A prime ideal in a polynomial ring whose intersection with the base ring is zero
$begingroup$
It seems very straightforward but I can't prove or disprove this statement :
Let $R $ be an integral domain and $P $ be a prime ideal of $R [x] $ such that $P cap R $ is zero. Then $P$ must be principal? Or is there any counterexamples?
What about $R $ is UFD?
Approach : we can localize $R [x] $ using $R^{times} $, then $P $ coincides with the prime ideal in this localized ring, that is, $K[x] $ where $K $ is a field of fraction of $R $. So in this ring, corresponding prime ideal is principal, but it seems not give a nice information.
abstract-algebra commutative-algebra ideals
asked Nov 30 '18 at 11:25
LWWLWW
337111
$endgroup$
add a comment |
$begingroup$
It seems very straightforward but I can't prove or disprove this statement :
Let $R $ be an integral domain and $P $ be a prime ideal of $R [x] $ such that $P cap R $ is zero. Then $P$ must be principal? Or is there any counterexamples?
What about $R $ is UFD?
Approach : we can localize $R [x] $ using $R^{times} $, then $P $ coincides with the prime ideal in this localized ring, that is, $K[x] $ where $K $ is a field of fraction of $R $. So in this ring, corresponding prime ideal is principal, but it seems not give a nice information.
abstract-algebra commutative-algebra ideals
asked Nov 30 '18 at 11:25
LWWLWW
337111
$endgroup$
$begingroup$
When $R$ is a UFD I think it's easy enough to show that $P$ is principal generated by the irreducible primitive generator of its image in $K[X]$. (Use Gauss Lemma.)
$endgroup$
– user26857
Nov 30 '18 at 15:58
$begingroup$
See also here: math.stackexchange.com/a/1343353/121097
$endgroup$
– user26857
Nov 30 '18 at 16:03
add a comment |
$begingroup$
It seems very straightforward but I can't prove or disprove this statement :
Let $R $ be an integral domain and $P $ be a prime ideal of $R [x] $ such that $P cap R $ is zero. Then $P$ must be principal? Or is there any counterexamples?
What about $R $ is UFD?
Approach : we can localize $R [x] $ using $R^{times} $, then $P $ coincides with the prime ideal in this localized ring, that is, $K[x] $ where $K $ is a field of fraction of $R $. So in this ring, corresponding prime ideal is principal, but it seems not give a nice information.
abstract-algebra commutative-algebra ideals
asked Nov 30 '18 at 11:25
LWWLWW
337111
$endgroup$
It seems very straightforward but I can't prove or disprove this statement :
Let $R $ be an integral domain and $P $ be a prime ideal of $R [x] $ such that $P cap R $ is zero. Then $P$ must be principal? Or is there any counterexamples?
What about $R $ is UFD?
Approach : we can localize $R [x] $ using $R^{times} $, then $P $ coincides with the prime ideal in this localized ring, that is, $K[x] $ where $K $ is a field of fraction of $R $. So in this ring, corresponding prime ideal is principal, but it seems not give a nice information.
abstract-algebra commutative-algebra ideals
abstract-algebra commutative-algebra ideals
asked Nov 30 '18 at 11:25
LWWLWW
337111
asked Nov 30 '18 at 11:25
LWWLWW
337111
edited Nov 30 '18 at 12:04
LWW
asked Nov 30 '18 at 11:25
LWWLWW
337111
asked Nov 30 '18 at 11:25
LWWLWW
337111
asked Nov 30 '18 at 11:25
LWWLWW
337111
337111
$begingroup$
When $R$ is a UFD I think it's easy enough to show that $P$ is principal generated by the irreducible primitive generator of its image in $K[X]$. (Use Gauss Lemma.)
$endgroup$
– user26857
Nov 30 '18 at 15:58
$begingroup$
See also here: math.stackexchange.com/a/1343353/121097
$endgroup$
– user26857
Nov 30 '18 at 16:03
add a comment |
$begingroup$
When $R$ is a UFD I think it's easy enough to show that $P$ is principal generated by the irreducible primitive generator of its image in $K[X]$. (Use Gauss Lemma.)
$endgroup$
– user26857
Nov 30 '18 at 15:58
$begingroup$
See also here: math.stackexchange.com/a/1343353/121097
$endgroup$
– user26857
Nov 30 '18 at 16:03
$begingroup$
When $R$ is a UFD I think it's easy enough to show that $P$ is principal generated by the irreducible primitive generator of its image in $K[X]$. (Use Gauss Lemma.)
