Fact checking: are these inclusion relations regarding algebraic varieties of polynomial ideals correct?

Question

I'm studying inclusion identities within polynomial ideals theory. More precisely, i'm interested in the correspondece of an ideal $I\subseteq \mathbb{F}[\vec{x}]$ and its associated affine variety $\mathcal{V}(I)$ for an arbitrary commutative field $\mathbb{F}$.

I've encountered some useful facts: mainly, that if $\{\mathfrak{a,b}\}\subseteq\mathbb{F}[\vec{x}]$ we have that: $$\mathfrak{a\cdot b\subseteq a\cap b\subseteq a+b\subseteq a\cup b}\quad (1)$$ These inclusion identities can be found by following some algebraic properties of ideals, while noting that every element $r\in\mathfrak{a\cup b}$ can be written in the form: $$r=f_1a+f_2b$$ Where $f_i\in\mathbb{F}[\vec{x}]$, $a\in\mathfrak{a}$ and $b\in\mathfrak{b}$. In a similar fashion, we can show: $$\mathcal{V}(\mathfrak{a\cup b})\subseteq\mathcal{V}(\mathfrak{a+b})\subseteq\mathcal{V}(\mathfrak{a\cap b})\subseteq\mathcal{V}(\mathfrak{a\cdot b})$$ So there seems to be a complete inclusion-reversing correspondence between ideals and their associated affine varieties (this is, provided that $\mathfrak{a\cup b}$ is actually an ideal). I have two questions:

1. Do these inclusion relations are always satisfied (again, $\mathfrak{a\cup b}$ is an ideal)?
2. Do $\mathfrak{a}$ and $\mathfrak{b}$ exist such that the inclusions in the inclusion tower described by (1) are strict?

Answer 1

The inclusions $\mathfrak{a} \cdot \mathfrak{b} \subseteq \mathfrak{a} \cap \mathfrak{b} \subseteq \mathfrak{a} + \mathfrak{b}$ are very well-known results; you can find them in any textbook on Commutative Algebra.

As an example where all inclusions are strict, consider, say, $\mathfrak{a} = \langle x^2y \rangle$ and $\mathfrak{b} = \langle xy^2 \rangle$. Then $\mathfrak{a} \cdot \mathfrak{b} = \langle x^3y^3 \rangle$, $\mathfrak{a} \cap \mathfrak{b} = \langle x^2y^2 \rangle$ and $\mathfrak{a} + \mathfrak{b} = \langle xy^2, x^2y \rangle$.

As for $\mathfrak{a} \cup \mathfrak{b}$, what do you mean by that? The basic set union will fail to be an ideal; if you mean the smallest ideal that contains $\mathfrak{a}$ and $\mathfrak{b}$, sometimes denoted as $\langle \mathfrak{a} \cup \mathfrak{b}\rangle$, we have, in fact, $\langle \mathfrak{a} \cup \mathfrak{b}\rangle = \mathfrak{a} + \mathfrak{b}$ (why?).