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darij grinberg
Birch & Swinnerton Dyer
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Location: Karlsruhe / Munich
Post Posted: Jul 20, 2009, 9:28 am • # 1 


Let A and B be two commutative rings with unity such that A\subseteq B. Let u\in B and v\in B^{\times}. Assume that u\in A\left[v\right]\cap A\left[v^{ - 1}\right].

(a) Prove that u is integral over A.

(This is J. S. Milne, Algebraic Number Theory, version 3.02, exercise 2-5, without the useless "integral domains" condition.)

(b) If P\in A\left[X\right] and Q\in A\left[X\right] are polynomials satisfying u = P\left(v\right) = Q\left(v^{ - 1}\right), then show that u is the root of a monic polynomial of degree \deg P + \deg Q over A. (Of course, this is a stronger version of part (a).)

(c) Modify the above statements to replace the condition v\in B^{\times} by v\in B.

(Remark. While (c) is easy, (b) seems to be harder than (a). I know two solutions for (a), but only one of them still shows (b).)

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EBz89
Hodge Conjecture

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Location: Glasgow/Cambridge
Post Posted: Jul 25, 2009, 1:15 am • # 2 


Let the degrees of the polynomials be n and m respectively. I think (b) can be proven reasonably easily by simply writing the two equations for u (clearing denominators in the second equation) then multiplying the first equation by v^{i} for all i<m, multiplying the second equation by v^{j} for all j<n and then writing the resulting n+m equations in matrix form. You get an equation of the form Mx=0 where M is an (n+m) by (n+m) matrix and x\in B^{n+m} is the vector with i-th entry v^{i-1}. The equation \det(M)=0 is the desired equation satisfied by u.
 
 
darij grinberg
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Post Posted: Jul 25, 2009, 8:46 am • # 3 


Nice - I know this solution (it is the same as in Milne, and the same as I gave in A few facts on integrality, as a proof to Corollary 3 - the real argument is in the proof of Theorem 2), but I didn't look at it from the same angle as you. I considered the main idea to be to show that there is a faithful A-submodule of B generated by n + m elements such that multiplication by u maps this module to itself. The way you wrote up the solution, it now looks different: The matrix M is the Sylvester matrix of the polynomials P\left(X\right) - u and Q\left(X^{ - 1}\right)X^m - uX^m, and its determinant is the resultant of these two polynomials.

For (c), all I did was just replacing the condition Q\left(v^{ - 1}\right) = u by Q\left(v^{ - 1}\right)v^m = uv^m, where m = \deg Q (note that Q\left(v^{ - 1}\right)v^m means the polynomial Q\left(X^{ - 1}\right)X^m evaluated at X = v, which is defined for every v and not just for invertible v). You seem to be aware of this generalization, too (as you say you clear denominators in the second equation), and all proofs I know for the v\in B^{\times} case also work in this one.

I mentioned a different proof of (a) that does not help showing (b). Here is an outline:

Let u = \sum_{i = 0}^n s_iv^i = \sum_{i = 0}^m t_iv^{ - i} for some elements s_0, s_1, ..., s_n, t_0, t_1, ..., t_n of A. Define n + m + 1 elements a_0, a_1, ..., a_{n + m + 1} of A by

a_i = \left\{\begin{array}{c} t_{m - i},\text{ if }i < m; \\ t_0 - s_0,\text{ if }i = m; \\ - s_{i - m},\text{ if }i > ... for all i\in\left\{0,1,...,n + m\right\}.

Then, \sum_{i = 0}^{n + m}a_iv^i = 0 and \sum_{i = 0}^n a_{i + m}v^i = u. So it is enough to prove the following fact:

Theorem 2. Let A and B be two rings (commutative, with unity) such that A\subseteq B. Let N\in\mathbb{N}. Let a_0, a_1, ..., a_N be N + 1 elements of A, and let v\in B be such that \sum_{i = 0}^N a_iv^i = 0. Then, \sum_{i = 0}^{N - k} a_{i + k}v^i is integral over A for every k\in\left\{0,1,...,N\right\}.

And this can be proven by backward induction over k:

For k = N, we have \sum_{i = 0}^{N - k} a_{i + k}v^i = a_N, which is obviously integral over A.

