In algebraic geometry one proves that the affine coordinate ring of $\mathbb{P}^n$ is trivial by using that the only polynomials $f$ satisfying $f(\lambda x_0,\ldots,\lambda x_n) = f(x_0,\ldots,x_n) $ are the constant polynomials. Geometrically this is obvious, if we're dealing with polynomials over $\mathbb{R}$ at least, but I'm also not sure how to prove it.
Question. How do we know that the only polynomials $f$ satisfying $f(\lambda x_0,\ldots,\lambda x_n) = f( x_0,\ldots,x_n) $ are the constant polynomials?
Remark. Robert Lewis solved this by assuming the identity was true for all $\lambda.$ However AFAIK, in the context of algebraic geometry we only know that it's true for all non-zero. Perhaps it's possible to adapt his proof somehow?
Best Answer
If
$f(\lambda x_0, \lambda x_1, \ldots, \lambda x_n) = f(x_0, x_1, \ldots, x_n) \tag 1$
for all $\lambda$ in some (base) field, then taking $\lambda = 0$, $(x_0, x_1, \ldots, x_n)$ arbitrary yields
$f(x_0, x_1, \ldots, x_n) = f(0 \cdot x_0, 0 \cdot x_1, \ldots, 0 \cdot x_n) = f(0, 0, \ldots, 0), \tag 2$
which looks pretty constant to me.