# Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements from $k[X_1,X_2,X_3,X_4]$?

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.

• It is worth to mention that $(X_1,X_2) \cap (X_3,X_4)=(X_1X_3,X_1X_4,X_2X_3,X_2X_4)$. – 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.
• 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.) – user26857 Mar 6 '14 at 22:58
• 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 $M$.) – user26857 Mar 7 '14 at 9:04
• 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. – zcn Mar 7 '14 at 9:14