I am reading the concept of mapping cone of a resolution. I need help with the following.
Let $I_1=\langle x_1^2-x_2 x_4, x_1 x_2-x_3 x_4, x_1 x_3-x_4^2,x_2^2-x_1 x_3, x_2 x_3-x_1 x_4,x_3^2-x_2 x_4 \rangle \subseteq R_1=\mathbb{K}[x_1,x_2,x_3,x_4]$
and $I_2= \langle x_5^2-x_6 x_8, x_5 x_6 -x_7 x_8, x_5 x_7-x_8^2,x_6^2-x_5 x_7,x_6 x_7-x_5 x_8,x_7^2-x_6 x_8 \rangle \subseteq R_2=\mathbb{K}[x_5,x_6,x_7,x_8]$
are two ideals with $R_i$-minimal free resolution of $I_i$ given by $\mathbf{F}_i$, for $i=1,2$ respectively.
Let $J=I_1 +I_2+\langle x_4-x_8 \rangle \subseteq R=\mathbb{K}[x_1,\dots,x_4,x_5,\dots,x_8]$ be an ideal. What will be the $R$-minimal free resolution of $J$?
I think using tensor product of free resolution $\mathbf{F}_1 \otimes \mathbf{F}_2$ and then from mapping cone we can obtained minimal free resolution of $J$. But, I don't able to prove this.
Any help would be appreciated. Thank you.
This is the perfect kind of question for a computer algebra package such as Macaulay2. https://macaulay2.com/ I highly recommend learning how to use this if you need to computations like this.
The main issue here is the emphasis on "minimal". What you propose sounds fine for producing a free resolution. But it may be subtle to prove that it is in fact minimal.