669: For Principal Integral Domain, Rectangle Matrix over Domain, and Invertible Square Matrix over Domain, Sum of Principal Ideals by Specified-Dimensional Subdeterminants of Product Is Sum of Principal Ideals by Same Dimensional Subdeterminants of Rectangle Matrix

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description/proof of that for principal integral domain, rectangle matrix over domain, and invertible square matrix over domain, sum of principal ideals by specified-dimensional subdeterminants of product is sum of principal ideals by same dimensional subdeterminants of rectangle matrix

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of principal integral domain.
• The reader admits the proposition that for any principal integral domain, any rectangle matrix over the domain, and any rectangular-matrix-columns-dimensional or rectangular-matrix-rows-dimensional square matrix over the domain, the sum of the principal ideals by the specified-dimensional subdeterminants of the product is contained in the sum of the principal ideals by the same-dimensional subdeterminants of the rectangle matrix.

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Target Context

• The reader will have a description and a proof of the proposition that for any principal integral domain, any rectangle matrix over the domain, and any invertible rectangular-matrix-columns-dimensional or rectangular-matrix-rows-dimensional square matrix over the domain, the sum of the principal ideals by the specified-dimensional subdeterminants of the product is the sum of the principal ideals by the same-dimensional subdeterminants of the rectangle matrix.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the principal integral domains }\}$
$M$: $\in \{\text{ the m x n matrices over }R\}$
$N$: $\in \{\text{ the invertible n x n matrices over }R\}$
$O$: $\in \{\text{ the invertible m x m matrices over }R\}$
$k$: $\in \mathbb{N}$, $1\le k\le min(m,n)$
$I_k(M)$: $=\text{ the sum of the principal ideals by the }k\text{ -dimensional subdeterminants of }M$
$I_k(MN)$: $=\text{ the sum of the principal ideals by the }k\text{ -dimensional subdeterminants of }MN$
$I_k(OM)$: $=\text{ the sum of the principal ideals by the }k\text{ -dimensional subdeterminants of }OM$
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Statements:
$I_k(MN)=I_k(M)$
$\wedge$
$I_k(OM)=I_k(M)$
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As an immediate corollary, $I_k(OMN)=I_k(M)$: $I_k(OMN)=I_k(MN)=I_k(M)$.

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2: Natural Language Description

For any principal integral domain, $R$, any $mxn$ matrix over $R$, $M$, any invertible $nxn$ matrix over $R$, $N$, any invertible $mxm$ matrix over $R$, $O$, any natural number, $k$, such that $1\le k\le min(m,n)$, the sum of the principal ideals by the $k$-dimensional subdeterminants of $M$, $I_k(M)$, the sum of the principal ideals by the $k$-dimensional subdeterminants of $MN$, $I_k(MN)$, and the sum of the principal ideals by the $k$-dimensional subdeterminants of $OM$, $I_k(OM)$, $I_k(MN)=I_k(M)$ and $I_k(OM)=I_k(M)$.

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3: Proof

Generally, for any matrix, $A$, $I_k(A)$ denotes the sum of the principal ideals by the $k$ -dimensional subdeterminants of $A$.

$I_k(MN)\subseteq I_k(M)$, by the proposition that for any principal integral domain, any rectangle matrix over the domain, and any rectangular-matrix-columns-dimensional or rectangular-matrix-rows-dimensional square matrix over the domain, the sum of the principal ideals by the specified-dimensional subdeterminants of the product is contained in the sum of the principal ideals by the same-dimensional subdeterminants of the rectangle matrix.

There is an inverse, $N^{-1}$.

$I_k(M)=I_k(MN{N}^{-1})\subseteq I_k(MN)$, by the proposition that for any principal integral domain, any rectangle matrix over the domain, and any rectangular-matrix-columns-dimensional or rectangular-matrix-rows-dimensional square matrix over the domain, the sum of the principal ideals by the specified-dimensional subdeterminants of the product is contained in the sum of the principal ideals by the same-dimensional subdeterminants of the rectangle matrix.

So, $I_k(MN)=I_k(M)$.

$I_k(OM)\subseteq I_k(M)$, by the proposition that for any principal integral domain, any rectangle matrix over the domain, and any rectangular-matrix-columns-dimensional or rectangular-matrix-rows-dimensional square matrix over the domain, the sum of the principal ideals by the specified-dimensional subdeterminants of the product is contained in the sum of the principal ideals by the same-dimensional subdeterminants of the rectangle matrix.

There is an inverse, $O^{-1}$.

$I_k(M)=I_k(O^{-1}OM)\subseteq I_k(OM)$, by the proposition that for any principal integral domain, any rectangle matrix over the domain, and any rectangular-matrix-columns-dimensional or rectangular-matrix-rows-dimensional square matrix over the domain, the sum of the principal ideals by the specified-dimensional subdeterminants of the product is contained in the sum of the principal ideals by the same-dimensional subdeterminants of the rectangle matrix.

So, $I_k(OM)=I_k(M)$.

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References

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