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freemind
Riemann Hypothesis
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Post Posted: Mar 03, 2008, 9:10 am • # 1 


A non-zero polynomial S\in\mathbb{R}[X,Y] is called homogeneous of degree d if there is a positive integer d so that S(\lambda x,\lambda y)=\lambda^dS(x,y) for any \lambda\in\mathbb{R}. Let P,Q\in\mathbb{R}[X,Y] so that Q is homogeneous and P divides Q (that is, P|Q). Prove that P is homogeneous too.

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olorin
Yang-Mills Theory

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Post Posted: Mar 03, 2008, 2:53 pm • # 2 


\mathbb{R}[X,Y] is a factorial ring.
So we find p_1(X,Y),p_2(X,Y),\ldots,p_n(X,Y)\in\mathbb{R}[X,Y] irreducible with
(*) Q(X,Y) = p_1(X,Y)p_2(X,Y)\cdots p_n(X,Y).

Then for d: = \deg Q(X,Y) and d_k: = \deg p_k(X,Y),k = 1,\ldots,n we have d = d_1 + \ldots + d_n and
Q(X,Y) = Y^dQ({X\over Y},1) \\ \hspace*{0.5in} = Y^dp_1({X\over Y},1)p_2({X\over Y},1)\cdots p_n({X\over Y},1) \\ \hspace*{0....
with q_k(X,Y): = Y^{d_k}p_k({X\over Y},1)\in\mathbb{R}[X,Y] homogeneous not constant for k = 1,\ldots,n.
But (*) is up to order and scalar multiples the only factorisation of Q(X,Y) in n not constant factors.
So for every k = 1,\ldots,n we find i = 1,\ldots,n and \lambda\in\mathbb{R}^* with p_k(X,Y) = \lambda q_i(X,Y) homogeneous.
So all irreducible factors of Q(X,Y) are homogeneous, and so are all other factors.
 
 
darij grinberg
Birch & Swinnerton Dyer
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Post Posted: Mar 03, 2008, 3:50 pm • # 3 


Isn't the problem trivial anyway? If p and q are two polynomials which are not both homogeneous, then pq cannot be homogeneous either; this holds in any R\left[X_1,X_2,...,X_n\right] with R being an integral domain. The simple reason is that the minimal degree of a monomial in pq is the sum of the minimal degree of a monomial in p and the minimal degree of a monomial in q, and same holds for the maximal degrees - thus, if at least one of the polynomials p and q is not homogeneous, it has monomials of different degrees, and so must pq.

darij

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