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Multiplicative group of $p$-adic numbers $mathbbQ_p$
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$DeclareMathOperatorKerKerDeclareMathOperatorUU$I'm reading Serre's text, "A course in arithmetic". I have some problems with this argument.
Let $U=mathbbZ_p^*$ the group of $p$-adic units.
For every $n le 1$ let $U_n=1+p^nmathbbZ_p$.
I don't understand how to prove that $U_n=Ker(f_n)$ with $f_n:U to (mathbbZ/p^nmathbbZ)^*$ morphism.
In particular, I see that $U_n subseteqKer(f_n)$ but not the inverse direction.
Then Serre says that the map $(1+p^nx) to x pmod p$ from $U_n/U_n+1$ to $mathbbZ/pmathbbZ$ is an isomorphism because$(1+p^nx)(1+p^ny)=1+p^n(x+y)$ modulo $p^n+1$.
Also this point isn't clear to me.
Is there anybody who has some suggestions?
Thanks!
abstract-algebra p-adic-number-theory
edited Jul 20 at 20:43
Fabio Lucchini
5,62411025
asked Jul 20 at 17:23
robbis
305
add a comment |Â
up vote
3
down vote
favorite
$DeclareMathOperatorKerKerDeclareMathOperatorUU$I'm reading Serre's text, "A course in arithmetic". I have some problems with this argument.
Let $U=mathbbZ_p^*$ the group of $p$-adic units.
For every $n le 1$ let $U_n=1+p^nmathbbZ_p$.
I don't understand how to prove that $U_n=Ker(f_n)$ with $f_n:U to (mathbbZ/p^nmathbbZ)^*$ morphism.
In particular, I see that $U_n subseteqKer(f_n)$ but not the inverse direction.
Then Serre says that the map $(1+p^nx) to x pmod p$ from $U_n/U_n+1$ to $mathbbZ/pmathbbZ$ is an isomorphism because$(1+p^nx)(1+p^ny)=1+p^n(x+y)$ modulo $p^n+1$.
Also this point isn't clear to me.
Is there anybody who has some suggestions?
Thanks!
abstract-algebra p-adic-number-theory
edited Jul 20 at 20:43
Fabio Lucchini
5,62411025
asked Jul 20 at 17:23
robbis
305
I'm not sure what the question is. That $U_n$ is the kernel of $f_n$ just comes from the definition of kernel. The second question: surely this means that $1+p^nxmapsto xpmod p$ is a homomorphism. Is being a homomorphism the problem, or is its bijectivity?
– Lord Shark the Unknown
Jul 20 at 17:26
For the first point: is $f_n$ the morphism that maps a $p$-adic unit $x$ to its $n$-th component, which is in $(mathbbZ/p^nmathbbZ)^*$? is it right?
– robbis
Jul 20 at 17:31
About the second question, I don't understand what the formula means. I think I need this formula to say that it is a morphism, because the formula says that in $U_n/U_n+1$ the product of the class of $x$ and $y$ is equal to the class of $x+y$, becuse of the congruence modulo $p^n+1$? But I'm not sure, because elements of $U_n+1$ are of the form $1+p^n+1mathbbZ_p$
– robbis
Jul 20 at 17:39
“Then Serre says†means that if you start with $xiin U_n/U_n+1$, choose an element $xin U_n$ representing $xi$, and form $fracx-1p^npmod pinBbb Z/pBbb Z$, then the result doesn’t depend on your choice of $x$, and the map is a homomorphism, even an isomorphism.
– Lubin
Jul 20 at 18:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$DeclareMathOperatorKerKerDeclareMathOperatorUU$I'm reading Serre's text, "A course in arithmetic". I have some problems with this argument.
Let $U=mathbbZ_p^*$ the group of $p$-adic units.
For every $n le 1$ let $U_n=1+p^nmathbbZ_p$.
I don't understand how to prove that $U_n=Ker(f_n)$ with $f_n:U to (mathbbZ/p^nmathbbZ)^*$ morphism.
In particular, I see that $U_n subseteqKer(f_n)$ but not the inverse direction.
