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Contraction of an ideal in the polynomial ring over the fraction field of a UFD
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Let $K$ be the fraction field of a UFD $R$. Let $0ne f(X)in R[X]$. Let $I=f(X)R[X]$ and $J= f(X)K[X] cap R[X]$. Then how to show that $I=aJ$ for some $0ne a in R$ ?
I think $a$ should be the gcd of the coefficients of $f(X)$, but I'm not sure. Please help.
ring-theory commutative-algebra ideals unique-factorization-domains
asked Jul 28 at 19:46
user521337
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Let $K$ be the fraction field of a UFD $R$. Let $0ne f(X)in R[X]$. Let $I=f(X)R[X]$ and $J= f(X)K[X] cap R[X]$. Then how to show that $I=aJ$ for some $0ne a in R$ ?
I think $a$ should be the gcd of the coefficients of $f(X)$, but I'm not sure. Please help.
ring-theory commutative-algebra ideals unique-factorization-domains
asked Jul 28 at 19:46
user521337
606
Your guess is (as far as I can tell) the correct one, and an approach should follow from some Gauss' lemma type ideas. (The specific approach I have in mind uses the idea of the ''content'' of a polynomial). Could you please indicate whether you are familiar with Gauss' lemma, and whether you know what the ''content'' of a polynomial in $K[X]$ means?
– Alex Wertheim
Jul 28 at 21:22
@AlexWertheim: yes I know Gauss' Lemma ... could you please elaborate your comment in an answer ?
– user521337
Jul 28 at 21:24
add a comment |Â
up vote
2
down vote
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up vote
2
down vote
favorite
Let $K$ be the fraction field of a UFD $R$. Let $0ne f(X)in R[X]$. Let $I=f(X)R[X]$ and $J= f(X)K[X] cap R[X]$. Then how to show that $I=aJ$ for some $0ne a in R$ ?
I think $a$ should be the gcd of the coefficients of $f(X)$, but I'm not sure. Please help.
ring-theory commutative-algebra ideals unique-factorization-domains
asked Jul 28 at 19:46
user521337
606
Let $K$ be the fraction field of a UFD $R$. Let $0ne f(X)in R[X]$. Let $I=f(X)R[X]$ and $J= f(X)K[X] cap R[X]$. Then how to show that $I=aJ$ for some $0ne a in R$ ?
I think $a$ should be the gcd of the coefficients of $f(X)$, but I'm not sure. Please help.
ring-theory commutative-algebra ideals unique-factorization-domains
asked Jul 28 at 19:46
user521337
606
asked Jul 28 at 19:46
user521337
606
asked Jul 28 at 19:46
user521337
606
asked Jul 28 at 19:46
user521337
606
606
Your guess is (as far as I can tell) the correct one, and an approach should follow from some Gauss' lemma type ideas. (The specific approach I have in mind uses the idea of the ''content'' of a polynomial). Could you please indicate whether you are familiar with Gauss' lemma, and whether you know what the ''content'' of a polynomial in $K[X]$ means?
– Alex Wertheim
Jul 28 at 21:22
@AlexWertheim: yes I know Gauss' Lemma ... could you please elaborate your comment in an answer ?
– user521337
Jul 28 at 21:24
add a comment |Â
Your guess is (as far as I can tell) the correct one, and an approach should follow from some Gauss' lemma type ideas. (The specific approach I have in mind uses the idea of the ''content'' of a polynomial). Could you please indicate whether you are familiar with Gauss' lemma, and whether you know what the ''content'' of a polynomial in $K[X]$ means?
– Alex Wertheim
Jul 28 at 21:22
@AlexWertheim: yes I know Gauss' Lemma ... could you please elaborate your comment in an answer ?
– user521337
Jul 28 at 21:24
Your guess is (as far as I can tell) the correct one, and an approach should follow from some Gauss' lemma type ideas. (The specific approach I have in mind uses the idea of the ''content'' of a polynomial). Could you please indicate whether you are familiar with Gauss' lemma, and whether you know what the ''content'' of a polynomial in $K[X]$ means?
– Alex Wertheim
Jul 28 at 21:22
Your guess is (as far as I can tell) the correct one, and an approach should follow from some Gauss' lemma type ideas. (The specific approach I have in mind uses the idea of the ''content'' of a polynomial). Could you please indicate whether you are familiar with Gauss' lemma, and whether you know what the ''content'' of a polynomial in $K[X]$ means?
– Alex Wertheim
Jul 28 at 21:22
@AlexWertheim: yes I know Gauss' Lemma ... could you please elaborate your comment in an answer ?
