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Notations: $R$- Noetherian graded ring and $I,J$ homogeneous ideals in $R$

Definition: The projective dimension of $R/I$, denoted $pd(R/I)$, is the length of a minimal free graded resolution of $R/I$: $0 \rightarrow \bigoplus_{j} R(-j)^ {\beta_{p,j}(R/I)}\rightarrow \bigoplus_{j} R(-j)^ {\beta_{p-1,j}(R/I)} \rightarrow \cdots \rightarrow \bigoplus_{j} R(-j)^ {\beta_{0,j}(R/I)} \rightarrow R/I \rightarrow 0$.

where $p \leq n$ and $R(-j)$ indicates the ring $R$ with the shifted graduation such that, for all $a \in \mathbb{Z}$, $R(-j)_{a}=R_{a-j}$ and $\beta_{i,j}(R/I)$ is the number of copies of $R(-j)$ appearing in the $i$-th module of the resolution, and is called graded Betti number degree $(i,j)$. The $i$-th Betti number is $\beta_{i}(R/I)=\sum \limits_{j \in \mathbb{Z}}\beta_{i,j}(R/I)$. The Castelnuovo-Mumford regularity (or simply regularity) of $R/I$ is

$\operatorname{reg}(R/I)=\max\{j-i\mid\beta_{i,j}(R/I)\neq 0\}$

Let $I$ be quadratic square free monomial ideal in $k[x_1,\ldots,x_n]$ and $J$ be quadratic square free monomial ideal in $k[y_1,\ldots,y_m]$ with $I\ncong J$. Suppose $\beta_{i,2i}(I) \neq 0$ and $\beta_{j,2j}(J) \neq 0$ for some $i,j$. Let $IJ$ be square free monomial in $k[x_1,\ldots,x_n,y_1,\ldots,y_m]$.

Can we say that $\beta_{(i+j+2),2(i+j+2)}(IJ)\neq 0$ ?

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I believe the answer to your question is no, we cannot conclude that $\beta_{(i+j+2), 2(i+j+2)}(IJ)\neq0$.

For example, let us take $n=m=4$ and $I$ and $J$ to be the following ideals $$I = \langle x_1x_2, x_2x_3\rangle \subset \mathbb{Q}[x_1, x_2, x_3, x_4]=R$$ $$J = \langle y_1y_2, y_3y_4\rangle \subset \mathbb{Q}[y_1, y_2, y_3, y_4]=S.$$ Both of these are generated by square-free monomials and by considering dimensions we see $R/I$ is not isomorphic to $S/J$. (I assume this is what you mean by $I\not\cong J$, sorry if I missed interpreted.) Using Macualay2 to compute the Betti tables of $I$, $J$, and $IJ$ we get:

$$b(I)= \begin{array}{c | c c c c c} - & 0 & 1 & 2 \\ \hline T & 1 & 2 & 1 \\ 0 & 1 & - & - \\ 1 & - & 2 & 1 \\ \end{array} \quad b(J)= \begin{array}{c | c c c c c} - & 0 & 1 & 2 \\ \hline T & 1 & 2 & 1 \\ 0 & 1 & - & -\\ 1 & - & 2 & - \\ 2 & - & - & 1 \end{array} \quad b(IJ)= \begin{array}{c | c c c c c} - & 0 & 1 & 2 & 3 \\ \hline T & 1 & 4 & 4 & 1 \\ 0 & 1 & - & - & - \\ 1 & - & - & - & - \\ 2 & - & - & - & - \\ 3 & - & 4 & 2 & - \\ 4 & - & - & 2 & 1 \end{array}$$

In particular, we see that: $$b_{1,1}(I)=\beta_{1,2}(I)=2, \quad \quad b_{1,1}(J)=\beta_{1,2}(J)=2, \quad \quad \text{and} \quad\quad b_{2,2}(J)=\beta_{2,4}(J)=1.$$ So we may take $i=1$ and $j=1,2$, but in either case: $$\beta_{1+1+2, 2(1+1+2)}(IJ)=b_{4,4}(IJ)=0 \quad \quad \text{and} \quad \quad \beta_{1+2+2, 2(1+2+2)}(IJ)=b_{5,5}(IJ)=0.$$ Out of curiosity what lead you to believe this might be true? (I don't have a lot of intuition for this, and found the above by just messing mourned in Macaulay2.)

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