How to show that non-homogeneous polynomials can't generate homogeneous ideals? Definitions: For each $d \ge 0$, define $R_d$ to be the set of all degree-$d$ homogeneous polynomials in $\mathbb{C}[x_0, \dots, x_n]$. By convention, $0 \in R_d$ for all $d \ge 0$.
(1) An ideal $I$ of $\mathbb{C}[x_0, \dots, x_n]$ is called a homogeneous ideal if $I = \oplus_{d \ge 0} (I \cap R_d)$.
(2) $I$ is generated by homogeneous polynomials.

2 Questions:
(i) How to correctly state the negation of (2), i.e.
$\quad \quad \quad \neg$($I$ is generated by homogeneous polynomials) ?
(ii) How to generalize from the case $I = \langle f \rangle$ ($f$ non-homogeneous) to show in general that:
$\quad \quad \quad \neg (2) \implies \neg (1)$?

Problem: I have already shown that $(2) \implies (1)$. I want to show that $(1) \implies (2)$, ideally by contraposition, i.e. that $\neg (2) \implies \neg(1)$. However, I am having some difficulties stating $\neg(2)$.
In the special case of $\neg (2)$ that the ideal $I$ is the principal ideal $\langle f \rangle$ generated by a non-homogeneous polynomial $f$, I have shown that (1) is not true (see below for details).
However, I have failed to generalize from this case (where $I = \langle f \rangle$ for $f$ non-homogeneous) because lots of different things can happen when $I$ is no longer assumed to be principal.
The problem comes down largely to the fact that non-homogeneous polynomials with field coefficients are closed under multiplication, but that they are not closed under addition.
EDIT: I have moved some of what I have done, in terms of attempting to find an answer, to a community wiki answer to this post, since the original post was too long to read easily.
Note: This question is based essentially on Exercises 5.2.7.(1), 5.2.7.(3), and 5.2.8. on p. 277 of Garrity et al's Algebraic Geometry: A Problem Solving Approach.
 A: It is actually much easier to prove $(1)\Rightarrow(2)$ directly than to prove the contrapositive.  Assuming (1), you want to prove there exists a set of homogeneous polynomials which generates $I$.  To make your life easy, you might as well pick the largest possible such set--namely, take the set $S$ of all homogeneous polynomials which are elements of $I$.  That is, take $S=\bigcup_d I\cap R_d$  It is then immediate from (1) that $S$ generates $I$, since $I$ is just the set of sums of elements of the sets $I\cap R_d$.
If $I$ happens to be finitely generated, you then immediately get that it is generated by finitely many homogeneous elements.  Namely, each of your generators of $I$ is a sum of finitely many elements of $S$, and so you can just take all of those elements of $S$ for each of your generators.
As for how to state the negation of (2), you need to keep your quantifiers straight.  The meaning of (2) is that "there exists a set of homogeneous polynomials which generates $I$".  So the negation is "there does not exist a set of homogeneous polynomials which generates $I$".  If you like, this is equivalent to "every set of generators for $I$ contains a non-homogeneous polynomial".  This, I think, makes it clear why the contrapositive is difficult to use: the negation is not a statement about a single set of non-homogeneous generators of $I$, but a statement about all possible sets of generators of $I$.  It is difficult to come up with a direct argument that makes use of the full strength of this assertion, since you need to somehow use all possible sets of generators at once.  On the other hand, to prove  $(1)\Rightarrow(2)$ directly, you just have to find a single set of homogeneous generators.
A: Update: As Eric Wofsey mentioned in the comments to his answer, there is a 3rd condition which is equivalent to (1) and (2):
(3) If $f = \sum_{d \ge 0} f_d \in I$, where for each $d$, $f_d \in R_d$, then $f_d \in I$ for each $d \ge 0$.
