Geometrically, we look at the ideal $I$ as a closed subscheme $$\mathrm{Spec}(R/I) \hookrightarrow \mathrm{Spec}(R),$$ and in this setting, Prime Avoidance says that if $\mathrm{Spec}(R/I)$ (or more precisely its image $V_{R}(I)$) contains the intersection of a finitely many irreducible closed subsets $Z_{1}, \dots, Z_{r}$ of $\mathrm{Spec}(R)$, then it must contain one $Z_{i}$ of the closed subsets.
When $R$ contains an infinite field, then this can be strengthened: $\mathfrak{p}_{i}$ do note need to be prime ideals, but just ideals. Geometrically, this means that we do not need to require $Z_{i}$ to be irreducible.
Prime Avoidance Lemma (over infinite fields). Let $k$ be an infinite field and $R$ a vector space over $k$ (e.g., $R$ can be a $k$-algebra). Let $I$ be a subspace of $R$. Given finitely many proper subspaces $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}$ of $R$, if $I$ is contained in the union of the subspaces, then there is a single $\mathfrak{p}_{i}$ containing $I$.
Proof. We follow Hochster's notes. Replacing $\mathfrak{p}_{i}$ with $\mathfrak{p}_{i} \cap I,$ we may assume that $\mathfrak{p}_{i} \subsetneq I.$ Then we show that $I = \mathfrak{p}_{i}$ for some $i$.
Suppose for contradiction that $I$ is not equal to any of $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}$. Under this assumption, we may choose $v_{i} \in I \setminus \mathfrak{p}_{i}$ for each $1 \leq i \leq r$. Thus, the $k$-subspace $J$ of $I$ generated by $v_{1}, \dots, v_{r}$ will not be equal to any $\mathfrak{p}_{1} \cap J, \dots, \mathfrak{p}_{r} \cap J$.
Focusing on $J$ reduces our statement to the case where $I$ is finite dimensional over $k.$ Hence, we may assume that $I = k^{n}$ for some $n \geq 0$. For each $\mathfrak{p}_{i}$, since it is not the whole $k^{n},$ there is a $k$-linear map $L_{i} : k^{n} \rightarrow k$ such that $L_{i}$ is not identically zero on $I = k^{n}$ but it is on $\mathfrak{p}_{i}$. Since $L_{i}$ is an $1 \times n$ row vector $(a_{i1}, \dots, a_{in})$, we may write $L_{i}(x_{1}, \dots, x_{n}) = a_{i1}x_{1} + \cdots + a_{in}x_{n}$. Now, consider the polynomial $$f(x_{1}, \dots, x_{n}) := \prod_{i=1}^{n}L_{i}(x_{1}, \dots, x_{n}).$$ We note that $f$ is a nonzero polynomial in $k[x_{1}, \dots, x_{n}]$ because none of $L_{i}$ are zero polynomials, but each $L_{i}$ vanishes on $\mathfrak{p}_{i}$, and the union of $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{n}$ is $k^{n}$. Hence, we must have $f(x_{1}, \dots, x_{n}) = 0$ for any evaluation of $x_{1}, \dots, x_{n} \in k$. Since $k$ is infinite, the only such polynomial is $f = 0$, a contradiction. $\Box$
Remark. Although we chose ring-friendly notations, the above statement is purely a statement about vector spaces over an infinite field $k.$ It says: any vector space over an infinite field cannot be covered by finitely many proper subspaces.
Avoidance in geometry. Now, suppose that we have finitely many nonzero vectors $(x_{i0}, \dots, x_{in}) \in k^{n+1}$ for $1 \leq i \leq r.$ For any fixed $i,$ the vectors $(a_{0}, \dots, a_{n}) \in k^{n+1}$ satsifying the equation $$a_{0}x_{i0} + \cdots + a_{n}x_{in} = 0$$ form a proper subsapce $V_{i}$ of $k^{n+1}.$ If $k$ is infinite, then $V_{1}, \dots, V_{r}$ cannot cover $k^{n+1},$ so there must be a vector $(a_{0}, \dots, a_{n}) \in k^{n+1}$ that does not satisfy any of the $r$ equations. This vector is necessarily nonzero because zero is in any vector space. This shows the following.
Proposition. Given any finitely many $k$-points of $\mathbb{P}^{n},$ there is a hyperplane avoiding all those points.
Q. Can we find a hyperplane in $\mathbb{P}^{n}$ (over an infinite field $k$) that avoids any given finitely many points (not just $k$-points)?
The answer is yes.
Proof. Let $\mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}$ be homogeneous prime ideals in $k[x_{0}, \dots, x_{n}]$ not containing $(x_{0}, \dots, x_{n})$, recalling that such prime ideals parametrize the points of $$\mathbb{P}^{n}_{k} = \mathrm{Proj}(k[x_{0}, \dots, x_{n}]).$$ Using Prime Avoidance, all we need to show that there is a nonzero vector $(a_{0}, \dots, a_{n}) \in k^{n+1}$ such that $a_{0}x_{0} + \cdots + a_{n}x_{n}$ is not in $$\mathfrak{p}_{1} \cup \cdots \cup \mathfrak{p}_{r}.$$ This has to be true because otherwise all of $x_{0}, \dots, x_{n}$ must be in the union so that $(x_{0}, \dots, x_{n})$ must be in the union, but this would mean that one of $\mathfrak{p}_{i}$ must be equal to $(x_{0}, \dots, x_{n}),$ a contradiction. This finishes the proof $\Box$
Q. What about a projective variety (instead of $\mathbb{P}^{n}$)?
The answer can be yes, but this is just a matter of definition.
Let $X$ be a projective $k$-scheme. This means that we can assume $$X = \text{Proj}(k[x_{0}, \dots, x_{n}]/I) \hookrightarrow \mathbb{P}^{n}_{k},$$ for a homogenous ideal $I$ in $k[x_{0}, \dots, x_{n}]$ not containing $(x_{0}, \dots, x_{n}).$ The line bundle $\mathscr{O}_{\mathbb{P}^{n}}(1)$ pulls back to $\mathscr{O}_{X}(1)$ under this inclusion. Thus, if $k$ is an infinite field, for any finitely many points $p_{1}, \dots, p_{r} \in X,$ we can find a hyperplane $H \subset \mathbb{P}^{n}$ not containing any of $p_{1}, \dots, p_{r}.$ If we say $L$ is a nonzero linear form in $k[x_{0}, \dots, x_{n}]$ (i.e., $L$ is a global section of $\mathscr{O}_{\mathbb{P}^{n}}(1)$) presenting $H,$ then $L$ does not vanish on any of the points $p_{1}, \dots, p_{r},$ so its pull back $L|_{X}$ does not either. Hence, the locus of $L|_{X}$ is the "hyperplane" of $X$ that does not pass through any of the points $p_{1}, \dots, p_{r}.$
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