Hilbert series of tensor product of quotient ring by squarefree monomial ideals

Question

Let $R=K[x_1,\dots,x_n]$ be a polynomial ring and $I,J$ be two distinct squarefree monomial ideals of $R$.

It is known that the Hilbert series of $R/I\otimes_K R/J$ is equal to the product of Hilbert series of $R/I$ and $R/J$, that is, $$HS(R/I\otimes_K R/J)=HS(R/I)HS(R/J).$$

In general, it is not true that $$HS(R/I\otimes_R R/J)=HS(R/I)HS(R/J).$$

Is it known a relation among the Hilbert series of $R/I\otimes_R R/J$, $R/I$ and $R/J$, where $I$ and $J$ are squarefree monomial ideals of $R=K[x_1,\dots,x_n]$?

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I did this example using a computational method, from which it is possible to find that the following formulas are not true in this case:

$$HS(R/I\otimes_R R/J)=\frac{HS(R/I)HS(R/J)}{HS(R)}$$

$$HS(R/I\otimes_R R/J)=HS(R/I)+HS(R/J)-HS(R/(I\cap J)$$

Remember that $R/I\otimes_R R/J$ is isomorphic to $R/(I+J)$. Consider $R=K[a,b,c,d,e]$ and $G=(ab,ac,ad,bd,ae)$.

i3 : reduceHilbert hilbertSeries G

               2    3           3
  ( 1 + 2T - 2T  + T ) / (1 - T)

Now we consider $I=(ab,ac,ad)$ and $J=(bd,ae)$, since $G=I+J$.

i5 : reduceHilbert hilbertSeries I

           2    3            4
 1 + T - 2T  + T  /   (1 - T)

i7 : reduceHilbert hilbertSeries J

           2          3
 1 + 2T + T  / (1 - T)

From the previous calculations we have $HS(R/I\otimes_R R/J)\neq \frac{HS(R/I)HS(R/J)}{HS(R)}$.

Consider $L=I\cap J$.

i9 : reduceHilbert hilbertSeries L

           2     3    4           4
  1 + T + T  - 3T  + T  /  (1 - T)

It follows that $HS(R/I\otimes_R R/J)\neq HS(R/I)+HS(R/J)-HS(R/(I\cap J)$.

Answer 1

Here is a general formula/fact:

Proposition (See Lemma 7 here): Let $k$ be a field, let $R$ be a positively graded $k$-algebra, and let $M$ and $N$ be graded $R$-modules. Suppose $M$ and $N$ have finite dimensional components and are nonzero in at most finitely many negative degrees. Then $$\displaystyle \sum_{i=0}^{\infty} (-1)^iHS(\operatorname{Tor}^R_i(M,N))=\dfrac{HS(M)HS(N)}{HS(R)}$$

Taking $M=R/I$ and $N=R/J$ and viewing them as graded $k$-vector spaces immediately recovers the formula $$HS(R/I \otimes_k R/J)=HS(R/I)HS(R/J),$$ since $\operatorname{Tor}^k_i(M,N)=0$ whenever $i>0$ and since $HS(k)=1$.

Working over more general rings (e.g. $k[x_1,\dots,x_n]$) instead, we will not have this $\operatorname{Tor}$-vanishing in general, so we cannot reduce the left hand side to $HS(R/I \otimes_R R/J)$.

As sketched in the comments, the formula $HS(R/I \otimes_R R/J)=HS(R/I)+HS(R/J)-HS(R/I \cap J)$ does hold. The example you computed with Macaulay2 evidences this as $$\dfrac{1+2t-2t^2+t^3}{(1-t)^3}=\dfrac{1+t-2t^2+t^3}{(1-t)^4}+\dfrac{1+2t+t^2}{(1-t)^3}-\dfrac{1+t+t^2-3t^3+t^4}{(1-t)^4}$$