nLab homogeneous polynomial

Contents

Context

Algebra

higher algebra

universal algebra

Contents

Idea

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

Definition

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

Euler identity

A specific property of homogeneous polynomials is the Euler identity:

Theorem

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

Proof

Suppose that $P \in R_{n}[X_{1},...,X_{q}]$ and write $P=\underset{\substack{0\le i_{1},...,i_{q} \le n\\ i_{1}+...+i_{q} = n}}{\sum} a_{i_{1},...,i_{q}}X_{1}^{i_{1}}...X_{q}^{i_{q}}$. Then $\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = \underset{1 \le k \le q}{\sum}\underset{\substack{0\le i_{1},...,i_{q} \le n\\ i_{1}+...+i_{q} = n}}{\sum} a_{i_{1},...,i_{q}}.i_{k}.X_{1}^{i_{1}}...X_{q}^{i_{q}}$

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

$= \underset{\substack{0\le i_{1},...,i_{q} \le n\\ i_{1}+...+i_{q} = n}}{\sum} a_{i_{1},...,i_{q}}.n.X_{1}^{i_{1}}...X_{q}^{i_{q}}$

$=n.P$

Reciprocally, suppose that $R$ is multiplicatively cancellable, that $P \in R[X_{1},...,X_{q}]$ and that $\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = n.P$. Write $P=\underset{0 \le i_{1},...,i_{q} \le n}{\sum} a_{i_{1},...,i_{q}}X_{1}^{i_{1}}...X_{q}^{i_{q}}$.

Then $\underset{1 \le k \le q}{\sum} \frac{\partial P}{\partial X_{k}}X_{k} = \underset{1 \le k \le q}{\sum}\underset{0 \le i_{1},...,i_{q} \le q}{\sum} a_{i_{1},...,i_{q}}.i_{k}.X_{1}^{i_{1}}...X_{q}^{i_{q}}$

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

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

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

which is equivalent to $a_{i_{1},...,i_{q}}.(i_{1}+...+i_{q}) = a_{i_{1},...,i_{q}}.n$ for all $0 \le i_{1},...,i_{q} \le n$ which is equivalent to $a_{i_{1},...,i_{q}}.(i_{1}+...+i_{q}) = a_{i_{1},...,i_{q}}.n$ for all $0 \le i_{1},...,i_{q} \le n$ such that $a_{i_{1},...,i_{q}} \neq 0$. $R$ is multiplicatively cancellable, thus it is equivalent to $i_{1}+...+i_{q} = n$ for all $0 \le i_{1},...,i_{q} \le n$ such that $a_{i_{1},...,i_{q}} \neq 0$ which is equivalent to $P\in R_{n}[X_{1},...,X_{q}]$.

Last revised on November 24, 2022 at 03:03:56. See the history of this page for a list of all contributions to it.