How should I go about this proof about homogeneous polynomials?

Question

Question

Let $f_1,…,f_s$ be homogeneous polynomials of total degrees $d_1<d_2\leq …\leq d_s$ and let $I=\langle f_1,\ldots,f_s\rangle\subseteq k$. Show that if $g$ is another homogeneous polynomial of degree $d_1$ in $I$, then $g$ must be a constant multiple of $f_1$.

Note: I have earlier proved that if $f=\sum_if_i$ and $g=\sum_ig_i$ are the expansions of two polynomials as the sums of their homogeneous components, then $f=g$ iff $f_i=g_i$ for all $i$. And that if $h_l$ is the homogeneous component of $h=f.g$, then $h_l=\sum_{i+j=l}f_i.g_j$. The hint provided in the exercise asks me to use these results to do the proof above, but I can’t seem to know how this hint shows that $g=cf_1$, where $c$ is constant. Obviously, there is something I can’t seem to get here.

Your help would be greatly appreciated.

Answer 1

Suppose that $g(x)$ is an arbitrary member of $I$ that is homogeneous of degree $d_1$. We may write $g(x) = \sum_{i=1}^s g_i(x)f_i(x)$ for some functions $g_i(x)$. Write $g_1(x) = g_1(0) + h_1(x)$ where $h_1(x)$ doesn't have a constant term. Then you have that $$g(x) = \sum_{i=1}^s g_i(x)f_i(x)$$ $$= g_1(0)f_1(x) + h_1(x)f_1(x) + \sum_{i=2}^s g_i(x) f_i(x)$$ Since the terms of $h_1(x)f_1(x) + \sum_{i=2}^s g_i(x) f_i(x)$ are monomials of degree $d_1 + 1$ or greater, by the first fact you proved these terms sum to zero. Thus you are left with $g(x) = g_1(0) f_1(x)$, which is of the desired form.