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mathwonk
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Well here is what I meant. Suppose I have not just any statement but an equation i want to show is true, like the one above, det(1+dX)=1+dtr(X) (mod squares of entries).
Since any equation involves only a finite number of entries, it suffices to show it is true in all finitely generated subrings of the given ring. Now every finitely generated ring is the image of a surjective ring homomorphism from a polynomial ring in a finite number of variables over the integers (oh I guessI I didn't need this but anyway).
Now the equation I am trying to prove involves only ring operations so remains true in any homomorphic image of a ring in which it is true. So if I can prove it in the ring of polynomials in sevberal .variables over ther integers, I will be done. That is my idea of a sort of "universal" ring.
Now such a polynomial ring is a domain of characteristic zero. Thus it has a quotient field of characteristic zero. hence if we can prove our equation holds in all fileds of characteristic zero then it is also true in all domains of characteristic zero and also in all homomorphic images of them, and hence in all rings, no matter what cahracteristic.
Is this of interest? This is what I came up with in the tub, after reading the citation to bourbaki. (I also declined to look it up.)
But since you guys have explained the direct argument to me, I see it now as well. I.e. I had never noticed it before, but if a term in a determinant expansion has all entries but one taken from the diagonal, then the last entry is also from the diagonal, because there is no other choice that does not repeat a column or a row.
Hence, if an entry in a determinant expansion has one entry not on the diagonal, then it has at least two such. Thus, as you both saw on reflection, all terms in a determinant expansion except the product of the diagonal entries, involve at least two non diagonal entries, i.e. is zero in the dual numbers in our case.
Since any equation involves only a finite number of entries, it suffices to show it is true in all finitely generated subrings of the given ring. Now every finitely generated ring is the image of a surjective ring homomorphism from a polynomial ring in a finite number of variables over the integers (oh I guessI I didn't need this but anyway).
Now the equation I am trying to prove involves only ring operations so remains true in any homomorphic image of a ring in which it is true. So if I can prove it in the ring of polynomials in sevberal .variables over ther integers, I will be done. That is my idea of a sort of "universal" ring.
Now such a polynomial ring is a domain of characteristic zero. Thus it has a quotient field of characteristic zero. hence if we can prove our equation holds in all fileds of characteristic zero then it is also true in all domains of characteristic zero and also in all homomorphic images of them, and hence in all rings, no matter what cahracteristic.
Is this of interest? This is what I came up with in the tub, after reading the citation to bourbaki. (I also declined to look it up.)
But since you guys have explained the direct argument to me, I see it now as well. I.e. I had never noticed it before, but if a term in a determinant expansion has all entries but one taken from the diagonal, then the last entry is also from the diagonal, because there is no other choice that does not repeat a column or a row.
Hence, if an entry in a determinant expansion has one entry not on the diagonal, then it has at least two such. Thus, as you both saw on reflection, all terms in a determinant expansion except the product of the diagonal entries, involve at least two non diagonal entries, i.e. is zero in the dual numbers in our case.