3

I want to show, that the intersection of two squarefree monomial ideals is again a squarefree monomial ideal. The definition of a squarefree monomial ideal I have is that the minimal set generating this ideal contains only squarefree monomial polynomials.

What I don't see is if $f \in G(I \cap J)$ (where I and J are squarefree monomial ideals and G(I) is the minimal generating set of ideal I) implies that $f\in G(I)$ or $f\in G(J)$. Than the statement would be obvious. So what is the relation of $G(I\cap J)$ and $G(I)\cap G(J)$?

Pedro
  • 122,002
krimas
  • 31
  • You’ve defined square-free monomials in terms of square-free monomials. This is very confusing. It appears you are meaning ideals of polynomial rings, but it isn’t clear. – Thomas Andrews Nov 07 '21 at 15:52
  • You can compute a minimal generating set by taking largest common multiples of the monomials generating you ideals. Can you try and prove this? – Pedro Nov 07 '21 at 19:36
  • If I understand correctly you mean this? Let I and J be monomial ideals with $G(I)={ u_1, \dots, u_n }$ and $G(J) = { v_1, \dots, v_n }$. Then $I \cap J = ({ lcm(u_i, v_j)})$. But my profesor said, that this doesnt necessarily mean this is the minimal generating set. – krimas Nov 07 '21 at 19:53

1 Answers1

0

Continuing from what you have $I\cap J=(\{lcm(u_i,v_j)\})$, you can find a minimal generating set of $I\cap J$ step by step removing redundant generators. The redundant ones in the generating set $(g_k)$ is whenever you have $g_{k_1}|g_{k_2}$ for some $k_1\neq k_2$. Continue this process until you cannot find any redundant generators anymore.

Therefore, if $I$ and $J$ are both square-free monomial ideals, then $I\cap J$ is necessarily a square-free monomial ideal.

For example, $$ I=\langle ab, bc, ca\rangle, \ J=\langle bd, cd, ad\rangle. $$ Then $$ I\cap J=\langle abd,abcd,abd,bcd,bcd,bcad,cabd,cad,cad \rangle. $$ In the above, redundant ones are the second $abcd$, the third $abd$, the fifth $bcd$, the sixth $bcad$, the seventh $cabd$, the last $cad$. Then the minimal generating set of $I\cap J$ is $$ I\cap J=\langle abd, bcd, cad\rangle. $$

As for the relation between $G(I\cap J)$ and $G(I)\cap G(J)$, we do not have any meaningful relationship. You can see from the above example, $G(I)\cap G(J)=\emptyset$.

Sungjin Kim
  • 20,102