Given $p,q in R$, when $(p,q) subseteq k[p,q]$?
$begingroup$
$k$ is a field of characteristic zero.
$R$ is a commutative $k$-algebra, which is an integral domain.
$Q(R)$ is the field of fractions of $R$.
$p,q in R^{times}$.
$I:=(p,q)$ (= the ideal generated by $p$ and $q$).
$S:=k[p,q] subseteq R$ (= the $k$-subalgebra of $R$ generated by $p$ and $q$).
An element of $I$ is of the form $Ap+Bq$, for some $A,B in R$.
(1) I am looking for an example in which $S subsetneq R$ and $I=(p,q) subseteq k[p,q]=S$. (Or an additional condition which guarantees that $I subseteq S$).
Notice that if $S=R$, then $I subseteq R=S$.
(2) Is something interesting can be said about the set $Q(I):={frac{u}{v}| 0 neq v,u in I}$? or the set
$frac{I}{S}:={frac{u}{v}| 0 neq v in S,u in I}$?
It seems that $Q(I)$ is not a fractional ideal, but just a $Q(R)$-submodule of $Q(R)$.
Also, $frac{I}{S}$ is not a fractional ideal,
and $frac{I}{v_0}:={frac{i}{v_0}|i in I}$ (for some fixed $v_0$) is not a fractional ideal (they are even not submodules).
Any hints are welcome!
algebraic-geometry ring-theory commutative-algebra modules
asked Jan 2 at 2:09
user237522user237522
2,1631617
$endgroup$
add a comment |
$begingroup$
$k$ is a field of characteristic zero.
$R$ is a commutative $k$-algebra, which is an integral domain.
$Q(R)$ is the field of fractions of $R$.
$p,q in R^{times}$.
$I:=(p,q)$ (= the ideal generated by $p$ and $q$).
$S:=k[p,q] subseteq R$ (= the $k$-subalgebra of $R$ generated by $p$ and $q$).
An element of $I$ is of the form $Ap+Bq$, for some $A,B in R$.
(1) I am looking for an example in which $S subsetneq R$ and $I=(p,q) subseteq k[p,q]=S$. (Or an additional condition which guarantees that $I subseteq S$).
Notice that if $S=R$, then $I subseteq R=S$.
(2) Is something interesting can be said about the set $Q(I):={frac{u}{v}| 0 neq v,u in I}$? or the set
$frac{I}{S}:={frac{u}{v}| 0 neq v in S,u in I}$?
It seems that $Q(I)$ is not a fractional ideal, but just a $Q(R)$-submodule of $Q(R)$.
Also, $frac{I}{S}$ is not a fractional ideal,
and $frac{I}{v_0}:={frac{i}{v_0}|i in I}$ (for some fixed $v_0$) is not a fractional ideal (they are even not submodules).
Any hints are welcome!
algebraic-geometry ring-theory commutative-algebra modules
asked Jan 2 at 2:09
user237522user237522
2,1631617
$endgroup$
$begingroup$
As written, this is, of course, impossible. If $p$ and $q$ are units, then $I=R$, so it cannot possibly be a subset of $S$ when $Ssubsetneq R$.
$endgroup$
– jgon
Jan 2 at 3:26
add a comment |
$begingroup$
$k$ is a field of characteristic zero.
$R$ is a commutative $k$-algebra, which is an integral domain.
$Q(R)$ is the field of fractions of $R$.
$p,q in R^{times}$.
$I:=(p,q)$ (= the ideal generated by $p$ and $q$).
$S:=k[p,q] subseteq R$ (= the $k$-subalgebra of $R$ generated by $p$ and $q$).
An element of $I$ is of the form $Ap+Bq$, for some $A,B in R$.
(1) I am looking for an example in which $S subsetneq R$ and $I=(p,q) subseteq k[p,q]=S$. (Or an additional condition which guarantees that $I subseteq S$).
Notice that if $S=R$, then $I subseteq R=S$.
(2) Is something interesting can be said about the set $Q(I):={frac{u}{v}| 0 neq v,u in I}$? or the set
$frac{I}{S}:={frac{u}{v}| 0 neq v in S,u in I}$?
It seems that $Q(I)$ is not a fractional ideal, but just a $Q(R)$-submodule of $Q(R)$.
Also, $frac{I}{S}$ is not a fractional ideal,
and $frac{I}{v_0}:={frac{i}{v_0}|i in I}$ (for some fixed $v_0$) is not a fractional ideal (they are even not submodules).
