0
$\begingroup$

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?

2 Answers 2

1

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.

  • 0
    I have given a counterexample to the general question in another, independent answer.2012-08-02
2

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).

  • 0
    I have addressed the case of poly$n$omial ri$n$gs i$n$ a$n$other, independent answer.2012-08-02