Then Serre says that the map $(1+p^nx) to x pmod p$ from $U_n/U_n+1$ to $mathbbZ/pmathbbZ$ is an isomorphism because$(1+p^nx)(1+p^ny)=1+p^n(x+y)$ modulo $p^n+1$.
Also this point isn't clear to me.
Is there anybody who has some suggestions?
Thanks!
abstract-algebra p-adic-number-theory
edited Jul 20 at 20:43
Fabio Lucchini
5,62411025
asked Jul 20 at 17:23
robbis
305
$DeclareMathOperatorKerKerDeclareMathOperatorUU$I'm reading Serre's text, "A course in arithmetic". I have some problems with this argument.
Let $U=mathbbZ_p^*$ the group of $p$-adic units.
For every $n le 1$ let $U_n=1+p^nmathbbZ_p$.
I don't understand how to prove that $U_n=Ker(f_n)$ with $f_n:U to (mathbbZ/p^nmathbbZ)^*$ morphism.
In particular, I see that $U_n subseteqKer(f_n)$ but not the inverse direction.
Then Serre says that the map $(1+p^nx) to x pmod p$ from $U_n/U_n+1$ to $mathbbZ/pmathbbZ$ is an isomorphism because$(1+p^nx)(1+p^ny)=1+p^n(x+y)$ modulo $p^n+1$.
Also this point isn't clear to me.
Is there anybody who has some suggestions?
Thanks!
abstract-algebra p-adic-number-theory
edited Jul 20 at 20:43
Fabio Lucchini
5,62411025
asked Jul 20 at 17:23
robbis
305
edited Jul 20 at 20:43
Fabio Lucchini
5,62411025
edited Jul 20 at 20:43
Fabio Lucchini
5,62411025
edited Jul 20 at 20:43
Fabio Lucchini
5,62411025
5,62411025
asked Jul 20 at 17:23
robbis
305
asked Jul 20 at 17:23
robbis
305
asked Jul 20 at 17:23
robbis
305
305
I'm not sure what the question is. That $U_n$ is the kernel of $f_n$ just comes from the definition of kernel. The second question: surely this means that $1+p^nxmapsto xpmod p$ is a homomorphism. Is being a homomorphism the problem, or is its bijectivity?
– Lord Shark the Unknown
Jul 20 at 17:26
For the first point: is $f_n$ the morphism that maps a $p$-adic unit $x$ to its $n$-th component, which is in $(mathbbZ/p^nmathbbZ)^*$? is it right?
– robbis
Jul 20 at 17:31
About the second question, I don't understand what the formula means. I think I need this formula to say that it is a morphism, because the formula says that in $U_n/U_n+1$ the product of the class of $x$ and $y$ is equal to the class of $x+y$, becuse of the congruence modulo $p^n+1$? But I'm not sure, because elements of $U_n+1$ are of the form $1+p^n+1mathbbZ_p$
– robbis
Jul 20 at 17:39
“Then Serre says†means that if you start with $xiin U_n/U_n+1$, choose an element $xin U_n$ representing $xi$, and form $fracx-1p^npmod pinBbb Z/pBbb Z$, then the result doesn’t depend on your choice of $x$, and the map is a homomorphism, even an isomorphism.
– Lubin
Jul 20 at 18:36
add a comment |Â
I'm not sure what the question is. That $U_n$ is the kernel of $f_n$ just comes from the definition of kernel. The second question: surely this means that $1+p^nxmapsto xpmod p$ is a homomorphism. Is being a homomorphism the problem, or is its bijectivity?
– Lord Shark the Unknown
Jul 20 at 17:26
For the first point: is $f_n$ the morphism that maps a $p$-adic unit $x$ to its $n$-th component, which is in $(mathbbZ/p^nmathbbZ)^*$? is it right?
– robbis
Jul 20 at 17:31
About the second question, I don't understand what the formula means. I think I need this formula to say that it is a morphism, because the formula says that in $U_n/U_n+1$ the product of the class of $x$ and $y$ is equal to the class of $x+y$, becuse of the congruence modulo $p^n+1$? But I'm not sure, because elements of $U_n+1$ are of the form $1+p^n+1mathbbZ_p$
– robbis
Jul 20 at 17:39
“Then Serre says†means that if you start with $xiin U_n/U_n+1$, choose an element $xin U_n$ representing $xi$, and form $fracx-1p^npmod pinBbb Z/pBbb Z$, then the result doesn’t depend on your choice of $x$, and the map is a homomorphism, even an isomorphism.
