When do we know a quotient of the polynomial ring over a local ring is torsion free

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Let $R$ be a local ring (e.g. the discrete valuation ring $mathbb C[[T]]$) and $mathfrakm$ its maximal ideal. Consider the polynomial ring $R[X_1,dots, X_n]$ in $n$ variables and a finitely generated ideal $I=(f_1,dots,f_l)$.



Question:



  1. What conditions can guarantee $A=R[X_1,dots,X_n]/I$ is torsion free (i.e. if $mathfrakm^N x=0$ for some positive integer $N$ and $xin A$ then $x=0$)? (it would be better if there is no Noetherian condition)

  2. How about the ring of formal power series, $R[[X_1,dots,X_n]]/I$ ?



abstract-algebra polynomials commutative-algebra formal-power-series torsion-groups




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  • Have you worked through a simple case? Say one variable and the localization of $mathbb Z$ at a prime?
    – hardmath
    Jul 15 at 14:24










  • I considered the case $R=mathbb C[[T]]$ and $mathfrakm=(T)$, which I prefer.
    – Hang
    Jul 15 at 14:37






  • 1




    If you're looking at valuation rings, then torsion free is equivalent to flat. And in that case, I think this paper of Jack Ohm will be your friend ams.org/journals/tran/1972-171-00/S0002-9947-1972-0306176-6/…
    – Badam Baplan
    Jul 15 at 16:26










  • @BadamBaplan Thanks. It seems that our notions of torsion-free are slightly different. Here we'd better to say adic torsion free.
    – Hang
    Jul 15 at 17:08















up vote
1
down vote

favorite












Let $R$ be a local ring (e.g. the discrete valuation ring $mathbb C[[T]]$) and $mathfrakm$ its maximal ideal. Consider the polynomial ring $R[X_1,dots, X_n]$ in $n$ variables and a finitely generated ideal $I=(f_1,dots,f_l)$.



Question:



  1. What conditions can guarantee $A=R[X_1,dots,X_n]/I$ is torsion free (i.e. if $mathfrakm^N x=0$ for some positive integer $N$ and $xin A$ then $x=0$)? (it would be better if there is no Noetherian condition)

  2. How about the ring of formal power series, $R[[X_1,dots,X_n]]/I$ ?



abstract-algebra polynomials commutative-algebra formal-power-series torsion-groups




share|cite|improve this question





















  • Have you worked through a simple case? Say one variable and the localization of $mathbb Z$ at a prime?
    – hardmath
    Jul 15 at 14:24










  • I considered the case $R=mathbb C[[T]]$ and $mathfrakm=(T)$, which I prefer.
    – Hang
    Jul 15 at 14:37






  • 1




    If you're looking at valuation rings, then torsion free is equivalent to flat. And in that case, I think this paper of Jack Ohm will be your friend ams.org/journals/tran/1972-171-00/S0002-9947-1972-0306176-6/…
    – Badam Baplan
    Jul 15 at 16:26










  • @BadamBaplan Thanks. It seems that our notions of torsion-free are slightly different. Here we'd better to say adic torsion free.
    – Hang
    Jul 15 at 17:08













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $R$ be a local ring (e.g. the discrete valuation ring $mathbb C[[T]]$) and $mathfrakm$ its maximal ideal. Consider the polynomial ring $R[X_1,dots, X_n]$ in $n$ variables and a finitely generated ideal $I=(f_1,dots,f_l)$.



Question:



  1. What conditions can guarantee $A=R[X_1,dots,X_n]/I$ is torsion free (i.e. if $mathfrakm^N x=0$ for some positive integer $N$ and $xin A$ then $x=0$)? (it would be better if there is no Noetherian condition)

  2. How about the ring of formal power series, $R[[X_1,dots,X_n]]/I$ ?



abstract-algebra polynomials commutative-algebra formal-power-series torsion-groups




share|cite|improve this question













Let $R$ be a local ring (e.g. the discrete valuation ring $mathbb C[[T]]$) and $mathfrakm$ its maximal ideal. Consider the polynomial ring $R[X_1,dots, X_n]$ in $n$ variables and a finitely generated ideal $I=(f_1,dots,f_l)$.



Question:



  1. What conditions can guarantee $A=R[X_1,dots,X_n]/I$ is torsion free (i.e. if $mathfrakm^N x=0$ for some positive integer $N$ and $xin A$ then $x=0$)? (it would be better if there is no Noetherian condition)

  2. How about the ring of formal power series, $R[[X_1,dots,X_n]]/I$ ?




abstract-algebra polynomials commutative-algebra formal-power-series torsion-groups





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 15 at 14:55
























asked Jul 15 at 14:18









Hang

395214




395214











  • Have you worked through a simple case? Say one variable and the localization of $mathbb Z$ at a prime?
    – hardmath
    Jul 15 at 14:24










  • I considered the case $R=mathbb C[[T]]$ and $mathfrakm=(T)$, which I prefer.
    – Hang
    Jul 15 at 14:37






  • 1




    If you're looking at valuation rings, then torsion free is equivalent to flat. And in that case, I think this paper of Jack Ohm will be your friend ams.org/journals/tran/1972-171-00/S0002-9947-1972-0306176-6/…
    – Badam Baplan
    Jul 15 at 16:26










  • @BadamBaplan Thanks. It seems that our notions of torsion-free are slightly different. Here we'd better to say adic torsion free.
    – Hang
    Jul 15 at 17:08

















  • Have you worked through a simple case? Say one variable and the localization of $mathbb Z$ at a prime?
    – hardmath
    Jul 15 at 14:24










  • I considered the case $R=mathbb C[[T]]$ and $mathfrakm=(T)$, which I prefer.
    – Hang
    Jul 15 at 14:37






  • 1




    If you're looking at valuation rings, then torsion free is equivalent to flat. And in that case, I think this paper of Jack Ohm will be your friend ams.org/journals/tran/1972-171-00/S0002-9947-1972-0306176-6/…
    – Badam Baplan
    Jul 15 at 16:26










  • @BadamBaplan Thanks. It seems that our notions of torsion-free are slightly different. Here we'd better to say adic torsion free.
    – Hang
    Jul 15 at 17:08
















Have you worked through a simple case? Say one variable and the localization of $mathbb Z$ at a prime?
– hardmath
Jul 15 at 14:24




Have you worked through a simple case? Say one variable and the localization of $mathbb Z$ at a prime?
– hardmath
Jul 15 at 14:24












I considered the case $R=mathbb C[[T]]$ and $mathfrakm=(T)$, which I prefer.
– Hang
Jul 15 at 14:37




I considered the case $R=mathbb C[[T]]$ and $mathfrakm=(T)$, which I prefer.
– Hang
Jul 15 at 14:37




1




1




If you're looking at valuation rings, then torsion free is equivalent to flat. And in that case, I think this paper of Jack Ohm will be your friend ams.org/journals/tran/1972-171-00/S0002-9947-1972-0306176-6/…
– Badam Baplan
Jul 15 at 16:26




If you're looking at valuation rings, then torsion free is equivalent to flat. And in that case, I think this paper of Jack Ohm will be your friend ams.org/journals/tran/1972-171-00/S0002-9947-1972-0306176-6/…
– Badam Baplan
Jul 15 at 16:26












@BadamBaplan Thanks. It seems that our notions of torsion-free are slightly different. Here we'd better to say adic torsion free.
– Hang
Jul 15 at 17:08





@BadamBaplan Thanks. It seems that our notions of torsion-free are slightly different. Here we'd better to say adic torsion free.
– Hang
Jul 15 at 17:08
















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