An ideal is homogeneous iff it is generated by homogeneous polynomials

Question

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.

Answer 1

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.