Distinct terms in Ascending or descending chain of ideals $k-$algebra R which is also a finite dimensional...
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Suppose the $k$-algebra $R$ is finite-dimensional as a vector space over $k$, for example, when $R = k[x]/<f(x)>$, where $f$ is any nonzero polynomial in $k[x]$. Then in particular $R$ is a
finitely generated $k$-algebra since a vector space basis also generates $R$ as a ring. In this case since ideals are also $k$-subspaces any ascending or descending chain of ideals has at most $mathbf{dim_k R + 1}$ distinct terms, hence $R$ satisfies both A.C.C and D.C.C- on ideals.
Doubt: How will any ideal have at most $dim_k R +1$ distinct terms? Shouldn't it be just $dim_k R$ and not $dim_k R +1$?
Suppose $f(x)=a_nx^n+cdots+a_0$, then $R={b_0+b_1x+cdots+b_{n-1}x^{n-1}+<f(x)>| ; b_0,cdots,b_{n-1} in k }$ is an $textbf{$n-$dimensional vector space}$ over $k$. So, any ideal might contain constant, linear etc terms which can be at most $n$.
In fact, for any $n$ dimensional vector space, it should be $n$ only.
P.S.: This is from Dummit and Foote, Abstract Algebra.
abstract-algebra finitely-generated
asked Jan 2 at 12:08
MUHMUH
427316
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add a comment |
$begingroup$
Suppose the $k$-algebra $R$ is finite-dimensional as a vector space over $k$, for example, when $R = k[x]/<f(x)>$, where $f$ is any nonzero polynomial in $k[x]$. Then in particular $R$ is a
finitely generated $k$-algebra since a vector space basis also generates $R$ as a ring. In this case since ideals are also $k$-subspaces any ascending or descending chain of ideals has at most $mathbf{dim_k R + 1}$ distinct terms, hence $R$ satisfies both A.C.C and D.C.C- on ideals.
Doubt: How will any ideal have at most $dim_k R +1$ distinct terms? Shouldn't it be just $dim_k R$ and not $dim_k R +1$?
Suppose $f(x)=a_nx^n+cdots+a_0$, then $R={b_0+b_1x+cdots+b_{n-1}x^{n-1}+<f(x)>| ; b_0,cdots,b_{n-1} in k }$ is an $textbf{$n-$dimensional vector space}$ over $k$. So, any ideal might contain constant, linear etc terms which can be at most $n$.
In fact, for any $n$ dimensional vector space, it should be $n$ only.
P.S.: This is from Dummit and Foote, Abstract Algebra.
abstract-algebra finitely-generated
asked Jan 2 at 12:08
MUHMUH
427316
$endgroup$
add a comment |
$begingroup$
Suppose the $k$-algebra $R$ is finite-dimensional as a vector space over $k$, for example, when $R = k[x]/<f(x)>$, where $f$ is any nonzero polynomial in $k[x]$. Then in particular $R$ is a
finitely generated $k$-algebra since a vector space basis also generates $R$ as a ring. In this case since ideals are also $k$-subspaces any ascending or descending chain of ideals has at most $mathbf{dim_k R + 1}$ distinct terms, hence $R$ satisfies both A.C.C and D.C.C- on ideals.
Doubt: How will any ideal have at most $dim_k R +1$ distinct terms? Shouldn't it be just $dim_k R$ and not $dim_k R +1$?
Suppose $f(x)=a_nx^n+cdots+a_0$, then $R={b_0+b_1x+cdots+b_{n-1}x^{n-1}+<f(x)>| ; b_0,cdots,b_{n-1} in k }$ is an $textbf{$n-$dimensional vector space}$ over $k$. So, any ideal might contain constant, linear etc terms which can be at most $n$.
In fact, for any $n$ dimensional vector space, it should be $n$ only.