Set \rho_k = \sum_{i = 0}^{N - k} a_{i + k}v^i for every k\in\left\{0,1,...,N\right\}. Then, \rho_{k - 1} = \sum_{i = 0}^{N - k} a_{i + k - 1}v^i satisfies \rho_{k - 1} = \rho_k v + a_{k - 1}. Now, \sum_{i = 0}^N a_iv^i = 0 yields

\sum_{i = 0}^{k - 1}a_i\rho_k^{k - 1 - i}\left(\rho_k v\right)^i + \left(\rho_k v\right)^k = \rho_k^{k - 1}\left(\sum_{i = 0}...
= \rho_k^{k - 1}\left(\sum_{i = 0}^{k - 1}a_iv^i + \sum_{i = 0}^{N - k} a_{i + k}v^i v^k\right) = \rho_k^{k - 1}\cdot\underbr...,

so that \rho_k v is integral over A\left[\rho_k\right]. Thus, \rho_{k - 1} = \rho_k v + a_{k - 1} is integral over A\left[\rho_k\right] as well. Hence, if \rho_k is integral over A, then so is \rho_{k - 1}, and we can proceed by induction.

If you try to get an estimate on the degree of \rho_k from this proof, you get something like \left(n - k\right)!...

darij

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EBz89
Hodge Conjecture

Posts: 61
Location: Glasgow/Cambridge
Post Posted: Jul 26, 2009, 2:28 am • # 4 


Nice notes, I learned a lot from them actually - thanks. (I like a\in A implies a is 1-integral in particular :P). Seriously though, thm 2 is rather nice.
 
 
darij grinberg
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Location: Karlsruhe / Munich
Post Posted: Jul 26, 2009, 2:56 am • # 5 


Well, I wrote them up just to have the proofs written out with every detail, so I can think about how they generalize to the \left(A,I\right)-integral case. This is the case when A\subseteq B is a ring extension, and I is an ideal of A, and we call some u\in B integral over \left(A,I\right) if and only if there exists some n\in\mathbb{N} and elements i_k\in I^k for every k\in\left\{1,2,...,n\right\} such that u^n + \sum_{k = 1}^ni_ku^{n - k} = 0. Note that this generalizes two well-known things: integrality over a ring (when I = A) and integrality over an ideal within the same ring (when B = A). If I have not made mistakes (what is quite a strong assumption), we have the following facts:

Analogue of Theorem 4. Let A and B be two rings such that A\subseteq B. Let v\in B and u\in B. Let m\in\mathbb{N} and n\in\mathbb{N}. Assume that v is m-integral over A (not necessarily over \left(A,I\right) !), and that u is n-integral over \left(A\left[v\right],IA\left[v\right]\right). Then, u is nm-integral over \left(A,I\right).

Analogue of Theorem 5. Let A and B be two rings such that A\subseteq B.
(a) Let a\in I. Then, a is 1-integral over \left(A,I\right).
(b) Let x\in B and y\in B. Let m\in \mathbb{N} and n\in \mathbb{N}. Assume that x is m-integral over \left(A,I\right), and that y is n-integral over \left(A,I\right). Then, x + y is integral (and the degree is nm) over \left(A,I\right).
(c) Let x\in B and y\in B. Let m\in \mathbb{N} and n\in \mathbb{N}. Assume that x is m-integral over \left(A,I\right), and that y is n-integral over A (not necessarily over \left(A,I\right) !). Then, xy is integral (and the degree is nm) over \left(A,I\right).

Note that Theorem 4 can be proven using resultants, but the analogue of Theorem 5 doesn't follow from it anymore as nicely as Theorem 5 followed from Theorem 4. But at least, the analogue of Theorem 4 allows you to reduce the general case to the case A = I and the case B = A. The first case is known, and the second one is thoroughly studied in Irena Swanson and Craig Huneke, Integral Closure of Ideals, Rings, and Modules, 2006-2009. Now what somewhat frightens me is that I don't see any literature about the general case. Anyway, if you are still reading this, I could as well tell you the most general case: Let A\subseteq B be a ring extension, and let \left(I_k\right)_{k\in\mathbb{N}} be a filtration of ideals in A (that is, I_1\subseteq I_2\subseteq I_3\subseteq ... are ideals of A satisfying I_aI_b\subseteq I_{a + b} for all a and b). An element u\in B will be called integral over \left(A,\left(I_k\right)_{k\in\mathbb{N}}\right) if and only if there exists some n\in\mathbb{N} and elements i_k\in I_k for every k\in\left\{1,2,...,n\right\} such that u^n + \sum_{k = 1}^ni_ku^{n - k} = 0. How do elements of integral over \left(A,\left(I_k\right)_{k\in\mathbb{N}}\right) behave with respect to addition, multiplication and integrality over integrality? See the new version of my integrality notes for some answers.

darij

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Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...
 
 
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