$endgroup$
– user26857
Nov 30 '18 at 15:58
$begingroup$
When $R$ is a UFD I think it's easy enough to show that $P$ is principal generated by the irreducible primitive generator of its image in $K[X]$. (Use Gauss Lemma.)
$endgroup$
– user26857
Nov 30 '18 at 15:58
$begingroup$
See also here: math.stackexchange.com/a/1343353/121097
$endgroup$
– user26857
Nov 30 '18 at 16:03
$begingroup$
See also here: math.stackexchange.com/a/1343353/121097
$endgroup$
– user26857
Nov 30 '18 at 16:03
add a comment |
1 Answer
1
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oldest
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$begingroup$
An answer to your question is given in Sharma, P.K., A note on Ideals in Polynomial Rings, Arch. Math. 37 (1981), 325-329.
The main theorem of the article is
Theorem 1: Let $R[x]$ be a polynomial ring over an integral domain $R$ and let $Pneq 0$ be a prime ideal in $R[x]$ such that $Pcap R=0$. Let
$$varphi(X)=a_{0}X^{d}+a_{1}X^{d-1}+ldots+a_{d-1}X+a_{d}$$
be a polynomial of least positive degree in $P$. Then $P=(varphi(X))$ if and only if there does not exist $tnotin (a_{0})$ such that $ta_{i}in (a_{0})$ for $1leq ileq d$.
One of the corollaries of the article is
Corollary 1: Let $R$ be a unique factorization domain or a valuation ring and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. Then there exists an irreducible polynomial $varphi(X)$ with least positive degree among non-zero elements of $P$ such that $P=(varphi(X))$.
To fully answer your question, define $operatorname{ct}(f(x))$ to be the ideal generated by the coefficients of $f(x)$ in $R$. Then we have,
Corollary: Let $R$ be an integral domain and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. If there exists a polynomial $varphi(X)in P$ such that $varphi(X)$ is of least positive degree in $P$ and $operatorname{ct}(varphi(X))=R$ then $P=(varphi(X))$.
Proof: Write $varphi(X)=a_{0}X^{d}+ldots+a_{d-1}X+a_{d}$. Let $tin R$ be given such that $ta_{i}in (a_{0})$ for $1leq ileq d$. Since $operatorname{ct}(varphi(X))=R$ there exists $b_{i}in R$ such that
$$sumlimits_{0leq ileq d}a_{i}b_{i}=1.$$
Write $ta_{i}=a_{0}r_{i}$ for each $1leq ileq d$. Then
$$t=tleft(sumlimits_{0leq ileq d}a_{i}b_{i}right)=a_{0}left(sumlimits_{1leq ileq d}r_{i}b_{i}+tb_{0}right)in(a_{0}).$$
Thus $P=(varphi(X))$ by the theorem.$$tag*{$blacksquare$}$$
answered Nov 30 '18 at 14:01
YumekuiMathYumekuiMath
34114
$endgroup$
$begingroup$
I still can't see a counterexample for the general case.
$endgroup$
– user26857
Nov 30 '18 at 16:08
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
An answer to your question is given in Sharma, P.K., A note on Ideals in Polynomial Rings, Arch. Math. 37 (1981), 325-329.
The main theorem of the article is
Theorem 1: Let $R[x]$ be a polynomial ring over an integral domain $R$ and let $Pneq 0$ be a prime ideal in $R[x]$ such that $Pcap R=0$. Let
$$varphi(X)=a_{0}X^{d}+a_{1}X^{d-1}+ldots+a_{d-1}X+a_{d}$$
be a polynomial of least positive degree in $P$. Then $P=(varphi(X))$ if and only if there does not exist $tnotin (a_{0})$ such that $ta_{i}in (a_{0})$ for $1leq ileq d$.
One of the corollaries of the article is
Corollary 1: Let $R$ be a unique factorization domain or a valuation ring and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. Then there exists an irreducible polynomial $varphi(X)$ with least positive degree among non-zero elements of $P$ such that $P=(varphi(X))$.
To fully answer your question, define $operatorname{ct}(f(x))$ to be the ideal generated by the coefficients of $f(x)$ in $R$. Then we have,
Corollary: Let $R$ be an integral domain and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. If there exists a polynomial $varphi(X)in P$ such that $varphi(X)$ is of least positive degree in $P$ and $operatorname{ct}(varphi(X))=R$ then $P=(varphi(X))$.