Since $\mathbb{C}[x_0, \dots, x_n] = \oplus_{d \ge 0} R_d$, every $f \in \mathbb{C}[x_0, \dots, x_n]$ is of the form $f = \sum_{d \ge 0} f_d$ where for each $d \ge 0$, $f_d \in R_d$. So (3) is equivalent to:
(3') If $f \in I$, then $f_d \in I$ for all $d \ge 0$.
Which is itself equivalent to:
(1') $I \subseteq \oplus_{d \ge 0} (I \cap R_d)$. (Since $f_d \in R_d$ by definition, so if it is also in $I$, then $f_d \in I \cap R_d $.)
(Skipping fewer steps, (3) is equivalent to the statement:
If $f \in I \cap ( +_{d \ge 0} R_d)$, then $f \in +_{d \ge 0} (I \cap R_d)$. I.e. intersection with $I$ distributes over the sum.
But anyway, we know that $+_{d \ge 0} R_d = \oplus_{d \ge 0} R_d = \mathbb{C}[x_0, \dots, x_n]$, so $I \cap (+_{d \ge 0} R_d) = I \cap (\oplus_{d \ge 0} R_d) = I \cap \mathbb{C}[x_0, \dots, x_n] = I$.
Moreover, $+_{d \ge 0} (I \cap R_d) = \oplus_{d \ge 0}(I \cap R_d)$, which follows from the direct sum decomposition of $\mathbb{C}[x_0, \dots, x_n]$, because $(\cap_{d \not= e} R_d) \cap R_e = \{0\} \implies I \cap [(\cap_{d \not= e} R_d) \cap R_e ] = \{0\} \iff [I \cap ( \cap_{d \not= e} R_d)] \cap [I \cap R_d] = \{0 \}$.
So [If $f \in I \cap ( +_{d \ge 0} R_d)$, then $f \in +_{d \ge 0} (I \cap R_d)$] $\iff$ [If $f \in I$, then $f_d \in I$ for all $d \ge 0$.])
Anyway the moral of this story is: it is easier to prove $\neg(2) \implies \neg (3)$.
First, let's state correctly $\neg(2)$, as Eric Wofsey explained how to do in his answer:
$\neg(2)$ Every set of generators for $I$ contains a non-homogeneous polynomial.
Let's state $\neg (3)$ (or something equivalent to it):
$\neg(3)$ There is some element of $I$ which has a homogeneous part which is not in $I$.
This is clearly true in the case that $I$ is the principal ideal generated by some non-homogeneous polynomial $f$, since no homogeneous polynomial is in $I$, so $f$ has (at least two) homogeneous parts which are not in $I$ (because the homogeneous parts are homogeneous polynomials, and no homogeneous polynomial is in $I$).
So proving $\neg(2) \implies \neg(3)$ would clearly generalize the argument from the special case that $I$ is a principal ideal generated by a non-homogeneous polynomial, which is what I wanted to do.
So let's show more generally that $\neg(2) \implies \neg (3)$. Assume that every generating set of $I$ has at least one non-homogeneous polynomial in it. Then assume by means of contradiction that (3) were true (that $I = \oplus_{d \ge 0} (I \cap R_d)$). Then $\cup_{d \ge 0}(I \cap R_d)$ would be a generating set for $I$ consisting exclusively of homogeneous polynomials. (Because any generating set for $I$ as an abelian group also has to be a generating set for $I$ as an ideal, since $I$ is by assumption closed under $\mathbb{C}[x_0, \dots, x_n]$ multiplication.) This would be a contradiction, since $\neg(2)$ says that every generating set of $I$ contains at least one non-homogeneous polynomial. Therefore (3) $\iff$ (1) is false, i.e. we have $\neg(3) \iff \neg(1)$. (This argument also seems to explain what Eric Wofsey meant in his comment when he said "I don't see a way to prove this is equivalent to $\neg(2)$ without basically going through (1)".
