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Assume that $I\subset k[x_1,\ldots,x_n]$ and $J\subset k[y_1,\ldots,y_m]$ are monomial ideals in different rings, and the minimal free resolution of $S/I$ and $S/J$, say $F_\cdot$ and $G_\cdot$, are both linear. I believe that $F\otimes G$ is a minimal free resolution for $S/I+J$. Does anyone have any comment for the proof, or any reference?

The tensor product of two chain complexes $(A,d_1)$ and $(B,d_2)$, say $A\otimes B$, is formed by taking all products $A_i \otimes B_j$ and letting $(A \otimes B)_k$ be the direct sum of $A_i\otimes B_j$ for $i+j=k$. The differential maps are defined as $\partial(a\otimes b) = d_ia \otimes b + (-1)^i a \otimes d_2b$ when $a\in A_i$. Then we have $\partial^2 = 0$.

So in order to prove the question we need to check the exactness of the complex, and how it resolves the resolution of $I+J$.

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    $\begingroup$ What is $S{}{}$? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 11, 2012 at 3:57
  • $\begingroup$ In any case, the tensor product of linear complexes is linear; if they are both exact and composed of free modules, their tensor product is exact. Since it is both exact and linear, it is minimal. But you should look for at few examples to see what exactly it resolves. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 11, 2012 at 4:03
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    $\begingroup$ If $I$ and $J$ are ideals in DIFFERENT rings then, as Mariano says, what is the ambient ring $S$ into which you are embedding $I$ and $J$? Also, what do you mean by "resolving the resolution" of $I+J$? $\endgroup$
    – Yemon Choi
    Commented Jan 11, 2012 at 4:38
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    $\begingroup$ @Yeman, $S=k[x_1,\ldots,x_n,y_1,\ldots,y_m]$. I mean is the $F_\cdot\otimes G_\cdot$ is the minimal free resolutiuon of $S/I+J$. The above example shows two ideals with linear resolution such that $I+J$ does not have linear resolution! $\endgroup$
    – today user
    Commented Jan 11, 2012 at 4:42
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    $\begingroup$ Ah. The thing is, when I say linear I mean the module in the $i$th position is generated by elements of degree $i$, which is the definition I am used to. If you only mean the degrees of the generators of the modules in the resolution grow by one, then of course it is not true that linear otimes linear is linear. But in any case, the argument I sketcked above (that is, the Künneth formula) shows that the tensor product of your resolutions is exact, and minimality follows from the definition of its differential and the minimality of the two resolutions you started with. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 11, 2012 at 6:25

2 Answers 2

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There are really two separate things being asked. (1) When is the complex $F\otimes G$ exact? (2) If it is exact, when is $F\otimes G$ a minimal free resolution?

The first question is computed by Tor. Namely $F\otimes G$ is exact if and only if $\text{Tor}_i(S/I,S/J)=0$ for all $i>0$

I believe that the second question is easier. Since the differential $\partial$ on $F\otimes G$ is defined in terms of differentials on $F$ and $G$ (which were assumed to be minimal free resolutions), we see that $\partial (F\otimes G)_i$ belongs to the maximal ideal times $(F\otimes G)_{i-1}$. Thus, $F\otimes G$ is a minimal free resolution if and only if it is exact.

Of course, in your example where $S=k[x_1,\dots,x_n,y_1,\dots,y_m]$, and $I$ only involves $x$-variables and $J$ and only involves $y$-variables, then the higher Tor's vanish and thus $F\otimes G$ is a minimal free resolution.

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  • $\begingroup$ I'm having a formatting issue in the third paragraph. If someone knows how to fix it, that'd be great. $\endgroup$
    – Daniel Erman
    Commented Jan 11, 2012 at 12:04
  • $\begingroup$ Thanks Daniel! Then do we also need to show that $I\otimes J$ is isomorphic to $I+J$? Since we have already just shown that $F\otimes G$ is a minimal free resolution of the tensor product of $I$ and $J$. $\endgroup$
    – today user
    Commented Jan 13, 2012 at 0:41
  • $\begingroup$ @today user: Be careful about conflating an ideal $I$ and the corresponding quotient ring $S/I$. The key fact is that $(S/I)\otimes_S (S/J)$ is isomorphic to $S/(I+J)$, which is always true. So I don't think there's anything more to show (after applying Mariano's argument, say). $\endgroup$
    – Daniel Erman
    Commented Jan 13, 2012 at 14:52
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I'll use Daniel's notation.

One way to see that $F\otimes G$ is exact is to notice the following. If $A$ and $B$ are the polynomial rings generated by the $x$s and the $y$s, respectively, then the minimal resolutions of $S/I$ and $S/J$ can be constructed first constructing the minimal resolutions $\bar F$ and $\bar G$ of $A/I$ and $B/J$ and then extending scalars to $S$. Therefore $F\otimes_S G$ can in fact be gotten as $\bar F\otimes_k\bar G$ as a complex of modules over $S=A\otimes_k B$. Now the Künneth formula tells us that the tensor product of two exact complexes of $k$-vector spaces is exact, and this applies to $\bar F\otimes_k\bar G$.

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