Factorization in the field of fractions of a local PID.
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Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^ncdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.
I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.
abstract-algebra ring-theory commutative-algebra
asked 18 hours ago
tangentbundle
350111
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up vote
0
down vote
favorite
Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^ncdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.
I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.
abstract-algebra ring-theory commutative-algebra
asked 18 hours ago
tangentbundle
350111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^ncdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.
I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.
abstract-algebra ring-theory commutative-algebra
asked 18 hours ago
tangentbundle
350111
Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^ncdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.
I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
asked 18 hours ago
tangentbundle
350111
asked 18 hours ago
tangentbundle
350111
asked 18 hours ago
tangentbundle
350111
asked 18 hours ago
tangentbundle
350111
asked 18 hours ago
tangentbundle
350111
350111
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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1
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accepted
Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.
If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
$$
frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
$$
answered 18 hours ago
egreg
173k1383197
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.
If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
$$
frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
$$
answered 18 hours ago
egreg
173k1383197
add a comment |
up vote
1
down vote
accepted
Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.
If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
$$
frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
$$
answered 18 hours ago
egreg
173k1383197
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.
If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
$$
frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
$$
answered 18 hours ago
egreg
173k1383197
Since $R$ is a local PID, every element $rin R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.
If $r,sin R$, with $r,sne0$, we have $r=up^n$ and $s=vp^m$, so
$$
frac{r}{s}=frac{up^n}{vp^m}=(uv^{-1})p^{n-m}
$$
answered 18 hours ago
egreg
173k1383197
answered 18 hours ago
egreg
173k1383197
answered 18 hours ago
egreg
173k1383197
answered 18 hours ago
egreg
173k1383197
173k1383197
add a comment |
add a comment |
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