Uniqueness in Weierstraß p-adic preparation theorem
I have given the Weierstraß p-adic preparation theorem. It is stated as follows: Let $f=a_0 + a_1T + ... in mathbb{Z}_p[[T]]$ for a prime $p$ such that $p mid a_0,...,a_{n-1}$ and $pnotmid a_n$. Then there exists a unique unit $Uinmathbb{Z}_p[[T]]^*$ and a distinguished polynomial P of degree $n$, i.e. $P= b_0 + b_1 T + ... + b_{n-1}T^{n-1}+T^n$ with $pmid b_0,...,b_{n-1}$ such that $f=PU$.
The existence is clear to me but the uniqueness not. Both supposedly follow from a prior theorem, division with rest. It is stated as follows:
Let $f=a_0 + a_1T + ... in mathbb{Z}_p[[T]]$ for a prime $p$ such that $p mid a_0,...,a_{n-1}$ and $pnotmid a_n$. Then, for all $ginmathbb{Z}_p[[T]]$ there is a unique expression $$ g = qf + r $$ with $ q in mathbb{Z}_p[[T]]$ and $ r $ a polynomial of degree $ deg(r) le n-1 $.
The uniqueness statement that I require is a rather strong one. I need that if $f=PU$ as in the first theorem and also $f=P'U'$ with $U'$ a unit in $mathbb{Z}_p[[T]]$ and $P'$ a polynomial, then $P=P'$ and $U=U'$. My problem is that P' might have a different degree than $P$, in which case I do not see how to apply division with rest.
I need this strong uniquement because I would like to conclude from this theorem that $mathbb{Z}_p[[T]]$ is a UFD.
Help would be greatly appreciated!
abstract-algebra number-theory commutative-algebra p-adic-number-theory formal-power-series
asked Nov 13 at 14:21
B. Alb
62
add a comment |
I have given the Weierstraß p-adic preparation theorem. It is stated as follows: Let $f=a_0 + a_1T + ... in mathbb{Z}_p[[T]]$ for a prime $p$ such that $p mid a_0,...,a_{n-1}$ and $pnotmid a_n$. Then there exists a unique unit $Uinmathbb{Z}_p[[T]]^*$ and a distinguished polynomial P of degree $n$, i.e. $P= b_0 + b_1 T + ... + b_{n-1}T^{n-1}+T^n$ with $pmid b_0,...,b_{n-1}$ such that $f=PU$.
The existence is clear to me but the uniqueness not. Both supposedly follow from a prior theorem, division with rest. It is stated as follows:
Let $f=a_0 + a_1T + ... in mathbb{Z}_p[[T]]$ for a prime $p$ such that $p mid a_0,...,a_{n-1}$ and $pnotmid a_n$. Then, for all $ginmathbb{Z}_p[[T]]$ there is a unique expression $$ g = qf + r $$ with $ q in mathbb{Z}_p[[T]]$ and $ r $ a polynomial of degree $ deg(r) le n-1 $.
The uniqueness statement that I require is a rather strong one. I need that if $f=PU$ as in the first theorem and also $f=P'U'$ with $U'$ a unit in $mathbb{Z}_p[[T]]$ and $P'$ a polynomial, then $P=P'$ and $U=U'$. My problem is that P' might have a different degree than $P$, in which case I do not see how to apply division with rest.
I need this strong uniquement because I would like to conclude from this theorem that $mathbb{Z}_p[[T]]$ is a UFD.
Help would be greatly appreciated!
abstract-algebra number-theory commutative-algebra p-adic-number-theory formal-power-series
asked Nov 13 at 14:21
B. Alb
62
1
If $P'=PU$ with $U$ a unit, go modulo $p$ to get an equation $T^n=T^mu$ where $u$ is $U$ modulo $p$. This is an equation in $mathbb{F}_p[[T]]$ and $u$ is a unit in this ring. Can you prove now that $m=n$?
– Mohan
Nov 14 at 2:01
1
It was not clear to me that OP intended for $P'$ to be a monic polynomial with nonleading coefficients divisible by $p$. Of course this stronger assertion of uniqueness, with $P'$ even required to be a monic polynomial, is never valid. Counterexamples like $f = (T + b_0)frac{1}{T+1} = (T^2 + (b_0 + 1)T + b_0) frac{1}{(T+1)^2}$ abound.
