More algebraic geometry questions

[Hurkyl]

one way to see det(1+dX)=1+dtr(X)

This turns out to be fairly straightforward if you recall the messy definition of determinant: the sum of all possible products taking one element from each column (with each product multiplied by 1 or -1 as appropriate).

The only product that doesn't contain two terms of the form df is the product of the diagonal entries, and it's fairly easy to see that their product is simply the RHS (plus some additional terms that have two factors of the form df)

besides it is a "universal" statement about matrcies

How does this work?

 

[mathwonk]

I just meant it only has to do with properties of addition and multiplications, true in any ring at all. and you just proved that. But to make it more rpecise you could prove it for matrices with variable entries in a polynomila ring with integer coefficients i guess and then it is true for any specialization of the values.

There is general principle of this type though. bourbaki calls it prolongation of identities. a polynomial over the integers which is zero when the arguments take values in any field of characteristic zero is in fact identically zero. see bourbaki, algebre, chapIV, paragraph 2, no.5.

 

[Hurkyl]

The reason I ask is because I have seen a proof that went something like:

"Statement <foo> is clearly true for all nice things of type <bar>. Since <foo> is a formal statement about <bar>s, it must be true for all <bar>."

I was with a couple people far smarter than I at the time, and none of them had any clue on this either... your statement that the property was "universal" for matrices beared an uncanny resemblance to this, so I was hoping you could have shed some light on it.

(Incidentally, the proof was of something in the appendix of Fulton's Intersection Theory... I've since returned it to the library so I can't dig up the exact statement)

 

[matt grime]

This turns out to be fairly straightforward if you recall the messy definition of determinant: the sum of all possible products taking one element from each column (with each product multiplied by 1 or -1 as appropriate).

The only product that doesn't contain two terms of the form df is the product of the diagonal entries, and it's fairly easy to see that their product is simply the RHS (plus some additional terms that have two factors of the form df)

Yep, that's what I meant about it becomes quite clear after a few minutes. I don't think I've ever actually formally shown this to be true.

 

[matt grime]

And as for the formal statement thing, I think the way to explain it might be (but I'm sort of guessing).

Suppose we have some statemtent along the lines of:

A matrix M with entries in some field is invertible iff it has non-zero determinant.
Now that isn't true if we replace field with ring. For the non-zero making it invertible it is important that we are in a field. However if we were to write it as
M (over a field) is invertibl iff the the determinant is invertible is true for all matrices (over a ring). We are not using all of the hypothesis so it's true in greater generality. Or at least that's how I read your statement.

 

[mathwonk]

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.

 

[mathwonk]

well fulton's appendix does contain an argument to the effect that the resultant of two polynomials f,g in one variable, where f is monic, equals the determinant of the endomorphism of A[T]/(f) defined by multiplication by g.

This is apprently an equation of exactly the type i was discussing and the same formal - universal type argument should indeed work. I am past the point of giving more details au moment.

Well you probably do not need this, but here is another detail. the argument I gavre above showed that any equation true in all fields of characteristic zero is true in all rings, but i could have taken the algebaric closure fo the quotient field oif the polynom,ial ring over Z and proved that an equation which is true over all algebraically clsoed fields of charac zero is true in all rings. This justifies Fulton's remark that it suffices to prove his statement when g factors into linear factors.

 

[mathwonk]

did this help? please tell me. I am getting so desperate i am reading my own posts!

 

[Hurkyl]

Yes, I follow, I think:

(1) If we have an algorithm that takes (finitely many) elements of a ring R generates polynomial identities.
(2) The algorithm is preserved by ring homomorphisms.
(3) The polynomial identities hold whenever R is a polynomial ring over the integers.

Then,

(4) The polynomial identities hold for any R at all.


Furthermore, to show property (3), it sufficies to select any ring S of characteristic zero and let R range over polynomial rings over S.

 

[mathwonk]

wow! very nice and clean. however this would not cover your "formal" example from fulton, where he uses the weaker form of your 3), i.e. it holds in the algebraic closure of a polynomial ring over the integers.

 

[Hurkyl]

Wasn't it the other way around? He proved it for polynomials over the algebraic closure of the integers?


To be happy, though, I still need to work out what "algorithm that generates polynomial identities" means.

 

[mathwonk]

that's what i thought too, but remember he said it sufficed to rpove it for a polynomial that factors completely, and since it is a polynomial over the ring of interest, we have to map a polynomial ring onto that coefficient ring, and then take the algebraic closure of that polynomial ring to get a proof that it works in the coefficient ring he was using.

what do you think?"

 

[Hurkyl]

I don't remember it well enough; I took the book back to the library since our group planning to go over the text never took off.

 

[mathwonk]

Fulton was trying to give two different expression for the resultant of two polynomials, but we should not need the book for this. He had two polynomials f,g over a ring A, and then he stated that two specific polynomials P,Q in the coefficients of f and g, were always equal, as long as they were equal when g is a polynomial that splits into linear factors.

so essentially he has two polynomials P,Q in variables representing the coefficients of f and g, and wants to show they are always equal, no matter what values the coefficient variables take in any ring. Essentially he states it is sufficient to let the variables (i.e. the coefficients of f and g) have values in an algebraically closed field.

to prove this, we observe that in every ring, any finitely generate subring is the homomorphic image of a domain (of characteristic zero), which is itself a subring of an algebraically closed field. Hence if the given two polynomials P,Q are equal when their variables (i.e. the coefficients of f and g) take values in every algebraically closed field of characteristic zero, then the polynomials are equal when their variables take values in any ring.

but since the domain we use to map onto an arbitrary ring is a polynomial ring over the integers, it is the algebraic closure of that ring we need to prove equality in.

that's my best shot anyway.