Any hints are welcome!
algebraic-geometry ring-theory commutative-algebra modules
asked Jan 2 at 2:09
user237522user237522
2,1631617
$endgroup$
$k$ is a field of characteristic zero.
$R$ is a commutative $k$-algebra, which is an integral domain.
$Q(R)$ is the field of fractions of $R$.
$p,q in R^{times}$.
$I:=(p,q)$ (= the ideal generated by $p$ and $q$).
$S:=k[p,q] subseteq R$ (= the $k$-subalgebra of $R$ generated by $p$ and $q$).
An element of $I$ is of the form $Ap+Bq$, for some $A,B in R$.
(1) I am looking for an example in which $S subsetneq R$ and $I=(p,q) subseteq k[p,q]=S$. (Or an additional condition which guarantees that $I subseteq S$).
Notice that if $S=R$, then $I subseteq R=S$.
(2) Is something interesting can be said about the set $Q(I):={frac{u}{v}| 0 neq v,u in I}$? or the set
$frac{I}{S}:={frac{u}{v}| 0 neq v in S,u in I}$?
It seems that $Q(I)$ is not a fractional ideal, but just a $Q(R)$-submodule of $Q(R)$.
Also, $frac{I}{S}$ is not a fractional ideal,
and $frac{I}{v_0}:={frac{i}{v_0}|i in I}$ (for some fixed $v_0$) is not a fractional ideal (they are even not submodules).
Any hints are welcome!
algebraic-geometry ring-theory commutative-algebra modules
algebraic-geometry ring-theory commutative-algebra modules
asked Jan 2 at 2:09
user237522user237522
2,1631617
asked Jan 2 at 2:09
user237522user237522
2,1631617
edited Jan 2 at 3:13
user237522
asked Jan 2 at 2:09
user237522user237522
2,1631617
asked Jan 2 at 2:09
user237522user237522
2,1631617
asked Jan 2 at 2:09
user237522user237522
2,1631617
2,1631617
$begingroup$
As written, this is, of course, impossible. If $p$ and $q$ are units, then $I=R$, so it cannot possibly be a subset of $S$ when $Ssubsetneq R$.
$endgroup$
– jgon
Jan 2 at 3:26
add a comment |
$begingroup$
As written, this is, of course, impossible. If $p$ and $q$ are units, then $I=R$, so it cannot possibly be a subset of $S$ when $Ssubsetneq R$.
$endgroup$
– jgon
Jan 2 at 3:26
$begingroup$
As written, this is, of course, impossible. If $p$ and $q$ are units, then $I=R$, so it cannot possibly be a subset of $S$ when $Ssubsetneq R$.
$endgroup$
– jgon
Jan 2 at 3:26
$begingroup$
As written, this is, of course, impossible. If $p$ and $q$ are units, then $I=R$, so it cannot possibly be a subset of $S$ when $Ssubsetneq R$.
$endgroup$
– jgon
Jan 2 at 3:26
add a comment |
1 Answer
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$begingroup$
Let's suppose $p$ and $q$ are nonunits, since otherwise this is impossible.
You don't ask about one generator, but it helps me think to try small cases:
Let's first see if we can solve this for one generator.
Is it possible to find $R$, a domain, and $pin R$ with $pRsubseteq k[p] subsetneq R$?
Assume it is possible. Choose $fin Rsetminus k[p]$, then $fpin k[p]$, so $fp=r(p)p+c$, with $c$ a constant, but then if $cne 0$, $(f-r(p))p = c$, so $cin pR$, so $pR=R$, contradiction. Otherwise $fp=r(p)p$, so $f=r(p)$ (since $R$ is a domain), contradicting the choice of $f$.
Now with two generators.
Assume we have $pR+qR subseteq k[p,q] subsetneq R$.
Let $I=pR+qR$. Let $J=pS+qS$.
Then $I=J$. Why? Well, first $Jsubseteq I$, clearly, and if
$iin I$, then $iin S$, so $i=j+c$ for some constant $cin k$ and $jin J$. But then $i-j=c$, so $cin I$, and $Isubsetneq R$, so $c=0$. Thus $i=jin J$.
On the other hand, if $I=J$, then $Isubseteq k[p,q]$ obviously.
Thus given $k[p,q]subsetneq R$, then $Isubseteq k[p,q]$ if and only if $I=J$.
Now consider $k[x^2,x^3]subsetneq k[x]$.
