOG assume this in your question.
– dalbouvet
3 hours ago
@dalbouvet : thank you. But I precisely did not assume that condition (under which the statement is indeed trivial).
– Alphonse
3 hours ago
yes indeed, I would also be interested if someone else can shed some light on this issue (and thus answers the set theoretic issues I'm concerned about).
– dalbouvet
3 hours ago
Such an $f$ has a root in $mathbbZ_p$ if and only if there exists a compatible sequence as in @dalbouvet's answer. As they said, this is really a question about understanding the definition of $mathbbZ_p$, not Hensel's lemma.
– RghtHndSd
2 hours ago
add a comment |Â
If $a_n equiv a_n-1 pmodp^n-1$ for all $n$ then the answer is yes, so it feels to me this is more of a "definition of $mathbbZ_p$" problem rather than something to do with Hensel's Lemma. But I am not familiar with the set theoretic issues to say whether or not we can WLOG assume this in your question.
– dalbouvet
3 hours ago
@dalbouvet : thank you. But I precisely did not assume that condition (under which the statement is indeed trivial).
– Alphonse
3 hours ago
yes indeed, I would also be interested if someone else can shed some light on this issue (and thus answers the set theoretic issues I'm concerned about).
– dalbouvet
3 hours ago
Such an $f$ has a root in $mathbbZ_p$ if and only if there exists a compatible sequence as in @dalbouvet's answer. As they said, this is really a question about understanding the definition of $mathbbZ_p$, not Hensel's lemma.
– RghtHndSd
2 hours ago
If $a_n equiv a_n-1 pmodp^n-1$ for all $n$ then the answer is yes, so it feels to me this is more of a "definition of $mathbbZ_p$" problem rather than something to do with Hensel's Lemma. But I am not familiar with the set theoretic issues to say whether or not we can WLOG assume this in your question.
– dalbouvet
3 hours ago
If $a_n equiv a_n-1 pmodp^n-1$ for all $n$ then the answer is yes, so it feels to me this is more of a "definition of $mathbbZ_p$" problem rather than something to do with Hensel's Lemma. But I am not familiar with the set theoretic issues to say whether or not we can WLOG assume this in your question.
– dalbouvet
3 hours ago
@dalbouvet : thank you. But I precisely did not assume that condition (under which the statement is indeed trivial).
– Alphonse
3 hours ago
@dalbouvet : thank you. But I precisely did not assume that condition (under which the statement is indeed trivial).
– Alphonse
3 hours ago
yes indeed, I would also be interested if someone else can shed some light on this issue (and thus answers the set theoretic issues I'm concerned about).
– dalbouvet
3 hours ago
yes indeed, I would also be interested if someone else can shed some light on this issue (and thus answers the set theoretic issues I'm concerned about).
– dalbouvet
3 hours ago
Such an $f$ has a root in $mathbbZ_p$ if and only if there exists a compatible sequence as in @dalbouvet's answer. As they said, this is really a question about understanding the definition of $mathbbZ_p$, not Hensel's lemma.
– RghtHndSd
2 hours ago
Such an $f$ has a root in $mathbbZ_p$ if and only if there exists a compatible sequence as in @dalbouvet's answer. As they said, this is really a question about understanding the definition of $mathbbZ_p$, not Hensel's lemma.
– RghtHndSd
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Yes, and this is not a generalization of Hensel's lemma.
We will repeatedly pass to subsequences as follows. First, the set $R_1 = a_n bmod p $ has finitely many elements, so there must be some residue $r_1 in R_1$ such that there are infinitely many $n$ with $a_n equiv r_1 bmod p$. Pass to this subsequence; that is, assume WLOG that every $a_n$ has this property (mostly in order to avoid having to come up with some annoying piece of notation for the new subsequence).
Next construct $R_2 = a_n bmod p^2 : n ge 2 $, which again is finite, so again there is some residue $r_2 in R_2$ such that there are infinitely many $n$ with $a_n equiv r_2 bmod p^2$. Again pass to this subsequence. Etc.
In this way we construct a sequence of residues $r_k in mathbbZ/p^kmathbbZ$ such that $r_k equiv r_k-1 bmod p^k-1$ and such that there exist infinitely many $a_n$ such that $a_n equiv r_k bmod p^k$. The $r_k$ define an element $r in mathbbZ_p$ which is a root of $f$. Note that we don't need to assume that $f$ is monic.
answered 2 hours ago
Qiaochu Yuan
273k32572910
2
Brilliant! This argument looks very much alike to proving a bounded sequence in $mathbbR$ has a convergent subsequence.
– dalbouvet
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Yes, and this is not a generalization of Hensel's lemma.
We will repeatedly pass to subsequences as follows. First, the set $R_1 = a_n bmod p $ has finitely many elements, so there must be some residue $r_1 in R_1$ such that there are infinitely many $n$ with $a_n equiv r_1 bmod p$. Pass to this subsequence; that is, assume WLOG that every $a_n$ has this property (mostly in order to avoid having to come up with some annoying piece of notation for the new subsequence).
Next construct $R_2 = a_n bmod p^2 : n ge 2 $, which again is finite, so again there is some residue $r_2 in R_2$ such that there are infinitely many $n$ with $a_n equiv r_2 bmod p^2$. Again pass to this subsequence. Etc.
In this way we construct a sequence of residues $r_k in mathbbZ/p^kmathbbZ$ such that $r_k equiv r_k-1 bmod p^k-1$ and such that there exist infinitely many $a_n$ such that $a_n equiv r_k bmod p^k$. The $r_k$ define an element $r in mathbbZ_p$ which is a root of $f$. Note that we don't need to assume that $f$ is monic.
answered 2 hours ago
Qiaochu Yuan
273k32572910
2
Brilliant! This argument looks very much alike to proving a bounded sequence in $mathbbR$ has a convergent subsequence.
