Comparison of saturated ideals and radical ideals

Question

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.)

Answer 1

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.