– Badam Baplan
Nov 14 at 5:24
add a comment |
I have given the Weierstraß p-adic preparation theorem. It is stated as follows: Let $f=a_0 + a_1T + ... in mathbb{Z}_p[[T]]$ for a prime $p$ such that $p mid a_0,...,a_{n-1}$ and $pnotmid a_n$. Then there exists a unique unit $Uinmathbb{Z}_p[[T]]^*$ and a distinguished polynomial P of degree $n$, i.e. $P= b_0 + b_1 T + ... + b_{n-1}T^{n-1}+T^n$ with $pmid b_0,...,b_{n-1}$ such that $f=PU$.
The existence is clear to me but the uniqueness not. Both supposedly follow from a prior theorem, division with rest. It is stated as follows:
Let $f=a_0 + a_1T + ... in mathbb{Z}_p[[T]]$ for a prime $p$ such that $p mid a_0,...,a_{n-1}$ and $pnotmid a_n$. Then, for all $ginmathbb{Z}_p[[T]]$ there is a unique expression $$ g = qf + r $$ with $ q in mathbb{Z}_p[[T]]$ and $ r $ a polynomial of degree $ deg(r) le n-1 $.
The uniqueness statement that I require is a rather strong one. I need that if $f=PU$ as in the first theorem and also $f=P'U'$ with $U'$ a unit in $mathbb{Z}_p[[T]]$ and $P'$ a polynomial, then $P=P'$ and $U=U'$. My problem is that P' might have a different degree than $P$, in which case I do not see how to apply division with rest.
I need this strong uniquement because I would like to conclude from this theorem that $mathbb{Z}_p[[T]]$ is a UFD.
Help would be greatly appreciated!
abstract-algebra number-theory commutative-algebra p-adic-number-theory formal-power-series
asked Nov 13 at 14:21
B. Alb
62
I have given the Weierstraß p-adic preparation theorem. It is stated as follows: Let $f=a_0 + a_1T + ... in mathbb{Z}_p[[T]]$ for a prime $p$ such that $p mid a_0,...,a_{n-1}$ and $pnotmid a_n$. Then there exists a unique unit $Uinmathbb{Z}_p[[T]]^*$ and a distinguished polynomial P of degree $n$, i.e. $P= b_0 + b_1 T + ... + b_{n-1}T^{n-1}+T^n$ with $pmid b_0,...,b_{n-1}$ such that $f=PU$.
The existence is clear to me but the uniqueness not. Both supposedly follow from a prior theorem, division with rest. It is stated as follows:
Let $f=a_0 + a_1T + ... in mathbb{Z}_p[[T]]$ for a prime $p$ such that $p mid a_0,...,a_{n-1}$ and $pnotmid a_n$. Then, for all $ginmathbb{Z}_p[[T]]$ there is a unique expression $$ g = qf + r $$ with $ q in mathbb{Z}_p[[T]]$ and $ r $ a polynomial of degree $ deg(r) le n-1 $.
The uniqueness statement that I require is a rather strong one. I need that if $f=PU$ as in the first theorem and also $f=P'U'$ with $U'$ a unit in $mathbb{Z}_p[[T]]$ and $P'$ a polynomial, then $P=P'$ and $U=U'$. My problem is that P' might have a different degree than $P$, in which case I do not see how to apply division with rest.
I need this strong uniquement because I would like to conclude from this theorem that $mathbb{Z}_p[[T]]$ is a UFD.
Help would be greatly appreciated!
abstract-algebra number-theory commutative-algebra p-adic-number-theory formal-power-series
abstract-algebra number-theory commutative-algebra p-adic-number-theory formal-power-series
asked Nov 13 at 14:21
B. Alb
62
asked Nov 13 at 14:21
B. Alb
62
asked Nov 13 at 14:21
B. Alb
62
asked Nov 13 at 14:21
B. Alb
62
asked Nov 13 at 14:21
B. Alb
62
62
1
If $P'=PU$ with $U$ a unit, go modulo $p$ to get an equation $T^n=T^mu$ where $u$ is $U$ modulo $p$. This is an equation in $mathbb{F}_p[[T]]$ and $u$ is a unit in this ring. Can you prove now that $m=n$?
– Mohan
Nov 14 at 2:01
1
It was not clear to me that OP intended for $P'$ to be a monic polynomial with nonleading coefficients divisible by $p$. Of course this stronger assertion of uniqueness, with $P'$ even required to be a monic polynomial, is never valid. Counterexamples like $f = (T + b_0)frac{1}{T+1} = (T^2 + (b_0 + 1)T + b_0) frac{1}{(T+1)^2}$ abound.