– Lubin
Jul 20 at 18:36
I'm not sure what the question is. That $U_n$ is the kernel of $f_n$ just comes from the definition of kernel. The second question: surely this means that $1+p^nxmapsto xpmod p$ is a homomorphism. Is being a homomorphism the problem, or is its bijectivity?
– Lord Shark the Unknown
Jul 20 at 17:26
I'm not sure what the question is. That $U_n$ is the kernel of $f_n$ just comes from the definition of kernel. The second question: surely this means that $1+p^nxmapsto xpmod p$ is a homomorphism. Is being a homomorphism the problem, or is its bijectivity?
– Lord Shark the Unknown
Jul 20 at 17:26
For the first point: is $f_n$ the morphism that maps a $p$-adic unit $x$ to its $n$-th component, which is in $(mathbbZ/p^nmathbbZ)^*$? is it right?
– robbis
Jul 20 at 17:31
For the first point: is $f_n$ the morphism that maps a $p$-adic unit $x$ to its $n$-th component, which is in $(mathbbZ/p^nmathbbZ)^*$? is it right?
– robbis
Jul 20 at 17:31
About the second question, I don't understand what the formula means. I think I need this formula to say that it is a morphism, because the formula says that in $U_n/U_n+1$ the product of the class of $x$ and $y$ is equal to the class of $x+y$, becuse of the congruence modulo $p^n+1$? But I'm not sure, because elements of $U_n+1$ are of the form $1+p^n+1mathbbZ_p$
– robbis
Jul 20 at 17:39
About the second question, I don't understand what the formula means. I think I need this formula to say that it is a morphism, because the formula says that in $U_n/U_n+1$ the product of the class of $x$ and $y$ is equal to the class of $x+y$, becuse of the congruence modulo $p^n+1$? But I'm not sure, because elements of $U_n+1$ are of the form $1+p^n+1mathbbZ_p$
– robbis
Jul 20 at 17:39
“Then Serre says†means that if you start with $xiin U_n/U_n+1$, choose an element $xin U_n$ representing $xi$, and form $fracx-1p^npmod pinBbb Z/pBbb Z$, then the result doesn’t depend on your choice of $x$, and the map is a homomorphism, even an isomorphism.
– Lubin
Jul 20 at 18:36
“Then Serre says†means that if you start with $xiin U_n/U_n+1$, choose an element $xin U_n$ representing $xi$, and form $fracx-1p^npmod pinBbb Z/pBbb Z$, then the result doesn’t depend on your choice of $x$, and the map is a homomorphism, even an isomorphism.
– Lubin
Jul 20 at 18:36
add a comment |Â
1 Answer
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$DeclareMathOperatorKerKerDeclareMathOperatorUU$To prove $Ker f_n=1+p^nBbb Z_p$, recall that the group homomorphism $f_n:Uto(Bbb Z/p^nBbb Z)^times$ is induced by the ring homomorphism $varepsilon_n:Bbb Z_ptoBbb Z/p^nBbb Z$ whose kernel is $Kervarepsilon_n=p^nBbb Z_p$ (cfr. proposition 1).
Consequently,
beginalign
Ker f_n
&=xinU:f_n(x)=1\
&=UcapxinBbb Z_p:varepsilon_n(x)=1\
&=Ucap (1+Kervarepsilon_n)\
&=(Bbb Z_psetminus pBbb Z_p)cap (1+p^nBbb Z_p)\
&=1+p^nBbb Z_p
endalign
where note that $U=Bbb Z_psetminus pBbb Z_p$ as stated in proposition 2.
For the second question, consider the function
beginalign
&varphi:Bbb Z_ptoU_n/U_n+1&
&xmapsto(1+p^nx)U_n+1
endalign
Then the formula $(1+p^nx)(1+p^ny)equiv 1+p^n(x+y)pmodp^n+1$ proves that $varphi$ is a group homomorphism from the additive group $Bbb Z_p$ to the multiplicative group $U_n/U_n+1$.
