Exercise 4.4.1 in Weibel's 'An Introduction to Homological Algebra'.












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I can solve this question on the assumption that the $x_i$s are not zero-divisors since $dim(R/(x)) = dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?



I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $mathfrak{m}$, but I'm not sure if this leads to anything.






commutative-algebra homological-algebra local-rings regular-rings







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$endgroup$

















    1












    $begingroup$


    enter image description here



    I can solve this question on the assumption that the $x_i$s are not zero-divisors since $dim(R/(x)) = dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?



    I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $mathfrak{m}$, but I'm not sure if this leads to anything.






    commutative-algebra homological-algebra local-rings regular-rings







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      enter image description here



      I can solve this question on the assumption that the $x_i$s are not zero-divisors since $dim(R/(x)) = dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?



      I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $mathfrak{m}$, but I'm not sure if this leads to anything.






      commutative-algebra homological-algebra local-rings regular-rings







      share|cite|improve this question











      $endgroup$




      enter image description here



      I can solve this question on the assumption that the $x_i$s are not zero-divisors since $dim(R/(x)) = dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?



      I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $mathfrak{m}$, but I'm not sure if this leads to anything.






      commutative-algebra homological-algebra local-rings regular-rings




      commutative-algebra homological-algebra local-rings regular-rings






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 28 at 21:39









      user26857

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      39.4k124183










      asked Jan 28 at 13:54









      Jehu314Jehu314

      1569




      1569






















          1 Answer
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          $begingroup$

          For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?



          EDIT:



          From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, if the $x_i$s are not zero divisors.
            $endgroup$
            – Jehu314
            Jan 28 at 14:11










          • $begingroup$
            OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
            $endgroup$
            – Ben
            Jan 28 at 16:39












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          $begingroup$

          For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?



          EDIT:



          From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, if the $x_i$s are not zero divisors.
            $endgroup$
            – Jehu314
            Jan 28 at 14:11










          • $begingroup$
            OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
            $endgroup$
            – Ben
            Jan 28 at 16:39
















          1












          $begingroup$

          For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?



          EDIT:



          From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes, if the $x_i$s are not zero divisors.
            $endgroup$
            – Jehu314
            Jan 28 at 14:11










          • $begingroup$
            OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
            $endgroup$
            – Ben
            Jan 28 at 16:39














          1












          1








          1





          $begingroup$

          For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?



          EDIT:



          From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.






          share|cite|improve this answer











          $endgroup$



          For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?



          EDIT:



          From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 28 at 16:38

























          answered Jan 28 at 14:06









          BenBen

          4,313617




          4,313617












          • $begingroup$
            Yes, if the $x_i$s are not zero divisors.
            $endgroup$
            – Jehu314
            Jan 28 at 14:11










          • $begingroup$
            OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
            $endgroup$
            – Ben
            Jan 28 at 16:39


















          • $begingroup$
            Yes, if the $x_i$s are not zero divisors.
            $endgroup$
            – Jehu314
            Jan 28 at 14:11










          • $begingroup$
            OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
            $endgroup$
            – Ben
            Jan 28 at 16:39
















          $begingroup$
          Yes, if the $x_i$s are not zero divisors.
          $endgroup$
          – Jehu314
          Jan 28 at 14:11




          $begingroup$
          Yes, if the $x_i$s are not zero divisors.
          $endgroup$
          – Jehu314
          Jan 28 at 14:11












          $begingroup$
          OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
          $endgroup$
          – Ben
          Jan 28 at 16:39




          $begingroup$
          OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
          $endgroup$
          – Ben
          Jan 28 at 16:39


















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