# Classes of Schubert Cycles form a basis

I am reading the not yet published book of Eisenbud and Harris about intersection theory (http://isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf) and I don't quite understand the following: In Proposition 4.31, page 113, they state that $deg(\sigma_a \sigma_b) = 1$ if $a_i + b_{k-i+1} = n-k$ for all i and $\sigma_a \sigma_b = 0$ otherwise, where $\sigma_a, \sigma_b$ are the classes of two Schubert cycles of complementary dimension.

1) The part that I don't understand is that they now conclude (in Corollary 4.32) that the Schubert Cycles form a Basis of the Chow Group A(G) of the Grassmannian G.

The "generating-part" is already clear by a different theorem, but I don't see how the "free-part" follows from the proposition above.

2) They also state that the "Schubert Classes form the dual bases of the intersection forms $A^m(G)\times A^{dimG-m} \rightarrow \mathrm Z$". I guess they mean that for a fixed cyle $\sigma_a$ we get an element of the dual of the (codimension m - part of the) Chow Group (viewed as a Z-module) by defining $\phi_a (\sigma_b) := deg(\sigma_a \sigma_b)$, and IF the $\sigma_a$'s would form a basis, than we had a dual basis that way by the proposition. But still, this does not proof that the Schuber Cycles really form a basis.

Do I get s.th. totally wrong? Am I blind? Thanks in advance.

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If I have the details right, this is a straightforward generalization of the familiar proof that an orthogonal family of vectors is linearly independent. –  Qiaochu Yuan Jun 20 '12 at 23:27
This solves the problem, thank you. –  Dalboz Jun 21 '12 at 8:23
You know there is already a newer version availible? You might try to find it or message me. –  sebigu Jun 21 '12 at 8:30
I already have a newer version, but for my purpose here the first google result was enough. Thanks anyway. –  Dalboz Jun 21 '12 at 8:33
Dear @Dalboz, I am sorry I have to dig in this old question, but I need your help! I see that the content of Corollary 4.32 in "3264 and all that" is exactly that the classes of the Schubert cycles form a free basis. Here is my question: isn't this already the content of Theorem 1.9 (which on my version is at page 10)? In fact I'm wondering if I can just use the existence of the affine stratification for the Grassmannian to conclude that its Chow ring has that free basis. Thank you! A. –  Brenin Nov 17 '12 at 18:50