Indeed, there exists some $winBbb Z_p$ such that
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))+p^n+1w$$
hence
$$z=frac w1+p^n(x+y)$$
belongs to $Bbb Z_p$ because $1+p^n(x+y)$ is invertible and satisfy
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))(1+p^n+1z)$$
Since $1+p^n+1zinU_n+1$, it follows that $1+p^n(x+y)$ and $(1+p^nx)(1+p^ny)$ determine the same coset in $U_n/U_n+1$, hence $varphi(x+y)=varphi(x)varphi(y)$.
Clearly $varphi$ is surjective and $Kervarphi=pBbb Z_p$ thus giving rise to a group isomorphism $barvarphi:Bbb Z_p/pBbb Z_ptoU_n/U_n+1$.
Finally, the ring homomorphism $varepsilon_1:Bbb Z_ptoBbb Z/pBbb Z$ is surjective and induce a (ring, hence) group isomorphism $barvarepsilon_1:Bbb Z_p/pBbb Z_ptoBbb Z/pBbb Z$.
Then the composition
$$U_n/U_n+1xrightarrow[sim]barvarphi^-1Bbb Z_p/pBbb Z_pxrightarrow[sim]barvarepsilon_1Bbb Z/pBbb Z$$
gives the required group isomorphism.
answered Jul 20 at 19:12
Fabio Lucchini
5,62411025
Thanks for your answer, it's very clear! I still have 2 questions. In proving that $Ker f_n=1+p^nmathbbZ_p$ I haven't seen the link with the ring homomorphism, but I don't understand how to relate $ker(epsilon_n)$ to $ker(f_n)$. If $x in ker(f_n)$, $f_n(x)=1$, right? So $f_n(x)-1 in ker(epsilon_n)$? How can I conclude?
– robbis
Jul 23 at 11:19
The second one is: the formula with $z$ says that $varphi(x)varphi(y)=varphi(x+y)$ because $(1+p^n+1z) in U_n+1$ so the class is the same, right? Thanks!
– robbis
Jul 23 at 11:22
@robbis: I have add some explanation to my answer.
– Fabio Lucchini
Jul 23 at 11:35
Now it is all clear to me, thanks for your help!
– robbis
Jul 23 at 11:43
Glad to help you!
– Fabio Lucchini
Jul 23 at 11:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$DeclareMathOperatorKerKerDeclareMathOperatorUU$To prove $Ker f_n=1+p^nBbb Z_p$, recall that the group homomorphism $f_n:Uto(Bbb Z/p^nBbb Z)^times$ is induced by the ring homomorphism $varepsilon_n:Bbb Z_ptoBbb Z/p^nBbb Z$ whose kernel is $Kervarepsilon_n=p^nBbb Z_p$ (cfr. proposition 1).
Consequently,
beginalign
Ker f_n
&=xinU:f_n(x)=1\
&=UcapxinBbb Z_p:varepsilon_n(x)=1\
&=Ucap (1+Kervarepsilon_n)\
&=(Bbb Z_psetminus pBbb Z_p)cap (1+p^nBbb Z_p)\
&=1+p^nBbb Z_p
endalign
where note that $U=Bbb Z_psetminus pBbb Z_p$ as stated in proposition 2.
For the second question, consider the function
beginalign
&varphi:Bbb Z_ptoU_n/U_n+1&
&xmapsto(1+p^nx)U_n+1
endalign
Then the formula $(1+p^nx)(1+p^ny)equiv 1+p^n(x+y)pmodp^n+1$ proves that $varphi$ is a group homomorphism from the additive group $Bbb Z_p$ to the multiplicative group $U_n/U_n+1$.
Indeed, there exists some $winBbb Z_p$ such that
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))+p^n+1w$$
hence
$$z=frac w1+p^n(x+y)$$
belongs to $Bbb Z_p$ because $1+p^n(x+y)$ is invertible and satisfy
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))(1+p^n+1z)$$
Since $1+p^n+1zinU_n+1$, it follows that $1+p^n(x+y)$ and $(1+p^nx)(1+p^ny)$ determine the same coset in $U_n/U_n+1$, hence $varphi(x+y)=varphi(x)varphi(y)$.