– user521337
Jul 28 at 21:24
@AlexWertheim: yes I know Gauss' Lemma ... could you please elaborate your comment in an answer ?
– user521337
Jul 28 at 21:24
add a comment |Â
1 Answer
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The approach that I outline uses the notation and a result from the wonderful notes of Paul Garrett, found here.
For any $g(X) in K[X]$, write $c(g)$ for the content of $g(X)$ (see the material from the paragraph preceding Proposition 2.0.1 to the sentence after it in the notes linked above). From the definition of content, it is clear that $c(g) in R$ if and only if $g(X) in R[X]$. Indeed, if $c(g)$ is the greatest common divisor of the coefficients of $g$, then for any coefficient $alpha in K$ of $g$, there is some $r in R$ such that $c(g) cdot r = alpha$. Since $c(g), r in R$, this implies $alpha in R$.
Now, let $f(X) in R[X]$, and let $I = f(X)R[X]$. Note that $f'(X) := (1/c(f))f(X) in R[X]$, and $f'(X)K[X] = f(X)K[X]$. I claim that, if $J := f(X)K[X] cap R[X] = f'(X)K[X] cap R[X]$, then $J = f'(X)R[X]$, whence
$$I = f(X)R[X] = c(f)(f'(X)R[X]) = c(f)J.$$
Clearly, $J supseteq f'(X)R[X]$, so it remains to show that $J subseteq f'(X)R[X]$. Suppose that $g(X) in J$. Then $g(X) = f'(X)p(X)$ for some $p(X) in K[X]$. By Lemma 2.0.2 of the notes above, we must have $c(g) = c(f')c(p)$. Since $g in R[X]$ and $c(f') = 1$, we get $c(p) = c(g) in R$, whence $p(X) in R[X]$, and the claim follows.
answered Jul 28 at 22:18
Alex Wertheim
15.6k22748
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The approach that I outline uses the notation and a result from the wonderful notes of Paul Garrett, found here.
For any $g(X) in K[X]$, write $c(g)$ for the content of $g(X)$ (see the material from the paragraph preceding Proposition 2.0.1 to the sentence after it in the notes linked above). From the definition of content, it is clear that $c(g) in R$ if and only if $g(X) in R[X]$. Indeed, if $c(g)$ is the greatest common divisor of the coefficients of $g$, then for any coefficient $alpha in K$ of $g$, there is some $r in R$ such that $c(g) cdot r = alpha$. Since $c(g), r in R$, this implies $alpha in R$.
Now, let $f(X) in R[X]$, and let $I = f(X)R[X]$. Note that $f'(X) := (1/c(f))f(X) in R[X]$, and $f'(X)K[X] = f(X)K[X]$. I claim that, if $J := f(X)K[X] cap R[X] = f'(X)K[X] cap R[X]$, then $J = f'(X)R[X]$, whence
$$I = f(X)R[X] = c(f)(f'(X)R[X]) = c(f)J.$$
Clearly, $J supseteq f'(X)R[X]$, so it remains to show that $J subseteq f'(X)R[X]$. Suppose that $g(X) in J$. Then $g(X) = f'(X)p(X)$ for some $p(X) in K[X]$. By Lemma 2.0.2 of the notes above, we must have $c(g) = c(f')c(p)$. Since $g in R[X]$ and $c(f') = 1$, we get $c(p) = c(g) in R$, whence $p(X) in R[X]$, and the claim follows.
answered Jul 28 at 22:18
Alex Wertheim
15.6k22748
add a comment |Â
up vote
0
down vote
The approach that I outline uses the notation and a result from the wonderful notes of Paul Garrett, found here.
For any $g(X) in K[X]$, write $c(g)$ for the content of $g(X)$ (see the material from the paragraph preceding Proposition 2.0.1 to the sentence after it in the notes linked above). From the definition of content, it is clear that $c(g) in R$ if and only if $g(X) in R[X]$. Indeed, if $c(g)$ is the greatest common divisor of the coefficients of $g$, then for any coefficient $alpha in K$ of $g$, there is some $r in R$ such that $c(g) cdot r = alpha$. Since $c(g), r in R$, this implies $alpha in R$.
