Trouble with a step in proof of "$R$ is a U.F.D. $implies R[x]$ is a U.F.D.

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I am reading a proof of this in Dummit & Foote (Chapter $9.3$, Thm. $7$).

The proof uses Gauss' Lemma:

Gauss' Lemma: Let $R$ be a U.F.D. and let $F$ be its field of fractions. If $p(x) in R[x]$ is reducible in $F[x]$, then $p(x)$ is also reducible in $F[x]$. More specifically, if $p(x) = A(x) B(x)$ where $A(x), B(x) in F[x]$, there exist $s, t in F$ such that $a(x) = sA(x)$ and $b(x) = tB(x)$ are in $R[x]$ and $p(x) = a(x)b(x)$.

---

Now on to the question. In this proof, we assume that the g.c.d of the coefficients of $p(x)$ is $1$, and that $deg p(x) ge 1$. The proof goes like this:

Since $F[x]$ is a U.F.D., $p(x)$ can be factored uniquely into irreducibles in $F[x]$. By Gauss' Lemma, such a factorization implies there is a factorization of $p(x)$ in $R[x]$ whose factors are $F$-multiples of the factors in $F[x]$. $colorred textSince the greatest common divisor of the coefficients of p(x) is 1, the g.c.d. of the$ $colorredtextcoefficients in each of these factors must be 1$

I fail to see how the last step in red is justified. This seems like a pretty big jump to me, which is weird since in other places D&F prove very trivial results. So I am thinking that this is something obvious I'm just not seeing?

abstract-algebra ring-theory unique-factorization-domains polynomial-rings

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edited Jul 22 at 1:02

asked Jul 22 at 0:38

Ovi

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I am reading a proof of this in Dummit & Foote (Chapter $9.3$, Thm. $7$).

The proof uses Gauss' Lemma:

Gauss' Lemma: Let $R$ be a U.F.D. and let $F$ be its field of fractions. If $p(x) in R[x]$ is reducible in $F[x]$, then $p(x)$ is also reducible in $F[x]$. More specifically, if $p(x) = A(x) B(x)$ where $A(x), B(x) in F[x]$, there exist $s, t in F$ such that $a(x) = sA(x)$ and $b(x) = tB(x)$ are in $R[x]$ and $p(x) = a(x)b(x)$.

---

Now on to the question. In this proof, we assume that the g.c.d of the coefficients of $p(x)$ is $1$, and that $deg p(x) ge 1$. The proof goes like this:

Since $F[x]$ is a U.F.D., $p(x)$ can be factored uniquely into irreducibles in $F[x]$. By Gauss' Lemma, such a factorization implies there is a factorization of $p(x)$ in $R[x]$ whose factors are $F$-multiples of the factors in $F[x]$. $colorred textSince the greatest common divisor of the coefficients of p(x) is 1, the g.c.d. of the$ $colorredtextcoefficients in each of these factors must be 1$

I fail to see how the last step in red is justified. This seems like a pretty big jump to me, which is weird since in other places D&F prove very trivial results. So I am thinking that this is something obvious I'm just not seeing?

abstract-algebra ring-theory unique-factorization-domains polynomial-rings

share|cite|improve this question

edited Jul 22 at 1:02

asked Jul 22 at 0:38

Ovi

11.3k935105

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I am reading a proof of this in Dummit & Foote (Chapter $9.3$, Thm. $7$).

The proof uses Gauss' Lemma:

Gauss' Lemma: Let $R$ be a U.F.D. and let $F$ be its field of fractions. If $p(x) in R[x]$ is reducible in $F[x]$, then $p(x)$ is also reducible in $F[x]$. More specifically, if $p(x) = A(x) B(x)$ where $A(x), B(x) in F[x]$, there exist $s, t in F$ such that $a(x) = sA(x)$ and $b(x) = tB(x)$ are in $R[x]$ and $p(x) = a(x)b(x)$.

