# Topoi (collected papers) by Lawvere F.W.

By Lawvere F.W.

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3) shows that the first row of B consists only of zeros. , 0. If n was 1, then the remaining element M is the unit matrix, and the proof is finished. Assume the assertion to be correct for n - 1 in place of n, and cross out the first and (n + I)th columns and rows of M. The remaining matrix M ( " - ' E ) r"-' or CY"", respectively. , 0, one arrives at a representation of M as desired, at least in the case of the group 0". In the case of the group r" one must still demonstrate that the matrix J1, arising from completion of J("-'), can be represented by the given generating elements.

LINEAR ALGEBRA By the finiteness criterion of §1,4 we thus know that a, is a finite i-module. It remains to show that it is of rank n. As all the a,, are of rank n there exists, for each place p , an exponent h such that phmiE ap. Because of property (b) we can take h = 0 almost everywhere. Letting a be the product of all the ph, we have am, E a, for all i, completing the proof. T In this connection we consider the dual space k"*. The complementary modules a,* to the a, form the components of a linear divisor a* of k"*; a and a* are said to be complementary.

N - 1 . 1 0 E, has v ones in the diagonal, and zeros otherwise. J: = E. d. d. , cnl). = min, Then multiply from the left with a suitable unimodular U , , so that the first column of C contains only zeros, with the possible exception of c1 If cI1# 0, continue to multiply from the left, with a T s . d. ( a r l )< IcI1l,which, however, contradicts the assumption made. Nevertheless the same argument holds with one exception: = ... But then the determinant is divisible by cI1,which must therefore be & 1.