Now, let $f(X) in R[X]$, and let $I = f(X)R[X]$. Note that $f'(X) := (1/c(f))f(X) in R[X]$, and $f'(X)K[X] = f(X)K[X]$. I claim that, if $J := f(X)K[X] cap R[X] = f'(X)K[X] cap R[X]$, then $J = f'(X)R[X]$, whence
$$I = f(X)R[X] = c(f)(f'(X)R[X]) = c(f)J.$$
Clearly, $J supseteq f'(X)R[X]$, so it remains to show that $J subseteq f'(X)R[X]$. Suppose that $g(X) in J$. Then $g(X) = f'(X)p(X)$ for some $p(X) in K[X]$. By Lemma 2.0.2 of the notes above, we must have $c(g) = c(f')c(p)$. Since $g in R[X]$ and $c(f') = 1$, we get $c(p) = c(g) in R$, whence $p(X) in R[X]$, and the claim follows.
answered Jul 28 at 22:18
Alex Wertheim
15.6k22748
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The approach that I outline uses the notation and a result from the wonderful notes of Paul Garrett, found here.
For any $g(X) in K[X]$, write $c(g)$ for the content of $g(X)$ (see the material from the paragraph preceding Proposition 2.0.1 to the sentence after it in the notes linked above). From the definition of content, it is clear that $c(g) in R$ if and only if $g(X) in R[X]$. Indeed, if $c(g)$ is the greatest common divisor of the coefficients of $g$, then for any coefficient $alpha in K$ of $g$, there is some $r in R$ such that $c(g) cdot r = alpha$. Since $c(g), r in R$, this implies $alpha in R$.
Now, let $f(X) in R[X]$, and let $I = f(X)R[X]$. Note that $f'(X) := (1/c(f))f(X) in R[X]$, and $f'(X)K[X] = f(X)K[X]$. I claim that, if $J := f(X)K[X] cap R[X] = f'(X)K[X] cap R[X]$, then $J = f'(X)R[X]$, whence
$$I = f(X)R[X] = c(f)(f'(X)R[X]) = c(f)J.$$
Clearly, $J supseteq f'(X)R[X]$, so it remains to show that $J subseteq f'(X)R[X]$. Suppose that $g(X) in J$. Then $g(X) = f'(X)p(X)$ for some $p(X) in K[X]$. By Lemma 2.0.2 of the notes above, we must have $c(g) = c(f')c(p)$. Since $g in R[X]$ and $c(f') = 1$, we get $c(p) = c(g) in R$, whence $p(X) in R[X]$, and the claim follows.
answered Jul 28 at 22:18
Alex Wertheim
15.6k22748
The approach that I outline uses the notation and a result from the wonderful notes of Paul Garrett, found here.
For any $g(X) in K[X]$, write $c(g)$ for the content of $g(X)$ (see the material from the paragraph preceding Proposition 2.0.1 to the sentence after it in the notes linked above). From the definition of content, it is clear that $c(g) in R$ if and only if $g(X) in R[X]$. Indeed, if $c(g)$ is the greatest common divisor of the coefficients of $g$, then for any coefficient $alpha in K$ of $g$, there is some $r in R$ such that $c(g) cdot r = alpha$. Since $c(g), r in R$, this implies $alpha in R$.
Now, let $f(X) in R[X]$, and let $I = f(X)R[X]$. Note that $f'(X) := (1/c(f))f(X) in R[X]$, and $f'(X)K[X] = f(X)K[X]$. I claim that, if $J := f(X)K[X] cap R[X] = f'(X)K[X] cap R[X]$, then $J = f'(X)R[X]$, whence
$$I = f(X)R[X] = c(f)(f'(X)R[X]) = c(f)J.$$
Clearly, $J supseteq f'(X)R[X]$, so it remains to show that $J subseteq f'(X)R[X]$. Suppose that $g(X) in J$. Then $g(X) = f'(X)p(X)$ for some $p(X) in K[X]$. By Lemma 2.0.2 of the notes above, we must have $c(g) = c(f')c(p)$. Since $g in R[X]$ and $c(f') = 1$, we get $c(p) = c(g) in R$, whence $p(X) in R[X]$, and the claim follows.
answered Jul 28 at 22:18
Alex Wertheim
15.6k22748
edited Jul 28 at 22:27
answered Jul 28 at 22:18
Alex Wertheim
15.6k22748
answered Jul 28 at 22:18
Alex Wertheim
15.6k22748
answered Jul 28 at 22:18
Alex Wertheim
15.6k22748
15.6k22748
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Your guess is (as far as I can tell) the correct one, and an approach should follow from some Gauss' lemma type ideas. (The specific approach I have in mind uses the idea of the ''content'' of a polynomial). Could you please indicate whether you are familiar with Gauss' lemma, and whether you know what the ''content'' of a polynomial in $K[X]$ means?
– Alex Wertheim
Jul 28 at 21:22
@AlexWertheim: yes I know Gauss' Lemma ... could you please elaborate your comment in an answer ?
– user521337
Jul 28 at 21:24