– Badam Baplan
Nov 14 at 5:24
add a comment |
1
If $P'=PU$ with $U$ a unit, go modulo $p$ to get an equation $T^n=T^mu$ where $u$ is $U$ modulo $p$. This is an equation in $mathbb{F}_p[[T]]$ and $u$ is a unit in this ring. Can you prove now that $m=n$?
– Mohan
Nov 14 at 2:01
1
It was not clear to me that OP intended for $P'$ to be a monic polynomial with nonleading coefficients divisible by $p$. Of course this stronger assertion of uniqueness, with $P'$ even required to be a monic polynomial, is never valid. Counterexamples like $f = (T + b_0)frac{1}{T+1} = (T^2 + (b_0 + 1)T + b_0) frac{1}{(T+1)^2}$ abound.
– Badam Baplan
Nov 14 at 5:24
1
1
If $P'=PU$ with $U$ a unit, go modulo $p$ to get an equation $T^n=T^mu$ where $u$ is $U$ modulo $p$. This is an equation in $mathbb{F}_p[[T]]$ and $u$ is a unit in this ring. Can you prove now that $m=n$?
– Mohan
Nov 14 at 2:01
If $P'=PU$ with $U$ a unit, go modulo $p$ to get an equation $T^n=T^mu$ where $u$ is $U$ modulo $p$. This is an equation in $mathbb{F}_p[[T]]$ and $u$ is a unit in this ring. Can you prove now that $m=n$?
– Mohan
Nov 14 at 2:01
1
1
It was not clear to me that OP intended for $P'$ to be a monic polynomial with nonleading coefficients divisible by $p$. Of course this stronger assertion of uniqueness, with $P'$ even required to be a monic polynomial, is never valid. Counterexamples like $f = (T + b_0)frac{1}{T+1} = (T^2 + (b_0 + 1)T + b_0) frac{1}{(T+1)^2}$ abound.
– Badam Baplan
Nov 14 at 5:24
It was not clear to me that OP intended for $P'$ to be a monic polynomial with nonleading coefficients divisible by $p$. Of course this stronger assertion of uniqueness, with $P'$ even required to be a monic polynomial, is never valid. Counterexamples like $f = (T + b_0)frac{1}{T+1} = (T^2 + (b_0 + 1)T + b_0) frac{1}{(T+1)^2}$ abound.
– Badam Baplan
Nov 14 at 5:24
add a comment |
1 Answer
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As @BadamBaplan says, your desired uniqueness does not hold. Consider, for instance, the polynomial $f=p+X^n+pX^{n+1}$. Then we have the factorization $U'=1$, $P'=f$, and the Weierstraß factorization $f=UP$, where indeed, $P$ and $U$ are given to you already by Hensel’s Lemma. If I understand your question aright, the factorization $f=U'P'$ satisfies your criteria.
answered Nov 24 at 21:33
Lubin
43.5k44485
What does "your criteria" in the last sentence mean? Is there some other uniqueness statement that holds for the factorization $U'P'$?
– aleph_two
yesterday
Oh, @aleph_two, “your criteria” refers to the paragraph in OP’s question that starts, “The uniqueness statement that I require…”.
– Lubin
yesterday
Thank you! So that sentence just means that your first factorization actually provides a counterexample.
– aleph_two
yesterday
Precisely, @aleph_two.
– Lubin
13 hours ago
add a comment |
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As @BadamBaplan says, your desired uniqueness does not hold. Consider, for instance, the polynomial $f=p+X^n+pX^{n+1}$. Then we have the factorization $U'=1$, $P'=f$, and the Weierstraß factorization $f=UP$, where indeed, $P$ and $U$ are given to you already by Hensel’s Lemma. If I understand your question aright, the factorization $f=U'P'$ satisfies your criteria.
answered Nov 24 at 21:33
Lubin
43.5k44485
What does "your criteria" in the last sentence mean? Is there some other uniqueness statement that holds for the factorization $U'P'$?
– aleph_two
yesterday
Oh, @aleph_two, “your criteria” refers to the paragraph in OP’s question that starts, “The uniqueness statement that I require…”.
– Lubin
yesterday
Thank you! So that sentence just means that your first factorization actually provides a counterexample.
– aleph_two
yesterday
Precisely, @aleph_two.
