Corollary of Gauss's Lemma (polynomials)
$begingroup$
I am trying to prove the following result. I have outlined my attempt at a proof but I get stuck.
Any help would be welcome!
Theorem:
Let $R$ be a UFD and let $K$ be its field of fractions.
Suppose that $f in R[X]$ is a monic polynomial.
If $f=gh$ where $g,h in K[X]$ and $g$ is monic, then $g in R[X].$
Proof Attempt:
Clearly we have $gh=(lambda cdot g_0)(mucdot h_0)$ for some $lambda,mu in K$ and $f_0, g_0 in R[X]$ primitive.
Write $lambda=a/b$ and $mu=c/d$ for some $a,b,c,d in R$.
Clearing denominators yields $(bd)cdot f = (ac)cdot g_0f_0$ where both sides belong to $R[X]$.
By Gauss's lemma $g_0f_0$ is primitive and so taking contents yields $lambdamu=1$.
This proves that $f=g_0h_0$ is a factorization in $R[X]$ but not necessarily that $g in R[X]$.
I can't seem to get any further than this - any help greatly appreciated?
abstract-algebra polynomials ring-theory
asked May 1 '15 at 20:22
user236080user236080
111
$endgroup$
add a comment |
$begingroup$
I am trying to prove the following result. I have outlined my attempt at a proof but I get stuck.
Any help would be welcome!
Theorem:
Let $R$ be a UFD and let $K$ be its field of fractions.
Suppose that $f in R[X]$ is a monic polynomial.
If $f=gh$ where $g,h in K[X]$ and $g$ is monic, then $g in R[X].$
Proof Attempt:
Clearly we have $gh=(lambda cdot g_0)(mucdot h_0)$ for some $lambda,mu in K$ and $f_0, g_0 in R[X]$ primitive.
Write $lambda=a/b$ and $mu=c/d$ for some $a,b,c,d in R$.
Clearing denominators yields $(bd)cdot f = (ac)cdot g_0f_0$ where both sides belong to $R[X]$.
By Gauss's lemma $g_0f_0$ is primitive and so taking contents yields $lambdamu=1$.
This proves that $f=g_0h_0$ is a factorization in $R[X]$ but not necessarily that $g in R[X]$.
I can't seem to get any further than this - any help greatly appreciated?
abstract-algebra polynomials ring-theory
asked May 1 '15 at 20:22
user236080user236080
111
$endgroup$
add a comment |
$begingroup$
I am trying to prove the following result. I have outlined my attempt at a proof but I get stuck.
Any help would be welcome!
Theorem:
Let $R$ be a UFD and let $K$ be its field of fractions.
Suppose that $f in R[X]$ is a monic polynomial.
If $f=gh$ where $g,h in K[X]$ and $g$ is monic, then $g in R[X].$
Proof Attempt:
Clearly we have $gh=(lambda cdot g_0)(mucdot h_0)$ for some $lambda,mu in K$ and $f_0, g_0 in R[X]$ primitive.
Write $lambda=a/b$ and $mu=c/d$ for some $a,b,c,d in R$.
Clearing denominators yields $(bd)cdot f = (ac)cdot g_0f_0$ where both sides belong to $R[X]$.
By Gauss's lemma $g_0f_0$ is primitive and so taking contents yields $lambdamu=1$.
This proves that $f=g_0h_0$ is a factorization in $R[X]$ but not necessarily that $g in R[X]$.
I can't seem to get any further than this - any help greatly appreciated?
abstract-algebra polynomials ring-theory
asked May 1 '15 at 20:22
user236080user236080
111
$endgroup$
I am trying to prove the following result. I have outlined my attempt at a proof but I get stuck.
Any help would be welcome!
Theorem:
Let $R$ be a UFD and let $K$ be its field of fractions.
Suppose that $f in R[X]$ is a monic polynomial.
If $f=gh$ where $g,h in K[X]$ and $g$ is monic, then $g in R[X].$
Proof Attempt:
Clearly we have $gh=(lambda cdot g_0)(mucdot h_0)$ for some $lambda,mu in K$ and $f_0, g_0 in R[X]$ primitive.
Write $lambda=a/b$ and $mu=c/d$ for some $a,b,c,d in R$.
Clearing denominators yields $(bd)cdot f = (ac)cdot g_0f_0$ where both sides belong to $R[X]$.
By Gauss's lemma $g_0f_0$ is primitive and so taking contents yields $lambdamu=1$.
This proves that $f=g_0h_0$ is a factorization in $R[X]$ but not necessarily that $g in R[X]$.
I can't seem to get any further than this - any help greatly appreciated?
abstract-algebra polynomials ring-theory
abstract-algebra polynomials ring-theory
asked May 1 '15 at 20:22
user236080user236080
111
asked May 1 '15 at 20:22
user236080user236080
111
asked May 1 '15 at 20:22
user236080user236080
111
asked May 1 '15 at 20:22
user236080user236080
111
asked May 1 '15 at 20:22
user236080user236080
111
111
add a comment |
add a comment |
2 Answers
2
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The leading term of $f$ is the product of the leading terms of $g_0$ and $h_0$, hence these are units. Hence the factor $lambda$ by which $g_0$ differs from $g$ is a unit.
answered May 5 '15 at 18:42
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
add a comment |
$begingroup$
why does $g$ and $h$ being monic imply that $k$ and $t$ are in $R$?
