Probably Known Property of homogeneous Polynomials in R[X,Y]

freemind

Probably Known Property of homogeneous Polynomials in R[X,Y]

**Posted:** Mon Mar 03, 2008 6:10 pm

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olorin · Yang-Mills Theory Offline Joined: 01 Jan 2007 Posts: 571

$\mathbb{R}[X,Y]$ 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.

**Posted:** Mon Mar 03, 2008 11:53 pm

darij grinberg

#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

**Posted:** Tue Mar 04, 2008 12:50 am

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