In particular, we would have at least one generating set with a non-homogeneous polynomial with at least one homogeneous part which is not generated by the homogeneous polynomials in $I$ -- this fact generalizes the example of $I = \langle x - yz, x^2 - yz \rangle$ which I gave earlier: $x \not\in I$, and since $x$ is the part of $x- yz \in I$ in $R_1$, we have that $x \not\in I \implies x-yz \not\in \oplus_{d \ge 0}(I \cap R_d) \implies \neg(3)$.
/Update
Attempt:
Because $\mathbb{C}$ is Noetherian, by Hilbert's Basis theorem, so is $\mathbb{C}[x_0, \dots, x_n]$. So (2) is equivalent to:
(2') $I$ is generated by a finite set of homogeneous polynomials.
Also, we have for all $d \ge 0$ that $(I \cap R_d) \subseteq I$ (trivially), and since $I$ is closed under addition, we must have as a result that $\oplus_{d \ge 0} (I \cap R_d) \subseteq I$. Therefore (1) is equivalent to:
(1') $I \subseteq \oplus_{d \ge 0} (I \cap R_d)$.
(I showed that the sum $+_{d \ge 0}(I \cap R_d)$ being a direct sum follows from the fact that the sum $\oplus_{d \ge 0} R_d = \mathbb{C}[x_0, \dots, x_n]$ is direct, which I also previously showed to be true.)
I showed that, for polynomials with field coefficients (more generally coefficients in any null-divisor-free ring) that the product $gf$ of any non-zero polynomial $g$ with a non-homogeneous polynomial $f$ is necessarily non-homogeneous. Thus, for any $d \ge 0$, $\langle f \rangle \cap R_d = \langle 0 \rangle$, with the result that $\oplus_{d \ge 0} (\langle f \rangle \cap R_d) = \langle 0 \rangle$. Clearly $\langle f \rangle \not\subseteq \langle 0 \rangle$, so (1') is false, and thus (1).
In the special case of $\neg (2)$ that $I = \langle x-yz, x^2 - yz \rangle = \langle yz, x^2 -x \rangle$, I have from the special case above that $\langle x^2 - x\rangle \cap R_d = \langle 0 \rangle$, but I'm not sure how to use this to show that $x^2 - x \not\in \oplus_{d \ge 0}(\langle yz, x^2 - x \rangle \cap R_d)$, since non-homogeneous polynomials are not closed under addition. I really want to say that, for all $d \ge 0$, $\langle yz, x^2 - x \rangle \cap R_d = \langle yz \rangle \cap R_d$, which would give me the result I want to show (since $yz$ doesn't divide $x^2 -x$), but (a) I'm not sure how to correctly show that, since non-homogeneous polynomials aren't closed under addition, and (b) even if I did show that, it wouldn't seem to generalize to all possible cases of $\neg (2)$, so it wouldn't be very useful line of thinking to pursue to show that, in general, $\neg (2) \implies \neg (1)$.
For example, if $f_1 = x - y^2$ and $f_2 = y^2 + z$, then $f_1 + f_2$ is homogeneous and non-zero even though both $f_1, f_2$ are non-zero and non-homogeneous, and clearly $f_1 + f_2 \in \langle f_1, f_2 \rangle$, so even though for all $d \ge 0$, $\langle f_1 \rangle \cap R_d = \langle 0 \rangle$ and $\langle f_2 \rangle \cap R_d = \langle 0 \rangle$, we do not have for all $d \ge 0$ that $\langle f_1, f_2 \rangle \cap R_d = \langle 0 \rangle$. Of course, in this case we also clearly have that $\langle f_1, f_2 \rangle = \langle x, z, y^2 \rangle$, which is generated by homogeneous polynomials, so this isn't actually a special case of $\neg (2)$. But even if $\neg (2)$ does exclude all problematic cases like this, I don't know how to state $\neg(2)$ correctly.
Showing $(1) \implies (2)$ directly might be easier, but the only way of proving it I have seen (see here or here) doesn't directly give a finite generating set for the ideal $I$. I would prefer a condition involving finite generating sets, to fully take advantage of the Noetherian assumption.