Clearly $varphi$ is surjective and $Kervarphi=pBbb Z_p$ thus giving rise to a group isomorphism $barvarphi:Bbb Z_p/pBbb Z_ptoU_n/U_n+1$.
Finally, the ring homomorphism $varepsilon_1:Bbb Z_ptoBbb Z/pBbb Z$ is surjective and induce a (ring, hence) group isomorphism $barvarepsilon_1:Bbb Z_p/pBbb Z_ptoBbb Z/pBbb Z$.
Then the composition
$$U_n/U_n+1xrightarrow[sim]barvarphi^-1Bbb Z_p/pBbb Z_pxrightarrow[sim]barvarepsilon_1Bbb Z/pBbb Z$$
gives the required group isomorphism.
answered Jul 20 at 19:12
Fabio Lucchini
5,62411025
Thanks for your answer, it's very clear! I still have 2 questions. In proving that $Ker f_n=1+p^nmathbbZ_p$ I haven't seen the link with the ring homomorphism, but I don't understand how to relate $ker(epsilon_n)$ to $ker(f_n)$. If $x in ker(f_n)$, $f_n(x)=1$, right? So $f_n(x)-1 in ker(epsilon_n)$? How can I conclude?
– robbis
Jul 23 at 11:19
The second one is: the formula with $z$ says that $varphi(x)varphi(y)=varphi(x+y)$ because $(1+p^n+1z) in U_n+1$ so the class is the same, right? Thanks!
– robbis
Jul 23 at 11:22
@robbis: I have add some explanation to my answer.
– Fabio Lucchini
Jul 23 at 11:35
Now it is all clear to me, thanks for your help!
– robbis
Jul 23 at 11:43
Glad to help you!
– Fabio Lucchini
Jul 23 at 11:44
add a comment |Â
up vote
1
down vote
accepted
$DeclareMathOperatorKerKerDeclareMathOperatorUU$To prove $Ker f_n=1+p^nBbb Z_p$, recall that the group homomorphism $f_n:Uto(Bbb Z/p^nBbb Z)^times$ is induced by the ring homomorphism $varepsilon_n:Bbb Z_ptoBbb Z/p^nBbb Z$ whose kernel is $Kervarepsilon_n=p^nBbb Z_p$ (cfr. proposition 1).
Consequently,
beginalign
Ker f_n
&=xinU:f_n(x)=1\
&=UcapxinBbb Z_p:varepsilon_n(x)=1\
&=Ucap (1+Kervarepsilon_n)\
&=(Bbb Z_psetminus pBbb Z_p)cap (1+p^nBbb Z_p)\
&=1+p^nBbb Z_p
endalign
where note that $U=Bbb Z_psetminus pBbb Z_p$ as stated in proposition 2.
For the second question, consider the function
beginalign
&varphi:Bbb Z_ptoU_n/U_n+1&
&xmapsto(1+p^nx)U_n+1
endalign
Then the formula $(1+p^nx)(1+p^ny)equiv 1+p^n(x+y)pmodp^n+1$ proves that $varphi$ is a group homomorphism from the additive group $Bbb Z_p$ to the multiplicative group $U_n/U_n+1$.
Indeed, there exists some $winBbb Z_p$ such that
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))+p^n+1w$$
hence
$$z=frac w1+p^n(x+y)$$
belongs to $Bbb Z_p$ because $1+p^n(x+y)$ is invertible and satisfy
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))(1+p^n+1z)$$
Since $1+p^n+1zinU_n+1$, it follows that $1+p^n(x+y)$ and $(1+p^nx)(1+p^ny)$ determine the same coset in $U_n/U_n+1$, hence $varphi(x+y)=varphi(x)varphi(y)$.
Clearly $varphi$ is surjective and $Kervarphi=pBbb Z_p$ thus giving rise to a group isomorphism $barvarphi:Bbb Z_p/pBbb Z_ptoU_n/U_n+1$.
