Where is "homogenuity" used in this proof?

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Theorem

Let $R$ be a ring and $P\in R[X_1,...,X_n]$ be a symmetric polynomial and $p_1,...,p_n$ be the elementary symmetric polynomials. Then, there exists a polynomial $Q\in R[X_1,...,X_n]$ such that $Q(p_1,...,p_n)=P$.

Proof

Let $\leq$ be the lexicographic monomial ordering on $R[X_1,...,X_n]$.

Let $P$ be a homogenous symmetric polynomial.

Let $\alpha$ be the multidegree of $P$. Since monomials are lexicographically ordered, $\alpha_1\geq ... \geq\alpha_n$.

Define $Q(X_1,...,X_n)=P(\alpha)X_1^{\alpha_1-\alpha_2}...X_{n-1}^{\alpha_{n-1}-\alpha_n} X_n^{\alpha_n}$.

Then, it can be directly shown that $LM(Q(p_1,...,p_n))= X_1^{\alpha_1}...X_n^{\alpha_n}=LM(P)$.

Thus, $R=P-Q(p_1,...,p_n)$ is a symmetric polynomial and its leading monomial is less than that of $P$.

Continuing this manner, $R$ must be $0$ at some step. Q.E.D

Where is homogenuity property used??

It's written in planetmath that "we first prove for the homogenous case, then prove in general", but I don't get why it assumes homogeneous condition first

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It is not directly used. However, it makes a certain argument which is implicit in the proof somewhat simpler. Namely, what exactly does "Continuing in this manner" mean? You can think of it as an algorithm where you start with a symmetric polynomial $P$ and, by subtracting certain polynomials of the form $Q\left(p_1, p_2, \ldots, p_n\right)$, you make its leading monomial smaller and smaller until there is no leading monomial anymore. The argument you have given shows a single step of this algorithm. But one thing that is left unsaid is: Why does this algorithm terminate? I.e., why cannot it go on forever, making the leading monomial smaller and smaller but never making it disappear?

If your polynomial is homogeneous of total degree $N$, then the reason for this is clear: There are only finitely many monomials of total degree $N$, and thus you cannot make the leading monomial smaller and smaller indefinitely (as it always stays of the same total degree $N$).

If, on the other hand, your polynomial is not homogeneous of total degree $N$, then you either have to argue that the leading monomial's total degree never rises (and this does not follow from it becoming smaller), or you have to argue that the lexicographic order is well-founded. These arguments are not exactly difficult (although the well-foundedness of the lexicographic order is proof-theoretically a bit more expensive than a simple induction, if I am not mistaken), but by restricting yourself to homogeneous polynomials, you can completely avoid them (at a very small price: reducing the general case to the case of homogeneous polynomials is easy).