Bourbaki

Exercise 25, § 8

Algebra, III, 644

Bourbaki marks this one as harder than the rest.

Let AA be a commutative ring containing the field Q\mathbf{Q} of rational numbers and let XX be a square matrix of even order 2n≥22n \geq 2 over AA. Let LL be the subset {1,2,…,n+1}\{1, 2, \ldots, n+1\} of {1,2,…,2n}\{1, 2, \ldots, 2n\} and L′L' its complement. Show the following formula ("Redei's identity"):

(*)

det⁡(XL,L)det⁡(XL′,L′)=∑H,K(−1)r1n(n−1)det⁡(XH,K)det⁡(XH′,K′) \det(X_{L,L}) \det(X_{L',L'}) = \sum_{H, K} (-1)^r \frac{1}{n(n-1)} \det(X_{H, K}) \det(X_{H', K'})

where HH runs through the set of subsets of nn elements of {1,2,…,2n}\{1, 2, \ldots, 2n\} containing 1 and KK the set of subsets of nn elements of LL and r=Card⁡(H∩L′)r = \operatorname{Card}(H \cap L'). (Evaluate the coefficient of a term xσ(1),1xσ(2),2…xσ(2n),2nx_{\sigma(1), 1} x_{\sigma(2), 2} \ldots x_{\sigma(2n), 2n} in the right-hand side of (*) for an arbitrary permutation σ∈S2n\sigma \in S_{2n}, noting that this coefficient can only be ≠0\neq 0 if H=σ(K)H = \sigma(K); show that it can only be ≠0\neq 0 if σ(L)=L\sigma(L) = L.)

Solution

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Transcribed from the printed volume. Not translated.