Finally, the ring homomorphism $varepsilon_1:Bbb Z_ptoBbb Z/pBbb Z$ is surjective and induce a (ring, hence) group isomorphism $barvarepsilon_1:Bbb Z_p/pBbb Z_ptoBbb Z/pBbb Z$.
Then the composition
$$U_n/U_n+1xrightarrow[sim]barvarphi^-1Bbb Z_p/pBbb Z_pxrightarrow[sim]barvarepsilon_1Bbb Z/pBbb Z$$
gives the required group isomorphism.
answered Jul 20 at 19:12
Fabio Lucchini
5,62411025
Thanks for your answer, it's very clear! I still have 2 questions. In proving that $Ker f_n=1+p^nmathbbZ_p$ I haven't seen the link with the ring homomorphism, but I don't understand how to relate $ker(epsilon_n)$ to $ker(f_n)$. If $x in ker(f_n)$, $f_n(x)=1$, right? So $f_n(x)-1 in ker(epsilon_n)$? How can I conclude?
– robbis
Jul 23 at 11:19
The second one is: the formula with $z$ says that $varphi(x)varphi(y)=varphi(x+y)$ because $(1+p^n+1z) in U_n+1$ so the class is the same, right? Thanks!
– robbis
Jul 23 at 11:22
@robbis: I have add some explanation to my answer.
– Fabio Lucchini
Jul 23 at 11:35
Now it is all clear to me, thanks for your help!
– robbis
Jul 23 at 11:43
Glad to help you!
– Fabio Lucchini
Jul 23 at 11:44
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$DeclareMathOperatorKerKerDeclareMathOperatorUU$To prove $Ker f_n=1+p^nBbb Z_p$, recall that the group homomorphism $f_n:Uto(Bbb Z/p^nBbb Z)^times$ is induced by the ring homomorphism $varepsilon_n:Bbb Z_ptoBbb Z/p^nBbb Z$ whose kernel is $Kervarepsilon_n=p^nBbb Z_p$ (cfr. proposition 1).
Consequently,
beginalign
Ker f_n
&=xinU:f_n(x)=1\
&=UcapxinBbb Z_p:varepsilon_n(x)=1\
&=Ucap (1+Kervarepsilon_n)\
&=(Bbb Z_psetminus pBbb Z_p)cap (1+p^nBbb Z_p)\
&=1+p^nBbb Z_p
endalign
where note that $U=Bbb Z_psetminus pBbb Z_p$ as stated in proposition 2.
For the second question, consider the function
beginalign
&varphi:Bbb Z_ptoU_n/U_n+1&
&xmapsto(1+p^nx)U_n+1
endalign
Then the formula $(1+p^nx)(1+p^ny)equiv 1+p^n(x+y)pmodp^n+1$ proves that $varphi$ is a group homomorphism from the additive group $Bbb Z_p$ to the multiplicative group $U_n/U_n+1$.
Indeed, there exists some $winBbb Z_p$ such that
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))+p^n+1w$$
hence
$$z=frac w1+p^n(x+y)$$
belongs to $Bbb Z_p$ because $1+p^n(x+y)$ is invertible and satisfy
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))(1+p^n+1z)$$
Since $1+p^n+1zinU_n+1$, it follows that $1+p^n(x+y)$ and $(1+p^nx)(1+p^ny)$ determine the same coset in $U_n/U_n+1$, hence $varphi(x+y)=varphi(x)varphi(y)$.
Clearly $varphi$ is surjective and $Kervarphi=pBbb Z_p$ thus giving rise to a group isomorphism $barvarphi:Bbb Z_p/pBbb Z_ptoU_n/U_n+1$.
Finally, the ring homomorphism $varepsilon_1:Bbb Z_ptoBbb Z/pBbb Z$ is surjective and induce a (ring, hence) group isomorphism $barvarepsilon_1:Bbb Z_p/pBbb Z_ptoBbb Z/pBbb Z$.