– dalbouvet
2 hours ago
add a comment |Â
up vote
4
down vote
accepted
Yes, and this is not a generalization of Hensel's lemma.
We will repeatedly pass to subsequences as follows. First, the set $R_1 = a_n bmod p $ has finitely many elements, so there must be some residue $r_1 in R_1$ such that there are infinitely many $n$ with $a_n equiv r_1 bmod p$. Pass to this subsequence; that is, assume WLOG that every $a_n$ has this property (mostly in order to avoid having to come up with some annoying piece of notation for the new subsequence).
Next construct $R_2 = a_n bmod p^2 : n ge 2 $, which again is finite, so again there is some residue $r_2 in R_2$ such that there are infinitely many $n$ with $a_n equiv r_2 bmod p^2$. Again pass to this subsequence. Etc.
In this way we construct a sequence of residues $r_k in mathbbZ/p^kmathbbZ$ such that $r_k equiv r_k-1 bmod p^k-1$ and such that there exist infinitely many $a_n$ such that $a_n equiv r_k bmod p^k$. The $r_k$ define an element $r in mathbbZ_p$ which is a root of $f$. Note that we don't need to assume that $f$ is monic.
answered 2 hours ago
Qiaochu Yuan
273k32572910
2
Brilliant! This argument looks very much alike to proving a bounded sequence in $mathbbR$ has a convergent subsequence.
– dalbouvet
2 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Yes, and this is not a generalization of Hensel's lemma.
We will repeatedly pass to subsequences as follows. First, the set $R_1 = a_n bmod p $ has finitely many elements, so there must be some residue $r_1 in R_1$ such that there are infinitely many $n$ with $a_n equiv r_1 bmod p$. Pass to this subsequence; that is, assume WLOG that every $a_n$ has this property (mostly in order to avoid having to come up with some annoying piece of notation for the new subsequence).
Next construct $R_2 = a_n bmod p^2 : n ge 2 $, which again is finite, so again there is some residue $r_2 in R_2$ such that there are infinitely many $n$ with $a_n equiv r_2 bmod p^2$. Again pass to this subsequence. Etc.
In this way we construct a sequence of residues $r_k in mathbbZ/p^kmathbbZ$ such that $r_k equiv r_k-1 bmod p^k-1$ and such that there exist infinitely many $a_n$ such that $a_n equiv r_k bmod p^k$. The $r_k$ define an element $r in mathbbZ_p$ which is a root of $f$. Note that we don't need to assume that $f$ is monic.
answered 2 hours ago
Qiaochu Yuan
273k32572910
Yes, and this is not a generalization of Hensel's lemma.
We will repeatedly pass to subsequences as follows. First, the set $R_1 = a_n bmod p $ has finitely many elements, so there must be some residue $r_1 in R_1$ such that there are infinitely many $n$ with $a_n equiv r_1 bmod p$. Pass to this subsequence; that is, assume WLOG that every $a_n$ has this property (mostly in order to avoid having to come up with some annoying piece of notation for the new subsequence).
Next construct $R_2 = a_n bmod p^2 : n ge 2 $, which again is finite, so again there is some residue $r_2 in R_2$ such that there are infinitely many $n$ with $a_n equiv r_2 bmod p^2$. Again pass to this subsequence. Etc.
In this way we construct a sequence of residues $r_k in mathbbZ/p^kmathbbZ$ such that $r_k equiv r_k-1 bmod p^k-1$ and such that there exist infinitely many $a_n$ such that $a_n equiv r_k bmod p^k$. The $r_k$ define an element $r in mathbbZ_p$ which is a root of $f$. Note that we don't need to assume that $f$ is monic.
answered 2 hours ago
Qiaochu Yuan
273k32572910
answered 2 hours ago
Qiaochu Yuan
273k32572910
answered 2 hours ago
Qiaochu Yuan
273k32572910
answered 2 hours ago
Qiaochu Yuan
273k32572910
273k32572910
2
Brilliant! This argument looks very much alike to proving a bounded sequence in $mathbbR$ has a convergent subsequence.
– dalbouvet
2 hours ago
add a comment |Â
2
Brilliant! This argument looks very much alike to proving a bounded sequence in $mathbbR$ has a convergent subsequence.
– dalbouvet
2 hours ago
2
2
Brilliant! This argument looks very much alike to proving a bounded sequence in $mathbbR$ has a convergent subsequence.
– dalbouvet
2 hours ago
Brilliant! This argument looks very much alike to proving a bounded sequence in $mathbbR$ has a convergent subsequence.
– dalbouvet
2 hours ago
add a comment |Â
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If $a_n equiv a_n-1 pmodp^n-1$ for all $n$ then the answer is yes, so it feels to me this is more of a "definition of $mathbbZ_p$" problem rather than something to do with Hensel's Lemma. But I am not familiar with the set theoretic issues to say whether or not we can WLOG assume this in your question.
– dalbouvet
3 hours ago
@dalbouvet : thank you. But I precisely did not assume that condition (under which the statement is indeed trivial).
– Alphonse
3 hours ago
yes indeed, I would also be interested if someone else can shed some light on this issue (and thus answers the set theoretic issues I'm concerned about).
– dalbouvet
3 hours ago
Such an $f$ has a root in $mathbbZ_p$ if and only if there exists a compatible sequence as in @dalbouvet's answer. As they said, this is really a question about understanding the definition of $mathbbZ_p$, not Hensel's lemma.
– RghtHndSd
2 hours ago