Observe that $I=x^2k[x]+x^3k[x]=x^2k[x]$, which is all polynomials with no linear or constant terms. Then $J$ contains $x^n$ for all $nge 2$, since it contains $x^2$, $x^3$, and by the Chicken McNugget theorem, for $n> 2cdot 3-2-3=1$, $n=2a+3b$ for some $a,bge 0$. Thus $J$ contains all monomials of degree $ge 2$, and hence $J=I$.
answered Jan 2 at 3:55
jgonjgon
14.9k32042
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let's suppose $p$ and $q$ are nonunits, since otherwise this is impossible.
You don't ask about one generator, but it helps me think to try small cases:
Let's first see if we can solve this for one generator.
Is it possible to find $R$, a domain, and $pin R$ with $pRsubseteq k[p] subsetneq R$?
Assume it is possible. Choose $fin Rsetminus k[p]$, then $fpin k[p]$, so $fp=r(p)p+c$, with $c$ a constant, but then if $cne 0$, $(f-r(p))p = c$, so $cin pR$, so $pR=R$, contradiction. Otherwise $fp=r(p)p$, so $f=r(p)$ (since $R$ is a domain), contradicting the choice of $f$.
Now with two generators.
Assume we have $pR+qR subseteq k[p,q] subsetneq R$.
Let $I=pR+qR$. Let $J=pS+qS$.
Then $I=J$. Why? Well, first $Jsubseteq I$, clearly, and if
$iin I$, then $iin S$, so $i=j+c$ for some constant $cin k$ and $jin J$. But then $i-j=c$, so $cin I$, and $Isubsetneq R$, so $c=0$. Thus $i=jin J$.
On the other hand, if $I=J$, then $Isubseteq k[p,q]$ obviously.
Thus given $k[p,q]subsetneq R$, then $Isubseteq k[p,q]$ if and only if $I=J$.
Now consider $k[x^2,x^3]subsetneq k[x]$.
Observe that $I=x^2k[x]+x^3k[x]=x^2k[x]$, which is all polynomials with no linear or constant terms. Then $J$ contains $x^n$ for all $nge 2$, since it contains $x^2$, $x^3$, and by the Chicken McNugget theorem, for $n> 2cdot 3-2-3=1$, $n=2a+3b$ for some $a,bge 0$. Thus $J$ contains all monomials of degree $ge 2$, and hence $J=I$.
answered Jan 2 at 3:55
jgonjgon
14.9k32042
$endgroup$
add a comment |
$begingroup$
Let's suppose $p$ and $q$ are nonunits, since otherwise this is impossible.
You don't ask about one generator, but it helps me think to try small cases:
Let's first see if we can solve this for one generator.
Is it possible to find $R$, a domain, and $pin R$ with $pRsubseteq k[p] subsetneq R$?
Assume it is possible. Choose $fin Rsetminus k[p]$, then $fpin k[p]$, so $fp=r(p)p+c$, with $c$ a constant, but then if $cne 0$, $(f-r(p))p = c$, so $cin pR$, so $pR=R$, contradiction. Otherwise $fp=r(p)p$, so $f=r(p)$ (since $R$ is a domain), contradicting the choice of $f$.
Now with two generators.
Assume we have $pR+qR subseteq k[p,q] subsetneq R$.
Let $I=pR+qR$. Let $J=pS+qS$.
Then $I=J$. Why? Well, first $Jsubseteq I$, clearly, and if
$iin I$, then $iin S$, so $i=j+c$ for some constant $cin k$ and $jin J$. But then $i-j=c$, so $cin I$, and $Isubsetneq R$, so $c=0$. Thus $i=jin J$.
On the other hand, if $I=J$, then $Isubseteq k[p,q]$ obviously.
Thus given $k[p,q]subsetneq R$, then $Isubseteq k[p,q]$ if and only if $I=J$.
Now consider $k[x^2,x^3]subsetneq k[x]$.
Observe that $I=x^2k[x]+x^3k[x]=x^2k[x]$, which is all polynomials with no linear or constant terms. Then $J$ contains $x^n$ for all $nge 2$, since it contains $x^2$, $x^3$, and by the Chicken McNugget theorem, for $n> 2cdot 3-2-3=1$, $n=2a+3b$ for some $a,bge 0$. Thus $J$ contains all monomials of degree $ge 2$, and hence $J=I$.
answered Jan 2 at 3:55
jgonjgon
14.9k32042
$endgroup$
add a comment |
$begingroup$
Let's suppose $p$ and $q$ are nonunits, since otherwise this is impossible.
