# Tensor product of minimal resolutions is a minimal resolution

I am reading http://www.math.cornell.edu/~irena/papers/regularity.pdf by J. McCullough and I. Peeva. Let $S=k[x_1,\dots,x_n]$ and let $I\subset S$ be a homogeneous ideal generated by $m$ forms. Let $\mathrm F$ be the minimal free resolution of $\mathrm{Syz}_1^S(I)$ as an $S$-module. Then let $$\overline T=k[x_1,\dots,x_n,y_1,\dots,y_m].$$ Finally, define $M$ and $N$ two homogeneous ideals in $\overline T$, which I don't know if we need to see explicitly in order to answer my question.

At page 15 they have $$\mathrm{Syz}_1^{\overline T}(M/(M\cap N))= (y_1,\dots,y_m)B_0+(\mathrm{Syz}^S_2(I)\otimes_S\overline T)$$ and they deduce that the minimal free resolution of $M/(M\cap N)$ is $$\mathbf B=\mathbf K_{\overline T}(y_1,\dots,y_m)\otimes_{\overline T}(\mathbf F(-1))\otimes_S\overline T),$$ where $\mathbf K_{\overline T}(y_1,\dots,y_m)$ is the Koszul complex on $y_1,\dots,y_m$ over $\overline T$ and $\mathbf F$ is the minimal $S$-free resolution of $\mathrm{Syz}^S_1(I)$.

I don't understand this passage, so my original question was:

Let $S=k[x_1,\dots,x_n]$. I have three $S$-modules $M$, $N$ and $P$. Let $\mathbf F$ be the minimal free resolution of $M$ and $\mathbf G$ the minimal free resolution of $N$. Suppose $$\mathrm{Syz}_1(P)=\mathrm{Syz}_1(M)\oplus\mathrm{Syz}_1(N).$$ Now I would like to deduce that the minimal free resolution of $P$ is the tensor product $\mathbf F\otimes\mathbf G$. Is it true?

By definition of tensor product of complexes I see that the first syzygy module is exactly the one I found, and I guess the resulting complex is minimal, since $\mathbf M$ and $\mathbf N$ are minimal and $$d^{\mathbf M\otimes\mathbf N}_k(m\otimes n):=d_i^{\mathbf M}(m)\otimes n+(-1)^im\otimes d_j^{\mathbf N}(n),$$ for all $m\in M_i$ and $n\in N_j$ with $i$ and $j$ such that $i+j=k$. Am I right? What puzzles me is: even if so far I am right, why is this a resolution of $P$? I don't even understand why it should be a resolution at all: I have been trying to prove the exactness, unsuccessfully.

Thank you very much in advance!

• The exactness of $F\otimes G$ follows from Kunneth's Formula (encyclopediaofmath.org/index.php/K%c3%bcnneth_formula). – Quimey Jul 6 '17 at 16:13
• Thank you very much, but from what I read I think Kunneth's Formula holds if one of the modules is flat and so is the image of the differential... Maybe it is better if I edit the question to study the particular case I am considering. – Miles Eagle Jul 6 '17 at 16:19
• I assumed the tensor product was over $k$, in which case all modules are flat... – Quimey Jul 6 '17 at 16:22
• I don't think the resolution is going to be minimal though because taking tensor product of complexes you add the lengths and you could go over $n$ (the maximum possible length of a minimal resolution). – Quimey Jul 6 '17 at 16:26
• Okay, I am an idiot. They wrote the reason why this is an exact complex. For what concers the minimality, though, I really think it is enough to take an element in some piece $Z_i$ of the tensor product, call it $\mathbf Z$ (because I don't know which letters to use) and see that it lands in $\mathfrak m Z_{i-1}$ because the two given resolutions are minimal. – Miles Eagle Jul 6 '17 at 16:48