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2. Square Matrices
                              18
                              x ij ’s are replaced by the scalars m ij .We observe that Det A does not really
                              depend on the ring A, in the sense that it is the image of Det ZZ through
                              the canonical ring homomorphism ZZ → A. For this reason, we shall simply
                              write Det. The polynomial Det may be viewed as the determinant of the
                              matrix X =(x ij ) 1≤i,j≤n ∈ M n (A[x 11 ,... ,x nn ]).
                              Theorem 2.1.1 The polynomial Det is irreducible in A[x 11 ,... ,x nn ].
                                Proof
                                We shall proceed by induction on the size n.If n = 1, there is nothing
                              to prove. Thus let us assume that n ≥ 2. We denote by D the ring of
                              polynomials in the x ij with (i, j)  =(1, 1), so that A[x 11 ,... ,x nn ]= D[x 11 ].
                              From the expansion with respect to the first row, we see that Det = x 11 P +
                              Q,with P, Q ∈ D. Since Det is of degree one as a polynomial in x 11 ,
                              any factorization must be of the form (x 11 R + S)T ,with R, S, T ∈ D.In
                              particular, RT = P.
                                By induction, and since P is the polynomial Det of (n − 1) × (n − 1)
                              matrices, it is irreducible in E, the ring of polynomials in the x ij ’s with
                              i, j > 1. Therefore, it is also irreducible in D,since D is the polynomial
                              ring E[x 12 ,... ,x 1n ,x 21 ,... ,x n1 ]. Therefore, we may assume that either R
                              or T equals 1.
                                If the factorization is nontrivial, then R =1 and T = P. It follows that
                              P divides Det. An expansion with respect to variousrowsshows similarly
                              that every minor of size n− 1, considered as an element of A[x 11 ,... ,x nn ],
                              divides Det. However, each such minor is irreducible, and they are pairwise
                              distinct, since they do not depend on the same set of x ij ’s. We conclude
                              that the product of all minors of size n − 1 divides Det. In particular, the
                                                                                2
                              degree n of Det is greater than or equal to the degree n (n − 1) of this
                              product, an obvious contradiction.



                              2.1.2 The Cauchy–Binet Formula

                              In the sequel, we shall use also the following result.
                              Proposition 2.1.2 Let B ∈ M n×m (A), C ∈ M m×l (A), and an integer
                              p ≤ n, l be given. Let 1 ≤ i 1 < ··· <i p ≤ n and 1 ≤ k 1 < ··· <k p ≤ l be
                              indices. Then the minor

                                                         i 1  i 2  ···  i p
                                                  (BC)
                                                         k 1  k 2  ···  k p
                              is given by the formula

                                                    i 1  i 2  ···  i p    j 1  j 2  ···  j p
                                               B                    · C                    .
                                                    j 1  j 2  ···  j p    k 1  k 2  ···  k p
                                 1≤j 1 <j 2 <···<j p ≤m
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