Proof: Write $varphi(X)=a_{0}X^{d}+ldots+a_{d-1}X+a_{d}$. Let $tin R$ be given such that $ta_{i}in (a_{0})$ for $1leq ileq d$. Since $operatorname{ct}(varphi(X))=R$ there exists $b_{i}in R$ such that
$$sumlimits_{0leq ileq d}a_{i}b_{i}=1.$$
Write $ta_{i}=a_{0}r_{i}$ for each $1leq ileq d$. Then
$$t=tleft(sumlimits_{0leq ileq d}a_{i}b_{i}right)=a_{0}left(sumlimits_{1leq ileq d}r_{i}b_{i}+tb_{0}right)in(a_{0}).$$
Thus $P=(varphi(X))$ by the theorem.$$tag*{$blacksquare$}$$
answered Nov 30 '18 at 14:01
YumekuiMathYumekuiMath
34114
$endgroup$
$begingroup$
I still can't see a counterexample for the general case.
$endgroup$
– user26857
Nov 30 '18 at 16:08
add a comment |
$begingroup$
An answer to your question is given in Sharma, P.K., A note on Ideals in Polynomial Rings, Arch. Math. 37 (1981), 325-329.
The main theorem of the article is
Theorem 1: Let $R[x]$ be a polynomial ring over an integral domain $R$ and let $Pneq 0$ be a prime ideal in $R[x]$ such that $Pcap R=0$. Let
$$varphi(X)=a_{0}X^{d}+a_{1}X^{d-1}+ldots+a_{d-1}X+a_{d}$$
be a polynomial of least positive degree in $P$. Then $P=(varphi(X))$ if and only if there does not exist $tnotin (a_{0})$ such that $ta_{i}in (a_{0})$ for $1leq ileq d$.
One of the corollaries of the article is
Corollary 1: Let $R$ be a unique factorization domain or a valuation ring and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. Then there exists an irreducible polynomial $varphi(X)$ with least positive degree among non-zero elements of $P$ such that $P=(varphi(X))$.
To fully answer your question, define $operatorname{ct}(f(x))$ to be the ideal generated by the coefficients of $f(x)$ in $R$. Then we have,
Corollary: Let $R$ be an integral domain and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. If there exists a polynomial $varphi(X)in P$ such that $varphi(X)$ is of least positive degree in $P$ and $operatorname{ct}(varphi(X))=R$ then $P=(varphi(X))$.
Proof: Write $varphi(X)=a_{0}X^{d}+ldots+a_{d-1}X+a_{d}$. Let $tin R$ be given such that $ta_{i}in (a_{0})$ for $1leq ileq d$. Since $operatorname{ct}(varphi(X))=R$ there exists $b_{i}in R$ such that
$$sumlimits_{0leq ileq d}a_{i}b_{i}=1.$$
Write $ta_{i}=a_{0}r_{i}$ for each $1leq ileq d$. Then
$$t=tleft(sumlimits_{0leq ileq d}a_{i}b_{i}right)=a_{0}left(sumlimits_{1leq ileq d}r_{i}b_{i}+tb_{0}right)in(a_{0}).$$
Thus $P=(varphi(X))$ by the theorem.$$tag*{$blacksquare$}$$
answered Nov 30 '18 at 14:01
YumekuiMathYumekuiMath
34114
$endgroup$
$begingroup$
I still can't see a counterexample for the general case.
$endgroup$
– user26857
Nov 30 '18 at 16:08
add a comment |
$begingroup$
An answer to your question is given in Sharma, P.K., A note on Ideals in Polynomial Rings, Arch. Math. 37 (1981), 325-329.
The main theorem of the article is
Theorem 1: Let $R[x]$ be a polynomial ring over an integral domain $R$ and let $Pneq 0$ be a prime ideal in $R[x]$ such that $Pcap R=0$. Let
$$varphi(X)=a_{0}X^{d}+a_{1}X^{d-1}+ldots+a_{d-1}X+a_{d}$$
be a polynomial of least positive degree in $P$. Then $P=(varphi(X))$ if and only if there does not exist $tnotin (a_{0})$ such that $ta_{i}in (a_{0})$ for $1leq ileq d$.
One of the corollaries of the article is
Corollary 1: Let $R$ be a unique factorization domain or a valuation ring and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. Then there exists an irreducible polynomial $varphi(X)$ with least positive degree among non-zero elements of $P$ such that $P=(varphi(X))$.
To fully answer your question, define $operatorname{ct}(f(x))$ to be the ideal generated by the coefficients of $f(x)$ in $R$. Then we have,
Corollary: Let $R$ be an integral domain and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. If there exists a polynomial $varphi(X)in P$ such that $varphi(X)$ is of least positive degree in $P$ and $operatorname{ct}(varphi(X))=R$ then $P=(varphi(X))$.