---

Now on to the question. In this proof, we assume that the g.c.d of the coefficients of $p(x)$ is $1$, and that $deg p(x) ge 1$. The proof goes like this:

Since $F[x]$ is a U.F.D., $p(x)$ can be factored uniquely into irreducibles in $F[x]$. By Gauss' Lemma, such a factorization implies there is a factorization of $p(x)$ in $R[x]$ whose factors are $F$-multiples of the factors in $F[x]$. $colorred textSince the greatest common divisor of the coefficients of p(x) is 1, the g.c.d. of the$ $colorredtextcoefficients in each of these factors must be 1$

I fail to see how the last step in red is justified. This seems like a pretty big jump to me, which is weird since in other places D&F prove very trivial results. So I am thinking that this is something obvious I'm just not seeing?

abstract-algebra ring-theory unique-factorization-domains polynomial-rings

share|cite|improve this question

edited Jul 22 at 1:02

asked Jul 22 at 0:38

Ovi

11.3k935105

I am reading a proof of this in Dummit & Foote (Chapter $9.3$, Thm. $7$).

The proof uses Gauss' Lemma:

Gauss' Lemma: Let $R$ be a U.F.D. and let $F$ be its field of fractions. If $p(x) in R[x]$ is reducible in $F[x]$, then $p(x)$ is also reducible in $F[x]$. More specifically, if $p(x) = A(x) B(x)$ where $A(x), B(x) in F[x]$, there exist $s, t in F$ such that $a(x) = sA(x)$ and $b(x) = tB(x)$ are in $R[x]$ and $p(x) = a(x)b(x)$.

---

Now on to the question. In this proof, we assume that the g.c.d of the coefficients of $p(x)$ is $1$, and that $deg p(x) ge 1$. The proof goes like this:

Since $F[x]$ is a U.F.D., $p(x)$ can be factored uniquely into irreducibles in $F[x]$. By Gauss' Lemma, such a factorization implies there is a factorization of $p(x)$ in $R[x]$ whose factors are $F$-multiples of the factors in $F[x]$. $colorred textSince the greatest common divisor of the coefficients of p(x) is 1, the g.c.d. of the$ $colorredtextcoefficients in each of these factors must be 1$

I fail to see how the last step in red is justified. This seems like a pretty big jump to me, which is weird since in other places D&F prove very trivial results. So I am thinking that this is something obvious I'm just not seeing?

abstract-algebra ring-theory unique-factorization-domains polynomial-rings

share|cite|improve this question

edited Jul 22 at 1:02

asked Jul 22 at 0:38

Ovi

11.3k935105

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 1:02

edited Jul 22 at 1:02

edited Jul 22 at 1:02

asked Jul 22 at 0:38

Ovi

11.3k935105

asked Jul 22 at 0:38

Ovi

11.3k935105

asked Jul 22 at 0:38

Ovi

11.3k935105

11.3k935105

add a comment | 

add a comment | 

2 Answers
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accepted

When you expand a product of polynomials $p=p_1p_2ldots p_k$, each coefficient in $p$ is a product of various coefficients from each polynomial $p_i$. Therefore, if one of the polynomials $p_i$ had some common divisor $d$ dividing all of its coefficients, that $d$ would also divide all of the coefficients of $p$.

Contrapositively, if the coefficients of $p$ are relatively prime, so must be the coefficients of each $p_i$.

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answered Jul 22 at 1:11

Alon Amit

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If in one of the factors the coefficients had a nontrivial $gcd$, say $d$, then $d$ would divide that factor, hence $d$ would divide the polynomial $p(x)$. I mean if $p(x)=f(x)g(x)$, and $f(x)=dh(x)$ for some $h(x)in R[x]$, then $p(x)=dh(x)g(x)$, so all coefficients of $p$ are divisible by $d$.

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answered Jul 22 at 1:10

A. Pongrácz

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

When you expand a product of polynomials $p=p_1p_2ldots p_k$, each coefficient in $p$ is a product of various coefficients from each polynomial $p_i$. Therefore, if one of the polynomials $p_i$ had some common divisor $d$ dividing all of its coefficients, that $d$ would also divide all of the coefficients of $p$.