P.S.: This is from Dummit and Foote, Abstract Algebra.
abstract-algebra finitely-generated
asked Jan 2 at 12:08
MUHMUH
427316
$endgroup$
Suppose the $k$-algebra $R$ is finite-dimensional as a vector space over $k$, for example, when $R = k[x]/<f(x)>$, where $f$ is any nonzero polynomial in $k[x]$. Then in particular $R$ is a
finitely generated $k$-algebra since a vector space basis also generates $R$ as a ring. In this case since ideals are also $k$-subspaces any ascending or descending chain of ideals has at most $mathbf{dim_k R + 1}$ distinct terms, hence $R$ satisfies both A.C.C and D.C.C- on ideals.
Doubt: How will any ideal have at most $dim_k R +1$ distinct terms? Shouldn't it be just $dim_k R$ and not $dim_k R +1$?
Suppose $f(x)=a_nx^n+cdots+a_0$, then $R={b_0+b_1x+cdots+b_{n-1}x^{n-1}+<f(x)>| ; b_0,cdots,b_{n-1} in k }$ is an $textbf{$n-$dimensional vector space}$ over $k$. So, any ideal might contain constant, linear etc terms which can be at most $n$.
In fact, for any $n$ dimensional vector space, it should be $n$ only.
P.S.: This is from Dummit and Foote, Abstract Algebra.
abstract-algebra finitely-generated
abstract-algebra finitely-generated
asked Jan 2 at 12:08
MUHMUH
427316
asked Jan 2 at 12:08
MUHMUH
427316
edited Jan 2 at 12:25
MUH
asked Jan 2 at 12:08
MUHMUH
427316
asked Jan 2 at 12:08
MUHMUH
427316
asked Jan 2 at 12:08
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if you agree that the ideals are the $k$-subspaces in an $n$-dimensional space, then the number of subspaces, i.e., $d$-dimensional subspace can take the value $d=0,dots,n$. Thus, it is $n+1$.
edited Jan 2 at 21:57
user26857
39.3k124183
answered Jan 2 at 13:54
bourbaki22034bourbaki22034
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$begingroup$
if you agree that the ideals are the $k$-subspaces in an $n$-dimensional space, then the number of subspaces, i.e., $d$-dimensional subspace can take the value $d=0,dots,n$. Thus, it is $n+1$.
edited Jan 2 at 21:57
user26857
39.3k124183
answered Jan 2 at 13:54
bourbaki22034bourbaki22034
614
$endgroup$
add a comment |
$begingroup$
if you agree that the ideals are the $k$-subspaces in an $n$-dimensional space, then the number of subspaces, i.e., $d$-dimensional subspace can take the value $d=0,dots,n$. Thus, it is $n+1$.
edited Jan 2 at 21:57
user26857
39.3k124183
answered Jan 2 at 13:54
bourbaki22034bourbaki22034
614
$endgroup$
add a comment |
$begingroup$
if you agree that the ideals are the $k$-subspaces in an $n$-dimensional space, then the number of subspaces, i.e., $d$-dimensional subspace can take the value $d=0,dots,n$. Thus, it is $n+1$.
edited Jan 2 at 21:57
user26857
39.3k124183
answered Jan 2 at 13:54
bourbaki22034bourbaki22034
614
$endgroup$
if you agree that the ideals are the $k$-subspaces in an $n$-dimensional space, then the number of subspaces, i.e., $d$-dimensional subspace can take the value $d=0,dots,n$. Thus, it is $n+1$.
edited Jan 2 at 21:57
user26857
39.3k124183
answered Jan 2 at 13:54
bourbaki22034bourbaki22034
614
edited Jan 2 at 21:57
user26857
39.3k124183
edited Jan 2 at 21:57
user26857
39.3k124183
edited Jan 2 at 21:57
user26857
39.3k124183
39.3k124183
answered Jan 2 at 13:54
bourbaki22034bourbaki22034
614
answered Jan 2 at 13:54
bourbaki22034bourbaki22034
614
answered Jan 2 at 13:54
bourbaki22034bourbaki22034
614
614
add a comment |
add a comment |
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