nLab homogeneous polynomial

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Contents

Idea

A polynomial all of whose monomial terms have the same degree. See also polynomial.

Definition

If RR is a commutative rig, then the set R n[X 1,...,X q]R_{n}[X_{1},...,X_{q}] of homogeneous polynomials of degree nn with variables X 1,...,X qX_{1},...,X_{q} is given by all the expressions of the form 0i 1,...,i qn i 1+...+i q=na i 1,...,i qX 1 i 1...X q i q\underset{\substack{0\le i_{1},...,i_{q} \le n\\ i_{1}+...+i_{q} = n}}{\sum} a_{i_{1},...,i_{q}}X_{1}^{i_{1}}...X_{q}^{i_{q}}.

Euler identity

A specific property of homogeneous polynomials is the Euler identity:

Theorem

If RR is a commutative rig and if PR n[X 1,...,X q]P \in R_{n}[X_{1},...,X_{q}] then 1kqPX kX k=n.P\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = n.P. Reciprocally, if RR is a commutative multiplicatively cancellable rig, if PR[X 1,...,X q]P \in R[X_{1},...,X_{q}] and 1kqPX kX k=n.P\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = n.P, then PR n[X 1,...,X q]P \in R_{n}[X_{1},...,X_{q}].

Proof

Suppose that PR n[X 1,...,X q]P \in R_{n}[X_{1},...,X_{q}] and write P=0i 1,...,i qn i 1+...+i q=na i 1,...,i qX 1 i 1...X q i qP=\underset{\substack{0\le i_{1},...,i_{q} \le n\\ i_{1}+...+i_{q} = n}}{\sum} a_{i_{1},...,i_{q}}X_{1}^{i_{1}}...X_{q}^{i_{q}}. Then 1kqPX kX k=1kq0i 1,...,i qn i 1+...+i q=na i 1,...,i q.i k.X 1 i 1...X q i q\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = \underset{1 \le k \le q}{\sum}\underset{\substack{0\le i_{1},...,i_{q} \le n\\ i_{1}+...+i_{q} = n}}{\sum} a_{i_{1},...,i_{q}}.i_{k}.X_{1}^{i_{1}}...X_{q}^{i_{q}}

=0i 1,...,i qn i 1+...+i q=na i 1,...,i q.(i 1+...+i q).X 1 i 1...X q i q = \underset{\substack{0\le i_{1},...,i_{q} \le n\\ i_{1}+...+i_{q} = n}}{\sum} a_{i_{1},...,i_{q}}.(i_{1}+...+i_{q}).X_{1}^{i_{1}}...X_{q}^{i_{q}}

=0i 1,...,i qn i 1+...+i q=na i 1,...,i q.n.X 1 i 1...X q i q = \underset{\substack{0\le i_{1},...,i_{q} \le n\\ i_{1}+...+i_{q} = n}}{\sum} a_{i_{1},...,i_{q}}.n.X_{1}^{i_{1}}...X_{q}^{i_{q}}

=n.P=n.P

Reciprocally, suppose that RR is multiplicatively cancellable, that PR[X 1,...,X q]P \in R[X_{1},...,X_{q}] and that 1kqPX kX k=n.P\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = n.P. Write P=0i 1,...,i qna i 1,...,i qX 1 i 1...X q i qP=\underset{0 \le i_{1},...,i_{q} \le n}{\sum} a_{i_{1},...,i_{q}}X_{1}^{i_{1}}...X_{q}^{i_{q}}.

Then 1kqPX kX k=1kq0i 1,...,i qqa i 1,...,i q.i k.X 1 i 1...X q i q\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = \underset{1 \le k \le q}{\sum}\underset{0 \le i_{1},...,i_{q} \le q}{\sum} a_{i_{1},...,i_{q}}.i_{k}.X_{1}^{i_{1}}...X_{q}^{i_{q}}

=0i 1,...,i qna i 1,...,i q.(i 1+...+i q).X 1 i 1...X q i q = \underset{0 \le i_{1},...,i_{q} \le n}{\sum} a_{i_{1},...,i_{q}}.(i_{1}+...+i_{q}).X_{1}^{i_{1}}...X_{q}^{i_{q}}.

Thus, if 1kqPX kX k=n.P\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = n.P, then

0i 1,...,i qna i 1,...,i q.n.X 1 i 1...X q i q=0i 1,...,i qna i 1,...,i q.(i 1+...+i q).X 1 i 1...X q i q\underset{0 \le i_{1},...,i_{q} \le n}{\sum} a_{i_{1},...,i_{q}}.n.X_{1}^{i_{1}}...X_{q}^{i_{q}} = \underset{0 \le i_{1},...,i_{q} \le n}{\sum} a_{i_{1},...,i_{q}}.(i_{1}+...+i_{q}).X_{1}^{i_{1}}...X_{q}^{i_{q}}

which is equivalent to a i 1,...,i q.(i 1+...+i q)=a i 1,...,i q.na_{i_{1},...,i_{q}}.(i_{1}+...+i_{q}) = a_{i_{1},...,i_{q}}.n for all 0i 1,...,i qn0 \le i_{1},...,i_{q} \le n which is equivalent to a i 1,...,i q.(i 1+...+i q)=a i 1,...,i q.na_{i_{1},...,i_{q}}.(i_{1}+...+i_{q}) = a_{i_{1},...,i_{q}}.n for all 0i 1,...,i qn0 \le i_{1},...,i_{q} \le n such that a i 1,...,i q0a_{i_{1},...,i_{q}} \neq 0. RR is multiplicatively cancellable, thus it is equivalent to i 1+...+i q=ni_{1}+...+i_{q} = n for all 0i 1,...,i qn0 \le i_{1},...,i_{q} \le n such that a i 1,...,i q0a_{i_{1},...,i_{q}} \neq 0 which is equivalent to PR n[X 1,...,X q]P\in R_{n}[X_{1},...,X_{q}].

Last revised on August 14, 2022 at 14:16:05. See the history of this page for a list of all contributions to it.