Then the composition
$$U_n/U_n+1xrightarrow[sim]barvarphi^-1Bbb Z_p/pBbb Z_pxrightarrow[sim]barvarepsilon_1Bbb Z/pBbb Z$$
gives the required group isomorphism.
answered Jul 20 at 19:12
Fabio Lucchini
5,62411025
$DeclareMathOperatorKerKerDeclareMathOperatorUU$To prove $Ker f_n=1+p^nBbb Z_p$, recall that the group homomorphism $f_n:Uto(Bbb Z/p^nBbb Z)^times$ is induced by the ring homomorphism $varepsilon_n:Bbb Z_ptoBbb Z/p^nBbb Z$ whose kernel is $Kervarepsilon_n=p^nBbb Z_p$ (cfr. proposition 1).
Consequently,
beginalign
Ker f_n
&=xinU:f_n(x)=1\
&=UcapxinBbb Z_p:varepsilon_n(x)=1\
&=Ucap (1+Kervarepsilon_n)\
&=(Bbb Z_psetminus pBbb Z_p)cap (1+p^nBbb Z_p)\
&=1+p^nBbb Z_p
endalign
where note that $U=Bbb Z_psetminus pBbb Z_p$ as stated in proposition 2.
For the second question, consider the function
beginalign
&varphi:Bbb Z_ptoU_n/U_n+1&
&xmapsto(1+p^nx)U_n+1
endalign
Then the formula $(1+p^nx)(1+p^ny)equiv 1+p^n(x+y)pmodp^n+1$ proves that $varphi$ is a group homomorphism from the additive group $Bbb Z_p$ to the multiplicative group $U_n/U_n+1$.
Indeed, there exists some $winBbb Z_p$ such that
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))+p^n+1w$$
hence
$$z=frac w1+p^n(x+y)$$
belongs to $Bbb Z_p$ because $1+p^n(x+y)$ is invertible and satisfy
$$(1+p^nx)(1+p^ny)=(1+p^n(x+y))(1+p^n+1z)$$
Since $1+p^n+1zinU_n+1$, it follows that $1+p^n(x+y)$ and $(1+p^nx)(1+p^ny)$ determine the same coset in $U_n/U_n+1$, hence $varphi(x+y)=varphi(x)varphi(y)$.
Clearly $varphi$ is surjective and $Kervarphi=pBbb Z_p$ thus giving rise to a group isomorphism $barvarphi:Bbb Z_p/pBbb Z_ptoU_n/U_n+1$.
Finally, the ring homomorphism $varepsilon_1:Bbb Z_ptoBbb Z/pBbb Z$ is surjective and induce a (ring, hence) group isomorphism $barvarepsilon_1:Bbb Z_p/pBbb Z_ptoBbb Z/pBbb Z$.
Then the composition
$$U_n/U_n+1xrightarrow[sim]barvarphi^-1Bbb Z_p/pBbb Z_pxrightarrow[sim]barvarepsilon_1Bbb Z/pBbb Z$$
gives the required group isomorphism.
answered Jul 20 at 19:12
Fabio Lucchini
5,62411025
edited Jul 23 at 11:35
answered Jul 20 at 19:12
Fabio Lucchini
5,62411025
answered Jul 20 at 19:12
Fabio Lucchini
5,62411025
answered Jul 20 at 19:12
Fabio Lucchini
5,62411025
5,62411025
Thanks for your answer, it's very clear! I still have 2 questions. In proving that $Ker f_n=1+p^nmathbbZ_p$ I haven't seen the link with the ring homomorphism, but I don't understand how to relate $ker(epsilon_n)$ to $ker(f_n)$. If $x in ker(f_n)$, $f_n(x)=1$, right? So $f_n(x)-1 in ker(epsilon_n)$? How can I conclude?
– robbis
Jul 23 at 11:19
The second one is: the formula with $z$ says that $varphi(x)varphi(y)=varphi(x+y)$ because $(1+p^n+1z) in U_n+1$ so the class is the same, right? Thanks!
– robbis
Jul 23 at 11:22
@robbis: I have add some explanation to my answer.
– Fabio Lucchini
Jul 23 at 11:35
Now it is all clear to me, thanks for your help!
– robbis
Jul 23 at 11:43
Glad to help you!
