Comparison of saturated ideals and radical ideals

Given a graded ring $B = A[x_0,\dots,x_n]$, $I$ a homogeneous ideal of $B$ not containing $B_+$. Then what are the relations between the racial ideal of $I$ and saturation of $I$?

As far as I know, there are the following results, indicating possible deeper relations between them:

1. Saturated ideals are not necessarily radical. Radical ideals are saturated.

2. There is a bijection between the closed subschemes of $\operatorname{Proj}(B)$ and the saturated homogeneous ideal of $B$ not containing $B_+$.

3. (Projective Nullstellensatz, from wiki): There is a bijection between homogeneous radical ideals not containing $B_+$ and subsets of $\mathbb{P}^n$ of the form $V(I):= \{x\in \mathbb{P}^n \mid f(x)=0 \text{ for all } f \in I\}.$

Is it true that saturated ideal indicate a scheme structure while radical ideal only indicates its topological property? Thanks in advance!

(Typos corrected based on the answer of KReiser.)

Not everything you say is correct. Let's fix the errors and discuss the implications.

1. This implication ("[s]aturated ideals are radical") is the wrong way around. The ideal $(x^2)\subset k[x,y]$ is saturated: any element $f\in(x^2)$ with $x^if$ and $y^jf$ in $(x^2)$ for some $i,j\geq 0$ must already be divisible by $x^2$. On the other hand, it is not radical: $x\cdot x\in (x^2)$ but $x\notin (x^2)$. It is true that any radical ideal is saturated, though: if $I\subset k[x_0,\cdots,x_n]$ is radical and $f$ is an element with $x_i^{j_i}f\in I$ for all large enough $j_i$, then for some $n\gg0$ one can write $f^n$ as a sum $\sum x_i^{l_i}f$ where $l_i\geq j_i$, and since $I$ is radical we have $f\in I$.
2. Yes, that's correct.
3. You're missing a "not containing $B_+$" in there (probably a typo), but the corrected statement is true.

Asking "[i]s it true that saturated ideal indicate a scheme structure while radical ideal only indicates its topological property" is not quite right. Any homogeneous ideal $I$ gives a scheme structure on $V(I)$, but there might be many $I$ which correspond to the same scheme structure. Saturation is the way to produce a unique largest ideal corresponding to a given scheme structure on a fixed closed subset. Taking the radical is the way to find largest ideal with the same vanishing set - this gives the reduced induced scheme structure, which is the "smallest" subscheme structure that you can put on a closed subset.