An ideal is homogeneous iff it is generated by homogeneous polynomials

Recall that $$f \in F[x_1, \cdots, x_n]$$ is homogeneous if all of its terms of monomials are of the same degree, i.e, if $$f = \sum_i a_ix_1^{d_{i_1}}\cdot ... x_n^{d_{i_n}}$$ then $$d_{i_1} + ... + d_{i_n}$$ is always the same for each $$i$$. And of course this means that $$f(ax_1, \cdots, ax_n) = a^mf(x_1, \cdots, x_n)$$ for some $$a$$.

An ideal $$I \subseteq F[x_1, \cdots, x_n]$$ is homogeneous if, for any $$p \in I$$, then the homogeneous components of $$I$$ (since every polynomial can be broken down into homogeneous components by simply selecting out the stuff of equal degree) is also in $$I$$.

I want to show:

An ideal is homogeneous iff it is generated by homogeneous polynomials.

$$\implies$$ is simple enough I think- the generating set is naturally the collection of all homogeneous polynomials in $$I$$.

$$\impliedby$$ I think an inductive argument should work. If I take a degree $$n+1$$ polynomial and $$f$$ and say that $$f = f' + g$$ where $$f'$$ is a homogeneous polynomial of degree $$n+1$$ and $$g$$ is the rest- I know that $$g$$ is generated by homogeneous polynomials by inductive assumption and $$f' = f-g \in I$$ and is a homogeneous polynomial so it is automatically generated by a homoegeneous polynomial in $$I$$, $$f'$$ itself? And together they generate $$f$$? Does that make sense? I am skeptical of this argument.

Your argument is fine, but I think the claim can be proven more directly. Let $$f_1,\cdots,f_m$$ be a finite collection of homogeneous poylnomials among the generators of $$I$$ and let $$g_1,\cdots,g_m$$ be collection of arbitrary polynomials. If $$h=\sum_i f_ig_i$$ is in $$I$$, then we can decompose each $$g_i=g_{i0}+g_{i1}+\cdots$$ in to homogeneous parts and rewrite $$h=\sum_i \sum_j f_ig_{ij}$$. Collecting terms of the same degree, we see that every homogeneous part of $$h$$ is in $$I$$, so $$I$$ is homogeneous.