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Let $f(X_1,\ldots,X_n)$ be a polynomial with integer coefficients, in $n$ variables. For every $\lambda\in\mathbb{C}$, i have $$f(\lambda X_1,\ldots,\lambda X_n)=\lambda^kf(X_1,\ldots,X_n)$$.

Can I conclude that $f$ is homogeneous of degree $k$? Why?

My definition of polynomial homogeneous of degree $k$ is: a polynomial whose all nonzero terms have the same degree $k$.

bateman
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1 Answers1

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You already know $f$ is a polynomial in $n$ variables so you can write it in the general form $$f(X_1,...,X_n)=\sum_{j_1,j_2,...j_n=0}^nc_kX_1^{j_1}X_2^{j_2}...X_n^{j_n}$$ for some $c_i\in \mathbb{Z}$, $j_i=0,...,n$, $i=0,...,n$. Calculate this with $\lambda X_i$ in the argument

$$f(\lambda X_1,...,\lambda X_n)=\sum_{j_1,j_2,...j_n=0}^nc_k(\lambda X_1)^{j_1}(\lambda X_2)^{j_2}...(\lambda X_n)^{j_n}$$ $$=\sum_{j_1,j_2,...j_n=0}^nc_k\lambda^{j_1+...+j_n} X_1^{j_1}X_2^{j_2}...X_n^{j_n}$$

Comparing this with $$\lambda^kf(X_1,...,X_n)=\lambda^k\sum_{j_1,j_2,...j_n=0}^nc_kX_1^{j_1}X_2^{j_2}...X_n^{j_n}$$ $$=\sum_{j_1,j_2,...j_n=0}^nc_k\lambda^k X_1^{j_1}X_2^{j_2}...X_n^{j_n}$$ for every $\lambda \in \mathbb{C}$ we can see that $$j_1+...+j_n=k$$ But this means precisely that $f$ is homogeneous of degree $k$

Here $$\sum_{j_1,j_2,...j_n=0}^n$$ denotes $$\sum_{j_1=0}^n\sum_{j_2=0}^n\cdots \sum_{j_n=0}^n$$

user89138
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  • Thanks! the only thing i don't understand is a little detail, namely: at the beginning, why do you say $j_i$ can go from $0$ to $n$? Couldn't they be independent, and have different ranges? – bateman Sep 12 '13 at 16:41
  • You're right, I assumed the polynomial had degree $n$ for some reason. The way to correct this should be to assume $f$ has a fixed degree $M$. Since the $j_1+...+j_n$ can never exceed $M$, and we agree that all $j_i>=0$, then all $j_i$ must be between $0$ and $M$. – user89138 Sep 12 '13 at 16:55
  • Also, yes, they can have different ranges, but that would mean more restrictions and more notation. All we needed to know is that they're bounded in $[0,M]$, but we don't have to worry how they behave in there. – user89138 Sep 12 '13 at 17:01