– Fabio Lucchini
Jul 23 at 11:44
add a comment |Â
Thanks for your answer, it's very clear! I still have 2 questions. In proving that $Ker f_n=1+p^nmathbbZ_p$ I haven't seen the link with the ring homomorphism, but I don't understand how to relate $ker(epsilon_n)$ to $ker(f_n)$. If $x in ker(f_n)$, $f_n(x)=1$, right? So $f_n(x)-1 in ker(epsilon_n)$? How can I conclude?
– robbis
Jul 23 at 11:19
The second one is: the formula with $z$ says that $varphi(x)varphi(y)=varphi(x+y)$ because $(1+p^n+1z) in U_n+1$ so the class is the same, right? Thanks!
– robbis
Jul 23 at 11:22
@robbis: I have add some explanation to my answer.
– Fabio Lucchini
Jul 23 at 11:35
Now it is all clear to me, thanks for your help!
– robbis
Jul 23 at 11:43
Glad to help you!
– Fabio Lucchini
Jul 23 at 11:44
Thanks for your answer, it's very clear! I still have 2 questions. In proving that $Ker f_n=1+p^nmathbbZ_p$ I haven't seen the link with the ring homomorphism, but I don't understand how to relate $ker(epsilon_n)$ to $ker(f_n)$. If $x in ker(f_n)$, $f_n(x)=1$, right? So $f_n(x)-1 in ker(epsilon_n)$? How can I conclude?
– robbis
Jul 23 at 11:19
Thanks for your answer, it's very clear! I still have 2 questions. In proving that $Ker f_n=1+p^nmathbbZ_p$ I haven't seen the link with the ring homomorphism, but I don't understand how to relate $ker(epsilon_n)$ to $ker(f_n)$. If $x in ker(f_n)$, $f_n(x)=1$, right? So $f_n(x)-1 in ker(epsilon_n)$? How can I conclude?
– robbis
Jul 23 at 11:19
The second one is: the formula with $z$ says that $varphi(x)varphi(y)=varphi(x+y)$ because $(1+p^n+1z) in U_n+1$ so the class is the same, right? Thanks!
– robbis
Jul 23 at 11:22
The second one is: the formula with $z$ says that $varphi(x)varphi(y)=varphi(x+y)$ because $(1+p^n+1z) in U_n+1$ so the class is the same, right? Thanks!
– robbis
Jul 23 at 11:22
@robbis: I have add some explanation to my answer.
– Fabio Lucchini
Jul 23 at 11:35
@robbis: I have add some explanation to my answer.
– Fabio Lucchini
Jul 23 at 11:35
Now it is all clear to me, thanks for your help!
– robbis
Jul 23 at 11:43
Now it is all clear to me, thanks for your help!
– robbis
Jul 23 at 11:43
Glad to help you!
– Fabio Lucchini
Jul 23 at 11:44
Glad to help you!
– Fabio Lucchini
Jul 23 at 11:44
add a comment |Â
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I'm not sure what the question is. That $U_n$ is the kernel of $f_n$ just comes from the definition of kernel. The second question: surely this means that $1+p^nxmapsto xpmod p$ is a homomorphism. Is being a homomorphism the problem, or is its bijectivity?
– Lord Shark the Unknown
Jul 20 at 17:26
For the first point: is $f_n$ the morphism that maps a $p$-adic unit $x$ to its $n$-th component, which is in $(mathbbZ/p^nmathbbZ)^*$? is it right?
– robbis
Jul 20 at 17:31
About the second question, I don't understand what the formula means. I think I need this formula to say that it is a morphism, because the formula says that in $U_n/U_n+1$ the product of the class of $x$ and $y$ is equal to the class of $x+y$, becuse of the congruence modulo $p^n+1$? But I'm not sure, because elements of $U_n+1$ are of the form $1+p^n+1mathbbZ_p$
– robbis
Jul 20 at 17:39
“Then Serre says†means that if you start with $xiin U_n/U_n+1$, choose an element $xin U_n$ representing $xi$, and form $fracx-1p^npmod pinBbb Z/pBbb Z$, then the result doesn’t depend on your choice of $x$, and the map is a homomorphism, even an isomorphism.
– Lubin
Jul 20 at 18:36