Is a complete intersection ring, which is a quotient of a maximal $A$-sequence, Artinian?

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Let $A$ be a noetherian regular local ring, $x_1,dots,x_n$ a regular $A$-sequence and $B = A / (x_1,dots,x_n)$. Then $B$ is a complete intersection ring by definition.



If $(x_1,dots,x_n)$ is a maximal $A$-sequence, does it follow that $B$ is Artinian?




abstract-algebra local-rings artinian regular-rings




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    Let $A$ be a noetherian regular local ring, $x_1,dots,x_n$ a regular $A$-sequence and $B = A / (x_1,dots,x_n)$. Then $B$ is a complete intersection ring by definition.



    If $(x_1,dots,x_n)$ is a maximal $A$-sequence, does it follow that $B$ is Artinian?




    abstract-algebra local-rings artinian regular-rings




    share|cite|improve this question






















      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      Let $A$ be a noetherian regular local ring, $x_1,dots,x_n$ a regular $A$-sequence and $B = A / (x_1,dots,x_n)$. Then $B$ is a complete intersection ring by definition.



      If $(x_1,dots,x_n)$ is a maximal $A$-sequence, does it follow that $B$ is Artinian?




      abstract-algebra local-rings artinian regular-rings




      share|cite|improve this question












      Let $A$ be a noetherian regular local ring, $x_1,dots,x_n$ a regular $A$-sequence and $B = A / (x_1,dots,x_n)$. Then $B$ is a complete intersection ring by definition.



      If $(x_1,dots,x_n)$ is a maximal $A$-sequence, does it follow that $B$ is Artinian?





      abstract-algebra local-rings artinian regular-rings





      share|cite|improve this question











      share|cite|improve this question




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      asked Aug 29 at 10:12









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          Yes it is. Because $A$ is regular, every maximal $A$-sequence has $textdim(A)$ elements, i.e. $textdim A = n$. By Krulls Hauptidealsatz we know, that every minimal prime over $x_i+1$ in $A/(x_1,dots,x_i)$ has height 0 or 1, and because $x_i+1$ is not a zero divisor, every minimal prime actually has height 1. Thus modding out $x_i+1$ reduces the dimension by at least $1$, so $B=A/(x_1,dots,x_n)$ has dimension 0. Noetherian, zero dimensional rings are artinian.






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            Found the answer myself:



            Yes it is. Because $A$ is regular, every maximal $A$-sequence has $textdim(A)$ elements, i.e. $textdim A = n$. By Krulls Hauptidealsatz we know, that every minimal prime over $x_i+1$ in $A/(x_1,dots,x_i)$ has height 0 or 1, and because $x_i+1$ is not a zero divisor, every minimal prime actually has height 1. Thus modding out $x_i+1$ reduces the dimension by at least $1$, so $B=A/(x_1,dots,x_n)$ has dimension 0. Noetherian, zero dimensional rings are artinian.






            share|cite|improve this answer


























              up vote
              0
              down vote



              accepted










              Found the answer myself:



              Yes it is. Because $A$ is regular, every maximal $A$-sequence has $textdim(A)$ elements, i.e. $textdim A = n$. By Krulls Hauptidealsatz we know, that every minimal prime over $x_i+1$ in $A/(x_1,dots,x_i)$ has height 0 or 1, and because $x_i+1$ is not a zero divisor, every minimal prime actually has height 1. Thus modding out $x_i+1$ reduces the dimension by at least $1$, so $B=A/(x_1,dots,x_n)$ has dimension 0. Noetherian, zero dimensional rings are artinian.






              share|cite|improve this answer
























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                Found the answer myself:



                Yes it is. Because $A$ is regular, every maximal $A$-sequence has $textdim(A)$ elements, i.e. $textdim A = n$. By Krulls Hauptidealsatz we know, that every minimal prime over $x_i+1$ in $A/(x_1,dots,x_i)$ has height 0 or 1, and because $x_i+1$ is not a zero divisor, every minimal prime actually has height 1. Thus modding out $x_i+1$ reduces the dimension by at least $1$, so $B=A/(x_1,dots,x_n)$ has dimension 0. Noetherian, zero dimensional rings are artinian.






                share|cite|improve this answer














                Found the answer myself:



                Yes it is. Because $A$ is regular, every maximal $A$-sequence has $textdim(A)$ elements, i.e. $textdim A = n$. By Krulls Hauptidealsatz we know, that every minimal prime over $x_i+1$ in $A/(x_1,dots,x_i)$ has height 0 or 1, and because $x_i+1$ is not a zero divisor, every minimal prime actually has height 1. Thus modding out $x_i+1$ reduces the dimension by at least $1$, so $B=A/(x_1,dots,x_n)$ has dimension 0. Noetherian, zero dimensional rings are artinian.







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                edited Sep 8 at 21:57

























                answered Aug 29 at 16:50









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