Because $kg$ and $th$ are primitive. In particular, they belong to $R[x]$. Since the highest coefficient of $kg$ is $k$, and the highest coefficient of $th$ is $h$, both $t$ and $h$ belong to $R$.
why does $k$ and $t$ being invertible in $R$ imply that $g$ and $h$ are in $R[x]$?
If the proof is still not clear to you, feel free to ask more questions.
Similarly, because $kgin R[x]$, then $g=k^{-1}kgin R[x]$ too. The same for $h$.
answered Dec 6 '18 at 15:52
Alex RavskyAlex Ravsky
41.1k32282
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The leading term of $f$ is the product of the leading terms of $g_0$ and $h_0$, hence these are units. Hence the factor $lambda$ by which $g_0$ differs from $g$ is a unit.
answered May 5 '15 at 18:42
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
add a comment |
$begingroup$
The leading term of $f$ is the product of the leading terms of $g_0$ and $h_0$, hence these are units. Hence the factor $lambda$ by which $g_0$ differs from $g$ is a unit.
answered May 5 '15 at 18:42
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
add a comment |
$begingroup$
The leading term of $f$ is the product of the leading terms of $g_0$ and $h_0$, hence these are units. Hence the factor $lambda$ by which $g_0$ differs from $g$ is a unit.
answered May 5 '15 at 18:42
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
The leading term of $f$ is the product of the leading terms of $g_0$ and $h_0$, hence these are units. Hence the factor $lambda$ by which $g_0$ differs from $g$ is a unit.
answered May 5 '15 at 18:42
Hagen von EitzenHagen von Eitzen
279k23271502
answered May 5 '15 at 18:42
Hagen von EitzenHagen von Eitzen
279k23271502
answered May 5 '15 at 18:42
Hagen von EitzenHagen von Eitzen
279k23271502
answered May 5 '15 at 18:42
Hagen von EitzenHagen von Eitzen
279k23271502
279k23271502
add a comment |
add a comment |
$begingroup$
why does $g$ and $h$ being monic imply that $k$ and $t$ are in $R$?
Because $kg$ and $th$ are primitive. In particular, they belong to $R[x]$. Since the highest coefficient of $kg$ is $k$, and the highest coefficient of $th$ is $h$, both $t$ and $h$ belong to $R$.
why does $k$ and $t$ being invertible in $R$ imply that $g$ and $h$ are in $R[x]$?
If the proof is still not clear to you, feel free to ask more questions.
Similarly, because $kgin R[x]$, then $g=k^{-1}kgin R[x]$ too. The same for $h$.
answered Dec 6 '18 at 15:52
Alex RavskyAlex Ravsky
41.1k32282
$endgroup$
add a comment |
$begingroup$
why does $g$ and $h$ being monic imply that $k$ and $t$ are in $R$?
Because $kg$ and $th$ are primitive. In particular, they belong to $R[x]$. Since the highest coefficient of $kg$ is $k$, and the highest coefficient of $th$ is $h$, both $t$ and $h$ belong to $R$.
why does $k$ and $t$ being invertible in $R$ imply that $g$ and $h$ are in $R[x]$?
If the proof is still not clear to you, feel free to ask more questions.
Similarly, because $kgin R[x]$, then $g=k^{-1}kgin R[x]$ too. The same for $h$.
answered Dec 6 '18 at 15:52
Alex RavskyAlex Ravsky
41.1k32282
$endgroup$
add a comment |
$begingroup$
why does $g$ and $h$ being monic imply that $k$ and $t$ are in $R$?
Because $kg$ and $th$ are primitive. In particular, they belong to $R[x]$. Since the highest coefficient of $kg$ is $k$, and the highest coefficient of $th$ is $h$, both $t$ and $h$ belong to $R$.
why does $k$ and $t$ being invertible in $R$ imply that $g$ and $h$ are in $R[x]$?
If the proof is still not clear to you, feel free to ask more questions.
Similarly, because $kgin R[x]$, then $g=k^{-1}kgin R[x]$ too. The same for $h$.
answered Dec 6 '18 at 15:52
Alex RavskyAlex Ravsky
41.1k32282
$endgroup$
why does $g$ and $h$ being monic imply that $k$ and $t$ are in $R$?
Because $kg$ and $th$ are primitive. In particular, they belong to $R[x]$. Since the highest coefficient of $kg$ is $k$, and the highest coefficient of $th$ is $h$, both $t$ and $h$ belong to $R$.
why does $k$ and $t$ being invertible in $R$ imply that $g$ and $h$ are in $R[x]$?
If the proof is still not clear to you, feel free to ask more questions.
Similarly, because $kgin R[x]$, then $g=k^{-1}kgin R[x]$ too. The same for $h$.
answered Dec 6 '18 at 15:52
Alex RavskyAlex Ravsky
41.1k32282
edited Dec 6 '18 at 15:59
answered Dec 6 '18 at 15:52
Alex RavskyAlex Ravsky
41.1k32282
answered Dec 6 '18 at 15:52
Alex RavskyAlex Ravsky
41.1k32282
answered Dec 6 '18 at 15:52
Alex RavskyAlex Ravsky
41.1k32282
41.1k32282
add a comment |
add a comment |
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