You don't ask about one generator, but it helps me think to try small cases:
Let's first see if we can solve this for one generator.
Is it possible to find $R$, a domain, and $pin R$ with $pRsubseteq k[p] subsetneq R$?
Assume it is possible. Choose $fin Rsetminus k[p]$, then $fpin k[p]$, so $fp=r(p)p+c$, with $c$ a constant, but then if $cne 0$, $(f-r(p))p = c$, so $cin pR$, so $pR=R$, contradiction. Otherwise $fp=r(p)p$, so $f=r(p)$ (since $R$ is a domain), contradicting the choice of $f$.
Now with two generators.
Assume we have $pR+qR subseteq k[p,q] subsetneq R$.
Let $I=pR+qR$. Let $J=pS+qS$.
Then $I=J$. Why? Well, first $Jsubseteq I$, clearly, and if
$iin I$, then $iin S$, so $i=j+c$ for some constant $cin k$ and $jin J$. But then $i-j=c$, so $cin I$, and $Isubsetneq R$, so $c=0$. Thus $i=jin J$.
On the other hand, if $I=J$, then $Isubseteq k[p,q]$ obviously.
Thus given $k[p,q]subsetneq R$, then $Isubseteq k[p,q]$ if and only if $I=J$.
Now consider $k[x^2,x^3]subsetneq k[x]$.
Observe that $I=x^2k[x]+x^3k[x]=x^2k[x]$, which is all polynomials with no linear or constant terms. Then $J$ contains $x^n$ for all $nge 2$, since it contains $x^2$, $x^3$, and by the Chicken McNugget theorem, for $n> 2cdot 3-2-3=1$, $n=2a+3b$ for some $a,bge 0$. Thus $J$ contains all monomials of degree $ge 2$, and hence $J=I$.
answered Jan 2 at 3:55
jgonjgon
14.9k32042
$endgroup$
Let's suppose $p$ and $q$ are nonunits, since otherwise this is impossible.
You don't ask about one generator, but it helps me think to try small cases:
Let's first see if we can solve this for one generator.
Is it possible to find $R$, a domain, and $pin R$ with $pRsubseteq k[p] subsetneq R$?
Assume it is possible. Choose $fin Rsetminus k[p]$, then $fpin k[p]$, so $fp=r(p)p+c$, with $c$ a constant, but then if $cne 0$, $(f-r(p))p = c$, so $cin pR$, so $pR=R$, contradiction. Otherwise $fp=r(p)p$, so $f=r(p)$ (since $R$ is a domain), contradicting the choice of $f$.
Now with two generators.
Assume we have $pR+qR subseteq k[p,q] subsetneq R$.
Let $I=pR+qR$. Let $J=pS+qS$.
Then $I=J$. Why? Well, first $Jsubseteq I$, clearly, and if
$iin I$, then $iin S$, so $i=j+c$ for some constant $cin k$ and $jin J$. But then $i-j=c$, so $cin I$, and $Isubsetneq R$, so $c=0$. Thus $i=jin J$.
On the other hand, if $I=J$, then $Isubseteq k[p,q]$ obviously.
Thus given $k[p,q]subsetneq R$, then $Isubseteq k[p,q]$ if and only if $I=J$.
Now consider $k[x^2,x^3]subsetneq k[x]$.
Observe that $I=x^2k[x]+x^3k[x]=x^2k[x]$, which is all polynomials with no linear or constant terms. Then $J$ contains $x^n$ for all $nge 2$, since it contains $x^2$, $x^3$, and by the Chicken McNugget theorem, for $n> 2cdot 3-2-3=1$, $n=2a+3b$ for some $a,bge 0$. Thus $J$ contains all monomials of degree $ge 2$, and hence $J=I$.
answered Jan 2 at 3:55
jgonjgon
14.9k32042
answered Jan 2 at 3:55
jgonjgon
14.9k32042
answered Jan 2 at 3:55
jgonjgon
14.9k32042
answered Jan 2 at 3:55
jgonjgon
14.9k32042
14.9k32042
add a comment |
add a comment |
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$begingroup$
As written, this is, of course, impossible. If $p$ and $q$ are units, then $I=R$, so it cannot possibly be a subset of $S$ when $Ssubsetneq R$.
$endgroup$
– jgon
Jan 2 at 3:26