Contrapositively, if the coefficients of $p$ are relatively prime, so must be the coefficients of each $p_i$.

share|cite|improve this answer

answered Jul 22 at 1:11

Alon Amit

10.2k3765

add a comment | 

up vote
3
down vote



accepted

When you expand a product of polynomials $p=p_1p_2ldots p_k$, each coefficient in $p$ is a product of various coefficients from each polynomial $p_i$. Therefore, if one of the polynomials $p_i$ had some common divisor $d$ dividing all of its coefficients, that $d$ would also divide all of the coefficients of $p$.

Contrapositively, if the coefficients of $p$ are relatively prime, so must be the coefficients of each $p_i$.

share|cite|improve this answer

answered Jul 22 at 1:11

Alon Amit

10.2k3765

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

When you expand a product of polynomials $p=p_1p_2ldots p_k$, each coefficient in $p$ is a product of various coefficients from each polynomial $p_i$. Therefore, if one of the polynomials $p_i$ had some common divisor $d$ dividing all of its coefficients, that $d$ would also divide all of the coefficients of $p$.

Contrapositively, if the coefficients of $p$ are relatively prime, so must be the coefficients of each $p_i$.

share|cite|improve this answer

answered Jul 22 at 1:11

Alon Amit

10.2k3765

When you expand a product of polynomials $p=p_1p_2ldots p_k$, each coefficient in $p$ is a product of various coefficients from each polynomial $p_i$. Therefore, if one of the polynomials $p_i$ had some common divisor $d$ dividing all of its coefficients, that $d$ would also divide all of the coefficients of $p$.

Contrapositively, if the coefficients of $p$ are relatively prime, so must be the coefficients of each $p_i$.

share|cite|improve this answer

answered Jul 22 at 1:11

Alon Amit

10.2k3765

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 1:11

Alon Amit

10.2k3765

answered Jul 22 at 1:11

Alon Amit

10.2k3765

answered Jul 22 at 1:11

Alon Amit

10.2k3765

10.2k3765

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add a comment | 

up vote
1
down vote

If in one of the factors the coefficients had a nontrivial $gcd$, say $d$, then $d$ would divide that factor, hence $d$ would divide the polynomial $p(x)$. I mean if $p(x)=f(x)g(x)$, and $f(x)=dh(x)$ for some $h(x)in R[x]$, then $p(x)=dh(x)g(x)$, so all coefficients of $p$ are divisible by $d$.

share|cite|improve this answer

answered Jul 22 at 1:10

A. Pongrácz

2,269221

add a comment | 

up vote
1
down vote

If in one of the factors the coefficients had a nontrivial $gcd$, say $d$, then $d$ would divide that factor, hence $d$ would divide the polynomial $p(x)$. I mean if $p(x)=f(x)g(x)$, and $f(x)=dh(x)$ for some $h(x)in R[x]$, then $p(x)=dh(x)g(x)$, so all coefficients of $p$ are divisible by $d$.

share|cite|improve this answer

answered Jul 22 at 1:10

A. Pongrácz

2,269221

add a comment | 

up vote
1
down vote

up vote
1
down vote

If in one of the factors the coefficients had a nontrivial $gcd$, say $d$, then $d$ would divide that factor, hence $d$ would divide the polynomial $p(x)$. I mean if $p(x)=f(x)g(x)$, and $f(x)=dh(x)$ for some $h(x)in R[x]$, then $p(x)=dh(x)g(x)$, so all coefficients of $p$ are divisible by $d$.

share|cite|improve this answer

answered Jul 22 at 1:10

A. Pongrácz

2,269221

If in one of the factors the coefficients had a nontrivial $gcd$, say $d$, then $d$ would divide that factor, hence $d$ would divide the polynomial $p(x)$. I mean if $p(x)=f(x)g(x)$, and $f(x)=dh(x)$ for some $h(x)in R[x]$, then $p(x)=dh(x)g(x)$, so all coefficients of $p$ are divisible by $d$.

share|cite|improve this answer

answered Jul 22 at 1:10

A. Pongrácz

2,269221

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 1:10

A. Pongrácz

2,269221

answered Jul 22 at 1:10

A. Pongrácz

2,269221

answered Jul 22 at 1:10

A. Pongrácz

2,269221

2,269221

add a comment | 

add a comment | 

 

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