– Lubin
13 hours ago
add a comment |
As @BadamBaplan says, your desired uniqueness does not hold. Consider, for instance, the polynomial $f=p+X^n+pX^{n+1}$. Then we have the factorization $U'=1$, $P'=f$, and the Weierstraß factorization $f=UP$, where indeed, $P$ and $U$ are given to you already by Hensel’s Lemma. If I understand your question aright, the factorization $f=U'P'$ satisfies your criteria.
answered Nov 24 at 21:33
Lubin
43.5k44485
What does "your criteria" in the last sentence mean? Is there some other uniqueness statement that holds for the factorization $U'P'$?
– aleph_two
yesterday
Oh, @aleph_two, “your criteria” refers to the paragraph in OP’s question that starts, “The uniqueness statement that I require…”.
– Lubin
yesterday
Thank you! So that sentence just means that your first factorization actually provides a counterexample.
– aleph_two
yesterday
Precisely, @aleph_two.
– Lubin
13 hours ago
add a comment |
As @BadamBaplan says, your desired uniqueness does not hold. Consider, for instance, the polynomial $f=p+X^n+pX^{n+1}$. Then we have the factorization $U'=1$, $P'=f$, and the Weierstraß factorization $f=UP$, where indeed, $P$ and $U$ are given to you already by Hensel’s Lemma. If I understand your question aright, the factorization $f=U'P'$ satisfies your criteria.
answered Nov 24 at 21:33
Lubin
43.5k44485
As @BadamBaplan says, your desired uniqueness does not hold. Consider, for instance, the polynomial $f=p+X^n+pX^{n+1}$. Then we have the factorization $U'=1$, $P'=f$, and the Weierstraß factorization $f=UP$, where indeed, $P$ and $U$ are given to you already by Hensel’s Lemma. If I understand your question aright, the factorization $f=U'P'$ satisfies your criteria.
answered Nov 24 at 21:33
Lubin
43.5k44485
answered Nov 24 at 21:33
Lubin
43.5k44485
answered Nov 24 at 21:33
Lubin
43.5k44485
answered Nov 24 at 21:33
Lubin
43.5k44485
43.5k44485
What does "your criteria" in the last sentence mean? Is there some other uniqueness statement that holds for the factorization $U'P'$?
– aleph_two
yesterday
Oh, @aleph_two, “your criteria” refers to the paragraph in OP’s question that starts, “The uniqueness statement that I require…”.
– Lubin
yesterday
Thank you! So that sentence just means that your first factorization actually provides a counterexample.
– aleph_two
yesterday
Precisely, @aleph_two.
– Lubin
13 hours ago
add a comment |
What does "your criteria" in the last sentence mean? Is there some other uniqueness statement that holds for the factorization $U'P'$?
– aleph_two
yesterday
Oh, @aleph_two, “your criteria” refers to the paragraph in OP’s question that starts, “The uniqueness statement that I require…”.
– Lubin
yesterday
Thank you! So that sentence just means that your first factorization actually provides a counterexample.
– aleph_two
yesterday
Precisely, @aleph_two.
– Lubin
13 hours ago
What does "your criteria" in the last sentence mean? Is there some other uniqueness statement that holds for the factorization $U'P'$?
– aleph_two
yesterday
What does "your criteria" in the last sentence mean? Is there some other uniqueness statement that holds for the factorization $U'P'$?
– aleph_two
yesterday
Oh, @aleph_two, “your criteria” refers to the paragraph in OP’s question that starts, “The uniqueness statement that I require…”.
– Lubin
yesterday
Oh, @aleph_two, “your criteria” refers to the paragraph in OP’s question that starts, “The uniqueness statement that I require…”.
– Lubin
yesterday
Thank you! So that sentence just means that your first factorization actually provides a counterexample.
– aleph_two
yesterday
Thank you! So that sentence just means that your first factorization actually provides a counterexample.
– aleph_two
yesterday
Precisely, @aleph_two.
– Lubin
13 hours ago
Precisely, @aleph_two.
– Lubin
13 hours ago
add a comment |
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1
If $P'=PU$ with $U$ a unit, go modulo $p$ to get an equation $T^n=T^mu$ where $u$ is $U$ modulo $p$. This is an equation in $mathbb{F}_p[[T]]$ and $u$ is a unit in this ring. Can you prove now that $m=n$?
– Mohan
Nov 14 at 2:01
1
It was not clear to me that OP intended for $P'$ to be a monic polynomial with nonleading coefficients divisible by $p$. Of course this stronger assertion of uniqueness, with $P'$ even required to be a monic polynomial, is never valid. Counterexamples like $f = (T + b_0)frac{1}{T+1} = (T^2 + (b_0 + 1)T + b_0) frac{1}{(T+1)^2}$ abound.
– Badam Baplan
Nov 14 at 5:24