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For the question, everything is over an algebraically closed field and by a scheme we mean a scheme of finite type. The theorem of the cube is the following: Let $X$ be a complete variety, $Y$ a scheme, and $L$ a line bundle on $X\times Y$. Then there is a unique closed subscheme $Y_1$ of $Y$ satisfying the following properties: (a) If $L_1$ is the restriction of $L$ to $X\times Y_1$, there is a line bundle $M_1$ on $Y_1$ and an isomorphism $p_2^*M_1\rightarrow L_1$ on $X\times Y_1$. (b), since (b) has little to do with the question, let us focus on (a). I am puzzled by the following thing: by the Künneth formula, if $p_2^*M_1\rightarrow L_1$ is an isomorphism (and $M_1$ is a line bundle on $Y_1$), then $M_1 \rightarrow {p_2}_*(L_1)$ is an isomorphism. Does anyone knows what is the content of the Künneth formula mentioned here and more importantly, why the statement is true? Thanks!

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up vote 3 down vote accepted

One uses the following trick.

By the projection formula we have $${p_2}_*(\mathcal{O}_{X\times Y_1}) \cong {p_2}_*(p_2^*M_1^{-1} \otimes L_1) \cong M_1^{-1} \otimes {p_2}_*(L_1)$$ and since $X$ is complete (and $k$ is algebraically closed) it follows from the Künneth formula that $${p_2}_*\mathcal{O}_{X\times Y_1} \cong \mathcal{O}_{Y_1}$$ which gives the desired isomorphism on $Y_1$ by uniqueness of inverses up to isomorphism in the Picard group.

In case I misunderstood and you were asking what the relevant version of the Künneth formula is in algebraic geometry there is an explanation here at the Encyclopedia of Mathematics (search for algebraic geometry to get to the relevant part).

To address Wayne's comment (where by the way it should be $M_1$ not its inverse): I think it is not necessarily true (well at least it isn't clear to me why it should be) that the adjunct of your original isomorphism will still be an isomorphism. The point (at least in the proof I know from Mumford's Abelian Varieties) is that the existence of such an isomorphism allows one to reduce to checking that ${p_2}_*L_1$ is a line bundle and that the counit of the adjunction is an isomorphism (which is good because this is a natural map which we get for free).

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Awesome, thank you! I think to make the question more precise, I should add "and $M_1$ is a line bundle on $Y_1$" (please look at line 6-7 in the question.) I do have a small question (and please forgive my fussiness): so we know $M_1^{-1}$ and $p_2{_*}(L_1)$ are isomorphic line bundles, but how do we know that the explicit morphism given in the question is an isomorphism? –  Wayne Nov 16 '09 at 5:03
    
No worries at all about the fussiness! I edited to try and answer your question in the comment. I hope this helps (in the proof I know nothing is asserted about what the isomorphisms actually are). –  Greg Stevenson Nov 16 '09 at 5:34
    
Thank you for pointing out the key point - it definitely helped me. Yes for the theorem of the cube I only need an abstract isomorphism to reduce the argument. –  Wayne Nov 16 '09 at 16:18
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