Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements in the ring $R=k[X_1,X_2,X_3,X_4]$? Can it be generated with three elements? (Here $k$ is a field.)

Thanks for any help.

  • 1
    $\begingroup$ It is worth to mention that $(X_1,X_2) \cap (X_3,X_4)=(X_1X_3,X_1X_4,X_2X_3,X_2X_4)$. $\endgroup$ – user26857 Oct 17 '14 at 16:31

Localize at the maximal ideal $m = (X_1, X_2, X_3, X_4)$. The number of generators can only drop, so it's enough to show that $I_m$ requires more than $2$ generators, where $I = (X_1, X_2) \cap (X_3, X_4)$ is your ideal. But a local ring has a well-defined notion of minimal number of generators, which is also a vector space dimension. Thus it's enough to find $3$ linearly independent elements in the $k$-vector space $I_m/mI_m$ ($\cong I/mI$).

Note: this assumes that $k$ is a field.

  • 2
    $\begingroup$ There is no need to localize: for a graded $R$-module like this the minimal number of generators equals the minimal number of homogeneous generators, that is, $\dim_KI/mI$. (For more details see here.) $\endgroup$ – user26857 Mar 6 '14 at 22:58
  • $\begingroup$ @user121097: Yes, but this approach works even in the ungraded case: one can choose a maximal ideal containing a nonhomogeneous ideal and the same reasoning applies $\endgroup$ – zcn Mar 7 '14 at 0:15
  • $\begingroup$ Actually your isomorphism shows exactly what I said: in practice we don't need to localize. Let me make more explicit what I mean: if $M$ is a f.g. $R$-module (here $R$ is a polynomial ring over a field $K$), then $\mu(M)\ge\mu(M_m)=\dim_{R_m/mR_m}M_m/mM_m=\dim_K M/mM$. What can say more is that if $M$ is graded, then $\dim_KM/mM$ equals the minimal number of homogeneous generators of $M$, hence equals $\mu(M)$. (By $\mu(M)$ I've denoted the minimal number of generators of $M$.) $\endgroup$ – user26857 Mar 7 '14 at 9:04
  • $\begingroup$ I agree with your statement in practice: if $m$ is maximal, one can effectively ignore the localization (at the end). For purposes of explanation though, I think it's good to explicitly localize - no harm in doing so, and $\mu(M)$ doesn't behave that well in the nonlocal case. $\endgroup$ – zcn Mar 7 '14 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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