A note on multiplicative property of the Smith normal form

M. Marcus, E. E. Underwood
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引用次数: 5

Abstract

In [2; p. 33]1 the following interesting result appears: If A and Bare n-square matrices over a principal ideal domain Rand g.c.d. (det (A), det (B»= 1 then S(AB)= S(A)S(B) where -SeA) is the Smith normal form of A. The purpose of this note is to present a simple proof of the result that uses elementary properties of compound matrices. LEMMA. Let Q=diag (q., ... , qn), P=diag (p., ... , Pn), qllqj' Pllpj,j=l, ... , nand g.c.d. (Pb qJ)= 1, i, j= 1, . .. , n. Let U be an n-square matrix with the property that g.c.d. (Ull' U2!, _ .. , Unl)= g.c.d. (Ull' U12, ... , Utn)= 1. Then the g.c.d. of all the entries in QUP is Plqt. PROOF. Obviously PlqlIQUP, i.e., Ptql divides every entry of QUP. Write QUP= PlqtD. Suppose that plD where p is a prime. It is simple to see that the first row and column of Dare respectively
关于史密斯范式的乘法性质的注记
在[2;在主理想域上,如果A和裸n方阵(det (A), det (B)»= 1,则S(AB)= S(A)S(B),其中-SeA)是A的史密斯范式。本文的目的是利用复合矩阵的初等性质给出一个简单的证明。引理。设Q=diag (Q .,…, qn), P=diag (P .,…, Pn), qllqj' Pllpj,j= 1,…, nand g.c.d. (pbqj)= 1, i, j= 1,…设U是一个n平方矩阵,其性质为gcd (U ' U2!),……(1), (1), (1), (2), (3), Utn)= 1。那么QUP中所有条目的gcd为Plqt。证明。显然PlqlIQUP,即Ptql除QUP的每一项。写QUP= PlqtD。假设plD中p是素数。很容易看出,Dare的第一行和第一列分别为
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