Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a ring $R$ and a graded ideal $I$, let $I_{\geqq p}= \oplus_{i\geqq p}I_i$.

If $R/I$ is an artinian, is $R/I_{\geqq p}$ artinian?

If it is false, then is it true in polynomial ring case?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Yes, in the case of polynomial rings, what you ask is true.

More precisely, if $R=k[X_1,...,X_n]$ is a polynomial ring over a field and $I\subset R$ is a graded ideal, then $R/I \;\text {artinian}\implies R/I_{\geqq p} \;\text {artinian}$.

Indeed, $R/I$ artinian means that the only prime ideals containing $I$ are maximal ideals . (Geometrically this says that the subscheme $V(I)\subset \mathbb A^n_k$ has dimension zero)
Now we have $ I_{\geqq p}\supset I^p $ so that any maximal ideal containing $I_{\geqq p}$ must contain $I$ and is thus maximal.
This proves that the ring $R/I_{\geqq p}$ is artinian too.

share|improve this answer
    
I have given a counterexample to the general question in another, independent answer. –  Georges Elencwajg Aug 2 '12 at 12:01

No, $R/I_{\geqq p}$ needn't be artinian.

Let $k$ be a field and define $$R=k[X_1,X_2,...,X_n,...]/\langle X_iX_j\mid i,j\geq 1\rangle=k[x_1,x_2,..., x_n,...]$$ This quotient of the polynomial ring in infinitely many indeterminates inherits a graded structure from the polynomial ring (with $\text {deg} \:x_i=1$) since we have factored out a homogeneous ideal.

Now if $I=\langle x_i\mid i\geq 1\rangle$, we have $R/I=k$ which is certainly artinian.
However $I_{\geqq 2}=0$ so that $R/I_{\geqq 2}=R$, which is not artinian (nor even noetherian).

share|improve this answer
    
thanks Georges, but I saw at Irena peeva "Graded syzygies" p.80 in this book it is true. but I don't reason. –  Sang Cheol Lee Aug 2 '12 at 9:40
    
Dear Sang, I am afraid I don't quite understand what you mean. What "is true" in that reference by Irena peeva ? And what do you mean by "but I don't reason" ? –  Georges Elencwajg Aug 2 '12 at 9:46
    
In Irena peeva book, it is true. but I don't know reason which is artinian. –  Sang Cheol Lee Aug 2 '12 at 10:09
    
I have addressed the case of polynomial rings in another, independent answer. –  Georges Elencwajg Aug 2 '12 at 10:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.