Proof: Write $varphi(X)=a_{0}X^{d}+ldots+a_{d-1}X+a_{d}$. Let $tin R$ be given such that $ta_{i}in (a_{0})$ for $1leq ileq d$. Since $operatorname{ct}(varphi(X))=R$ there exists $b_{i}in R$ such that
$$sumlimits_{0leq ileq d}a_{i}b_{i}=1.$$
Write $ta_{i}=a_{0}r_{i}$ for each $1leq ileq d$. Then
$$t=tleft(sumlimits_{0leq ileq d}a_{i}b_{i}right)=a_{0}left(sumlimits_{1leq ileq d}r_{i}b_{i}+tb_{0}right)in(a_{0}).$$
Thus $P=(varphi(X))$ by the theorem.$$tag*{$blacksquare$}$$
answered Nov 30 '18 at 14:01
YumekuiMathYumekuiMath
34114
$endgroup$
An answer to your question is given in Sharma, P.K., A note on Ideals in Polynomial Rings, Arch. Math. 37 (1981), 325-329.
The main theorem of the article is
Theorem 1: Let $R[x]$ be a polynomial ring over an integral domain $R$ and let $Pneq 0$ be a prime ideal in $R[x]$ such that $Pcap R=0$. Let
$$varphi(X)=a_{0}X^{d}+a_{1}X^{d-1}+ldots+a_{d-1}X+a_{d}$$
be a polynomial of least positive degree in $P$. Then $P=(varphi(X))$ if and only if there does not exist $tnotin (a_{0})$ such that $ta_{i}in (a_{0})$ for $1leq ileq d$.
One of the corollaries of the article is
Corollary 1: Let $R$ be a unique factorization domain or a valuation ring and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. Then there exists an irreducible polynomial $varphi(X)$ with least positive degree among non-zero elements of $P$ such that $P=(varphi(X))$.
To fully answer your question, define $operatorname{ct}(f(x))$ to be the ideal generated by the coefficients of $f(x)$ in $R$. Then we have,
Corollary: Let $R$ be an integral domain and $Pneq 0$ a prime ideal in $R[x]$ with $Pcap R=0$. If there exists a polynomial $varphi(X)in P$ such that $varphi(X)$ is of least positive degree in $P$ and $operatorname{ct}(varphi(X))=R$ then $P=(varphi(X))$.
Proof: Write $varphi(X)=a_{0}X^{d}+ldots+a_{d-1}X+a_{d}$. Let $tin R$ be given such that $ta_{i}in (a_{0})$ for $1leq ileq d$. Since $operatorname{ct}(varphi(X))=R$ there exists $b_{i}in R$ such that
$$sumlimits_{0leq ileq d}a_{i}b_{i}=1.$$
Write $ta_{i}=a_{0}r_{i}$ for each $1leq ileq d$. Then
$$t=tleft(sumlimits_{0leq ileq d}a_{i}b_{i}right)=a_{0}left(sumlimits_{1leq ileq d}r_{i}b_{i}+tb_{0}right)in(a_{0}).$$
Thus $P=(varphi(X))$ by the theorem.$$tag*{$blacksquare$}$$
answered Nov 30 '18 at 14:01
YumekuiMathYumekuiMath
34114
answered Nov 30 '18 at 14:01
YumekuiMathYumekuiMath
34114
answered Nov 30 '18 at 14:01
YumekuiMathYumekuiMath
34114
answered Nov 30 '18 at 14:01
YumekuiMathYumekuiMath
34114
34114
$begingroup$
I still can't see a counterexample for the general case.
$endgroup$
– user26857
Nov 30 '18 at 16:08
add a comment |
$begingroup$
I still can't see a counterexample for the general case.
$endgroup$
– user26857
Nov 30 '18 at 16:08
$begingroup$
I still can't see a counterexample for the general case.
$endgroup$
– user26857
Nov 30 '18 at 16:08
$begingroup$
I still can't see a counterexample for the general case.
$endgroup$
– user26857
Nov 30 '18 at 16:08
add a comment |
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$begingroup$
When $R$ is a UFD I think it's easy enough to show that $P$ is principal generated by the irreducible primitive generator of its image in $K[X]$. (Use Gauss Lemma.)
$endgroup$
– user26857
Nov 30 '18 at 15:58
$begingroup$
See also here: math.stackexchange.com/a/1343353/121097
$endgroup$
– user26857
Nov 30 '18 at 16:03