Proof: Let R=Z[x1,...] and I=(x1,...). It suffices to find an alternating n-multilinear map φn:I×...×I→Z (n factors) such that φn(x1,...,xn)=1 for all n. To that end, define φn as follows,φn:I×...×I→Zφn(∑a1,ixi,...,∑an,ixi)=det (a′ij)1≤i,j≤nwhere a′ij is the constant term of aij. Note that ∑aixi=∑bixi implies that a′i=b′i for all i, so that (a′ij)1≤i,j≤n is uniquely determined and φn is well defined. This map is multilinear alternating on components of I×...×I just as det is multilinear alternating on matrix rows. Here (x1,...,xn) represents the identity matrix, and as such φn(x1,...,xn)=1.◻
Tuesday, July 30, 2013
Exterior Algebras and Fraction Fields (11.5.8c)
Dummit and Foote Abstract Algebra, section 11.5, exercise 8(c):
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(c) Give an example of an integral domain R with fraction field F and ideal I⊆F considered as an R-module such that ⋀nI≠0 for all n.
Proof: Let R=Z[x1,...] and I=(x1,...). It suffices to find an alternating n-multilinear map φn:I×...×I→Z (n factors) such that φn(x1,...,xn)=1 for all n. To that end, define φn as follows,φn:I×...×I→Zφn(∑a1,ixi,...,∑an,ixi)=det (a′ij)1≤i,j≤nwhere a′ij is the constant term of aij. Note that ∑aixi=∑bixi implies that a′i=b′i for all i, so that (a′ij)1≤i,j≤n is uniquely determined and φn is well defined. This map is multilinear alternating on components of I×...×I just as det is multilinear alternating on matrix rows. Here (x1,...,xn) represents the identity matrix, and as such φn(x1,...,xn)=1.◻
Proof: Let R=Z[x1,...] and I=(x1,...). It suffices to find an alternating n-multilinear map φn:I×...×I→Z (n factors) such that φn(x1,...,xn)=1 for all n. To that end, define φn as follows,φn:I×...×I→Zφn(∑a1,ixi,...,∑an,ixi)=det (a′ij)1≤i,j≤nwhere a′ij is the constant term of aij. Note that ∑aixi=∑bixi implies that a′i=b′i for all i, so that (a′ij)1≤i,j≤n is uniquely determined and φn is well defined. This map is multilinear alternating on components of I×...×I just as det is multilinear alternating on matrix rows. Here (x1,...,xn) represents the identity matrix, and as such φn(x1,...,xn)=1.◻
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